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Exercise 1.1

Question 1:

Which of the following are sets? Justify our answer.

(i) The collection of all months of a year beginning with the letter J.

(ii) The collection of ten most talented writers of India.

(iii) A team of eleven best-cricket batsmen of the world.

(iv) The collection of all boys in your class.

(v) The collection of all natural numbers less than 100.

(vi) A collection of novels written by the writer Munshi Prem Chand.

(vii) The collection of all even integers.

(viii) The collection of questions in this Chapter.

(ix) A collection of most dangerous animals of the world.

Answer

(i) The collection of all months of a year beginning with the letter J is a well-defined

collection of objects because one can definitely identify a month that belongs to this

collection.

Hence, this collection is a set.

(ii) The collection of ten most talented writers of India is not a well-defined collection

because the criteria for determining a writer’s talent may vary from person to person.

Hence, this collection is not a set.

(iii) A team of eleven best cricket batsmen of the world is not a well-defined collection

because the criteria for determining a batsman’s talent may vary from person to person.


(iv) The collection of all boys in your class is a well-defined collection because you can

definitely identify a boy who belongs to this collection.


(v) The collection of all natural numbers less than 100 is a well-defined collection

because one can definitely identify a number that belongs to this collection.


(vi) A collection of novels written by the writer Munshi Prem Chand is a well-defined

collection because one can definitely identify a book that belongs to this collection.

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(vii) The collection of all even integers is a well-defined collection because one can

definitely identify an even integer that belongs to this collection.


(viii) The collection of questions in this chapter is a well-defined collection because one

can definitely identify a question that belongs to this chapter.


(ix) The collection of most dangerous animals of the world is not a well-defined

collection because the criteria for determining the dangerousness of an animal can vary

from person to person.


Question 2:

Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈or ∉ in the blank spaces:

(i) 5…A (ii) 8…A (iii) 0…A

(iv) 4…A (v) 2…A (vi) 10…A

Answer

(i) 5 ∈ A

(ii) 8 ∉ A

(iii) 0 ∉ A

(iv) 4 ∈ A

(v) 2 ∈ A

(vi) 10 ∉ A

Question 3:

Write the following sets in roster form:

(i) A = {x: x is an integer and –3 < x < 7}.

(ii) B = {x: x is a natural number less than 6}.

(iii) C = {x: x is a two-digit natural number such that the sum of its digits is 8}

(iv) D = {x: x is a prime number which is divisor of 60}.

(v) E = The set of all letters in the word TRIGONOMETRY.

(vi) F = The set of all letters in the word BETTER.

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Answer

(i) A = {x: x is an integer and –3 < x < 7}

The elements of this set are –2, –1, 0, 1, 2, 3, 4, 5, and 6 only.

Therefore, the given set can be written in roster form as

A = {–2, –1, 0, 1, 2, 3, 4, 5, 6}

(ii) B = {x: x is a natural number less than 6}

The elements of this set are 1, 2, 3, 4, and 5 only.

Therefore, the given set can be written in roster form as

B = {1, 2, 3, 4, 5}

(iii) C = {x: x is a two-digit natural number such that the sum of its digits is 8}

The elements of this set are 17, 26, 35, 44, 53, 62, 71, and 80 only.

Therefore, this set can be written in roster form as

C = {17, 26, 35, 44, 53, 62, 71, 80}

(iv) D = {x: x is a prime number which is a divisor of 60}

2 60

2 30

3 15

5

∴60 = 2 × 2 × 3 × 5

The elements of this set are 2, 3, and 5 only.

Therefore, this set can be written in roster form as D = {2, 3, 5}.

(v) E = The set of all letters in the word TRIGONOMETRY

There are 12 letters in the word TRIGONOMETRY, out of which letters T, R, and O are

repeated.


E = {T, R, I, G, O, N, M, E, Y}

(vi) F = The set of all letters in the word BETTER

There are 6 letters in the word BETTER, out of which letters E and T are repeated.


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F = {B, E, T, R}

Question 4:

Write the following sets in the set-builder form:

(i) (3, 6, 9, 12) (ii) {2, 4, 8, 16, 32}

(iii) {5, 25, 125, 625} (iv) {2, 4, 6 …}

(v) {1, 4, 9 … 100}

Answer

(i) {3, 6, 9, 12} = {x: x = 3n, n∈ N and 1 ≤ n ≤ 4}

(ii) {2, 4, 8, 16, 32}

It can be seen that 2 = 21, 4 = 22, 8 = 23, 16 = 24, and 32 = 25.

∴ {2, 4, 8, 16, 32} = {x: x = 2n, n∈ N and 1 ≤ n ≤ 5}

(iii) {5, 25, 125, 625}

It can be seen that 5 = 51, 25 = 52, 125 = 53, and 625 = 54.

∴ {5, 25, 125, 625} = {x: x = 5n, n∈N and 1 ≤ n ≤ 4}

(iv) {2, 4, 6 …}

It is a set of all even natural numbers.

∴ {2, 4, 6 …} = {x: x is an even natural number}

(v) {1, 4, 9 … 100}

It can be seen that 1 = 12, 4 = 22, 9 = 32 …100 = 102.

∴ {1, 4, 9… 100} = {x: x = n2, n∈N and 1 ≤ n ≤ 10}

Question 5:

List all the elements of the following sets:

(i) A = {x: x is an odd natural number}

(ii) B = {x: x is an integer, }

(iii) C = {x: x is an integer, }

(iv) D = {x: x is a letter in the word “LOYAL”}

(v) E = {x: x is a month of a year not having 31 days}

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(vi) F = {x: x is a consonant in the English alphabet which proceeds k}.

Answer

(i) A = {x: x is an odd natural number} = {1, 3, 5, 7, 9 …}

(ii) B = {x: x is an integer; }

It can be seen that and

∴ B

(iii) C = {x: x is an integer; }

It can be seen that

(–1)2 = 1 ≤ 4; (–2)2 = 4 ≤ 4; (–3)2 = 9 > 4

02 = 0 ≤ 4

12 = 1 ≤ 4

22 = 4 ≤ 4

32 = 9 > 4

∴C = {–2, –1, 0, 1, 2}

(iv) D = (x: x is a letter in the word “LOYAL”) = {L, O, Y, A}

(v) E = {x: x is a month of a year not having 31 days}

= {February, April, June, September, November}

(vi) F = {x: x is a consonant in the English alphabet which precedes k}

= {b, c, d, f, g, h, j}

Question 6:

Match each of the set on the left in the roster form with the same set on the right

described in set-builder form:

(i) {1, 2, 3, 6} (a) {x: x is a prime number and a divisor of 6}

(ii) {2, 3} (b) {x: x is an odd natural number less than 10}

(iii) {M, A,T, H, E, I,C, S} (c) {x: x is natural number and divisor of 6}

(iv) {1, 3, 5, 7, 9} (d) {x: x is a letter of the word MATHEMATICS}

Answer

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(i) All the elements of this set are natural numbers as well as the divisors of 6.

Therefore, (i) matches with (c).

(ii) It can be seen that 2 and 3 are prime numbers. They are also the divisors of 6.

Therefore, (ii) matches with (a).

(iii) All the elements of this set are letters of the word MATHEMATICS. Therefore, (iii)

matches with (d).

(iv) All the elements of this set are odd natural numbers less than 10. Therefore, (iv)

matches with (b).

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Exercise 1.2

Question 1:

Which of the following are examples of the null set

(i) Set of odd natural numbers divisible by 2

(ii) Set of even prime numbers

(iii) {x:x is a natural numbers, x < 5 and x > 7 }

(iv) {y:y is a point common to any two parallel lines}

Answer

(i) A set of odd natural numbers divisible by 2 is a null set because no odd number is

divisible by 2.

(ii) A set of even prime numbers is not a null set because 2 is an even prime number.

(iii) {x: x is a natural number, x < 5 and x > 7} is a null set because a number cannot

be simultaneously less than 5 and greater than 7.

(iv) {y: y is a point common to any two parallel lines} is a null set because parallel lines

do not intersect. Hence, they have no common point.

Question 2:

Which of the following sets are finite or infinite

(i) The set of months of a year

(ii) {1, 2, 3 ...}

(iii) {1, 2, 3 ... 99, 100}

(iv) The set of positive integers greater than 100

(v) The set of prime numbers less than 99

Answer

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(i) The set of months of a year is a finite set because it has 12 elements.

(ii) {1, 2, 3 …} is an infinite set as it has infinite number of natural numbers.

(iii) {1, 2, 3 …99, 100} is a finite set because the numbers from 1 to 100 are finite in

number.

(iv) The set of positive integers greater than 100 is an infinite set because positive

integers greater than 100 are infinite in number.

(v) The set of prime numbers less than 99 is a finite set because prime numbers less

than 99 are finite in number.

Question 3:

State whether each of the following set is finite or infinite:

(i) The set of lines which are parallel to the x-axis

(ii) The set of letters in the English alphabet

(iii) The set of numbers which are multiple of 5

(iv) The set of animals living on the earth

(v) The set of circles passing through the origin (0, 0)

Answer

(i) The set of lines which are parallel to the x-axis is an infinite set because lines

parallel to the x-axis are infinite in number.

(ii) The set of letters in the English alphabet is a finite set because it has 26 elements.

(iii) The set of numbers which are multiple of 5 is an infinite set because multiples of 5

are infinite in number.

(iv) The set of animals living on the earth is a finite set because the number of animals

living on the earth is finite (although it is quite a big number).

(v) The set of circles passing through the origin (0, 0) is an infinite set because infinite

number of circles can pass through the origin.

Question 4:

In the following, state whether A = B or not:

(i) A = {a, b, c, d}; B = {d, c, b, a}

(ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18}

(iii) A = {2, 4, 6, 8, 10}; B = {x: x is positive even integer and x ≤ 10}

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(iv) A = {x: x is a multiple of 10}; B = {10, 15, 20, 25, 30 ...}

Answer

(i) A = {a, b, c, d}; B = {d, c, b, a}

The order in which the elements of a set are listed is not significant.

∴A = B

(ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18}

It can be seen that 12 ∈ A but 12 ∉ B.

∴A ≠ B

(iii) A = {2, 4, 6, 8, 10}

B = {x: x is a positive even integer and x ≤ 10}

= {2, 4, 6, 8, 10}

∴A = B

(iv) A = {x: x is a multiple of 10}

B = {10, 15, 20, 25, 30 …}

It can be seen that 15 ∈ B but 15 ∉ A.

∴A ≠ B

Question 5:

Are the following pair of sets equal? Give reasons.

(i) A = {2, 3}; B = {x: x is solution of x2 + 5x + 6 = 0}

(ii) A = {x: x is a letter in the word FOLLOW}; B = {y: y is a letter in the word WOLF}

Answer

(i) A = {2, 3}; B = {x: x is a solution of x2 + 5x + 6 = 0}

The equation x2 + 5x + 6 = 0 can be solved as:

x(x + 3) + 2(x + 3) = 0

(x + 2)(x + 3) = 0

x = –2 or x = –3

∴A = {2, 3}; B = {–2, –3}

∴A ≠ B

(ii) A = {x: x is a letter in the word FOLLOW} = {F, O, L, W}

B = {y: y is a letter in the word WOLF} = {W, O, L, F}

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∴A = B

Question 6:

From the sets given below, select equal sets:

A = {2, 4, 8, 12}, B = {1, 2, 3, 4}, C = {4, 8, 12, 14}, D = {3, 1, 4, 2}

E = {–1, 1}, F = {0, a}, G = {1, –1}, H = {0, 1}

Answer

A = {2, 4, 8, 12}; B = {1, 2, 3, 4}; C = {4, 8, 12, 14}

D = {3, 1, 4, 2}; E = {–1, 1}; F = {0, a}

G = {1, –1}; A = {0, 1}

It can be seen that

8 ∈ A, 8 ∉ B, 8 ∉ D, 8 ∉ E, 8 ∉ F, 8 ∉ G, 8 ∉ H

⇒ A ≠ B, A ≠ D, A ≠ E, A ≠ F, A ≠ G, A ≠ H

Also, 2 ∈ A, 2 ∉ C

∴ A ≠ C

3 ∈ B, 3 ∉ C, 3 ∉ E, 3 ∉ F, 3 ∉ G, 3 ∉ H

∴ B ≠ C, B ≠ E, B ≠ F, B ≠ G, B ≠ H

12 ∈ C, 12 ∉ D, 12 ∉ E, 12 ∉ F, 12 ∉ G, 12 ∉ H

∴ C ≠ D, C ≠ E, C ≠ F, C ≠ G, C ≠ H

4 ∈ D, 4 ∉ E, 4 ∉ F, 4 ∉ G, 4 ∉ H

∴ D ≠ E, D ≠ F, D ≠ G, D ≠ H

Similarly, E ≠ F, E ≠ G, E ≠ H

F ≠ G, F ≠ H, G ≠ H


∴ B = D and E = G

Hence, among the given sets, B = D and E = G.

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Exercise 1.3

Question 1:

Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces:

(i) {2, 3, 4} … {1, 2, 3, 4, 5}

(ii) {a, b, c} … {b, c, d}

(iii) {x: x is a student of Class XI of your school} … {x: x student of your school}

(iv) {x: x is a circle in the plane} … {x: x is a circle in the same plane with radius 1

unit}

(v) {x: x is a triangle in a plane}…{x: x is a rectangle in the plane}

(vi) {x: x is an equilateral triangle in a plane}… {x: x is a triangle in the same plane}

(vii) {x: x is an even natural number} … {x: x is an integer}

Answer

(i)

(ii)

(iii) {x: x is a student of class XI of your school}⊂ {x: x is student of your school}

(iv) {x: x is a circle in the plane} ⊄ {x: x is a circle in the same plane with radius 1

unit}

(v) {x: x is a triangle in a plane} ⊄ {x: x is a rectangle in the plane}

(vi) {x: x is an equilateral triangle in a plane}⊂ {x: x in a triangle in the same plane}

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(vii) {x: x is an even natural number} ⊂ {x: x is an integer}

Question 2:

Examine whether the following statements are true or false:

(i) {a, b} ⊄ {b, c, a}

(ii) {a, e} ⊂ {x: x is a vowel in the English alphabet}

(iii) {1, 2, 3} ⊂{1, 3, 5}

(iv) {a} ⊂ {a. b, c}

(v) {a} ∈ (a, b, c)

(vi) {x: x is an even natural number less than 6} ⊂ {x: x is a natural number which

divides 36}

Answer

(i) False. Each element of {a, b} is also an element of {b, c, a}.

(ii) True. a, e are two vowels of the English alphabet.

(iii) False. 2∈{1, 2, 3}; however, 2∉{1, 3, 5}

(iv) True. Each element of {a} is also an element of {a, b, c}.

(v) False. The elements of {a, b, c} are a, b, c. Therefore, {a}⊂{a, b, c}

(vi) True. {x:x is an even natural number less than 6} = {2, 4}

{x:x is a natural number which divides 36}= {1, 2, 3, 4, 6, 9, 12, 18, 36}

Question 3:

Let A= {1, 2, {3, 4,}, 5}. Which of the following statements are incorrect and why?

(i) {3, 4}⊂ A

(ii) {3, 4}}∈ A

(iii) {{3, 4}}⊂ A

(iv) 1∈ A

(v) 1⊂ A

(vi) {1, 2, 5} ⊂ A

(vii) {1, 2, 5} ∈ A

(viii) {1, 2, 3} ⊂ A

(ix) Φ ∈ A

(x) Φ ⊂ A

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(xi) {Φ} ⊂ A

Answer

A = {1, 2, {3, 4}, 5}

(i) The statement {3, 4} ⊂ A is incorrect because 3 ∈ {3, 4}; however, 3∉A.

(ii) The statement {3, 4} ∈A is correct because {3, 4} is an element of A.

(iii) The statement {{3, 4}} ⊂ A is correct because {3, 4} ∈ {{3, 4}} and {3, 4} ∈ A.

(iv) The statement 1∈A is correct because 1 is an element of A.

(v) The statement 1⊂ A is incorrect because an element of a set can never be a subset

of itself.

(vi) The statement {1, 2, 5} ⊂ A is correct because each element of {1, 2, 5} is also an

element of A.

(vii) The statement {1, 2, 5} ∈ A is incorrect because {1, 2, 5} is not an element of A.

(viii) The statement {1, 2, 3} ⊂ A is incorrect because 3 ∈ {1, 2, 3}; however, 3 ∉ A.

(ix) The statement Φ ∈ A is incorrect because Φ is not an element of A.

(x) The statement Φ ⊂ A is correct because Φ is a subset of every set.

(xi) The statement {Φ} ⊂ A is incorrect because Φ∈ {Φ}; however, Φ ∈ A.

Question 4:

Write down all the subsets of the following sets:

(i) {a}

(ii) {a, b}

(iii) {1, 2, 3}

(iv) Φ

Answer

(i) The subsets of {a} are Φ and {a}.

(ii) The subsets of {a, b} areΦ, {a}, {b}, and {a, b}.

(iii) The subsets of {1, 2, 3} areΦ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and

{1, 2, 3}

(iv) The only subset of Φ isΦ.

Question 5:

How many elements has P(A), if A = Φ?

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Answer

We know that if A is a set with m elements i.e., n(A) = m, then n[P(A)] = 2m.

If A = Φ, then n(A) = 0.

∴ n[P(A)] = 20 = 1

Hence, P(A) has one element.

Question 6:

Write the following as intervals:

(i) {x: x ∈ R, –4 < x ≤ 6}

(ii) {x: x ∈ R, –12 < x < –10}

(iii) {x: x ∈ R, 0 ≤ x < 7}

(iv) {x: x ∈ R, 3 ≤ x ≤ 4}

Answer

(i) {x: x ∈ R, –4 < x ≤ 6} = (–4, 6]

(ii) {x: x ∈ R, –12 < x < –10} = (–12, –10)

(iii) {x: x ∈ R, 0 ≤ x < 7} = [0, 7)

(iv) {x: x ∈ R, 3 ≤ x ≤ 4} = [3, 4]

Question 7:

Write the following intervals in set-builder form:

(i) (–3, 0)

(ii) [6, 12]

(iii) (6, 12]

(iv) [–23, 5)

Answer

(i) (–3, 0) = {x: x ∈ R, –3 < x < 0}

(ii) [6, 12] = {x: x ∈ R, 6 ≤ x ≤ 12}

(iii) (6, 12] ={x: x ∈ R, 6 < x ≤ 12}

(iv) [–23, 5) = {x: x ∈ R, –23 ≤ x < 5}

Question 8:

What universal set (s) would you propose for each of the following:

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(i) The set of right triangles

(ii) The set of isosceles triangles

Answer

(i) For the set of right triangles, the universal set can be the set of triangles or the set

of polygons.

(ii) For the set of isosceles triangles, the universal set can be the set of triangles or the

set of polygons or the set of two-dimensional figures.

Question 9:

Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the

following may be considered as universals set (s) for all the three sets A, B and C

(i) {0, 1, 2, 3, 4, 5, 6}

(ii) Φ

(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(iv) {1, 2, 3, 4, 5, 6, 7, 8}

Answer

(i) It can be seen that A ⊂ {0, 1, 2, 3, 4, 5, 6}

B ⊂ {0, 1, 2, 3, 4, 5, 6}

However, C ⊄ {0, 1, 2, 3, 4, 5, 6}

Therefore, the set {0, 1, 2, 3, 4, 5, 6} cannot be the universal set for the sets A, B, and

C.

(ii) A ⊄ Φ, B ⊄ Φ, C ⊄ Φ

Therefore, Φ cannot be the universal set for the sets A, B, and C.

(iii) A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

C ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Therefore, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the sets A, B,

and C.

(iv) A ⊂ {1, 2, 3, 4, 5, 6, 7, 8}

B ⊂ {1, 2, 3, 4, 5, 6, 7, 8}

However, C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}

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Therefore, the set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets A, B,

and C.

Exercise 1.4

Question 1:

Find the union of each of the following pairs of sets:

(i) X = {1, 3, 5} Y = {1, 2, 3}

(ii) A = {a, e, i, o, u} B = {a, b, c}

(iii) A = {x: x is a natural number and multiple of 3}

B = {x: x is a natural number less than 6}

(iv) A = {x: x is a natural number and 1 < x ≤ 6}

B = {x: x is a natural number and 6 < x < 10}

(v) A = {1, 2, 3}, B = Φ

Answer

(i) X = {1, 3, 5} Y = {1, 2, 3}

X∪ Y= {1, 2, 3, 5}

(ii) A = {a, e, i, o, u} B = {a, b, c}

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A∪ B = {a, b, c, e, i, o, u}

(iii) A = {x: x is a natural number and multiple of 3} = {3, 6, 9 …}

As B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5, 6}

A ∪ B = {1, 2, 4, 5, 3, 6, 9, 12 …}

∴ A ∪ B = {x: x = 1, 2, 4, 5 or a multiple of 3}

(iv) A = {x: x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}

B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9}

A∪ B = {2, 3, 4, 5, 6, 7, 8, 9}

∴ A∪ B = {x: x ∈ N and 1 < x < 10}

(v) A = {1, 2, 3}, B = Φ

A∪ B = {1, 2, 3}

Question 2:

Let A = {a, b}, B = {a, b, c}. Is A ⊂ B? What is A ∪ B?

Answer

Here, A = {a, b} and B = {a, b, c}

Yes, A ⊂ B.

A∪ B = {a, b, c} = B

Question 3:

If A and B are two sets such that A ⊂ B, then what is A ∪ B?

Answer

If A and B are two sets such that A ⊂ B, then A ∪ B = B.

Question 4:

If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}; find

(i) A ∪ B

(ii) A ∪ C

(iii) B ∪ C

(iv) B ∪ D

(v) A ∪ B ∪ C

(vi) A ∪ B ∪ D

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(vii) B ∪ C ∪ D

Answer

A = {1, 2, 3, 4], B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}

(i) A ∪ B = {1, 2, 3, 4, 5, 6}

(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(iii) B ∪ C = {3, 4, 5, 6, 7, 8}

(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

Question 5:

Find the intersection of each pair of sets:

(i) X = {1, 3, 5} Y = {1, 2, 3}

(ii) A = {a, e, i, o, u} B = {a, b, c}

(iii) A = {x: x is a natural number and multiple of 3}

B = {x: x is a natural number less than 6}

(iv) A = {x: x is a natural number and 1 < x ≤ 6}

B = {x: x is a natural number and 6 < x < 10}

(v) A = {1, 2, 3}, B = Φ

Answer

(i) X = {1, 3, 5}, Y = {1, 2, 3}

X ∩ Y = {1, 3}

(ii) A = {a, e, i, o, u}, B = {a, b, c}

A ∩ B = {a}

(iii) A = {x: x is a natural number and multiple of 3} = (3, 6, 9 …}

B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5}

∴ A ∩ B = {3}

(iv) A = {x: x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}

B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9}

A ∩ B = Φ

(v) A = {1, 2, 3}, B = Φ

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A ∩ B = Φ

Question 6:

If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find

(i) A ∩ B

(ii) B ∩ C

(iii) A ∩ C ∩ D

(iv) A ∩ C

(v) B ∩ D

(vi) A ∩ (B ∪ C)

(vii) A ∩ D

(viii) A ∩ (B ∪ D)

(ix) (A ∩ B) ∩ (B ∪ C)

(x) (A ∪ D) ∩ (B ∪ C)

Answer

(i) A ∩ B = {7, 9, 11}

(ii) B ∩ C = {11, 13}

(iii) A ∩ C ∩ D = { A ∩ C} ∩ D = {11} ∩ {15, 17} = Φ

(iv) A ∩ C = {11}

(v) B ∩ D = Φ

(vi) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

= {7, 9, 11} ∪ {11} = {7, 9, 11}

(vii) A ∩ D = Φ

(viii) A ∩ (B ∪ D) = (A ∩ B) ∪ (A ∩ D)

= {7, 9, 11} ∪ Φ = {7, 9, 11}

(ix) (A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}

(x) (A ∪ D) ∩ (B ∪ C) = {3, 5, 7, 9, 11, 15, 17) ∩ {7, 9, 11, 13, 15}

= {7, 9, 11, 15}

Question 7:

If A = {x: x is a natural number}, B ={x: x is an even natural number}

C = {x: x is an odd natural number} and D = {x: x is a prime number}, find

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(i) A ∩ B

(ii) A ∩ C

(iii) A ∩ D

(iv) B ∩ C

(v) B ∩ D

(vi) C ∩ D

Answer

A = {x: x is a natural number} = {1, 2, 3, 4, 5 …}

B ={x: x is an even natural number} = {2, 4, 6, 8 …}

C = {x: x is an odd natural number} = {1, 3, 5, 7, 9 …}

D = {x: x is a prime number} = {2, 3, 5, 7 …}

(i) A ∩B = {x: x is a even natural number} = B

(ii) A ∩ C = {x: x is an odd natural number} = C

(iii) A ∩ D = {x: x is a prime number} = D

(iv) B ∩ C = Φ

(v) B ∩ D = {2}

(vi) C ∩ D = {x: x is odd prime number}

Question 8:

Which of the following pairs of sets are disjoint

(i) {1, 2, 3, 4} and {x: x is a natural number and 4 ≤ x ≤ 6}

(ii) {a, e, i, o, u}and {c, d, e, f}

(iii) {x: x is an even integer} and {x: x is an odd integer}

Answer

(i) {1, 2, 3, 4}

{x: x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}

Now, {1, 2, 3, 4} ∩ {4, 5, 6} = {4}

Therefore, this pair of sets is not disjoint.

(ii) {a, e, i, o, u} ∩ (c, d, e, f} = {e}

Therefore, {a, e, i, o, u} and (c, d, e, f} are not disjoint.

(iii) {x: x is an even integer} ∩ {x: x is an odd integer} = Φ

Therefore, this pair of sets is disjoint.

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Question 9:

If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20},

C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}; find

(i) A – B

(ii) A – C

(iii) A – D

(iv) B – A

(v) C – A

(vi) D – A

(vii) B – C

(viii) B – D

(ix) C – B

(x) D – B

(xi) C – D

(xii) D – C

Answer

(i) A – B = {3, 6, 9, 15, 18, 21}

(ii) A – C = {3, 9, 15, 18, 21}

(iii) A – D = {3, 6, 9, 12, 18, 21}

(iv) B – A = {4, 8, 16, 20}

(v) C – A = {2, 4, 8, 10, 14, 16}

(vi) D – A = {5, 10, 20}

(vii) B – C = {20}

(viii) B – D = {4, 8, 12, 16}

(ix) C – B = {2, 6, 10, 14}

(x) D – B = {5, 10, 15}

(xi) C – D = {2, 4, 6, 8, 12, 14, 16}

(xii) D – C = {5, 15, 20}

Question 10:

If X = {a, b, c, d} and Y = {f, b, d, g}, find

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(i) X – Y

(ii) Y – X

(iii) X ∩ Y

Answer

(i) X – Y = {a, c}

(ii) Y – X = {f, g}

(iii) X ∩ Y = {b, d}

Question 11:

If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?

Answer

R: set of real numbers

Q: set of rational numbers

Therefore, R – Q is a set of irrational numbers.

Question 12:

State whether each of the following statement is true or false. Justify your answer.

(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.

(ii) {a, e, i, o, u } and {a, b, c, d} are disjoint sets.

(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.

(iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.

Answer

(i) False

As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}

⇒ {2, 3, 4, 5} ∩ {3, 6} = {3}

(ii) False

As a ∈ {a, e, i, o, u}, a ∈ {a, b, c, d}

⇒ {a, e, i, o, u } ∩ {a, b, c, d} = {a}

(iii) True

As {2, 6, 10, 14} ∩ {3, 7, 11, 15} = Φ

(iv) True

As {2, 6, 10} ∩ {3, 7, 11} = Φ

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Exercise 1.5

Question 1:

Let U ={1, 2, 3; 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5,

6}. Find

(i)

(ii)

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(iii)

(iv)

(v)

(vi)

Answer

U ={1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {1, 2, 3, 4}

B = {2, 4, 6, 8}

C = {3, 4, 5, 6}

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Question 2:

If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets:

(i) A = {a, b, c}

(ii) B = {d, e, f, g}

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(iii) C = {a, c, e, g}

(iv) D = {f, g, h, a}

Answer

U = {a, b, c, d, e, f, g, h}

(i) A = {a, b, c}

(ii) B = {d, e, f, g}

(iii) C = {a, c, e, g}

(iv) D = {f, g, h, a}

Question 3:

Taking the set of natural numbers as the universal set, write down the complements of

the following sets:

(i) {x: x is an even natural number}

(ii) {x: x is an odd natural number}

(iii) {x: x is a positive multiple of 3}

(iv) {x: x is a prime number}

(v) {x: x is a natural number divisible by 3 and 5}

(vi) {x: x is a perfect square}

(vii) {x: x is perfect cube}

(viii) {x: x + 5 = 8}

(ix) {x: 2x + 5 = 9}

(x) {x: x ≥ 7}

(xi) {x: x ∈ N and 2x + 1 > 10}

Answer

U = N: Set of natural numbers

(i) {x: x is an even natural number}´ = {x: x is an odd natural number}

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(ii) {x: x is an odd natural number}´ = {x: x is an even natural number}

(iii) {x: x is a positive multiple of 3}´ = {x: x ∈ N and x is not a multiple of 3}

(iv) {x: x is a prime number}´ ={x: x is a positive composite number and x = 1}

(v) {x: x is a natural number divisible by 3 and 5}´ = {x: x is a natural number that is

not divisible by 3 or 5}

(vi) {x: x is a perfect square}´ = {x: x ∈ N and x is not a perfect square}

(vii) {x: x is a perfect cube}´ = {x: x ∈ N and x is not a perfect cube}

(viii) {x: x + 5 = 8}´ = {x: x ∈ N and x ≠ 3}

(ix) {x: 2x + 5 = 9}´ = {x: x ∈ N and x ≠ 2}

(x) {x: x ≥ 7}´ = {x: x ∈ N and x < 7}

(xi) {x: x ∈ N and 2x + 1 > 10}´ = {x: x ∈ N and x ≤ 9/2}

Question 4:

If U = {1, 2, 3, 4, 5,6,7,8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that

(i) (ii)

Answer

U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {2, 4, 6, 8}, B = {2, 3, 5, 7}

(i)

(ii)

Question 5:

Draw appropriate Venn diagram for each of the following:

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(i)

(ii)

(iii)

(iv)

Answer

(i)

(ii)

(iii)

(iv)

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Question 6:

Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one

angle different from 60°, what is ?

Answer

is the set of all equilateral triangles.

Question 7:

Fill in the blanks to make each of the following a true statement:

(i)

(ii) Φ′ ∩ A = …

(iii)

(iv)

Answer

(i)

(ii) Φ′ ∩ A = U ∩ A = A

∴ Φ′ ∩ A = A

(iii) A ∩ A′ = Φ

(iv) U′ ∩ A = Φ ∩ A = Φ

∴ U′ ∩ A = Φ

Exercise 1.6

Question 1:

If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩Y).

Answer

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It is given that:

n(X) = 17, n(Y) = 23, n(X ∪ Y) = 38

n(X ∩ Y) = ?

We know that:

Question 2:

If X and Y are two sets such that X ∪Y has 18 elements, X has 8 elements and Y has 15

elements; how many elements does X ∩Y have?

Answer

It is given that:

n(X ï�f�? Y) = ?

We know that:

Question 3:

In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many

people can speak both Hindi and English?

Answer

Let H be the set of people who speak Hindi, and

E be the set of people who speak English

∴ n(H ∪ E) = 400, n(H) = 250, n(E) = 200

n(H ∩ E) = ?

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We know that:

n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

∴ 400 = 250 + 200 – n(H ∩ E)

⇒ 400 = 450 – n(H ∩ E)

⇒ n(H ∩ E) = 450 – 400

∴ n(H ∩ E) = 50

Thus, 50 people can speak both Hindi and English.

Question 4:

If S and T are two sets such that S has 21 elements, T has 32 elements, and

S ∩ T has 11 elements, how many elements does S ∪ T have?

Answer

It is given that:

n(S) = 21, n(T) = 32, n(S ∩ T) = 11

We know that:

n (S ∪ T) = n (S) + n (T) – n (S ∩ T)

∴ n (S ∪ T) = 21 + 32 – 11 = 42

Thus, the set (S ∪ T) has 42 elements.

Question 5:

If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements and X ∩Y has

10 elements, how many elements does Y have?

Answer

It is given that:

n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10

We know that:

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

∴ 60 = 40 + n(Y) – 10

∴ n(Y) = 60 – (40 – 10) = 30

Thus, the set Y has 30 elements.

Question 6:

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In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of

the two drinks. How many people like both coffee and tea?

Answer

Let C denote the set of people who like coffee, and

T denote the set of people who like tea

n(C ∪ T) = 70, n(C) = 37, n(T) = 52

We know that:

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

∴ 70 = 37 + 52 – n(C ∩ T)

⇒ 70 = 89 – n(C ∩ T)

⇒ n(C ∩ T) = 89 – 70 = 19

Thus, 19 people like both coffee and tea.

Question 7:

In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like

tennis only and not cricket? How many like tennis?

Answer

Let C denote the set of people who like cricket, and

T denote the set of people who like tennis

∴ n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10

We know that:

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

∴ 65 = 40 + n(T) – 10

⇒ 65 = 30 + n(T)

⇒ n(T) = 65 – 30 = 35

Therefore, 35 people like tennis.

Now,

(T – C) ∪ (T ∩ C) = T

Also,

(T – C) ∩ (T ∩ C) = Φ

∴ n (T) = n (T – C) + n (T ∩ C)

⇒ 35 = n (T – C) + 10

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⇒ n (T – C) = 35 – 10 = 25

Thus, 25 people like only tennis.

Question 8:

In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish

and French. How many speak at least one of these two languages?

Answer

Let F be the set of people in the committee who speak French, and

S be the set of people in the committee who speak Spanish

∴ n(F) = 50, n(S) = 20, n(S ∩ F) = 10

We know that:

n(S ∪ F) = n(S) + n(F) – n(S ∩ F)

= 20 + 50 – 10

= 70 – 10 = 60

Thus, 60 people in the committee speak at least one of the two languages.

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Question 1:

Decide, among the following sets, which sets are subsets of one and another:

A = {x: x ∈ R and x satisfy x2 – 8x + 12 = 0},

B = {2, 4, 6}, C = {2, 4, 6, 8…}, D = {6}.

Answer

A = {x: x ∈ R and x satisfies x2 – 8x + 12 = 0}

2 and 6 are the only solutions of x2 – 8x + 12 = 0.

∴ A = {2, 6}

B = {2, 4, 6}, C = {2, 4, 6, 8 …}, D = {6}

∴ D ⊂ A ⊂ B ⊂ C

Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C

Question 2:

In each of the following, determine whether the statement is true or false. If it is true,

prove it. If it is false, give an example.

(i) If x ∈ A and A ∈ B, then x ∈ B

(ii) If A ⊂ B and B ∈ C, then A ∈ C

(iii) If A ⊂ B and B ⊂ C, then A ⊂ C

(iv) If A ⊄ B and B ⊄ C, then A ⊄ C

(v) If x ∈ A and A ⊄ B, then x ∈ B

(vi) If A ⊂ B and x ∉ B, then x ∉ A

Answer

(i) False

Let A = {1, 2} and B = {1, {1, 2}, {3}}

Now,

∴ A ∈ B

However,

(ii) False

Let

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As A ⊂ B

B ∈ C

However,

(iii) True

Let A ⊂ B and B ⊂ C.

Let x ∈ A

∴ A ⊂ C

(iv) False

Let

Accordingly, and .

However, A ⊂ C

(v) False

Let A = {3, 5, 7} and B = {3, 4, 6}

Now, 5 ∈ A and A ⊄ B

However, 5 ∉ B

(vi) True

Let A ⊂ B and x ∉ B.

To show: x ∉ A

If possible, suppose x ∈ A.

Then, x ∈ B, which is a contradiction as x ∉ B

∴x ∉ A

Question 3:

Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. show that B = C.

Answer

Let, A, B and C be the sets such that and .

To show: B = C

Let x ∈ B

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Case I

x ∈ A

Also, x ∈ B

∴

∴ x ∈ A and x ∈ C

∴ x ∈ C

∴ B ⊂ C

Similarly, we can show that C ⊂ B.

∴ B = C

Question 4:

Show that the following four conditions are equivalent:

(i) A ⊂ B (ii) A – B = Φ

(iii) A ∪ B = B (iv) A ∩ B = A

Answer

First, we have to show that (i) ⇔ (ii).

Let A ⊂ B

To show: A – B ≠ Φ

If possible, suppose A – B ≠ Φ

This means that there exists x ∈ A, x ≠ B, which is not possible as A ⊂ B.

∴ A – B = Φ

∴ A ⊂ B ⇒ A – B = Φ

Let A – B = Φ

To show: A ⊂ B

Let x ∈ A

Clearly, x ∈ B because if x ∉ B, then A – B ≠ Φ

∴ A – B = Φ ⇒ A ⊂ B

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∴ (i) ⇔ (ii)

Let A ⊂ B

To show:

Clearly,

Let

Case I: x ∈ A

∴

Case II: x ∈ B

Then,

Conversely, let

Let x ∈ A

∴ A ⊂ B

Hence, (i) ⇔ (iii)

Now, we have to show that (i) ⇔ (iv).

Let A ⊂ B

Clearly

Let x ∈ A

We have to show that

As A ⊂ B, x ∈ B

∴

∴

Hence, A = A ∩ B

Conversely, suppose A ∩ B = A

Let x ∈ A

⇒

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⇒ x ∈ A and x ∈ B

⇒ x ∈ B

∴ A ⊂ B

Hence, (i) ⇔ (iv).

Question 5:

Show that if A ⊂ B, then C – B ⊂ C – A.

Answer

Let A ⊂ B

To show: C – B ⊂ C – A

Let x ∈ C – B

⇒ x ∈ C and x ∉ B

⇒ x ∈ C and x ∉ A [A ⊂ B]

⇒ x ∈ C – A

∴ C – B ⊂ C – A

Question 6:

Assume that P (A) = P (B). Show that A = B.

Answer

Let P(A) = P(B)

To show: A = B

Let x ∈ A

A ∈ P(A) = P(B)

∴ x ∈ C, for some C ∈ P(B)

Now, C ⊂ B

∴ x ∈ B

∴ A ⊂ B

Similarly, B ⊂ A

∴ A = B

Question 7:

Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer.

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Answer

False

Let A = {0, 1} and B = {1, 2}

∴ A ∪ B = {0, 1, 2}

P(A) = {Φ, {0}, {1}, {0, 1}}

P(B) = {Φ, {1}, {2}, {1, 2}}

P(A ∪ B) = {Φ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}

P(A) ∪ P(B) = {Φ, {0}, {1}, {0, 1}, {2}, {1, 2}}

∴ P(A) ∪ P(B) ≠ P(A ∪ B)

Question 8:

Show that for any sets A and B,

A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)

Answer

To show: A = (A ∩ B) ∪ (A – B)

Let x ∈ A

We have to show that x ∈ (A ∩ B) ∪ (A – B)

Case I

x ∈ A ∩ B

Then, x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)

Case II

x ∉ A ∩ B

⇒ x ∉ A or x ∉ B

∴ x ∉ B [x ∉ A]

∴ x ∉ A – B ⊂ (A ∪ B) ∪ (A – B)

∴ A ⊂ (A ∩ B) ∪ (A – B) … (1)

It is clear that

A ∩ B ⊂ A and (A – B) ⊂ A

∴ (A ∩ B) ∪ (A – B) ⊂ A … (2)

From (1) and (2), we obtain

A = (A ∩ B) ∪ (A – B)

To prove: A ∪ (B – A) ⊂ A ∪ B

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Let x ∈ A ∪ (B – A)

⇒ x ∈ A or x ∈ (B – A)

⇒ x ∈ A or (x ∈ B and x ∉ A)

⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∉ A)

⇒ x ∈ (A ∪ B)

∴ A ∪ (B – A) ⊂ (A ∪ B) … (3)

Next, we show that (A ∪ B) ⊂ A ∪ (B – A).

Let y ∈ A ∪ B

⇒ y ∈ A or y ∈ B

⇒ (y ∈ A or y ∈ B) and (y ∈ A or y ∉ A)

⇒ y ∈ A or (y ∈ B and y ∉ A)

⇒ y ∈ A ∪ (B – A)

∴ A ∪ B ⊂ A ∪ (B – A) … (4)

Hence, from (3) and (4), we obtain A ∪ (B – A) = A ∪B.

Question 9:

Using properties of sets show that

(i) A ∪ (A ∩ B) = A (ii) A ∩ (A ∪ B) = A.

Answer

(i) To show: A ∪ (A ∩ B) = A

We know that

A ⊂ A

A ∩ B ⊂ A

∴ A ∪ (A ∩ B) ⊂ A … (1)

Also, A ⊂ A ∪ (A ∩ B) … (2)

∴ From (1) and (2), A ∪ (A ∩ B) = A

(ii) To show: A ∩ (A ∪ B) = A

A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)

= A ∪ (A ∩ B)

= A {from (1)}

Question 10:

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Show that A ∩ B = A ∩ C need not imply B = C.

Answer

Let A = {0, 1}, B = {0, 2, 3}, and C = {0, 4, 5}

Accordingly, A ∩ B = {0} and A ∩ C = {0}

Here, A ∩ B = A ∩ C = {0}

However, B ≠ C [2 ∈ B and 2 ∉ C]

Question 11:

Let A and B be sets. If A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X, show that A

= B.

(Hints A = A ∩ (A ∪ X), B = B ∩ (B ∪ X) and use distributive law)

Answer

Let A and B be two sets such that A ∩ X = B ∩ X = f and A ∪ X = B ∪ X for some set X.

To show: A = B

It can be seen that

A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ∪ X = B ∪ X]

= (A ∩ B) ∪ (A ∩ X) [Distributive law]

= (A ∩ B) ∪ Φ [A ∩ X = Φ]

= A ∩ B … (1)

Now, B = B ∩ (B ∪ X)

= B ∩ (A ∪ X) [A ∪ X = B ∪ X]

= (B ∩ A) ∪ (B ∩ X) [Distributive law]

= (B ∩ A) ∪ Φ [B ∩ X = Φ]

= B ∩ A

= A ∩ B … (2)

Hence, from (1) and (2), we obtain A = B.

Question 12:

Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C

= Φ.

Answer

Let A = {0, 1}, B = {1, 2}, and C = {2, 0}.

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Accordingly, A ∩ B = {1}, B ∩ C = {2}, and A ∩ C = {0}.

∴ A ∩ B, B ∩ C, and A ∩ C are non-empty.

However, A ∩ B ∩ C = Φ

Question 13:

In a survey of 600 students in a school, 150 students were found to be taking tea and

225 taking coffee, 100 were taking both tea and coffee. Find how many students were

taking neither tea nor coffee?

Answer

Let U be the set of all students who took part in the survey.

Let T be the set of students taking tea.

Let C be the set of students taking coffee.

Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100

To find: Number of student taking neither tea nor coffee i.e., we have to find n(T' ∩ C').

n(T' ∩ C') = n(T ∪ C)'

= n(U) – n(T ∪ C)

= n(U) – [n(T) + n(C) – n(T ∩ C)]

= 600 – [150 + 225 – 100]

= 600 – 275

= 325

Hence, 325 students were taking neither tea nor coffee.

Question 14:

In a group of students 100 students know Hindi, 50 know English and 25 know both.

Each of the students knows either Hindi or English. How many students are there in the

group?

Answer

Let U be the set of all students in the group.

Let E be the set of all students who know English.

Let H be the set of all students who know Hindi.

∴ H ∪ E = U

Accordingly, n(H) = 100 and n(E) = 50

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= 25

n(U) = n(H) + – n(H ∩ E)

= 100 + 50 – 25

= 125

Hence, there are 125 students in the group.

Question 15:

In a survey of 60 people, it was found that 25 people read newspaper H, 26 read

newspaper T, 26 read newspaper I, 9 read both H and I,11 read both H and T, 8 read

both T and I, 3 read all three newspapers. Find:

(i) the number of people who read at least one of the newspapers.

(ii) the number of people who read exactly one newspaper.

Answer

Let A be the set of people who read newspaper H.

Let B be the set of people who read newspaper T.

Let C be the set of people who read newspaper I.

Accordingly, n(A) = 25, n(B) = 26, and n(C) = 26

n(A ∩ C) = 9, n(A ∩ B) = 11, and n(B ∩ C) = 8

n(A ∩ B ∩ C) = 3

Let U be the set of people who took part in the survey.

(i) Accordingly,

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)

= 25 + 26 + 26 – 11 – 8 – 9 + 3

= 52

Hence, 52 people read at least one of the newspapers.

(ii) Let a be the number of people who read newspapers H and T only.

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Let b denote the number of people who read newspapers I and H only.

Let c denote the number of people who read newspapers T and I only.

Let d denote the number of people who read all three newspapers.

Accordingly, d = n(A ∩ B ∩ C) = 3

Now, n(A ∩ B) = a + d

n(B ∩ C) = c + d

n(C ∩ A) = b + d

∴ a + d + c + d + b + d = 11 + 8 + 9 = 28

⇒ a + b + c + d = 28 – 2d = 28 – 6 = 22

Hence, (52 – 22) = 30 people read exactly one newspaper.

Question 16:

In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked

product C. If 14 people liked products A and B, 12 people liked products C and A, 14

people liked products B and C and 8 liked all the three products. Find how many liked

product C only.

Answer

Let A, B, and C be the set of people who like product A, product B, and product C

respectively.

Accordingly, n(A) = 21, n(B) = 26, n(C) = 29, n(A ∩ B) = 14, n(C ∩ A) = 12,

n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8

The Venn diagram for the given problem can be drawn as

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It can be seen that number of people who like product C only is

{29 – (4 + 8 + 6)} = 11

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Exercise 2.1

Question 1:

If , find the values of x and y.

Answer

It is given that .

Since the ordered pairs are equal, the corresponding elements will also be equal.

Therefore, and .

∴ x = 2 and y = 1

Question 2:

If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements

in (A × B)?

Answer

It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.

⇒ Number of elements in set B = 3

Number of elements in (A × B)

= (Number of elements in A) × (Number of elements in B)

= 3 × 3 = 9

Thus, the number of elements in (A × B) is 9.

Question 3:

If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

Answer

G = {7, 8} and H = {5, 4, 2}

We know that the Cartesian product P × Q of two non-empty sets P and Q is defined as

P × Q = {(p, q): p∈ P, q ∈ Q}

∴G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

Question 4:

State whether each of the following statement are true or false. If the statement is false,

rewrite the given statement correctly.

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(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y)

such that x ∈ A and y ∈ B.

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Φ) = Φ.

Answer

(i) False

If P = {m, n} and Q = {n, m}, then

P × Q = {(m, m), (m, n), (n, m), (n, n)}

(ii) True

(iii) True

Question 5:

If A = {–1, 1}, find A × A × A.

Answer

It is known that for any non-empty set A, A × A × A is defined as

A × A × A = {(a, b, c): a, b, c ∈ A}

It is given that A = {–1, 1}

∴ A × A × A = {(–1, –1, –1), (–1, –1, 1), (–1, 1, –1), (–1, 1, 1),

(1, –1, –1), (1, –1, 1), (1, 1, –1), (1, 1, 1)}

Question 6:

If A × B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.

Answer

It is given that A × B = {(a, x), (a, y), (b, x), (b, y)}

We know that the Cartesian product of two non-empty sets P and Q is defined as P × Q

= {(p, q): p ∈ P, q ∈ Q}

∴ A is the set of all first elements and B is the set of all second elements.

Thus, A = {a, b} and B = {x, y}

Question 7:

Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that (i) A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) A × C is a subset of B × D

Answer (i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

∴L.H.S. = A × (B ∩ C) = A × Φ = Φ

A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

∴ R.H.S. = (A × B) ∩ (A × C) = Φ

∴L.H.S. = R.H.S

Hence, A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To verify: A × C is a subset of B × D

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7),

(3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

We can observe that all the elements of set A × C are the elements of set B × D.

Therefore, A × C is a subset of B × D.

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Question 8:

Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List

them.

Answer

A = {1, 2} and B = {3, 4}

∴A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

⇒ n(A × B) = 4

We know that if C is a set with n(C) = m, then n[P(C)] = 2m.

Therefore, the set A × B has 24 = 16 subsets. These are

Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)},

{(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)},

{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)},

{(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

Question 9:

Let A and B be two sets such that n(A) = 3 and n (B) = 2. If (x, 1), (y, 2), (z, 1) are in A

× B, find A and B, where x, y and z are distinct elements.

Answer

It is given that n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.

We know that A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B.

∴ x, y, and z are the elements of A; and 1 and 2 are the elements of B.

Since n(A) = 3 and n(B) = 2, it is clear that A = {x, y, z} and B = {1, 2}.

Question 10:

The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1).

Find the set A and the remaining elements of A × A.

Answer

We know that if n(A) = p and n(B) = q, then n(A × B) = pq.

∴ n(A × A) = n(A) × n(A)

It is given that n(A × A) = 9

∴ n(A) × n(A) = 9

⇒ n(A) = 3

The ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.

We know that A × A = {(a, a): a ∈ A}. Therefore, –1, 0, and 1 are elements of A.

Since n(A) = 3, it is clear that A = {–1, 0, 1}.

The remaining elements of set A × A are (–1, –1), (–1, 1), (0, –1), (0, 0),

(1, –1), (1, 0), and (1, 1)

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Exercise 2.2

Question 1:

Let A = {1, 2, 3, … , 14}. Define a relation R from A to A by R = {(x, y): 3x – y = 0,

where x, y ∈ A}. Write down its domain, codomain and range.

Answer

The relation R from A to A is given as

R = {(x, y): 3x – y = 0, where x, y ∈ A}

i.e., R = {(x, y): 3x = y, where x, y ∈ A}

∴R = {(1, 3), (2, 6), (3, 9), (4, 12)}

The domain of R is the set of all first elements of the ordered pairs in the relation.

∴Domain of R = {1, 2, 3, 4}

The whole set A is the codomainof the relation R.

∴Codomain of R = A = {1, 2, 3, …, 14}

The range of R is the set of all second elements of the ordered pairs in the relation.

∴Range of R = {3, 6, 9, 12}

Question 2:

Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a

natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write

down the domain and the range.

Answer

R = {(x, y): y = x + 5, x is a natural number less than 4, x, y ∈ N}

The natural numbers less than 4 are 1, 2, and 3.

∴R = {(1, 6), (2, 7), (3, 8)}

The domain of R is the set of all first elements of the ordered pairs in the relation.

∴ Domain of R = {1, 2, 3}

The range of R is the set of all second elements of the ordered pairs in the relation.

∴ Range of R = {6, 7, 8}

Question 3:

A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the

difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

Answer

A = {1, 2, 3, 5} and B = {4, 6, 9}

R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}

∴R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

Question 4:

The given figure shows a relationship between the sets P and Q. write this relation

(i) in set-builder form (ii) in roster form.

What is its domain and range?

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Answer

According to the given figure, P = {5, 6, 7}, Q = {3, 4, 5}

(i) R = {(x, y): y = x – 2; x ∈ P} or R = {(x, y): y = x – 2 for x = 5, 6, 7}

(ii) R = {(5, 3), (6, 4), (7, 5)}

Domain of R = {5, 6, 7}

Range of R = {3, 4, 5}

Question 5:

Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by

{(a, b): a, b ∈ A, b is exactly divisible by a}.

(i) Write R in roster form

(ii) Find the domain of R

(iii) Find the range of R.

Answer

A = {1, 2, 3, 4, 6}, R = {(a, b): a, b ∈ A, b is exactly divisible by a}

(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4),

(6, 6)}

(ii) Domain of R = {1, 2, 3, 4, 6}

(iii) Range of R = {1, 2, 3, 4, 6}

Question 6:

Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1,

2, 3, 4, 5}}.

Answer

R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}

∴ R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

∴Domain of R = {0, 1, 2, 3, 4, 5}

Range of R = {5, 6, 7, 8, 9, 10}

Question 7:

Write the relation R = {(x, x3): x is a prime number less than 10} in roster form.

Answer

R = {(x, x3): x is a prime number less than 10}

The prime numbers less than 10 are 2, 3, 5, and 7.

∴R = {(2, 8), (3, 27), (5, 125), (7, 343)}

Question 8:

Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Answer

It is given that A = {x, y, z} and B = {1, 2}.

∴ A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}

Since n(A × B) = 6, the number of subsets of A × B is 26.

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Therefore, the number of relations from A to B is 26.

Question 9:

Let R be the relation on Z defined by R = {(a, b): a, b ∈ Z, a – b is an integer}. Find the

domain and range of R.

Answer

R = {(a, b): a, b ∈ Z, a – b is an integer}

It is known that the difference between any two integers is always an integer.

∴Domain of R = Z

Range of R = Z

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Exercise 2.3

Question 1:

Which of the following relations are functions? Give reasons. If it is a function, determine

its domain and range.

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

(iii) {(1, 3), (1, 5), (2, 5)}

Answer

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having

their unique images, this relation is a function.

Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation

having their unique images, this relation is a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1, 3), (1, 5), (2, 5)}

Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5,

this relation is not a function.

Question 2:

Find the domain and range of the following real function:

(i) f(x) = –|x| (ii)

Answer

(i) f(x) = –|x|, x ∈ R

We know that |x| =

Since f(x) is defined for x ∈ R, the domain of f is R.

It can be observed that the range of f(x) = –|x| is all real numbers except positive real

numbers.

∴The range of f is (– , 0].

(ii)

Since is defined for all real numbers that are greater than or equal to –3 and less

than or equal to 3, the domain of f(x) is {x : –3 ≤ x ≤ 3} or [–3, 3].

For any value of x such that –3 ≤ x ≤ 3, the value of f(x) will lie between 0 and 3.

∴The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].

Question 3:

A function f is defined by f(x) = 2x – 5. Write down the values of

(i) f(0), (ii) f(7), (iii) f(–3)

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Answer

The given function is f(x) = 2x – 5.

Therefore,

(i) f(0) = 2 × 0 – 5 = 0 – 5 = –5

(ii) f(7) = 2 × 7 – 5 = 14 – 5 = 9

(iii) f(–3) = 2 × (–3) – 5 = – 6 – 5 = –11

Question 4:

The function ‘t’ which maps temperature in degree Celsius into temperature in degree

Fahrenheit is defined by .

Find (i) t (0) (ii) t (28) (iii) t (–10) (iv) The value of C, when t(C) = 212

Answer

The given function is .

Therefore,

(i)

(ii)

(iii)

(iv) It is given that t(C) = 212

Thus, the value of t, when t(C) = 212, is 100.

Question 5:

Find the range of each of the following functions.

(i) f(x) = 2 – 3x, x ∈ R, x > 0.

(ii) f(x) = x2 + 2, x, is a real number.

(iii) f(x) = x, x is a real number

Answer

(i) f(x) = 2 – 3x, x ∈ R, x > 0

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The values of f(x) for various values of real numbers x > 0 can be written in the tabular

form as

x 0.01 0.1 0.9 1 2 2.5 4 5 …

f(x) 1.97 1.7 –0.7 –1 –4 –5.5 –10 –13 …

Thus, it can be clearly observed that the range of f is the set of all real numbers less

than 2.

i.e., range of f = (– , 2)

Alter:

Let x > 0

⇒ 3x > 0

⇒ 2 –3x < 2

⇒ f(x) < 2

∴Range of f = (– , 2)

(ii) f(x) = x2 + 2, x, is a real number

The values of f(x) for various values of real numbers x can be written in the tabular form

as

x 0 ±0.3 ±0.8 ±1 ±2 ±3 …

f(x) 2 2.09 2.64 3 6 11 …..

Thus, it can be clearly observed that the range of f is the set of all real numbers greater

than 2.

i.e., range of f = [2, )

Alter:

Let x be any real number.

Accordingly,

x2 ≥ 0

⇒ x2 + 2 ≥ 0 + 2

⇒ x2 + 2 ≥ 2

⇒ f(x) ≥ 2

∴ Range of f = [2, )

(iii) f(x) = x, x is a real number

It is clear that the range of f is the set of all real numbers.

∴ Range of f = R

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Question 1:

The relation f is defined by

The relation g is defined by

Show that f is a function and g is not a function.

Answer

The relation f is defined as

It is observed that for

0 ≤ x < 3, f(x) = x2

3 < x ≤ 10, f(x) = 3x

Also, at x = 3, f(x) = 32 = 9 or f(x) = 3 × 3 = 9

i.e., at x = 3, f(x) = 9

Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique.

Thus, the given relation is a function.

The relation g is defined as

It can be observed that for x = 2, g(x) = 22 = 4 and g(x) = 3 × 2 = 6

Hence, element 2 of the domain of the relation g corresponds to two different images

i.e., 4 and 6. Hence, this relation is not a function.

Question 2:

If f(x) = x2, find .

Answer

Question 3:

Find the domain of the function

Answer


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Mathematics______________________________

________________________________________________________________________________________Page 10 of 14

It can be seen that function f is defined for all real numbers except at x = 6 and x = 2.

Hence, the domain of f is R – {2, 6}.

Question 4:

Find the domain and the range of the real function f defined by .

Answer

The given real function is .

It can be seen that is defined for (x – 1) ≥ 0.

i.e., is defined for x ≥ 1.

Therefore, the domain of f is the set of all real numbers greater than or equal to 1 i.e.,

the domain of f = [1, ).

As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒

Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e., the

range of f = [0, ).

Question 5:

Find the domain and the range of the real function f defined by f (x) = |x – 1|.

Answer

The given real function is f (x) = |x – 1|.

It is clear that |x – 1| is defined for all real numbers.

∴Domain of f = R

Also, for x ∈ R, |x – 1| assumes all real numbers.

Hence, the range of f is the set of all non-negative real numbers.

Question 6:

Let be a function from R into R. Determine the range of f.

Answer

The range of f is the set of all second elements. It can be observed that all these

elements are greater than or equal to 0 but less than 1.

[Denominator is greater numerator]

Thus, range of f = [0, 1)

Question 7:

Class XI______________________________


Mathematics______________________________

________________________________________________________________________________________Page 11 of 14

Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g

and .

Answer f, g: R → R is defined as f(x) = x + 1, g(x) = 2x – 3

(f + g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2

∴(f + g) (x) = 3x – 2

(f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4

∴ (f – g) (x) = –x + 4

Question 8:

Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax

+ b, for some integers a, b. Determine a, b.

Answer

f = {(1, 1), (2, 3), (0, –1), (–1, –3)}

f(x) = ax + b

(1, 1) ∈ f

⇒ f(1) = 1

⇒ a × 1 + b = 1

⇒ a + b = 1

(0, –1) ∈ f

⇒ f(0) = –1

⇒ a × 0 + b = –1

⇒ b = –1

On substituting b = –1 in a + b = 1, we obtain a + (–1) = 1 ⇒ a = 1 + 1 = 2.

Thus, the respective values of a and b are 2 and –1.

Question 9:

Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b2}. Are the

following true?

(i) (a, a) ∈ R, for all a ∈ N (ii) (a, b) ∈ R, implies (b, a) ∈ R

(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.

Justify your answer in each case.

Answer

R = {(a, b): a, b ∈ N and a = b2}

(i) It can be seen that 2 ∈ N;however, 2 ≠ 22 = 4.

Therefore, the statement “(a, a) ∈ R, for all a ∈ N” is not true.

(ii) It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 32.

Now, 3 ≠ 92 = 81; therefore, (3, 9) ∉ N

Class XI______________________________


Mathematics______________________________

________________________________________________________________________________________Page 12 of 14

Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.

(iii) It can be seen that (9, 3) ∈ R, (16, 4) ∈ R because 9, 3, 16, 4 ∈ N and 9 = 32 and 16

= 42.

Now, 9 ≠ 42 = 16; therefore, (9, 4) ∉ N

Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.

Question 10:

Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2,

11)}. Are the following true?

(i) f is a relation from A to B (ii) f is a function from A to B.

Justify your answer in each case.

Answer

A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

∴A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11),

(2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4,

11), (4, 15), (4, 16)}

It is given that f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian

product A × B.

It is observed that f is a subset of A × B.

Thus, f is a relation from A to B.

(ii) Since the same first element i.e., 2 corresponds to two different images i.e., 9 and

11, relation f is not a function.

Question 11:

Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z

to Z: justify your answer.

Answer

The relation f is defined as f = {(ab, a + b): a, b ∈ Z}

We know that a relation f from a set A to a set B is said to be a function if every element

of set A has unique images in set B.

Since 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f

i.e., (12, 8), (12, –8) ∈ f

It can be seen that the same first element i.e., 12 corresponds to two different images

i.e., 8 and –8. Thus, relation f is not a function.

Question 12:

Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime

factor of n. Find the range of f.

Answer

A = {9, 10, 11, 12, 13} f: A → N is defined as

f(n) = The highest prime factor of n

Prime factor of 9 = 3

Prime factors of 10 = 2, 5


Prime factors of 12 = 2, 3


∴f(9) = The highest prime factor of 9 = 3

Class XI______________________________


Mathematics______________________________

________________________________________________________________________________________Page 13 of 14

f(10) = The highest prime factor of 10 = 5




The range of f is the set of all f(n), where n ∈ A.

∴Range of f = {3, 5, 11, 13}

Class XI______________________________


Mathematics______________________________

________________________________________________________________________________________Page 14 of 14

Class XI Chapter 3 – Trigonometric Functions Maths

Exercise 3.1

Question 1:

Find the radian measures corresponding to the following degree measures:

(i) 25° (ii) – 47° 30' (iii) 240° (iv) 520°

Answer

(i) 25°

We know that 180° = π radian

(ii) –47° 30'

–47° 30' = degree [1° = 60']

degree

Since 180° = π radian

(iii) 240°


(iv) 520°


Question 2:

Find the degree measures corresponding to the following radian measures

.


Page 1 of 44


Page 2 of 44

(i) (ii) – 4 (iii) (iv)

Answer

(i)

We know that π radian = 180°

(ii) – 4


(iii)



Page 3 of 44

(iv)


Question 3:

A wheel makes 360 revolutions in one minute. Through how many radians does it turn in

one second?

Answer

Number of revolutions made by the wheel in 1 minute = 360

∴Number of revolutions made by the wheel in 1 second =

In one complete revolution, the wheel turns an angle of 2π radian.

Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e.,

12 π radian

Thus, in one second, the wheel turns an angle of 12π radian.

Question 4:

Find the degree measure of the angle subtended at the centre of a circle of radius 100

cm by an arc of length 22 cm .

Answer

We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ

radian at the centre, then

Therefore, forr = 100 cm, l = 22 cm, we have

Thus, the required angle is 12°36′.


Page 4 of 44

Question 5:

In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor

arc of the chord.

Answer

Diameter of the circle = 40 cm

∴Radius (r) of the circle =

Let AB be a chord (length = 20 cm) of the circle.

In ∆OAB, OA = OB = Radius of circle = 20 cm

Also, AB = 20 cm

Thus, ∆OAB is an equilateral triangle.

∴θ = 60° =


radian at the centre, then .

Thus, the length of the minor arc of the chord is .

Question 6:

If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find

the ratio of their radii.

Answer


Page 5 of 44

Let the radii of the two circles be and . Let an arc of length l subtend an angle of 60°

at the centre of the circle of radius r1, while let an arc of length l subtend an angle of 75°

at the centre of the circle of radius r2.

Now, 60° = and 75° =



Thus, the ratio of the radii is 5:4.

Question 7:

Find the angle in radian though which a pendulum swings if its length is 75 cm and the

tip describes an arc of length

(i) 10 cm (ii) 15 cm (iii) 21 cm

Answer



It is given that r = 75 cm

(i) Here, l = 10 cm

(ii) Here, l = 15 cm


Page 6 of 44

(iii) Here, l = 21 cm


Page 7 of 44

Exercise 3.2

Question 1:

Find the values of other five trigonometric functions if , x lies in third

quadrant.

Answer

Since x lies in the 3rd quadrant, the value of sin x will be negative.


Page 8 of 44

Question 2:

Find the values of other five trigonometric functions if , x lies in second

quadrant.

Answer

Since x lies in the 2nd quadrant, the value of cos x will be negative


Page 9 of 44

Question 3:

Find the values of other five trigonometric functions if , x lies in third quadrant.

Answer

Since x lies in the 3rd quadrant, the value of sec x will be negative.


Page 10 of 44

Question 4:

Find the values of other five trigonometric functions if , x lies in fourth

quadrant.

Answer

Since x lies in the 4th quadrant, the value of sin x will be negative.


Page 11 of 44

Question 5:

Find the values of other five trigonometric functions if , x lies in second

quadrant.

Answer


Page 12 of 44

Since x lies in the 2nd quadrant, the value of sec x will be negative.

∴sec x =

Question 6:

Find the value of the trigonometric function sin 765°

Answer

It is known that the values of sin x repeat after an interval of 2π or 360°.

Question 7:

Find the value of the trigonometric function cosec (–1410°)

Answer

It is known that the values of cosec x repeat after an interval of 2π or 360°.

Question 8:


Page 13 of 44

Find the value of the trigonometric function

Answer

It is known that the values of tan x repeat after an interval of π or 180°.

Question 9:


Answer

It is known that the values of sin x repeat after an interval of 2π or 360°.

Question 10:


Answer

It is known that the values of cot x repeat after an interval of π or 180°.


Page 14 of 44

Exercise 3.3

Question 1:

Answer

L.H.S. =

Question 2:

Prove that


Page 15 of 44

Answer

L.H.S. =

Question 3:

Prove that

Answer

L.H.S. =

Question 4:

Prove that

Answer

L.H.S =


Page 16 of 44

Question 5:

Find the value of:

(i) sin 75°

(ii) tan 15°

Answer

(i) sin 75° = sin (45° + 30°)

= sin 45° cos 30° + cos 45° sin 30°

[sin (x + y) = sin x cos y + cos x sin y]

(ii) tan 15° = tan (45° – 30°)


Page 17 of 44

Question 6:

Prove that:

Answer


Page 18 of 44

Question 7:

Prove that:

Answer

It is known that

�L.H.S. =

Question 8:

Prove that

Answer

Question 9:


Page 19 of 44

Answer

L.H.S. =

Question 10:

Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x

Answer

L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x

Question 11:

Prove that

Answer


Page 20 of 44

It is known that .

∴L.H.S. =

Question 12:

Prove that sin2 6x – sin2 4x = sin 2x sin 10x

Answer

It is known that

∴L.H.S. = sin26x – sin24x

= (sin 6x + sin 4x) (sin 6x – sin 4x)

= (2 sin 5x cos x) (2 cos 5x sin x)

= (2 sin 5x cos 5x) (2 sin x cos x)

= sin 10x sin 2x


Page 21 of 44

Website: www.vidhyarjan.com Email: [email protected] Mobile: 9999 249717

Head Office: 1/3-H-A-2, Street # 6, East Azad Nagar, Delhi-110051

(One Km from ‘Welcome’ Metro Station)

= R.H.S.

Question 13:

Prove that cos2 2x – cos2 6x = sin 4x sin 8x

Answer

It is known that

∴L.H.S. = cos2 2x – cos2 6x

= (cos 2x + cos 6x) (cos 2x – 6x)

= [2 cos 4x cos 2x] [–2 sin 4x (–sin 2x)]

= (2 sin 4x cos 4x) (2 sin 2x cos 2x)

= sin 8x sin 4x

= R.H.S.

Question 14:

Prove that sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x

Answer

L.H.S. = sin 2x + 2 sin 4x + sin 6x

= [sin 2x + sin 6x] + 2 sin 4x

= 2 sin 4x cos (– 2x) + 2 sin 4x

= 2 sin 4x cos 2x + 2 sin 4x

= 2 sin 4x (cos 2x + 1)

= 2 sin 4x (2 cos2 x – 1 + 1)

= 2 sin 4x (2 cos2 x)


Page 22 of 44

= 4cos2 x sin 4x

= R.H.S.

Question 15:

Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Answer

L.H.S = cot 4x (sin 5x + sin 3x)

= 2 cos 4x cos x

R.H.S. = cot x (sin 5x – sin 3x)

= 2 cos 4x. cos x

L.H.S. = R.H.S.

Question 16:

Prove that

Answer

It is known that


Page 23 of 44

∴L.H.S =

Question 17:

Prove that

Answer

It is known that

∴L.H.S. =

Question 18:

Prove that


Page 24 of 44

Answer

It is known that

∴L.H.S. =

Question 19:

Prove that

Answer

It is known that

∴L.H.S. =


Page 25 of 44

Question 20:

Prove that

Answer

It is known that

∴L.H.S. =

Question 21:

Prove that

Answer

L.H.S. =


Page 26 of 44

Question 22:

Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Answer

L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – cot (2x + x) (cot 2x + cot x)

= cot x cot 2x – (cot 2x cot x – 1)

= 1 = R.H.S.

Question 23:

Prove that

Answer

It is known that .


Page 27 of 44

∴L.H.S. = tan 4x = tan 2(2x)

Question 24:

Prove that cos 4x = 1 – 8sin2 x cos2 x

Answer

L.H.S. = cos 4x

= cos 2(2x)

= 1 – 2 sin2 2x [cos 2A = 1 – 2 sin2 A]

= 1 – 2(2 sin x cos x)2 [sin2A = 2sin A cosA]

= 1 – 8 sin2x cos2x

= R.H.S.


Page 28 of 44

Question 25:

Prove that: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1

Answer

L.H.S. = cos 6x

= cos 3(2x)

= 4 cos3 2x – 3 cos 2x [cos 3A = 4 cos3 A – 3 cos A]

= 4 [(2 cos2 x – 1)3 – 3 (2 cos2 x – 1) [cos 2x = 2 cos2 x – 1]

= 4 [(2 cos2 x)3 – (1)3 – 3 (2 cos2 x)2 + 3 (2 cos2 x)] – 6cos2 x + 3

= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3

= 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3

= 32 cos6x – 48 cos4x + 18 cos2x – 1

= R.H.S.


Page 29 of 44

Exercise 3.4

Question 1:

Find the principal and general solutions of the equation

Answer

Therefore, the principal solutions are x = and .

Therefore, the general solution is

Question 2:


Answer



Page 30 of 44

Therefore, the general solution is , where n ∈ Z

Question 3:


Answer



Question 4:

Find the general solution of cosec x = –2

Answer

cosec x = –2


Page 31 of 44

Therefore, the principal solutions are x = .


Question 5:

Find the general solution of the equation

Answer


Page 32 of 44

Question 6:


Answer

Question 7:


Answer

Therefore, the general solution is .


Page 33 of 44

Question 8:


Answer

Therefore, the general solution is .

Question 9:


Answer


Page 34 of 44



Page 35 of 44

NCERT Miscellaneous Solution

Question 1: Prove that:

Answer

L.H.S.

= 0 = R.H.S

Question 2:

Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

Answer

L.H.S.

= (sin 3x + sin x) sin x + (cos 3x – cos x) cos x


Page 36 of 44

= RH.S.

Question 3:

Prove that:

Answer

L.H.S. =

Question 4:

Prove that:

Answer

L.H.S. =


Page 37 of 44

Question 5:

Prove that:

Answer

It is known that .

�L.H.S. =

Question 6:

Prove that:

Answer


Page 38 of 44

It is known that

.

L.H.S. =

= tan 6x

= R.H.S.

Question 7:

Prove that:

Answer

L.H.S. =


Page 39 of 44

Question 8:

, x in quadrant II

Answer

Here, x is in quadrant II.

i.e.,

Therefore, are all positive.


Page 40 of 44

As x is in quadrant II, cosx is negative.

�


Page 41 of 44

Thus, the respective values of are .

Question 9:

Find for , x in quadrant III

Answer

Here, x is in quadrant III.

Therefore, and are negative, whereas is positive.


Page 42 of 44

Now,


Question 10:

Find for , x in quadrant II

Answer

Here, x is in quadrant II.

Therefore, , and are all positive.


Page 43 of 44

[cosx is negative in quadrant II]


Page 44 of 44


Class XI Chapter 4 – Principle of Mathematical Induction Maths

Exercise 4.1

Question 1:

Prove the following by using the principle of mathematical induction for all n ∈ N:

Answer

Let the given statement be P(n), i.e.,

P(n): 1 + 3 + 32 + …+ 3n–1 =

For n = 1, we have

P(1): 1 = , which is true.

Let P(k) be true for some positive integer k, i.e.,

We shall now prove that P(k + 1) is true.

Consider

1 + 3 + 32 + … + 3k–1 + 3(k+1) – 1

= (1 + 3 + 32 +… + 3k–1) + 3k


Page 1 of 27


Page 2 of 27

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural

numbers i.e., n.

Question 2:


Answer


P(n):

For n = 1, we have

P(1): 13 = 1 = , which is true.



Consider

13 + 23 + 33 + … + k3 + (k + 1)3


Page 3 of 27

= (13 + 23 + 33 + …. + k3) + (k + 1)3



numbers i.e., n.

Question 3:


Answer


P(n):

For n = 1, we have

P(1): 1 = which is true.



Page 4 of 27


Consider

Question 4:

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2.3

+ 2.3.4 + … + n(n + 1) (n + 2) =

Answer


P(n): 1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) =

For n = 1, we have

P(1): 1.2.3 = 6 = , which is true.


1.2.3 + 2.3.4 + … + k(k + 1) (k + 2)


Page 5 of 27


Consider

1.2.3 + 2.3.4 + … + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)

= {1.2.3 + 2.3.4 + … + k(k + 1) (k + 2)} + (k + 1) (k + 2) (k + 3)



numbers i.e., n.

Question 5:


Answer


P(n) :

For n = 1, we have

P(1): 1.3 = 3 , which is true.




Page 6 of 27

Consider

1.3 + 2.32 + 3.33 + … + k3k+ (k + 1) 3k+1

= (1.3 + 2.32 + 3.33 + …+ k.3k) + (k + 1) 3k+1



numbers i.e., n.

Question 6:


Answer


P(n):

For n = 1, we have

P(1): , which is true.


Page 7 of 27



Consider

1.2 + 2.3 + 3.4 + … + k.(k + 1) + (k + 1).(k + 2)

= [1.2 + 2.3 + 3.4 + … + k.(k + 1)] + (k + 1).(k + 2)



numbers i.e., n.

Question 7:


Answer


P(n):

For n = 1, we have

, which is true.



Page 8 of 27


Consider

(1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + {2(k + 1) – 1}{2(k + 1) + 1}



numbers i.e., n.


Page 9 of 27

Question 8:

Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 +

2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2

Answer


P(n): 1.2 + 2.22 + 3.22 + … + n.2n = (n – 1) 2n+1 + 2

For n = 1, we have

P(1): 1.2 = 2 = (1 – 1) 21+1 + 2 = 0 + 2 = 2, which is true.


1.2 + 2.22 + 3.22 + … + k.2k = (k – 1) 2k + 1 + 2 … (i)


Consider



numbers i.e., n.

Question 9:


Answer


P(n):


Page 10 of 27

For n = 1, we have

P(1): , which is true.



Consider



numbers i.e., n.

Question 10:


Answer


P(n):

For n = 1, we have


Page 11 of 27

, which is true.



Consider



numbers i.e., n.


Page 12 of 27

Question 11:


Answer


P(n):

For n = 1, we have

, which is true.



Consider


Page 13 of 27



numbers i.e., n.


Page 14 of 27

Question 12:


Answer


For n = 1, we have

, which is true.



Consider



Page 15 of 27


numbers i.e., n.

Question 13:


Answer


For n = 1, we have



Consider



Page 16 of 27


numbers i.e., n.

Question 14:


Answer


For n = 1, we have

, which is true.



Consider



numbers i.e., n.


Page 17 of 27

Question 15:


Answer




Consider


Page 18 of 27



numbers i.e., n.

Question 16:


Answer




Consider


Page 19 of 27



numbers i.e., n.

Question 17:


Answer


For n = 1, we have


Page 20 of 27

, which is true.



Consider



numbers i.e., n.

Question 18:


Page 21 of 27


Answer


It can be noted that P(n) is true for n = 1 since .


We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

Hence,



numbers i.e., n.

Question 19:


Page 22 of 27

Prove the following by using the principle of mathematical induction for all n ∈ N: n (n +

1) (n + 5) is a multiple of 3.

Answer


P(n): n (n + 1) (n + 5), which is a multiple of 3.

It can be noted that P(n) is true for n = 1 since 1 (1 + 1) (1 + 5) = 12, which is a

multiple of 3.


k (k + 1) (k + 5) is a multiple of 3.

∴k (k + 1) (k + 5) = 3m, where m ∈ N … (1)


Consider



numbers i.e., n.

Question 20:

Prove the following by using the principle of mathematical induction for all n ∈ N: 102n – 1

+ 1 is divisible by 11.

Answer



Page 23 of 27

P(n): 102n – 1 + 1 is divisible by 11.

It can be observed that P(n) is true for n = 1 since P(1) = 102.1 – 1 + 1 = 11, which is

divisible by 11.


102k – 1 + 1 is divisible by 11.

∴102k – 1 + 1 = 11m, where m ∈ N … (1)


Consider



numbers i.e., n.

Question 21:

Prove the following by using the principle of mathematical induction for all n ∈ N: x2n –

y2n is divisible by x + y.

Answer


P(n): x2n – y2n is divisible by x + y.

It can be observed that P(n) is true for n = 1.

This is so because x2 × 1 – y2 × 1 = x2 – y2 = (x + y) (x – y) is divisible by (x + y).



Page 24 of 27

x2k – y2k is divisible by x + y.

∴x2k – y2k = m (x + y), where m ∈ N … (1)


Consider



numbers i.e., n.

Question 22:

Prove the following by using the principle of mathematical induction for all n ∈ N: 32n + 2

– 8n – 9 is divisible by 8.

Answer


P(n): 32n + 2 – 8n – 9 is divisible by 8.

It can be observed that P(n) is true for n = 1 since 32 × 1 + 2 – 8 × 1 – 9 = 64, which is

divisible by 8.


32k + 2 – 8k – 9 is divisible by 8.

∴32k + 2 – 8k – 9 = 8m; where m ∈ N … (1)


Consider


Page 25 of 27



numbers i.e., n.

Question 23:

Prove the following by using the principle of mathematical induction for all n ∈ N: 41n –

14n is a multiple of 27.

Answer


P(n):41n – 14nis a multiple of 27.

It can be observed that P(n) is true for n = 1 since , which is a multiple of

27.


41k – 14kis a multiple of 27

∴41k – 14k = 27m, where m ∈ N … (1)


Consider


Page 26 of 27



numbers i.e., n.

Question 24:

Prove the following by using the principle of mathematical induction for all

(2n +7) < (n + 3)2

Answer


P(n): (2n +7) < (n + 3)2

It can be observed that P(n) is true for n = 1 since 2.1 + 7 = 9 < (1 + 3)2 = 16, which is

true.


(2k + 7) < (k + 3)2 … (1)


Consider


Page 27 of 27



numbers i.e., n.

Class XI Chapter 5 – Complex Numbers and Quadratic Equations Maths

Exercise 5.1

Question 1:

Express the given complex number in the form a + ib:

Answer

Question 2:

Express the given complex number in the form a + ib: i9 + i19

Answer

Question 3:

Express the given complex number in the form a + ib: i–39

Answer


Page 1 of 34


Page 2 of 34

Question 4:

Express the given complex number in the form a + ib: 3(7 + i7) + i(7 + i7)

Answer

Question 5:

Express the given complex number in the form a + ib: (1 – i) – (–1 + i6)

Answer

Question 6:


Answer

Question 7:


Answer


Page 3 of 34

Question 8:

Express the given complex number in the form a + ib: (1 – i)4

Answer

Question 9:


Answer


Page 4 of 34

Question 10:


Answer

Question 11:

Find the multiplicative inverse of the complex number 4 – 3i


Page 5 of 34

Answer

Let z = 4 – 3i

Then, = 4 + 3i and

Therefore, the multiplicative inverse of 4 – 3i is given by

Question 12:

Find the multiplicative inverse of the complex number

Answer

Let z =

Therefore, the multiplicative inverse of is given by

Question 13:

Find the multiplicative inverse of the complex number –i

Answer

Let z = –i

Therefore, the multiplicative inverse of –i is given by

Question 14:


Page 6 of 34

Express the following expression in the form of a + ib.

Answer


Page 7 of 34

Exercise 5.2

Question 1:

Find the modulus and the argument of the complex number

Answer

On squaring and adding, we obtain

Since both the values of sin θ and cos θ are negative and sinθ and cosθ are negative in

III quadrant,

Thus, the modulus and argument of the complex number are 2 and

respectively.

Question 2:


Page 8 of 34

Find the modulus and the argument of the complex number

Answer


Thus, the modulus and argument of the complex number are 2 and

respectively.

Question 3:

Convert the given complex number in polar form: 1 – i

Answer

1 – i

Let r cos θ = 1 and r sin θ = –1



Page 9 of 34

This is

the required polar form.

Question 4:

Convert the given complex number in polar form: – 1 + i

Answer

– 1 + i

Let r cos θ = –1 and r sin θ = 1


It can be written,


Page 10 of 34

This is the required polar form.

Question 5:

Convert the given complex number in polar form: – 1 – i

Answer

– 1 – i

Let r cos θ = –1 and r sin θ = –1


This is the

required polar form.

Question 6:

Convert the given complex number in polar form: –3

Answer

–3




Page 11 of 34


Question 7:

Convert the given complex number in polar form:

Answer

Let r cos θ = and r sin θ = 1



Page 12 of 34


Question 8:

Convert the given complex number in polar form: i

Answer

i

Let r cosθ = 0 and r sin θ = 1




Page 13 of 34

Exercise 5.3

Question 1:

Solve the equation x2 + 3 = 0

Answer

The given quadratic equation is x2 + 3 = 0

On comparing the given equation with ax2 + bx + c = 0, we obtain

a = 1, b = 0, and c = 3

Therefore, the discriminant of the given equation is

D = b2 – 4ac = 02 – 4 × 1 × 3 = –12

Therefore, the required solutions are

Question 2:

Solve the equation 2x2 + x + 1 = 0

Answer

The given quadratic equation is 2x2 + x + 1 = 0


a = 2, b = 1, and c = 1


D = b2 – 4ac = 12 – 4 × 2 × 1 = 1 – 8 = –7



Page 14 of 34

Question 3:

Solve the equation x2 + 3x + 9 = 0

Answer

The given quadratic equation is x2 + 3x + 9 = 0


a = 1, b = 3, and c = 9


D = b2 – 4ac = 32 – 4 × 1 × 9 = 9 – 36 = –27


Question 4:

Solve the equation –x2 + x – 2 = 0

Answer

The given quadratic equation is –x2 + x – 2 = 0


a = –1, b = 1, and c = –2


D = b2 – 4ac = 12 – 4 × (–1) × (–2) = 1 – 8 = –7


Question 5:

Solve the equation x2 + 3x + 5 = 0

Answer

The given quadratic equation is x2 + 3x + 5 = 0



Page 15 of 34

a = 1, b = 3, and c = 5


D = b2 – 4ac = 32 – 4 × 1 × 5 =9 – 20 = –11


Question 6:

Solve the equation x2 – x + 2 = 0

Answer

The given quadratic equation is x2 – x + 2 = 0


a = 1, b = –1, and c = 2


D = b2 – 4ac = (–1)2 – 4 × 1 × 2 = 1 – 8 = –7


Question 7:

Solve the equation

Answer

The given quadratic equation is


a = , b = 1, and c =


D = b2 – 4ac = 12 – = 1 – 8 = –7



Page 16 of 34

Question 8:

Solve the equation

Answer



a = , b = , and c =


D = b2 – 4ac =


Question 9:

Solve the equation

Answer


This equation can also be written as

On comparing this equation with ax2 + bx + c = 0, we obtain

a = , b = , and c = 1



Page 17 of 34

Question 10:

Solve the equation

Answer




a = , b = 1, and c =



Page 18 of 34

NCERT Miscellaneous Solutions

Question 1:

Evaluate:

Answer


Page 19 of 34

Question 2:

For any two complex numbers z1 and z2, prove that

Re (z1z2) = Re z1 Re z2 – Im z1 Im z2

Answer


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Question 3:

Reduce to the standard form.

Answer


Page 21 of 34

Question 4:

If x – iy = prove that .

Answer


Page 22 of 34

Question 5:

Convert the following in the polar form:

(i) , (ii)

Answer

(i) Here,



Page 23 of 34


r2 (cos2 θ + sin2 θ) = 1 + 1

⇒ r2 (cos2 θ + sin2 θ) = 2

⇒ r2 = 2 [cos2 θ + sin2 θ = 1]

∴z = r cos θ + i r sin θ


(ii) Here,



r2 (cos2 θ + sin2 θ) = 1 + 1

⇒r2 (cos2 θ + sin2 θ) = 2

⇒ r2 = 2 [cos2 θ + sin2 θ = 1]


Page 24 of 34

∴z = r cos θ + i r sin θ


Question 6:

Solve the equation

Answer




a = 9, b = –12, and c = 20


D = b2 – 4ac = (–12)2 – 4 × 9 × 20 = 144 – 720 = –576


Question 7:

Solve the equation

Answer


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a = 2, b = –4, and c = 3


D = b2 – 4ac = (–4)2 – 4 × 2 × 3 = 16 – 24 = –8


Question 8:

Solve the equation 27x2 – 10x + 1 = 0

Answer

The given quadratic equation is 27x2 – 10x + 1 = 0


a = 27, b = –10, and c = 1


D = b2 – 4ac = (–10)2 – 4 × 27 × 1 = 100 – 108 = –8


Question 9:

Solve the equation 21x2 – 28x + 10 = 0

Answer

The given quadratic equation is 21x2 – 28x + 10 = 0



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a = 21, b = –28, and c = 10


D = b2 – 4ac = (–28)2 – 4 × 21 × 10 = 784 – 840 = –56


Question 10:

If find .

Answer

Question 11:


Page 27 of 34

If a + ib = , prove that a2 + b2 =

Answer

On comparing real and imaginary parts, we obtain

Hence, proved.

Question 12:

Let . Find


Page 28 of 34

(i) , (ii)

Answer

(i)

On multiplying numerator and denominator by (2 – i), we obtain

On comparing real parts, we obtain

(ii)

On comparing imaginary parts, we obtain

Question 13:

Find the modulus and argument of the complex number .

Answer

Let , then


Page 29 of 34


Therefore, the modulus and argument of the given complex number are

respectively.

Question 14:

Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.

Answer

Let

It is given that,


Page 30 of 34

Equating real and imaginary parts, we obtain

Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain

Putting the value of x in equation (i), we obtain

Thus, the values of x and y are 3 and –3 respectively.

Question 15:

Find the modulus of .

Answer

Question 16:


Page 31 of 34

If (x + iy)3 = u + iv, then show that .

Answer

On equating real and imaginary parts, we obtain

Hence, proved.

Question 17:

If α and β are different complex numbers with = 1, then find .

Answer

Let α = a + ib and β = x + iy

It is given that,


Page 32 of 34

Question 18:

Find the number of non-zero integral solutions of the equation .

Answer


Page 33 of 34

Thus, 0 is the only integral solution of the given equation. Therefore, the number of non-

zero integral solutions of the given equation is 0.

Question 19:

If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that

(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2.

Answer

On squaring both sides, we obtain

(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2

Hence, proved.

Question 20:

If , then find the least positive integral value of m.

Answer


Page 34 of 34

Therefore, the least positive integer is 1.

Thus, the least positive integral value of m is 4 (= 4 × 1).

Class XI Chapter 6 – Linear Inequalities Maths

Exercise 6.1

Question 1:

Solve 24x < 100, when (i) x is a natural number (ii) x is an integer

Answer

The given inequality is 24x < 100.

(i) It is evident that 1, 2, 3, and 4 are the only natural numbers less than .

Thus, when x is a natural number, the solutions of the given inequality are 1, 2, 3, and

4.

Hence, in this case, the solution set is {1, 2, 3, 4}.

(ii) The integers less than are …–3, –2, –1, 0, 1, 2, 3, 4.

Thus, when x is an integer, the solutions of the given inequality are

…–3, –2, –1, 0, 1, 2, 3, 4.

Hence, in this case, the solution set is {…–3, –2, –1, 0, 1, 2, 3, 4}.

Question 2:

Solve –12x > 30, when

(i) x is a natural number (ii) x is an integer

Answer

The given inequality is –12x > 30.

(i) There is no natural number less than .


Page 1 of 48


Page 2 of 48

Thus, when x is a natural number, there is no solution of the given inequality.

(ii) The integers less than are …, –5, –4, –3.


…, –5, –4, –3.

Hence, in this case, the solution set is {…, –5, –4, –3}.

Question 3:

Solve 5x– 3 < 7, when

(i) x is an integer (ii) x is a real number

Answer

The given inequality is 5x– 3 < 7.

(i) The integers less than 2 are …, –4, –3, –2, –1, 0, 1.


…, –4, –3, –2, –1, 0, 1.

Hence, in this case, the solution set is {…, –4, –3, –2, –1, 0, 1}.

(ii) When x is a real number, the solutions of the given inequality are given by x < 2,

that is, all real numbers x which are less than 2.

Thus, the solution set of the given inequality is x ∈ (–∞, 2).

Question 4:

Solve 3x + 8 > 2, when

(i) x is an integer (ii) x is a real number

Answer

The given inequality is 3x + 8 > 2.


Page 3 of 48

(i) The integers greater than –2 are –1, 0, 1, 2, …


–1, 0, 1, 2 …

Hence, in this case, the solution set is {–1, 0, 1, 2, …}.

(ii) When x is a real number, the solutions of the given inequality are all the real

numbers, which are greater than –2.

Thus, in this case, the solution set is (– 2, ∞).

Question 5:

Solve the given inequality for real x: 4x + 3 < 5x + 7

Answer

4x + 3 < 5x + 7

⇒ 4x + 3 – 7 < 5x + 7 – 7

⇒ 4x – 4 < 5x

⇒ 4x – 4 – 4x < 5x – 4x

⇒ –4 < x

Thus, all real numbers x,which are greater than –4, are the solutions of the given

inequality.

Hence, the solution set of the given inequality is (–4, ∞).

Question 6:

Solve the given inequality for real x: 3x – 7 > 5x – 1

Answer

3x – 7 > 5x – 1

⇒ 3x – 7 + 7 > 5x – 1 + 7

⇒ 3x > 5x + 6

⇒ 3x – 5x > 5x + 6 – 5x


Page 4 of 48

⇒ – 2x > 6

Thus, all real numbers x,which are less than –3, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–∞, –3).

Question 7:

Solve the given inequality for real x: 3(x – 1) ≤ 2 (x – 3)

Answer

3(x – 1) ≤ 2(x – 3)

⇒ 3x – 3 ≤ 2x – 6

⇒ 3x – 3 + 3 ≤ 2x – 6 + 3

⇒ 3x ≤ 2x – 3

⇒ 3x – 2x ≤ 2x – 3 – 2x

⇒ x ≤ – 3

Thus, all real numbers x,which are less than or equal to –3, are the solutions of the

given inequality.

Hence, the solution set of the given inequality is (–∞, –3].

Question 8:

Solve the given inequality for real x: 3(2 – x) ≥ 2(1 – x)

Answer

3(2 – x) ≥ 2(1 – x)

⇒ 6 – 3x ≥ 2 – 2x

⇒ 6 – 3x + 2x ≥ 2 – 2x + 2x

⇒ 6 – x ≥ 2

⇒ 6 – x – 6 ≥ 2 – 6

⇒ –x ≥ –4

⇒ x ≤ 4

Thus, all real numbers x,which are less than or equal to 4, are the solutions of the given

inequality.

Hence, the solution set of the given inequality is (–∞, 4].


Page 5 of 48

Question 9:

Solve the given inequality for real x:

Answer

Thus, all real numbers x,which are less than 6, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–∞, 6).

Question 10:


Answer


Page 6 of 48

Thus, all real numbers x,which are less than –6, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–∞, –6).

Question 11:


Answer


inequality.


Question 12:


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Answer

Thus, all real numbers x,which are less than or equal to 120, are the solutions of the

given inequality.


Question 13:

Solve the given inequality for real x: 2(2x + 3) – 10 < 6 (x – 2)

Answer

Thus, all real numbers x,which are greater than or equal to 4, are the solutions of the

given inequality.

Hence, the solution set of the given inequality is [4, ∞).

Question 14:

Solve the given inequality for real x: 37 – (3x + 5) ≥ 9x – 8(x – 3)

Answer


Page 8 of 48


inequality.


Question 15:


Answer

Thus, all real numbers x,which are greater than 4, are the solutions of the given

inequality.

Hence, the solution set of the given inequality is (4, ∞).

Question 16:



Page 9 of 48

Answer


inequality.


Question 17:

Solve the given inequality and show the graph of the solution on number line: 3x – 2 <

2x +1

Answer

3x – 2 < 2x +1

⇒ 3x – 2x < 1 + 2

⇒ x < 3

The graphical representation of the solutions of the given inequality is as follows.

Question 18:

Solve the given inequality and show the graph of the solution on number line: 5x – 3 ≥

3x – 5

Answer


Page 10 of 48


Question 19:

Solve the given inequality and show the graph of the solution on number line: 3(1 – x) <

2 (x + 4)

Answer


Question 20:

Solve the given inequality and show the graph of the solution on number line:

Answer


Page 11 of 48


Question 21:

Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should

get in the third test to have an average of at least 60 marks.

Answer

Let x be the marks obtained by Ravi in the third unit test.

Since the student should have an average of at least 60 marks,

Thus, the student must obtain a minimum of 35 marks to have an average of at least 60

marks.


Page 12 of 48

Question 22:

To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five

examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87,

92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get

grade ‘A’ in the course.

Answer

Let x be the marks obtained by Sunita in the fifth examination.

In order to receive grade ‘A’ in the course, she must obtain an average of 90 marks or

more in five examinations.

Therefore,

Thus, Sunita must obtain greater than or equal to 82 marks in the fifth examination.

Question 23:

Find all pairs of consecutive odd positive integers both of which are smaller than 10 such

that their sum is more than 11.

Answer

Let x be the smaller of the two consecutive odd positive integers. Then, the other integer

is x + 2.

Since both the integers are smaller than 10,

x + 2 < 10

⇒ x < 10 – 2

⇒ x < 8 … (i)

Also, the sum of the two integers is more than 11.

∴x + (x + 2) > 11

⇒ 2x + 2 > 11

⇒ 2x > 11 – 2


Page 13 of 48

⇒ 2x > 9

From (i) and (ii), we obtain .

Since x is an odd number, x can take the values, 5 and 7.

Thus, the required possible pairs are (5, 7) and (7, 9).

Question 24:

Find all pairs of consecutive even positive integers, both of which are larger than 5 such

that their sum is less than 23.

Answer

Let x be the smaller of the two consecutive even positive integers. Then, the other

integer is x + 2.

Since both the integers are larger than 5,

x > 5 ... (1)

Also, the sum of the two integers is less than 23.

x + (x + 2) < 23

⇒ 2x + 2 < 23

⇒ 2x < 23 – 2

⇒ 2x < 21

From (1) and (2), we obtain 5 < x < 10.5.

Since x is an even number, x can take the values, 6, 8, and 10.

Thus, the required possible pairs are (6, 8), (8, 10), and (10, 12).

Question 25:

The longest side of a triangle is 3 times the shortest side and the third side is 2 cm

shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the

minimum length of the shortest side.

Answer


Page 14 of 48

Let the length of the shortest side of the triangle be x cm.

Then, length of the longest side = 3x cm

Length of the third side = (3x – 2) cm

Since the perimeter of the triangle is at least 61 cm,

Thus, the minimum length of the shortest side is 9 cm.

Question 26:

A man wants to cut three lengths from a single piece of board of length 91 cm. The

second length is to be 3 cm longer than the shortest and the third length is to be twice

as long as the shortest. What are the possible lengths of the shortest board if the third

piece is to be at least 5 cm longer than the second?

[Hint: If x is the length of the shortest board, then x, (x + 3) and 2x are the lengths of

the second and third piece, respectively. Thus, x = (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3)

+ 5]

Answer

Let the length of the shortest piece be x cm. Then, length of the second piece and the

third piece are (x + 3) cm and 2x cm respectively.

Since the three lengths are to be cut from a single piece of board of length 91 cm,

x cm + (x + 3) cm + 2x cm ≤ 91 cm

⇒ 4x + 3 ≤ 91

⇒ 4x ≤ 91 – 3

⇒ 4x ≤ 88

Also, the third piece is at least 5 cm longer than the second piece.


Page 15 of 48

∴2x ≥ (x + 3) + 5

⇒ 2x ≥ x + 8

⇒ x ≥ 8 … (2)


8 ≤ x ≤ 22

Thus, the possible length of the shortest board is greater than or equal to 8 cm but less

than or equal to 22 cm.


Page 16 of 48

Exercise 6.2

Question 1:

Solve the given inequality graphically in two-dimensional plane: x + y < 5

Answer

The graphical representation of x + y = 5 is given as dotted line in the figure below.

This line divides the xy-plane in two half planes, I and II.

Select a point (not on the line), which lies in one of the half planes, to determine

whether the point satisfies the given inequality or not.

We select the point as (0, 0).

It is observed that,

0 + 0 < 5 or, 0 < 5, which is true

Therefore, half plane II is not the solution region of the given inequality. Also, it is

evident that any point on the line does not satisfy the given strict inequality.

Thus, the solution region of the given inequality is the shaded half plane I excluding the

points on the line.

This can be represented as follows.

Question 2:

Solve the given inequality graphically in two-dimensional plane: 2x + y ≥ 6

Answer

The graphical representation of 2x + y = 6 is given in the figure below.



Page 17 of 48





2(0) + 0 ≥ 6 or 0 ≥ 6, which is false

Therefore, half plane I is not the solution region of the given inequality. Also, it is

evident that any point on the line satisfies the given inequality.

Thus, the solution region of the given inequality is the shaded half plane II including the

points on the line.


Question 3:

Solve the given inequality graphically in two-dimensional plane: 3x + 4y ≤ 12

Answer

3x + 4y ≤ 12

The graphical representation of 3x + 4y = 12 is given in the figure below.






3(0) + 4(0) ≤ 12 or 0 ≤ 12, which is true

Therefore, half plane II is not the solution region of the given inequality. Also, it is



Page 18 of 48

Thus, the solution region of the given inequality is the shaded half plane I including the

points on the line.


Question 4:

Solve the given inequality graphically in two-dimensional plane: y + 8 ≥ 2x

Answer

The graphical representation of y + 8 = 2x is given in the figure below.

This line divides the xy-plane in two half planes.





0 + 8 ≥ 2(0) or 8 ≥ 0, which is true

Therefore, lower half plane is not the solution region of the given inequality. Also, it is


Thus, the solution region of the given inequality is the half plane containing the point (0,

0) including the line.

The solution region is represented by the shaded region as follows.


Page 19 of 48

Question 5:

Solve the given inequality graphically in two-dimensional plane: x – y ≤ 2

Answer

The graphical representation of x – y = 2 is given in the figure below.






0 – 0 ≤ 2 or 0 ≤ 2, which is true

Therefore, the lower half plane is not the solution region of the given inequality. Also, it

is clear that any point on the line satisfies the given inequality.





Page 20 of 48

Question 6:

Solve the given inequality graphically in two-dimensional plane: 2x – 3y > 6

Answer

The graphical representation of 2x – 3y = 6 is given as dotted line in the figure below.






2(0) – 3(0) > 6 or 0 > 6, which is false

Therefore, the upper half plane is not the solution region of the given inequality. Also, it

is clear that any point on the line does not satisfy the given inequality.

Thus, the solution region of the given inequality is the half plane that does not contain

the point (0, 0) excluding the line.



Page 21 of 48

Question 7:

Solve the given inequality graphically in two-dimensional plane: –3x + 2y ≥ –6

Answer

The graphical representation of – 3x + 2y = – 6 is given in the figure below.






– 3(0) + 2(0) ≥ – 6 or 0 ≥ –6, which is true

Therefore, the lower half plane is not the solution region of the given inequality. Also, it

is evident that any point on the line satisfies the given inequality.




Question 8:

Solve the given inequality graphically in two-dimensional plane: 3y – 5x < 30

Answer

The graphical representation of 3y – 5x = 30 is given as dotted line in the figure below.



Page 22 of 48





3(0) – 5(0) < 30 or 0 < 30, which is true

Therefore, the upper half plane is not the solution region of the given inequality. Also, it

is evident that any point on the line does not satisfy the given inequality.


0) excluding the line.


Question 9:

Solve the given inequality graphically in two-dimensional plane: y < –2

Answer

The graphical representation of y = –2 is given as dotted line in the figure below. This

line divides the xy-plane in two half planes.






Page 23 of 48

0 < –2, which is false

Also, it is evident that any point on the line does not satisfy the given inequality.

Hence, every point below the line, y = –2 (excluding all the points on the line),

determines the solution of the given inequality.


Question 10:

Solve the given inequality graphically in two-dimensional plane: x > –3

Answer

The graphical representation of x = –3 is given as dotted line in the figure below. This

line divides the xy-plane in two half planes.





0 > –3, which is true

Also, it is evident that any point on the line does not satisfy the given inequality.

Hence, every point on the right side of the line, x = –3 (excluding all the points on the

line), determines the solution of the given inequality.



Page 24 of 48


Page 25 of 48

Exercise 6.3

Question 1:

Solve the following system of inequalities graphically: x ≥ 3, y ≥ 2

Answer

x ≥ 3 … (1)

y ≥ 2 … (2)

The graph of the lines, x = 3 and y = 2, are drawn in the figure below.

Inequality (1) represents the region on the right hand side of the line, x = 3 (including

the line x = 3), and inequality (2) represents the region above the line, y = 2 (including

the line y = 2).

Hence, the solution of the given system of linear inequalities is represented by the

common shaded region including the points on the respective lines as follows.

Question 2:

Solve the following system of inequalities graphically: 3x + 2y ≤ 12, x ≥ 1, y ≥ 2

Answer

3x + 2y ≤ 12 … (1)

x ≥ 1 … (2)

y ≥ 2 … (3)

The graphs of the lines, 3x + 2y = 12, x = 1, and y = 2, are drawn in the figure below.


Page 26 of 48

Inequality (1) represents the region below the line, 3x + 2y = 12 (including the line 3x +

2y = 12). Inequality (2) represents the region on the right side of the line, x = 1

(including the line x = 1). Inequality (3) represents the region above the line, y = 2

(including the line y = 2).



Question 3:

Solve the following system of inequalities graphically: 2x + y≥ 6, 3x + 4y ≤ 12

Answer

2x + y≥ 6 … (1)

3x + 4y ≤ 12 … (2)

The graph of the lines, 2x + y= 6 and 3x + 4y = 12, are drawn in the figure below.

Inequality (1) represents the region above the line, 2x + y= 6 (including the line 2x +

y= 6), and inequality (2) represents the region below the line, 3x + 4y =12 (including

the line 3x + 4y =12).




Page 27 of 48

Question 4:

Solve the following system of inequalities graphically: x + y≥ 4, 2x – y > 0

Answer

x + y≥ 4 … (1)

2x – y > 0 … (2)

The graph of the lines, x + y = 4 and 2x – y = 0, are drawn in the figure below.

Inequality (1) represents the region above the line, x + y = 4 (including the line x + y =

4).

It is observed that (1, 0) satisfies the inequality, 2x – y > 0. [2(1) – 0 = 2 > 0]

Therefore, inequality (2) represents the half plane corresponding to the line, 2x – y = 0,

containing the point (1, 0) [excluding the line 2x – y > 0].


common shaded region including the points on line x + y = 4 and excluding the points

on line 2x – y = 0 as follows.


Page 28 of 48

Question 5:

Solve the following system of inequalities graphically: 2x – y > 1, x – 2y < –1

Answer

2x – y > 1 … (1)

x – 2y < –1 … (2)

The graph of the lines, 2x – y = 1 and x – 2y = –1, are drawn in the figure below.

Inequality (1) represents the region below the line, 2x – y = 1 (excluding the line 2x – y

= 1), and inequality (2) represents the region above the line, x – 2y = –1 (excluding the

line x – 2y = –1).


common shaded region excluding the points on the respective lines as follows.


Page 29 of 48

Question 6:

Solve the following system of inequalities graphically: x + y ≤ 6, x + y ≥ 4

Answer

x + y ≤ 6 … (1)

x + y ≥ 4 … (2)

The graph of the lines, x + y = 6 and x + y = 4, are drawn in the figure below.

Inequality (1) represents the region below the line, x + y = 6 (including the line x + y =

6), and inequality (2) represents the region above the line, x + y = 4 (including the line

x + y = 4).




Page 30 of 48

Question 7:

Solve the following system of inequalities graphically: 2x + y≥ 8, x + 2y ≥ 10

Answer

2x + y= 8 … (1)

x + 2y = 10 … (2)

The graph of the lines, 2x + y= 8 and x + 2y = 10, are drawn in the figure below.

Inequality (1) represents the region above the line, 2x + y = 8, and inequality (2)

represents the region above the line, x + 2y = 10.




Page 31 of 48

Question 8:

Solve the following system of inequalities graphically: x + y ≤ 9, y > x, x ≥ 0

Answer

x + y ≤ 9 ... (1)

y > x ... (2)

x ≥ 0 ... (3)

The graph of the lines, x + y= 9 and y = x, are drawn in the figure below.

Inequality (1) represents the region below the line, x + y = 9 (including the line x + y =

9).

It is observed that (0, 1) satisfies the inequality, y > x. [1 > 0]

Therefore, inequality (2) represents the half plane corresponding to the line, y = x,

containing the point (0, 1) [excluding the line y = x].

Inequality (3) represents the region on the right hand side of the line, x = 0 or y-axis

(including y-axis).


common shaded region including the points on the lines, x + y = 9 and x = 0, and

excluding the points on line y = x as follows.


Page 32 of 48

Question 9:

Solve the following system of inequalities graphically: 5x + 4y ≤ 20, x ≥ 1, y ≥ 2

Answer

5x + 4y ≤ 20 … (1)

x ≥ 1 … (2)

y ≥ 2 … (3)

The graph of the lines, 5x + 4y = 20, x = 1, and y = 2, are drawn in the figure below.


4y = 20). Inequality (2) represents the region on the right hand side of the line, x = 1

(including the line x = 1). Inequality (3) represents the region above the line, y = 2

(including the line y = 2).




Page 33 of 48

Question 10:

Solve the following system of inequalities graphically: 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0,

y ≥ 0

Answer

3x + 4y ≤ 60 … (1)

x + 3y ≤ 30 … (2)

The graph of the lines, 3x + 4y = 60 and x + 3y = 30, are drawn in the figure below.


4y = 60), and inequality (2) represents the region below the line, x + 3y = 30 (including

the line x + 3y = 30).

Since x ≥ 0 and y ≥ 0, every point in the common shaded region in the first quadrant

including the points on the respective line and the axes represents the solution of the

given system of linear inequalities.


Page 34 of 48

Question 11:

Solve the following system of inequalities graphically: 2x + y≥ 4, x + y ≤ 3, 2x – 3y ≤ 6

Answer

2x + y≥ 4 … (1)

x + y ≤ 3 … (2)

2x – 3y ≤ 6 … (3)

The graph of the lines, 2x + y= 4, x + y = 3, and 2x – 3y = 6, are drawn in the figure

below.

Inequality (1) represents the region above the line, 2x + y= 4 (including the line 2x +

y= 4). Inequality (2) represents the region below the line,

x + y = 3 (including the line x + y = 3). Inequality (3) represents the region above the

line, 2x – 3y = 6 (including the line 2x – 3y = 6).




Page 35 of 48

Question 12:

Solve the following system of inequalities graphically:

x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1

Answer

x – 2y ≤ 3 … (1)

3x + 4y ≥ 12 … (2)

y ≥ 1 … (3)

The graph of the lines, x – 2y = 3, 3x + 4y = 12, and y = 1, are drawn in the figure

below.

Inequality (1) represents the region above the line, x – 2y = 3 (including the line x – 2y

= 3). Inequality (2) represents the region above the line, 3x + 4y = 12 (including the

line 3x + 4y = 12). Inequality (3) represents the region above the line, y = 1 (including

the line y = 1).

The inequality, x ≥ 0, represents the region on the right hand side of y-axis (including y-

axis).


common shaded region including the points on the respective lines and y- axis as

follows.


Page 36 of 48

Question 13:

[[Q]]

Solve the following system of inequalities graphically:

4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0

Answer

4x + 3y ≤ 60 … (1)

y ≥ 2x … (2)

x ≥ 3 … (3)

The graph of the lines, 4x + 3y = 60, y = 2x, and x = 3, are drawn in the figure below.


3y = 60). Inequality (2) represents the region above the line, y = 2x (including the line

y = 2x). Inequality (3) represents the region on the right hand side of the line, x = 3

(including the line x = 3).




Page 37 of 48

Question 14:

Solve the following system of inequalities graphically: 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤

15, y ≥ 0, x ≥ 0

Answer

3x + 2y ≤ 150 … (1)

x + 4y ≤ 80 … (2)

x ≤ 15 … (3)

The graph of the lines, 3x + 2y = 150, x + 4y = 80, and x = 15, are drawn in the figure

below.

Inequality (1) represents the region below the line, 3x + 2y = 150 (including the line 3x

+ 2y = 150). Inequality (2) represents the region below the line, x + 4y = 80 (including

the line x + 4y = 80). Inequality (3) represents the region on the left hand side of the

line, x = 15 (including the line x = 15).


Page 38 of 48


including the points on the respective lines and the axes represents the solution of the


Question 15:

Solve the following system of inequalities graphically: x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0,

x ≥ 0, y ≥ 0

Answer

x + 2y ≤ 10 … (1)

x + y ≥ 1 … (2)

x – y ≤ 0 … (3)

The graph of the lines, x + 2y = 10, x + y = 1, and x – y = 0, are drawn in the figure

below.

Inequality (1) represents the region below the line, x + 2y = 10 (including the line x +

2y = 10). Inequality (2) represents the region above the line, x + y = 1 (including the

line x + y = 1). Inequality (3) represents the region above the line, x – y = 0 (including

the line x – y = 0).


Page 39 of 48


including the points on the respective lines and the axes represents the solution of the



Page 40 of 48

NCERT Miscellaneous Solution

Question 1:

Solve the inequality 2 ≤ 3x – 4 ≤ 5

Answer

2 ≤ 3x – 4 ≤ 5

⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4

⇒ 6 ≤ 3x ≤ 9

⇒ 2 ≤ x ≤ 3

Thus, all the real numbers, x, which are greater than or equal to 2 but less than or equal

to 3, are the solutions of the given inequality. The solution set for the given inequalityis

[2, 3].

Question 2:

Solve the inequality 6 ≤ –3(2x – 4) < 12

Answer

6 ≤ – 3(2x – 4) < 12

⇒ 2 ≤ –(2x – 4) < 4

⇒ –2 ≥ 2x – 4 > –4

⇒ 4 – 2 ≥ 2x > 4 – 4

⇒ 2 ≥ 2x > 0

⇒1 ≥ x > 0

Thus, the solution set for the given inequalityis (0, 1].

Question 3:

Solve the inequality

Answer


Page 41 of 48

Thus, the solution set for the given inequalityis [–4, 2].

Question 4:


Answer

⇒ –75 < 3(x – 2) ≤ 0

⇒ –25 < x – 2 ≤ 0

⇒ – 25 + 2 < x ≤ 2

⇒ –23 < x ≤ 2

Thus, the solution set for the given inequalityis (–23, 2].

Question 5:


Answer


Page 42 of 48

Thus, the solution set for the given inequalityis .

Question 6:


Answer

Thus, the solution set for the given inequalityis .

Question 7:

Solve the inequalities and represent the solution graphically on number line: 5x + 1 > –

24, 5x – 1 < 24

Answer

5x + 1 > –24

⇒ 5x > –25


Page 43 of 48

⇒ x > –5 … (1)

5x – 1 < 24

⇒ 5x < 25

⇒ x < 5 … (2)

From (1) and (2), it can be concluded that the solution set for the given system of

inequalities is (–5, 5). The solution of the given system of inequalities can be represented

on number line as

Question 8:

Solve the inequalities and represent the solution graphically on number line: 2(x – 1) < x

+ 5, 3(x + 2) > 2 – x

Answer

2(x – 1) < x + 5

⇒ 2x – 2 < x + 5

⇒ 2x – x < 5 + 2

⇒ x < 7 … (1)

3(x + 2) > 2 – x

⇒ 3x + 6 > 2 – x

⇒ 3x + x > 2 – 6

⇒ 4x > – 4

⇒ x > – 1 … (2)


inequalities is (–1, 7). The solution of the given system of inequalities can be represented

on number line as

Question 9:

Solve the following inequalities and represent the solution graphically on number line:

3x – 7 > 2(x – 6), 6 – x > 11 – 2x

Answer:


Page 44 of 48

3x – 7 > 2(x – 6)

⇒ 3x – 7 > 2x – 12

⇒ 3x – 2x > – 12 + 7

⇒ x > –5 … (1)

6 – x > 11 – 2x

⇒ –x + 2x > 11 – 6

⇒ x > 5 … (2)


inequalities is . The solution of the given system of inequalities can be represented

on number line as

Question 10:

Solve the inequalities and represent the solution graphically on number line: 5(2x – 7) –

3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47

Answer

5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

⇒ 4x ≤ 44

⇒ x ≤ 11 … (1)

2x + 19 ≤ 6x + 47

⇒ 19 – 47 ≤ 6x – 2x

⇒ –28 ≤ 4x

⇒ –7 ≤ x … (2)


inequalities is [–7, 11]. The solution of the given system of inequalities can be

represented on number line as


Page 45 of 48

Question 11:

A solution is to be kept between 68°F and 77°F. What is the range in temperature in

degree Celsius (C) if the Celsius/Fahrenheit (F) conversion formula is given by

Answer

Since the solution is to be kept between 68°F and 77°F,

68 < F < 77

Putting we obtain

Thus, the required range of temperature in degree Celsius is between 20°C and 25°C.

Question 12:

A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The

resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres

of the 8% solution, how many litres of the 2% solution will have to be added?

Answer

Let x litres of 2% boric acid solution is required to be added.

Then, total mixture = (x + 640) litres

This resulting mixture is to be more than 4% but less than 6% boric acid.

∴2%x + 8% of 640 > 4% of (x + 640)

And, 2% x + 8% of 640 < 6% of (x + 640)

2%x + 8% of 640 > 4% of (x + 640)


Page 46 of 48

⇒ 2x + 5120 > 4x + 2560

⇒ 5120 – 2560 > 4x – 2x

⇒ 5120 – 2560 > 2x

⇒ 2560 > 2x

⇒ 1280 > x

2% x + 8% of 640 < 6% of (x + 640)

⇒ 2x + 5120 < 6x + 3840

⇒ 5120 – 3840 < 6x – 2x

⇒ 1280 < 4x

⇒ 320 < x

∴320 < x < 1280

Thus, the number of litres of 2% of boric acid solution that is to be added will have to be

more than 320 litres but less than 1280 litres.

Question 13:

How many litres of water will have to be added to 1125 litres of the 45% solution of acid

so that the resulting mixture will contain more than 25% but less than 30% acid content?

Answer

Let x litres of water is required to be added.

Then, total mixture = (x + 1125) litres

It is evident that the amount of acid contained in the resulting mixture is 45% of 1125

litres.

This resulting mixture will contain more than 25% but less than 30% acid content.

∴30% of (1125 + x) > 45% of 1125

And, 25% of (1125 + x) < 45% of 1125

30% of (1125 + x) > 45% of 1125


Page 47 of 48

25% of (1125 + x) < 45% of 1125

∴562.5 < x < 900

Thus, the required number of litres of water that is to be added will have to be more than

562.5 but less than 900.

Question 14:

IQ of a person is given by the formula

Where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12

years old children, find the range of their mental age.

Answer

It is given that for a group of 12 years old children, 80 ≤ IQ ≤ 140 … (i)

For a group of 12 years old children, CA = 12 years

Putting this value of IQ in (i), we obtain


Page 48 of 48

Thus, the range of mental age of the group of 12 years old children is .

Class XI Chapter 7 – Permutations and Combinations Maths

Exercise 7.1

Question 1:

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that

(i) repetition of the digits is allowed?

(ii) repetition of the digits is not allowed?

Answer

(i)

There will be as many ways as there are ways of filling 3 vacant places

in succession by the given five digits. In this case, repetition of digits is

allowed. Therefore, the units place can be filled in by any of the given five digits.

Similarly, tens and hundreds digits can be filled in by any of the given five digits.

Thus, by the multiplication principle, the number of ways in which three-digit numbers

can be formed from the given digits is 5 × 5 × 5 = 125

(ii)

In this case, repetition of digits is not allowed. Here, if units place is filled in first, then it

can be filled by any of the given five digits. Therefore, the number of ways of filling the

units place of the three-digit number is 5.

Then, the tens place can be filled with any of the remaining four digits and the hundreds

place can be filled with any of the remaining three digits.

Thus, by the multiplication principle, the number of ways in which three-digit numbers

can be formed without repeating the given digits is 5 × 4 × 3 = 60

Question 2:

How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the

digits can be repeated?

Answer

There will be as many ways as there are ways of filling 3 vacant places

in succession by the given six digits. In this case, the units place can be filled

by 2 or 4 or 6 only i.e., the units place can be filled in 3 ways. The tens place can be

filled by any of the 6 digits in 6 different ways and also the hundreds place can be filled

by any of the 6 digits in 6 different ways, as the digits can be repeated.


Page 2 of 26


Page 2 of 26

Therefore, by multiplication principle, the required number of three digit even numbers is

3 × 6 × 6 = 108

Question 3:

How many 4-letter code can be formed using the first 10 letters of the English alphabet,

if no letter can be repeated?

Answer

There are as many codes as there are ways of filling 4 vacant places in

succession by the first 10 letters of the English alphabet, keeping in mind that the

repetition of letters is not allowed.

The first place can be filled in 10 different ways by any of the first 10 letters of the

English alphabet following which, the second place can be filled in by any of the

remaining letters in 9 different ways. The third place can be filled in by any of the

remaining 8 letters in 8 different ways and the fourth place can be filled in by any of the

remaining 7 letters in 7 different ways.

Therefore, by multiplication principle, the required numbers of ways in which 4 vacant

places can be filled is 10 × 9 × 8 × 7 = 5040

Hence, 5040 four-letter codes can be formed using the first 10 letters of the English

alphabet, if no letter is repeated.

Question 4:

How many 5–digit telephone numbers can be constructed using the digits 0 to 9 if each

number starts with 67 and no digit appears more than once?

Answer

It is given that the 5-digit telephone numbers always start with 67.

Therefore, there will be as many phone numbers as there are ways of filling 3 vacant

places by the digits 0 – 9, keeping in mind that the digits cannot be

repeated.

The units place can be filled by any of the digits from 0 – 9, except digits 6 and 7.

Therefore, the units place can be filled in 8 different ways following which, the tens place

can be filled in by any of the remaining 7 digits in 7 different ways, and the hundreds

place can be filled in by any of the remaining 6 digits in 6 different ways.


Page 3 of 26

Therefore, by multiplication principle, the required number of ways in which 5-digit

telephone numbers can be constructed is 8 × 7 × 6 = 336

Question 5:

A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes

are there?

Answer

When a coin is tossed once, the number of outcomes is 2 (Head and tail) i.e., in each

throw, the number of ways of showing a different face is 2.

Thus, by multiplication principle, the required number of possible outcomes is 2 × 2 × 2

= 8

Question 6:

Given 5 flags of different colours, how many different signals can be generated if each

signal requires the use of 2 flags, one below the other?

Answer

Each signal requires the use of 2 flags.

There will be as many flags as there are ways of filling in 2 vacant places in

succession by the given 5 flags of different colours.

The upper vacant place can be filled in 5 different ways by any one of the 5 flags

following which, the lower vacant place can be filled in 4 different ways by any one of the

remaining 4 different flags.

Thus, by multiplication principle, the number of different signals that can be generated is

5 × 4 = 20


Page 4 of 26

Exercise 7.2

Question 1:

Evaluate

(i) 8! (ii) 4! – 3!

Answer

(i) 8! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 = 40320

(ii) 4! = 1 × 2 × 3 × 4 = 24

3! = 1 × 2 × 3 = 6

∴4! – 3! = 24 – 6 = 18

Question 2:

Is 3! + 4! = 7!?

Answer

3! = 1 × 2 × 3 = 6

4! = 1 × 2 × 3 × 4 = 24

∴3! + 4! = 6 + 24 = 30

7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040

∴ 3! + 4! ≠ 7!

Question 3:

Compute

Answer

Question 4:

If , find x.

Answer


Page 5 of 26

Question 5:

Evaluate , when

(i) n = 6, r = 2 (ii) n = 9, r = 5

Answer

(i) When n = 6, r = 2,

(ii) When n = 9, r = 5,


Page 6 of 26

Exercise 7.3

Question 1:

How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is

repeated?

Answer

3-digit numbers have to be formed using the digits 1 to 9.

Here, the order of the digits matters.

Therefore, there will be as many 3-digit numbers as there are permutations of 9

different digits taken 3 at a time.

Therefore, required number of 3-digit numbers

Question 2:

How many 4-digit numbers are there with no digit repeated?

Answer

The thousands place of the 4-digit number is to be filled with any of the digits from 1 to

9 as the digit 0 cannot be included. Therefore, the number of ways in which thousands

place can be filled is 9.

The hundreds, tens, and units place can be filled by any of the digits from 0 to 9.

However, the digits cannot be repeated in the 4-digit numbers and thousands place is

already occupied with a digit. The hundreds, tens, and units place is to be filled by the

remaining 9 digits.

Therefore, there will be as many such 3-digit numbers as there are permutations of 9

different digits taken 3 at a time.

Number of such 3-digit numbers

Thus, by multiplication principle, the required number of 4-digit numbers is

9 × 504 = 4536


Page 7 of 26

Question 3:

How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit

is repeated?

Answer

3-digit even numbers are to be formed using the given six digits, 1, 2, 3, 4, 6, and 7,

without repeating the digits.

Then, units digits can be filled in 3 ways by any of the digits, 2, 4, or 6.

Since the digits cannot be repeated in the 3-digit numbers and units place is already

occupied with a digit (which is even), the hundreds and tens place is to be filled by the

remaining 5 digits.

Therefore, the number of ways in which hundreds and tens place can be filled with the

remaining 5 digits is the permutation of 5 different digits taken 2 at a time.

Number of ways of filling hundreds and tens place

Thus, by multiplication principle, the required number of 3-digit numbers is

3 × 20 = 60

Question 4:

Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no

digit is repeated. How many of these will be even?

Answer

4-digit numbers are to be formed using the digits, 1, 2, 3, 4, and 5.

There will be as many 4-digit numbers as there are permutations of 5 different digits

taken 4 at a time.

Therefore, required number of 4 digit numbers =

Among the 4-digit numbers formed by using the digits, 1, 2, 3, 4, 5, even numbers end

with either 2 or 4.


Page 8 of 26

The number of ways in which units place is filled with digits is 2.

Since the digits are not repeated and the units place is already occupied with a digit

(which is even), the remaining places are to be filled by the remaining 4 digits.

Therefore, the number of ways in which the remaining places can be filled is the

permutation of 4 different digits taken 3 at a time.

Number of ways of filling the remaining places

= 4 × 3 × 2 × 1 = 24

Thus, by multiplication principle, the required number of even numbers is

24 × 2 = 48

Question 5:

From a committee of 8 persons, in how many ways can we choose a chairman and a vice

chairman assuming one person cannot hold more than one position?

Answer

From a committee of 8 persons, a chairman and a vice chairman are to be chosen in

such a way that one person cannot hold more than one position.

Here, the number of ways of choosing a chairman and a vice chairman is the

permutation of 8 different objects taken 2 at a time.

Thus, required number of ways =

Question 6:

Find n if

Answer


Page 9 of 26

Question 7:

Find r if (i) (ii) .

Answer

(i)


Page 10 of 26

It is known that,

∴0 ≤ r ≤ 5

Hence, r ≠ 10

∴r = 3

(ii)


Page 11 of 26

It is known that,

∴0 ≤ r ≤ 5

Hence, r ≠ 9

∴r = 4

Question 8:

How many words, with or without meaning, can be formed using all the letters of the

word EQUATION, using each letter exactly once?

Answer

There are 8 different letters in the word EQUATION.

Therefore, the number of words that can be formed using all the letters of the word

EQUATION, using each letter exactly once, is the number of permutations of 8 different

objects taken 8 at a time, which is .

Thus, required number of words that can be formed = 8! = 40320

Question 9:

How many words, with or without meaning can be made from the letters of the word

MONDAY, assuming that no letter is repeated, if

(i) 4 letters are used at a time, (ii) all letters are used at a time,

(iii) all letters are used but first letter is a vowel?

Answer

There are 6 different letters in the word MONDAY.

(i) Number of 4-letter words that can be formed from the letters of the word MONDAY,

without repetition of letters, is the number of permutations of 6 different objects taken 4

at a time, which is .

Thus, required number of words that can be formed using 4 letters at a time is


Page 12 of 26

(ii) Number of words that can be formed by using all the letters of the word MONDAY at

a time is the number of permutations of 6 different objects taken 6 at a time, which is

.

Thus, required number of words that can be formed when all letters are used at a time =

6! = 6 × 5 × 4 × 3 × 2 ×1 = 720

(iii) In the given word, there are 2 different vowels, which have to occupy the rightmost

place of the words formed. This can be done only in 2 ways.

Since the letters cannot be repeated and the rightmost place is already occupied with a

letter (which is a vowel), the remaining five places are to be filled by the remaining 5

letters. This can be done in 5! ways.

Thus, in this case, required number of words that can be formed is

5! × 2 = 120 × 2 = 240

Question 10:

In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s

not come together?

Answer

In the given word MISSISSIPPI, I appears 4 times, S appears 4 times, P appears 2

times, and M appears just once.

Therefore, number of distinct permutations of the letters in the given word

There are 4 Is in the given word. When they occur together, they are treated as a single

object for the time being. This single object together with the remaining 7 objects

will account for 8 objects.


Page 13 of 26

These 8 objects in which there are 4 Ss and 2 Ps can be arranged in ways i.e., 840

ways.

Number of arrangements where all Is occur together = 840

Thus, number of distinct permutations of the letters in MISSISSIPPI in which four Is do

not come together = 34650 – 840 = 33810

Question 11:

In how many ways can the letters of the word PERMUTATIONS be arranged if the

(i) words start with P and end with S, (ii) vowels are all together,

(ii) there are always 4 letters between P and S?

Answer

In the word PERMUTATIONS, there are 2 Ts and all the other letters appear only once.

(i) If P and S are fixed at the extreme ends (P at the left end and S at the right end),

then 10 letters are left.

Hence, in this case, required number of arrangements

(ii) There are 5 vowels in the given word, each appearing only once.

Since they have to always occur together, they are treated as a single object for the

time being. This single object together with the remaining 7 objects will account for 8

objects. These 8 objects in which there are 2 Ts can be arranged in .

Corresponding to each of these arrangements, the 5 different vowels can be arranged in

5! ways.

Therefore, by multiplication principle, required number of arrangements in this case

(iii) The letters have to be arranged in such a way that there are always 4 letters

between P and S.

Therefore, in a way, the places of P and S are fixed. The remaining 10 letters in which

there are 2 Ts can be arranged in .


Page 14 of 26

Also, the letters P and S can be placed such that there are 4 letters between them in 2 ×

7 = 14 ways.

Therefore, by multiplication principle, required number of arrangements in this case


Page 15 of 26

Exercise 7.4

Question 1:

If , find .

Answer

It is known that,

Therefore,

Question 2:

Determine n if

(i) (ii)

Answer

(i)


Page 16 of 26

(ii)


Page 17 of 26

Question 3:

How many chords can be drawn through 21 points on a circle?

Answer

For drawing one chord on a circle, only 2 points are required.

To know the number of chords that can be drawn through the given 21 points on a

circle, the number of combinations have to be counted.

Therefore, there will be as many chords as there are combinations of 21 points taken 2

at a time.

Thus, required number of chords =

Question 4:

In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Answer

A team of 3 boys and 3 girls is to be selected from 5 boys and 4 girls.

3 boys can be selected from 5 boys in ways.

3 girls can be selected from 4 girls in ways.

Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3

girls can be selected

Question 5:

Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue

balls if each selection consists of 3 balls of each colour.

Answer

There are a total of 6 red balls, 5 white balls, and 5 blue balls.


Page 18 of 26

9 balls have to be selected in such a way that each selection consists of 3 balls of each

colour.

Here,

3 balls can be selected from 6 red balls in ways.

3 balls can be selected from 5 white balls in ways.

3 balls can be selected from 5 blue balls in ways.

Thus, by multiplication principle, required number of ways of selecting 9 balls

Question 6:

Determine the number of 5 card combinations out of a deck of 52 cards if there is

exactly one ace in each combination.

Answer

In a deck of 52 cards, there are 4 aces. A combination of 5 cards have to be made in

which there is exactly one ace.

Then, one ace can be selected in ways and the remaining 4 cards can be selected

out of the 48 cards in ways.

Thus, by multiplication principle, required number of 5 card combinations

Question 7:

In how many ways can one select a cricket team of eleven from 17 players in which only

5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?


Page 19 of 26

Answer

Out of 17 players, 5 players are bowlers.

A cricket team of 11 players is to be selected in such a way that there are exactly 4

bowlers.

4 bowlers can be selected in ways and the remaining 7 players can be selected out

of the 12 players in ways.

Thus, by multiplication principle, required number of ways of selecting cricket team

Question 8:

A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black

and 3 red balls can be selected.

Answer

There are 5 black and 6 red balls in the bag.

2 black balls can be selected out of 5 black balls in ways and 3 red balls can be

selected out of 6 red balls in ways.

Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red

balls

Question 9:

In how many ways can a student choose a programme of 5 courses if 9 courses are

available and 2 specific courses are compulsory for every student?

Answer

There are 9 courses available out of which, 2 specific courses are compulsory for every

student.

Therefore, every student has to choose 3 courses out of the remaining 7 courses. This

can be chosen in ways.

Thus, required number of ways of choosing the programme


Page 20 of 26


Question 1:

How many words, with or without meaning, each of 2 vowels and 3 consonants can be

formed from the letters of the word DAUGHTER?

Answer

In the word DAUGHTER, there are 3 vowels namely, A, U, and E, and 5 consonants

namely, D, G, H, T, and R.

Number of ways of selecting 2 vowels out of 3 vowels =

Number of ways of selecting 3 consonants out of 5 consonants =

Therefore, number of combinations of 2 vowels and 3 consonants = 3 × 10 = 30

Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among

themselves in 5! ways.

Hence, required number of different words = 30 × 5! = 3600

Question 2:

How many words, with or without meaning, can be formed using all the letters of the

word EQUATION at a time so that the vowels and consonants occur together?

Answer

In the word EQUATION, there are 5 vowels, namely, A, E, I, O, and U, and 3 consonants,

namely, Q, T, and N.

Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN)

can be assumed as single objects. Then, the permutations of these 2 objects taken all at

a time are counted. This number would be

Corresponding to each of these permutations, there are 5! permutations of the five

vowels taken all at a time and 3! permutations of the 3 consonants taken all at a time.

Hence, by multiplication principle, required number of words = 2! × 5! × 3!

= 1440

Question 3:


Page 21 of 26

A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this

be done when the committee consists of:

(i) exactly 3 girls? (ii) atleast 3 girls? (iii) atmost 3 girls?

Answer

A committee of 7 has to be formed from 9 boys and 4 girls.

i. Since exactly 3 girls are to be there in every committee, each committee must

consist of (7 – 3) = 4 boys only.

Thus, in this case, required number of ways =

(ii) Since at least 3 girls are to be there in every committee, the committee can consist

of

(a) 3 girls and 4 boys or (b) 4 girls and 3 boys

3 girls and 4 boys can be selected in ways.


Therefore, in this case, required number of ways =

(iii) Since atmost 3 girls are to be there in every committee, the committee can consist

of

(a) 3 girls and 4 boys (b) 2 girls and 5 boys

(c) 1 girl and 6 boys (d) No girl and 7 boys



1 girl and 6 boys can be selected in ways.

No girl and 7 boys can be selected in ways.

Therefore, in this case, required number of ways


Page 22 of 26

Question 4:

If the different permutations of all the letter of the word EXAMINATION are listed as in a

dictionary, how many words are there in this list before the first word starting with E?

Answer

In the given word EXAMINATION, there are 11 letters out of which, A, I, and N appear 2

times and all the other letters appear only once.

The words that will be listed before the words starting with E in a dictionary will be the

words that start with A only.

Therefore, to get the number of words starting with A, the letter A is fixed at the

extreme left position, and then the remaining 10 letters taken all at a time are

rearranged.

Since there are 2 Is and 2 Ns in the remaining 10 letters,

Number of words starting with A =

Thus, the required numbers of words is 907200.

Question 5:

How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are

divisible by 10 and no digit is repeated?

Answer

A number is divisible by 10 if its units digits is 0.

Therefore, 0 is fixed at the units place.

Therefore, there will be as many ways as there are ways of filling 5 vacant places

in succession by the remaining 5 digits (i.e., 1, 3, 5, 7 and 9).

The 5 vacant places can be filled in 5! ways.

Hence, required number of 6-digit numbers = 5! = 120


Page 23 of 26

Question 6:

The English alphabet has 5 vowels and 21 consonants. How many words with two

different vowels and 2 different consonants can be formed from the alphabet?

Answer

2 different vowels and 2 different consonants are to be selected from the English

alphabet.

Since there are 5 vowels in the English alphabet, number of ways of selecting 2 different

vowels from the alphabet =

Since there are 21 consonants in the English alphabet, number of ways of selecting 2

different consonants from the alphabet

Therefore, number of combinations of 2 different vowels and 2 different consonants = 10

× 210 = 2100

Each of these 2100 combinations has 4 letters, which can be arranged among

themselves in 4! ways.

Therefore, required number of words = 2100 × 4! = 50400

Question 7:

In an examination, a question paper consists of 12 questions divided into two parts i.e.,

Part I and Part II, containing 5 and 7 questions, respectively. A student is required to

attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a

student select the questions?

Answer

It is given that the question paper consists of 12 questions divided into two parts – Part

I and Part II, containing 5 and 7 questions, respectively.

A student has to attempt 8 questions, selecting at least 3 from each part.

This can be done as follows.

(a) 3 questions from part I and 5 questions from part II

(b) 4 questions from part I and 4 questions from part II

(c) 5 questions from part I and 3 questions from part II


Page 24 of 26

3 questions from part I and 5 questions from part II can be selected in ways.



Thus, required number of ways of selecting questions

Question 8:

Determine the number of 5-card combinations out of a deck of 52 cards if each selection

of 5 cards has exactly one king.

Answer

From a deck of 52 cards, 5-card combinations have to be made in such a way that in

each selection of 5 cards, there is exactly one king.

In a deck of 52 cards, there are 4 kings.

1 king can be selected out of 4 kings in ways.

4 cards out of the remaining 48 cards can be selected in ways.

Thus, the required number of 5-card combinations is .

Question 9:

It is required to seat 5 men and 4 women in a row so that the women occupy the even

places. How many such arrangements are possible?

Answer

5 men and 4 women are to be seated in a row such that the women occupy the even

places.

The 5 men can be seated in 5! ways. For each arrangement, the 4 women can be seated

only at the cross marked places (so that women occupy the even places).

Therefore, the women can be seated in 4! ways.


Page 25 of 26

Thus, possible number of arrangements = 4! × 5! = 24 × 120 = 2880

Question 10:

From a class of 25 students, 10 are to be chosen for an excursion party. There are 3

students who decide that either all of them will join or none of them will join. In how

many ways can the excursion party be chosen?

Answer

From the class of 25 students, 10 are to be chosen for an excursion party.

Since there are 3 students who decide that either all of them will join or none of them

will join, there are two cases.

Case I: All the three students join.

Then, the remaining 7 students can be chosen from the remaining 22 students in

ways.

Case II: None of the three students join.

Then, 10 students can be chosen from the remaining 22 students in ways.

Thus, required number of ways of choosing the excursion party is .

Question 11:

In how many ways can the letters of the word ASSASSINATION be arranged so that all

the S’s are together?

Answer

In the given word ASSASSINATION, the letter A appears 3 times, S appears 4 times, I

appears 2 times, N appears 2 times, and all the other letters appear only once.

Since all the words have to be arranged in such a way that all the Ss are together, SSSS

is treated as a single object for the time being. This single object together with the

remaining 9 objects will account for 10 objects.

These 10 objects in which there are 3 As, 2 Is, and 2 Ns can be arranged in

ways.

Thus, required number of ways of arranging the letters of the given word

Class XI Chapter 8 – Binomial Theorem Maths

Exercise 8.1

Question 1:

Expand the expression (1– 2x)5

Answer

By using Binomial Theorem, the expression (1– 2x)5 can be expanded as

Question 2:

Expand the expression

Answer

By using Binomial Theorem, the expression can be expanded as

Question 3:

Expand the expression (2x – 3)6

Answer

By using Binomial Theorem, the expression (2x – 3)6 can be expanded as


Page 2 of 25


Page 2 of 25

Question 4:

Expand the expression

Answer


Question 5:

Expand

Answer



Page 3 of 25

Question 6:

Using Binomial Theorem, evaluate (96)3

Answer

96 can be expressed as the sum or difference of two numbers whose powers are easier

to calculate and then, binomial theorem can be applied.

It can be written that, 96 = 100 – 4

Question 7:


Answer


to calculate and then, Binomial Theorem can be applied.

It can be written that, 102 = 100 + 2


Page 4 of 25

Question 8:


Answer


to calculate and then, Binomial Theorem can be applied.

It can be written that, 101 = 100 + 1

Question 9:


Answer

99 can be written as the sum or difference of two numbers whose powers are easier to

calculate and then, Binomial Theorem can be applied.

It can be written that, 99 = 100 – 1


Page 5 of 25

Question 10:

Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.

Answer

By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1)10000 can

be obtained as

Question 11:

Find (a + b)4 – (a – b)4. Hence, evaluate .

Answer

Using Binomial Theorem, the expressions, (a + b)4 and (a – b)4, can be expanded as


Page 6 of 25

Question 12:

Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate .

Answer

Using Binomial Theorem, the expressions, (x + 1)6 and (x – 1)6, can be expanded as

By putting , we obtain


Page 7 of 25

Question 13:

Show that is divisible by 64, whenever n is a positive integer.

Answer

In order to show that is divisible by 64, it has to be proved that,

, where k is some natural number

By Binomial Theorem,

For a = 8 and m = n + 1, we obtain

Thus, is divisible by 64, whenever n is a positive integer.

Question 14:

Prove that .

Answer

By Binomial Theorem,

By putting b = 3 and a = 1 in the above equation, we obtain

Hence, proved.


Page 8 of 25

Exercise 8.2

Question 1:

Find the coefficient of x5 in (x + 3)8

Answer

It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by

.

Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8, we obtain

Comparing the indices of x in x5 and in Tr +1, we obtain

r = 3

Thus, the coefficient of x5 is

Question 2:

Find the coefficient of a5b7 in (a – 2b)12

Answer


.

Assuming that a5b7 occurs in the (r + 1)th term of the expansion (a – 2b)12, we obtain

Comparing the indices of a and b in a5 b7 and in Tr +1, we obtain

r = 7

Thus, the coefficient of a5b7 is

Question 3:

Write the general term in the expansion of (x2 – y)6

Answer


Page 9 of 25

It is known that the general term Tr+1 {which is the (r + 1)th term} in the binomial

expansion of (a + b)n is given by .

Thus, the general term in the expansion of (x2 – y6) is

Question 4:

Write the general term in the expansion of (x2 – yx)12, x ≠ 0

Answer

It is known that the general term Tr+1 {which is the (r + 1)th term} in the binomial

expansion of (a + b)n is given by .

Thus, the general term in the expansion of(x2 – yx)12 is

Question 5:

Find the 4th term in the expansion of (x – 2y)12 .

Answer


.

Thus, the 4th term in the expansion of (x – 2y)12 is

Question 6:

Find the 13th term in the expansion of .

Answer


.


Page 10 of 25

Thus, 13th term in the expansion of is

Question 7:

Find the middle terms in the expansions of

Answer

It is known that in the expansion of (a + b)n, if n is odd, then there are two middle

terms, namely, term and term.

Therefore, the middle terms in the expansion of are term and

term


Page 11 of 25

Thus, the middle terms in the expansion of are .

Question 8:

Find the middle terms in the expansions of

Answer

It is known that in the expansion (a + b)n, if n is even, then the middle term is

term.

Therefore, the middle term in the expansion of is term

Thus, the middle term in the expansion of is 61236 x5y5.

Question 9:

In the expansion of (1 + a)m + n, prove that coefficients of am and an are equal.

Answer


.

Assuming that am occurs in the (r + 1)th term of the expansion (1 + a)m + n, we obtain

Comparing the indices of a in am and in Tr + 1, we obtain

r = m


Page 12 of 25

Therefore, the coefficient of am is

Assuming that an occurs in the (k + 1)th term of the expansion (1 + a)m+n, we obtain

Comparing the indices of a in an and in Tk + 1, we obtain

k = n

Therefore, the coefficient of an is

Thus, from (1) and (2), it can be observed that the coefficients of am and an in the

expansion of (1 + a)m + n are equal.

Question 10:

The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of

(x + 1)n are in the ratio 1:3:5. Find n and r.

Answer

It is known that (k + 1)th term, (Tk+1), in the binomial expansion of (a + b)n is given by

.

Therefore, (r – 1)th term in the expansion of (x + 1)n is

r th term in the expansion of (x + 1)n is

(r + 1)th term in the expansion of (x + 1)n is

Therefore, the coefficients of the (r – 1)th, rth, and (r + 1)th terms in the expansion of (x

+ 1)n are respectively. Since these coefficients are in the ratio

1:3:5, we obtain


Page 13 of 25

Multiplying (1) by 3 and subtracting it from (2), we obtain

4r – 12 = 0

⇒ r = 3

Putting the value of r in (1), we obtain

n – 12 + 5 = 0

⇒ n = 7

Thus, n = 7 and r = 3

Question 11:

Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn

in the expansion of (1 + x)2n–1 .

Answer


.


Page 14 of 25

Assuming that xn occurs in the (r + 1)th term of the expansion of (1 + x)2n, we obtain

Comparing the indices of x in xn and in Tr + 1, we obtain

r = n

Therefore, the coefficient of xn in the expansion of (1 + x)2n is

Assuming that xn occurs in the (k +1)th term of the expansion (1 + x)2n – 1, we obtain

Comparing the indices of x in xn and Tk + 1, we obtain

k = n

Therefore, the coefficient of xn in the expansion of (1 + x)2n –1 is

From (1) and (2), it is observed that

Therefore, the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn

in the expansion of (1 + x)2n–1.

Hence, proved.

Question 12:

Find a positive value of m for which the coefficient of x2 in the expansion

(1 + x)m is 6.

Answer


.


Page 15 of 25

Assuming that x2 occurs in the (r + 1)th term of the expansion (1 +x)m, we obtain

Comparing the indices of x in x2 and in Tr + 1, we obtain

r = 2

Therefore, the coefficient of x2 is .

It is given that the coefficient of x2 in the expansion (1 + x)m is 6.

Thus, the positive value of m, for which the coefficient of x2 in the expansion

(1 + x)m is 6, is 4.


Page 16 of 25


Question 1:

Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are

729, 7290 and 30375, respectively.

Answer


.

The first three terms of the expansion are given as 729, 7290, and 30375 respectively.

Therefore, we obtain

Dividing (2) by (1), we obtain



Page 17 of 25


Substituting n = 6 in equation (1), we obtain

a6 = 729

From (5), we obtain

Thus, a = 3, b = 5, and n = 6.

Question 2:

Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.

Answer


.

Assuming that x2 occurs in the (r + 1)th term in the expansion of (3 + ax)9, we obtain

Comparing the indices of x in x2 and in Tr + 1, we obtain


Page 18 of 25

r = 2


Assuming that x3 occurs in the (k + 1)th term in the expansion of (3 + ax)9, we obtain

Comparing the indices of x in x3 and in Tk+ 1, we obtain

k = 3


It is given that the coefficients of x2 and x3 are the same.

Thus, the required value of a is .

Question 3:

Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.

Answer

Using Binomial Theorem, the expressions, (1 + 2x)6 and (1 – x)7, can be expanded as


Page 19 of 25

The complete multiplication of the two brackets is not required to be carried out. Only

those terms, which involve x5, are required.

The terms containing x5 are

Thus, the coefficient of x5 in the given product is 171.

Question 4:

If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a

positive integer.

[Hint: write an = (a – b + b)n and expand]

Answer

In order to prove that (a – b) is a factor of (an – bn), it has to be proved that

an – bn = k (a – b), where k is some natural number

It can be written that, a = a – b + b

This shows that (a – b) is a factor of (an – bn), where n is a positive integer.


Page 20 of 25

Question 5:

Evaluate .

Answer

Firstly, the expression (a + b)6 – (a – b)6 is simplified by using Binomial Theorem.

This can be done as

Question 6:

Find the value of .

Answer

Firstly, the expression (x + y)4 + (x – y)4 is simplified by using Binomial Theorem.

This can be done as


Page 21 of 25

Question 7:

Find an approximation of (0.99)5 using the first three terms of its expansion.

Answer

0.99 = 1 – 0.01

Thus, the value of (0.99)5 is approximately 0.951.

Question 8:


Page 22 of 25

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in

the expansion of

Answer

In the expansion, ,

Fifth term from the beginning

Fifth term from the end

Therefore, it is evident that in the expansion of , the fifth term from the

beginning is and the fifth term from the end is .

It is given that the ratio of the fifth term from the beginning to the fifth term from the

end is . Therefore, from (1) and (2), we obtain


Page 23 of 25

Thus, the value of n is 10.

Question 9:

Expand using Binomial Theorem .

Answer

Using Binomial Theorem, the given expression can be expanded as


Page 24 of 25

Again by using Binomial Theorem, we obtain

From (1), (2), and (3), we obtain

Question 10:


Page 25 of 25

Find the expansion of using binomial theorem.

Answer

Using Binomial Theorem, the given expression can be expanded as

Again by using Binomial Theorem, we obtain


Class XI Chapter 9 – Sequences and Series Maths

Exercise 9.1

Question 1:

Write the first five terms of the sequences whose nth term is

Answer

Substituting n = 1, 2, 3, 4, and 5, we obtain

Therefore, the required terms are 3, 8, 15, 24, and 35.

Question 2:


Answer

Substituting n = 1, 2, 3, 4, 5, we obtain

Therefore, the required terms are .

Question 3:

Write the first five terms of the sequences whose nth term is an = 2n

Answer

an = 2n



Page 2 of 80


Page 2 of 80

Therefore, the required terms are 2, 4, 8, 16, and 32.

Question 4:


Answer


Therefore, the required terms are .

Question 5:


Answer



Page 3 of 80

Therefore, the required terms are 25, –125, 625, –3125, and 15625.

Question 6:


Answer


Therefore, the required terms are

Question 7:

Find the 17th term in the following sequence whose nth term is

Answer

Substituting n = 17, we obtain



Page 4 of 80

Question 8:


Answer


Question 9:


Answer


Question 10:


Answer


Question 11:

Write the first five terms of the following sequence and obtain the corresponding series:

Answer


Page 5 of 80

Hence, the first five terms of the sequence are 3, 11, 35, 107, and 323.

The corresponding series is 3 + 11 + 35 + 107 + 323 + …

Question 12:


Answer

Hence, the first five terms of the sequence are

The corresponding series is

Question 13:


Answer


Page 6 of 80

Hence, the first five terms of the sequence are 2, 2, 1, 0, and –1.

The corresponding series is 2 + 2 + 1 + 0 + (–1) + …

Question 14:

The Fibonacci sequence is defined by

Find

Answer


Page 7 of 80

Exercise 9.2

Question 1:

Find the sum of odd integers from 1 to 2001.

Answer

The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

This sequence forms an A.P.

Here, first term, a = 1

Common difference, d = 2

Thus, the sum of odd numbers from 1 to 2001 is 1002001.

Question 2:

Find the sum of all natural numbers lying between 100 and 1000, which are multiples of

5.

Answer

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105,

110, … 995.


Page 8 of 80

Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples

of 5, is 98450.

Question 3:

In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next

five terms. Show that 20th term is –112.

Answer

First term = 2

Let d be the common difference of the A.P.

Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d, …

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

According to the given condition,


Page 9 of 80

Thus, the 20th term of the A.P. is –112.

Question 4:

How many terms of the A.P. are needed to give the sum –25?

Answer

Let the sum of n terms of the given A.P. be –25.

It is known that, , where n = number of terms, a = first term, and

d = common difference

Here, a = –6

Therefore, we obtain

Question 5:

In an A.P., if pth term is and qth term is , prove that the sum of first pq terms is


Page 10 of 80

Answer

It is known that the general term of an A.P. is an = a + (n – 1)d

∴ According to the given information,

Subtracting (2) from (1), we obtain

Putting the value of d in (1), we obtain

Thus, the sum of first pq terms of the A.P. is .


Page 11 of 80

Question 6:

If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last

term

Answer

Let the sum of n terms of the given A.P. be 116.

Here, a = 25 and d = 22 – 25 = – 3

However, n cannot be equal to . Therefore, n = 8

∴ a8 = Last term = a + (n – 1)d = 25 + (8 – 1) (– 3)

= 25 + (7) (– 3) = 25 – 21

= 4

Thus, the last term of the A.P. is 4.

Question 7:

Find the sum to n terms of the A.P., whose kth term is 5k + 1.

Answer

It is given that the kth term of the A.P. is 5k + 1.

kth term = ak = a + (k – 1)d

∴ a + (k – 1)d = 5k + 1

a + kd – d = 5k + 1


Page 12 of 80

Comparing the coefficient of k, we obtain d = 5

a – d = 1

⇒ a – 5 = 1

⇒ a = 6

Question 8:

If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the

common difference.

Answer

It is known that,


Comparing the coefficients of n2 on both sides, we obtain

∴ d = 2 q

Thus, the common difference of the A.P. is 2q.


Page 13 of 80

Question 9:

The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find

the ratio of their 18th terms.

Answer

Let a1, a2, and d1, d2 be the first terms and the common difference of the first and

second arithmetic progression respectively.


Substituting n = 35 in (1), we obtain


Thus, the ratio of 18th term of both the A.P.s is 179: 321.

Question 10:

If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find

the sum of the first (p + q) terms.

Answer

Let a and d be the first term and the common difference of the A.P. respectively.

Here,


Page 14 of 80


Thus, the sum of the first (p + q) terms of the A.P. is 0.

Question 11:

Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.

Prove that

Answer

Let a1 and d be the first term and the common difference of the A.P. respectively.

According to the given information,


Page 15 of 80



Equating both the values of d obtained in (4) and (5), we obtain


Page 16 of 80

Dividing both sides by pqr, we obtain

Thus, the given result is proved.

Question 12:

The ratio of the sums of m and n terms of an A.P. is m2: n2. Show that the ratio of mth

and nth term is (2m – 1): (2n – 1).

Answer

Let a and b be the first term and the common difference of the A.P. respectively.


Putting m = 2m – 1 and n = 2n – 1 in (1), we obtain


Page 17 of 80



Question 13:

If the sum of n terms of an A.P. is and its mth term is 164, find the value of m.

Answer


am = a + (m – 1)d = 164 … (1)

Sum of n terms,

Here,

Comparing the coefficient of n2 on both sides, we obtain

Comparing the coefficient of n on both sides, we obtain


Page 18 of 80

Therefore, from (1), we obtain

8 + (m – 1) 6 = 164

⇒ (m – 1) 6 = 164 – 8 = 156

⇒ m – 1 = 26

⇒ m = 27

Thus, the value of m is 27.

Question 14:

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Answer

Let A1, A2, A3, A4, and A5 be five numbers between 8 and 26 such that

8, A1, A2, A3, A4, A5, 26 is an A.P.

Here, a = 8, b = 26, n = 7

Therefore, 26 = 8 + (7 – 1) d

⇒ 6d = 26 – 8 = 18

⇒ d = 3

A1 = a + d = 8 + 3 = 11

A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14

A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17

A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20

A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23

Thus, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

Question 15:

If is the A.M. between a and b, then find the value of n.

Answer

A.M. of a and b



Page 19 of 80

Question 16:

Between 1 and 31, m numbers have been inserted in such a way that the resulting

sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5:9. Find the value of

m.

Answer

Let A1, A2, … Am be m numbers such that 1, A1, A2, … Am, 31 is an A.P.

Here, a = 1, b = 31, n = m + 2

∴ 31 = 1 + (m + 2 – 1) (d)

⇒ 30 = (m + 1) d

A1 = a + d

A2 = a + 2d

A3 = a + 3d …

∴ A7 = a + 7d

Am–1 = a + (m – 1) d



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Thus, the value of m is 14.

Question 17:

A man starts repaying a loan as first installment of Rs. 100. If he increases the

installment by Rs 5 every month, what amount he will pay in the 30th installment?

Answer

The first installment of the loan is Rs 100.

The second installment of the loan is Rs 105 and so on.

The amount that the man repays every month forms an A.P.

The A.P. is 100, 105, 110, …

First term, a = 100


A30 = a + (30 – 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Thus, the amount to be paid in the 30th installment is Rs 245.


Page 21 of 80

Question 18:

The difference between any two consecutive interior angles of a polygon is 5°. If the

smallest angle is 120°, find the number of the sides of the polygon.

Answer

The angles of the polygon will form an A.P. with common difference d as 5° and first

term a as 120°.

It is known that the sum of all angles of a polygon with n sides is 180° (n – 2).


Page 22 of 80

Exercise 9.3

Question 1:

Find the 20th and nthterms of the G.P.

Answer

The given G.P. is

Here, a = First term =

r = Common ratio =

Question 2:

Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

Answer

Common ratio, r = 2

Let a be the first term of the G.P.

∴ a8 = ar 8–1 = ar7

⇒ ar7 = 192

a(2)7 = 192

a(2)7 = (2)6 (3)


Page 23 of 80

Question 3:

The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.

Answer

Let a be the first term and r be the common ratio of the G.P.


a5 = a r5–1 = a r4 = p … (1)

a8 = a r8–1 = a r7 = q … (2)

a11 = a r11–1 = a r10 = s … (3)

Dividing equation (2) by (1), we obtain


Equating the values of r3 obtained in (4) and (5), we obtain


Question 4:

The 4th term of a G.P. is square of its second term, and the first term is –3. Determine

its 7th term.


Page 24 of 80

Answer


∴ a = –3

It is known that, an = arn–1

∴a4 = ar3 = (–3) r3

a2 = a r1 = (–3) r


(–3) r3 = [(–3) r]2

⇒ –3r3 = 9 r2

⇒ r = –3

a7 = a r 7–1 = a r6 = (–3) (–3)6 = – (3)7 = –2187

Thus, the seventh term of the G.P. is –2187.

Question 5:

Which term of the following sequences:

(a) (b) (c)

Answer

(a) The given sequence is

Here, a = 2 and r =

Let the nth term of the given sequence be 128.


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Thus, the 13th term of the given sequence is 128.

(b) The given sequence is

Here,

Let the nth term of the given sequence be 729.

Thus, the 12th term of the given sequence is 729.

(c) The given sequence is


Page 26 of 80

Here,

Let the nth term of the given sequence be .

Thus, the 9th term of the given sequence is .

Question 6:

For what values of x, the numbers are in G.P?

Answer

The given numbers are .

Common ratio =

Also, common ratio =


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Thus, for x = ± 1, the given numbers will be in G.P.

Question 7:

Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …

Answer

The given G.P. is 0.15, 0.015, 0.00015, …

Here, a = 0.15 and

Question 8:

Find the sum to n terms in the geometric progression

Answer

The given G.P. is

Here,


Page 28 of 80

Question 9:


Answer

The given G.P. is

Here, first term = a1 = 1

Common ratio = r = – a

Question 10:


Answer


Page 29 of 80

The given G.P. is

Here, a = x3 and r = x2

Question 11:

Evaluate

Answer

The terms of this sequence 3, 32, 33, … forms a G.P.

Substituting this value in equation (1), we obtain

Question 12:

The sum of first three terms of a G.P. is and their product is 1. Find the common

ratio and the terms.

Answer


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Let be the first three terms of the G.P.

From (2), we obtain

a3 = 1

⇒ a = 1 (Considering real roots only)

Substituting a = 1 in equation (1), we obtain

Thus, the three terms of G.P. are .

Question 13:

How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?

Answer

The given G.P. is 3, 32, 33, …

Let n terms of this G.P. be required to obtain the sum as 120.

Here, a = 3 and r = 3


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∴ n = 4

Thus, four terms of the given G.P. are required to obtain the sum as 120.

Question 14:

The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128.

Determine the first term, the common ratio and the sum to n terms of the G.P.

Answer

Let the G.P. be a, ar, ar2, ar3, …


a + ar + ar2 = 16 and ar3 + ar4 + ar5 = 128

⇒ a (1 + r + r2) = 16 … (1)

ar3(1 + r + r2) = 128 … (2)


Substituting r = 2 in (1), we obtain

a (1 + 2 + 4) = 16

⇒ a (7) = 16


Page 32 of 80

Question 15:

Given a G.P. with a = 729 and 7th term 64, determine S7.

Answer

a = 729

a7 = 64

Let r be the common ratio of the G.P.

It is known that, an = a rn–1

a7 = ar7–1 = (729)r6

⇒ 64 = 729 r6

Also, it is known that,


Page 33 of 80

Question 16:

Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the

third term.

Answer


According to the given conditions,

a5 = 4 × a3

ar4 = 4ar2

⇒ r2 = 4

∴ r = ± 2

From (1), we obtain


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Thus, the required G.P. is

4, –8, 16, –32, …

Question 17:

If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are

in G.P.

Answer



a4 = a r3 = x … (1)

a10 = a r9 = y … (2)

a16 = a r15 = z … (3)




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∴

Thus, x, y, z are in G. P.

Question 18:

Find the sum to n terms of the sequence, 8, 88, 888, 8888…

Answer

The given sequence is 8, 88, 888, 8888…

This sequence is not a G.P. However, it can be changed to G.P. by writing the terms as

Sn = 8 + 88 + 888 + 8888 + …………….. to n terms

Question 19:

Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32

and 128, 32, 8, 2, .

Answer

Required sum =


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Here, 4, 2, 1, is a G.P.

First term, a = 4

Common ratio, r =

It is known that,

∴Required sum =

Question 20:

Show that the products of the corresponding terms of the sequences

form a G.P, and find the common ratio.

Answer

It has to be proved that the sequence, aA, arAR, ar2AR2, …arn–1ARn–1, forms a G.P.

Thus, the above sequence forms a G.P. and the common ratio is rR.

Question 21:


Page 37 of 80

Find four numbers forming a geometric progression in which third term is greater than

the first term by 9, and the second term is greater than the 4th by 18.

Answer


a1 = a, a2 = ar, a3 = ar2, a4 = ar

3

By the given condition,

a3 = a1 + 9

⇒ ar2 = a + 9 … (1)

a2 = a4 + 18

⇒ ar = ar3 + 18 … (2)


a(r2 – 1) = 9 … (3)

ar (1– r2) = 18 … (4)


Substituting the value of r in (1), we obtain

4a = a + 9

⇒ 3a = 9

∴ a = 3

Thus, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)2, and 3(–2)3 i.e., 3¸–6, 12,

and –24.

Question 22:

If the terms of a G.P. are a, b and c, respectively. Prove that

Answer

Let A be the first term and R be the common ratio of the G.P.


ARp–1 = a


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ARq–1 = b

ARr–1 = c

aq–r br–p cp–q

= Aq–r × R(p–1) (q–r) × Ar–p × R(q–1) (r-p) × Ap–q × R(r –1)(p–q)

= Aq – r + r – p + p – q × R (pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)

= A0 × R0

= 1


Question 23:

If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n

terms, prove that P2 = (ab)n.

Answer

The first term of the G.P is a and the last term is b.

Therefore, the G.P. is a, ar, ar2, ar3, … arn–1, where r is the common ratio.

b = arn–1 … (1)

P = Product of n terms

= (a) (ar) (ar2) … (arn–1)

= (a × a ×…a) (r × r2 × …rn–1)

= an r 1 + 2 +…(n–1) … (2)

Here, 1, 2, …(n – 1) is an A.P.

∴1 + 2 + ……….+ (n – 1)



Page 39 of 80

Question 24:

Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from

.

Answer


Since there are n terms from (n +1)th to (2n)th term,

Sum of terms from(n + 1)th to (2n)th term

a n +1 = ar n + 1 – 1 = arn

Thus, required ratio =

Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to

(2n)th term is .

Question 25:

If a, b, c and d are in G.P. show that .

Answer

a, b, c, d are in G.P.

Therefore,

bc = ad … (1)

b2 = ac … (2)

c2 = bd … (3)

It has to be proved that,

(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc – cd)2

R.H.S.

= (ab + bc + cd)2


Page 40 of 80

= (ab + ad + cd)2 [Using (1)]

= [ab + d (a + c)]2

= a2b2 + 2abd (a + c) + d2 (a + c)2

= a2b2 +2a2bd + 2acbd + d2(a2 + 2ac + c2)

= a2b2 + 2a2c2 + 2b2c2 + d2a2 + 2d2b2 + d2c2 [Using (1) and (2)]

= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2

= a2b2 + a2c2 + a2d2 + b2 × b2 + b2c2 + b2d2 + c2b2 + c2 × c2 + c2d2

[Using (2) and (3) and rearranging terms]

= a2(b2 + c2 + d2) + b2 (b2 + c2 + d2) + c2 (b2+ c2 + d2)

= (a2 + b2 + c2) (b2 + c2 + d2)

= L.H.S.

∴ L.H.S. = R.H.S.

∴

Question 26:

Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Answer

Let G1 and G2 be two numbers between 3 and 81 such that the series, 3, G1, G2, 81,

forms a G.P.


∴81 = (3) (r)3

⇒ r3 = 27

∴ r = 3 (Taking real roots only)

For r = 3,

G1 = ar = (3) (3) = 9

G2 = ar2 = (3) (3)2 = 27

Thus, the required two numbers are 9 and 27.

Question 27:

Find the value of n so that may be the geometric mean between a and b.

Answer


Page 41 of 80

G. M. of a and b is .

By the given condition,

Squaring both sides, we obtain

Question 28:

The sum of two numbers is 6 times their geometric mean, show that numbers are in the

ratio .

Answer

Let the two numbers be a and b.

G.M. =


Also,


Page 42 of 80

Adding (1) and (2), we obtain

Substituting the value of a in (1), we obtain

Thus, the required ratio is .

Question 29:

If A and G be A.M. and G.M., respectively between two positive numbers, prove that the

numbers are .

Answer

It is given that A and G are A.M. and G.M. between two positive numbers. Let these two

positive numbers be a and b.


a + b = 2A … (3)

ab = G2 … (4)

Substituting the value of a and b from (3) and (4) in the identity (a – b)2 = (a + b)2 –

4ab, we obtain

(a – b)2 = 4A2 – 4G2 = 4 (A2–G2)


Page 43 of 80

(a – b)2 = 4 (A + G) (A – G)



Thus, the two numbers are .

Question 30:

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria

present in the culture originally, how many bacteria will be present at the end of 2nd

hour, 4th hour and nth hour?

Answer

It is given that the number of bacteria doubles every hour. Therefore, the number of

bacteria after every hour will form a G.P.

Here, a = 30 and r = 2

∴ a3 = ar2 = (30) (2)2 = 120

Therefore, the number of bacteria at the end of 2nd hour will be 120.

a5 = ar4 = (30) (2)4 = 480

The number of bacteria at the end of 4th hour will be 480.

an +1 = arn = (30) 2n

Thus, number of bacteria at the end of nth hour will be 30(2)n.

Question 31:

What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual

interest rate of 10% compounded annually?

Answer

The amount deposited in the bank is Rs 500.


Page 44 of 80

At the end of first year, amount = = Rs 500 (1.1)

At the end of 2nd year, amount = Rs 500 (1.1) (1.1)

At the end of 3rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on

∴Amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)

= Rs 500(1.1)10

Question 32:

If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain

the quadratic equation.

Answer

Let the root of the quadratic equation be a and b.


The quadratic equation is given by,

x2– x (Sum of roots) + (Product of roots) = 0

x2 – x (a + b) + (ab) = 0

x2 – 16x + 25 = 0 [Using (1) and (2)]

Thus, the required quadratic equation is x2 – 16x + 25 = 0


Page 45 of 80

Exercise 9.4

Question 1:

Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Answer

The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

nth term, an = n ( n + 1)

Question 2:

Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

Answer

The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

nth term, an = n ( n + 1) ( n + 2)

= (n2 + n) (n + 2)

= n3 + 3n2 + 2n


Page 46 of 80

Question 3:

Find the sum to n terms of the series 3 × 12 + 5 × 22 + 7 × 32 + …

Answer

The given series is 3 ×12 + 5 × 22 + 7 × 32 + …

nth term, an = ( 2n + 1) n2 = 2n3 + n2


Page 47 of 80

Question 4:

Find the sum to n terms of the series

Answer

The given series is

nth term, an =


Page 48 of 80

Adding the above terms column wise, we obtain

Question 5:

Find the sum to n terms of the series

Answer

The given series is 52 + 62 + 72 + … + 202

nth term, an = ( n + 4)2 = n2 + 8n + 16

16th term is (16 + 4)2 = 2022


Page 49 of 80

Question 6:

Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…

Answer

The given series is 3 × 8 + 6 × 11 + 9 × 14 + …

an = (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14, …)

= (3n) (3n + 5)

= 9n2 + 15n

Question 7:

Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …


Page 50 of 80

Answer

The given series is 12 + (12 + 22) + (12 + 22 + 33 ) + …

an = (12 + 22 + 33 +…….+ n2)

Question 8:

Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).


Page 51 of 80

Answer

an = n (n + 1) (n + 4) = n(n2 + 5n + 4) = n3 + 5n2 + 4n

Question 9:

Find the sum to n terms of the series whose nth terms is given by n2 + 2n

Answer

an = n2 + 2n

Consider

The above series 2, 22, 23, … is a G.P. with both the first term and common ratio equal

to 2.

Therefore, from (1) and (2), we obtain


Page 52 of 80

Question 10:

Find the sum to n terms of the series whose nth terms is given by (2n – 1)2

Answer

an = (2n – 1)2 = 4n2 – 4n + 1


Page 53 of 80


Question 1:

Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth

term.

Answer

Let a and d be the first term and the common difference of the A.P. respectively.

It is known that the kth term of an A. P. is given by

ak = a + (k –1) d

∴ am + n = a + (m + n –1) d

am – n = a + (m – n –1) d

am = a + (m –1) d

∴ am + n + am – n = a + (m + n –1) d + a + (m – n –1) d

= 2a + (m + n –1 + m – n –1) d

= 2a + (2m – 2) d

= 2a + 2 (m – 1) d

=2 [a + (m – 1) d]

= 2am

Thus, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

Question 2:

If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Answer

Let the three numbers in A.P. be a – d, a, and a + d.


(a – d) + (a) + (a + d) = 24 … (1)

⇒ 3a = 24

∴ a = 8

(a – d) a (a + d) = 440 … (2)

⇒ (8 – d) (8) (8 + d) = 440

⇒ (8 – d) (8 + d) = 55


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⇒ 64 – d2 = 55

⇒ d2 = 64 – 55 = 9

⇒ d = ± 3

Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are

11, 8, and 5.

Thus, the three numbers are 5, 8, and 11.

Question 3:

Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 =

3 (S2– S1)

Answer


Therefore,


Hence, the given result is proved.

Question 4:

Find the sum of all numbers between 200 and 400 which are divisible by 7.


Page 55 of 80

Answer

The numbers lying between 200 and 400, which are divisible by 7, are

203, 210, 217, … 399

∴First term, a = 203

Last term, l = 399


Let the number of terms of the A.P. be n.

∴ an = 399 = a + (n –1) d

⇒ 399 = 203 + (n –1) 7

⇒ 7 (n –1) = 196

⇒ n –1 = 28

⇒ n = 29

Thus, the required sum is 8729.

Question 5:

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Answer

The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.

This forms an A.P. with both the first term and common difference equal to 2.

⇒100 = 2 + (n –1) 2

⇒ n = 50


Page 56 of 80

The integers from 1 to 100, which are divisible by 5, are 5, 10… 100.

This forms an A.P. with both the first term and common difference equal to 5.

∴100 = 5 + (n –1) 5

⇒ 5n = 100

⇒ n = 20

The integers, which are divisible by both 2 and 5, are 10, 20, … 100.

This also forms an A.P. with both the first term and common difference equal to 10.

∴100 = 10 + (n –1) (10)

⇒ 100 = 10n

⇒ n = 10

∴Required sum = 2550 + 1050 – 550 = 3050

Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.

Question 6:

Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Answer

The two-digit numbers, which when divided by 4, yield 1 as remainder, are

13, 17, … 97.

This series forms an A.P. with first term 13 and common difference 4.

Let n be the number of terms of the A.P.

It is known that the nth term of an A.P. is given by, an = a + (n –1) d

∴97 = 13 + (n –1) (4)

⇒ 4 (n –1) = 84

⇒ n – 1 = 21


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⇒ n = 22

Sum of n terms of an A.P. is given by,

Thus, the required sum is 1210.

Question 7:

If f is a function satisfying such that

, find the value of n.

Answer

It is given that,

f (x + y) = f (x) × f (y) for all x, y ∈ N … (1)

f (1) = 3

Taking x = y = 1 in (1), we obtain

f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9

Similarly,

f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) f (2) = 3 × 9 = 27

f (4) = f (1 + 3) = f (1) f (3) = 3 × 27 = 81

∴ f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with both the first term and

common ratio equal to 3.

It is known that,

It is given that,


Page 58 of 80

Thus, the value of n is 4.

Question 8:

The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and

2, respectively. Find the last term and the number of terms.

Answer

Let the sum of n terms of the G.P. be 315.

It is known that,

It is given that the first term a is 5 and common ratio r is 2.

∴Last term of the G.P = 6th term = ar6 – 1 = (5)(2)5 = (5)(32) = 160

Thus, the last term of the G.P. is 160.

Question 9:

The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the

common ratio of G.P.

Answer

Let a and r be the first term and the common ratio of the G.P. respectively.

∴ a = 1

a3 = ar2 = r2


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a5 = ar4 = r4

∴ r2 + r4 = 90

⇒ r4 + r2 – 90 = 0

Thus, the common ratio of the G.P. is ±3.

Question 10:

The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in

that order, we obtain an arithmetic progression. Find the numbers.

Answer

Let the three numbers in G.P. be a, ar, and ar2.

From the given condition, a + ar + ar2 = 56

⇒ a (1 + r + r2) = 56

… (1)

a – 1, ar – 7, ar2 – 21 forms an A.P.

∴(ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)

⇒ ar – a – 6 = ar2 – ar – 14

⇒ar2 – 2ar + a = 8

⇒ar2 – ar – ar + a = 8

⇒a(r2 + 1 – 2r) = 8

⇒ a (r – 1)2 = 8 … (2)

⇒7(r2 – 2r + 1) = 1 + r + r2

⇒7r2 – 14 r + 7 – 1 – r – r2 = 0

⇒ 6r2 – 15r + 6 = 0

⇒ 6r2 – 12r – 3r + 6 = 0

⇒ 6r (r – 2) – 3 (r – 2) = 0

⇒ (6r – 3) (r – 2) = 0


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When r = 2, a = 8

When

Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.

When , the three numbers in G.P. are 32, 16, and 8.

Thus, in either case, the three required numbers are 8, 16, and 32.

Question 11:

A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the

sum of terms occupying odd places, then find its common ratio.

Answer

Let the G.P. be T1, T2, T3, T4, … T2n.

Number of terms = 2n


T1 + T2 + T3 + …+ T2n = 5 [T1 + T3 + … +T2n–1]

⇒ T1 + T2 + T3 + … + T2n – 5 [T1 + T3 + … + T2n–1] = 0

⇒ T2 + T4 + … + T2n = 4 [T1 + T3 + … + T2n–1]

Let the G.P. be a, ar, ar2, ar3, …

Thus, the common ratio of the G.P. is 4.

Question 12:

The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If

its first term is 11, then find the number of terms.

Answer

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.

Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d


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Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d]

+ [a + n – 1) d]

= 4a + (4n – 10) d


4a + 6d = 56

⇒ 4(11) + 6d = 56 [Since a = 11 (given)]

⇒ 6d = 12

⇒ d = 2

∴ 4a + (4n –10) d = 112

⇒ 4(11) + (4n – 10)2 = 112

⇒ (4n – 10)2 = 68

⇒ 4n – 10 = 34

⇒ 4n = 44

⇒ n = 11

Thus, the number of terms of the A.P. is 11.

Question 13:

If , then show that a, b, c and d are in G.P.

Answer

It is given that,


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Thus, a, b, c, and d are in G.P.

Question 14:

Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove

that P2Rn = Sn

Answer

Let the G.P. be a, ar, ar2, ar3, … arn – 1…



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Hence, P2 Rn = Sn

Question 15:

The pth, qth and rth terms of an A.P. are a, b, c respectively. Show that

Answer

Let t and d be the first term and the common difference of the A.P. respectively.

The nth term of an A.P. is given by, an = t + (n – 1) d

Therefore,

ap = t + (p – 1) d = a … (1)


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aq = t + (q – 1)d = b … (2)

ar = t + (r – 1) d = c … (3)

Subtracting equation (2) from (1), we obtain

(p – 1 – q + 1) d = a – b

⇒ (p – q) d = a – b

Subtracting equation (3) from (2), we obtain

(q – 1 – r + 1) d = b – c

⇒ (q – r) d = b – c

Equating both the values of d obtained in (4) and (5), we obtain


Question 16:

If a are in A.P., prove that a, b, c are in A.P.

Answer

It is given that a are in A.P.


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Thus, a, b, and c are in A.P.

Question 17:

If a, b, c, d are in G.P, prove that are in G.P.

Answer

It is given that a, b, c,and d are in G.P.

∴b2 = ac … (1)

c2 = bd … (2)

ad = bc … (3)

It has to be proved that (an + bn), (bn + cn), (cn + dn) are in G.P. i.e.,

(bn + cn)2 = (an + bn) (cn + dn)

Consider L.H.S.

(bn + cn)2 = b2n + 2bncn + c2n

= (b2)n+ 2bncn + (c2) n

= (ac)n + 2bncn + (bd)n [Using (1) and (2)]

= an cn + bncn+ bn cn + bn dn

= an cn + bncn+ an dn + bn dn [Using (3)]

= cn (an + bn) + dn (an + bn)

= (an + bn) (cn + dn)

= R.H.S.

∴ (bn + cn)2 = (an + bn) (cn + dn)


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Thus, (an + bn), (bn + cn), and (cn + dn) are in G.P.

Question 18:

If a and b are the roots of are roots of , where a,

b, c, d, form a G.P. Prove that (q + p): (q – p) = 17:15.

Answer

It is given that a and b are the roots of x2 – 3x + p = 0

∴ a + b = 3 and ab = p … (1)

Also, c and d are the roots of

∴c + d = 12 and cd = q … (2)

It is given that a, b, c, d are in G.P.

Let a = x, b = xr, c = xr2, d = xr3


x + xr = 3

⇒ x (1 + r) = 3

xr2 + xr3 =12

⇒ xr2 (1 + r) = 12

On dividing, we obtain

Case I:

When r = 2 and x =1,

ab = x2r = 2

cd = x2r5 = 32


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Case II:

When r = –2, x = –3,

ab = x2r = –18

cd = x2r5 = – 288

Thus, in both the cases, we obtain (q + p): (q – p) = 17:15

Question 19:

The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that

.

Answer

Let the two numbers be a and b.

A.M and G.M. =


Using this in the identity (a – b)2 = (a + b)2 – 4ab, we obtain


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Adding (1) and (2), we obtain


Question 20:

If a, b, c are in A.P,; b, c, d are in G.P and are in A.P. prove that a, c, e are in

G.P.

Answer

It is given that a, b, c are in A.P.

∴ b – a = c – b … (1)

It is given that b, c, d, are in G.P.

∴ c2 = bd … (2)


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Also, are in A.P.

It has to be proved that a, c, e are in G.P. i.e., c2 = ae

From (1), we obtain

From (2), we obtain

Substituting these values in (3), we obtain

Thus, a, c, and e are in G.P.

Question 21:

Find the sum of the following series up to n terms:

(i) 5 + 55 + 555 + … (ii) .6 +.66 +. 666 +…

Answer

(i) 5 + 55 + 555 + …


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Let Sn = 5 + 55 + 555 + ….. to n terms

(ii) .6 +.66 +. 666 +…

Let Sn = 06. + 0.66 + 0.666 + … to n terms

Question 22:


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Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.

Answer

The given series is 2 × 4 + 4 × 6 + 6 × 8 + … n terms

∴ nth term = an = 2n × (2n + 2) = 4n2 + 4n

a20 = 4 (20)2 + 4(20) = 4 (400) + 80 = 1600 + 80 = 1680

Thus, the 20th term of the series is 1680.

Question 23:

Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …

Answer

The given series is 3 + 7 + 13 + 21 + 31 + …

S = 3 + 7 + 13 + 21 + 31 + …+ an–1 + an

S = 3 + 7 + 13 + 21 + …. + an – 2 + an – 1 + an

On subtracting both the equations, we obtain

S – S = [3 + (7 + 13 + 21 + 31 + …+ an–1 + an)] – [(3 + 7 + 13 + 21 + 31 + …+ an–1)

+ an]

S – S = 3 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (an – an–1)] – an

0 = 3 + [4 + 6 + 8 + … (n –1) terms] – an

an = 3 + [4 + 6 + 8 + … (n –1) terms]


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Question 24:

If S1, S2, S3 are the sum of first n natural numbers, their squares and their cubes,

respectively, show that

Answer

From the given information,


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Thus, from (1) and (2), we obtain

Question 25:

Find the sum of the following series up to n terms:

Answer

The nth term of the given series is


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Question 26:

Show that

Answer

nth term of the numerator = n(n + 1)2 = n3 + 2n2 + n

nth term of the denominator = n2(n + 1) = n3 + n2


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Page 76 of 80



Question 27:

A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the

balance in annual installments of Rs 500 plus 12% interest on the unpaid amount. How

much will be the tractor cost him?

Answer

It is given that the farmer pays Rs 6000 in cash.

Therefore, unpaid amount = Rs 12000 – Rs 6000 = Rs 6000

According to the given condition, the interest paid annually is

12% of 6000, 12% of 5500, 12% of 5000, …, 12% of 500

Thus, total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + … +

12% of 500

= 12% of (6000 + 5500 + 5000 + … + 500)


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= 12% of (500 + 1000 + 1500 + … + 6000)

Now, the series 500, 1000, 1500 … 6000 is an A.P. with both the first term and common

difference equal to 500.

Let the number of terms of the A.P. be n.

∴ 6000 = 500 + (n – 1) 500

⇒ 1 + (n – 1) = 12

⇒ n = 12

∴Sum of the A.P

Thus, total interest to be paid = 12% of (500 + 1000 + 1500 + … + 6000)

= 12% of 39000 = Rs 4680

Thus, cost of tractor = (Rs 12000 + Rs 4680) = Rs 16680

Question 28:

Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the

balance in annual installment of Rs 1000 plus 10% interest on the unpaid amount. How

much will the scooter cost him?

Answer

It is given that Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.

∴Unpaid amount = Rs 22000 – Rs 4000 = Rs 18000

According to the given condition, the interest paid annually is

10% of 18000, 10% of 17000, 10% of 16000 … 10% of 1000

Thus, total interest to be paid = 10% of 18000 + 10% of 17000 + 10% of 16000 + … +

10% of 1000

= 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of (1000 + 2000 + 3000 + … + 18000)

Here, 1000, 2000, 3000 … 18000 forms an A.P. with first term and common difference

both equal to 1000.

Let the number of terms be n.

∴ 18000 = 1000 + (n – 1) (1000)

⇒ n = 18


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∴ Total interest paid = 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of Rs 171000 = Rs 17100

∴Cost of scooter = Rs 22000 + Rs 17100 = Rs 39100

Question 29:

A person writes a letter to four of his friends. He asks each one of them to copy the

letter and mail to four different persons with instruction that they move the chain

similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one

letter. Find the amount spent on the postage when 8th set of letter is mailed.

Answer

The numbers of letters mailed forms a G.P.: 4, 42, … 48

First term = 4

Common ratio = 4

Number of terms = 8

It is known that the sum of n terms of a G.P. is given by

It is given that the cost to mail one letter is 50 paisa.

∴Cost of mailing 87380 letters = Rs 43690

Thus, the amount spent when 8th set of letter is mailed is Rs 43690.

Question 30:

A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the

amount in 15th year since he deposited the amount and also calculate the total amount

after 20 years.


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Answer

It is given that the man deposited Rs 10000 in a bank at the rate of 5% simple interest

annually.

∴ Interest in first year

∴Amount in 15th year = Rs

= Rs 10000 + 14 × Rs 500

= Rs 10000 + Rs 7000

= Rs 17000

Amount after 20 years =

= Rs 10000 + 20 × Rs 500

= Rs 10000 + Rs 10000

= Rs 20000

Question 31:

A manufacturer reckons that the value of a machine, which costs him Rs 15625, will

depreciate each year by 20%. Find the estimated value at the end of 5 years.

Answer

Cost of machine = Rs 15625

Machine depreciates by 20% every year.

Therefore, its value after every year is 80% of the original cost i.e., of the original

cost.

∴ Value at the end of 5 years = = 5 × 1024 = 5120

Thus, the value of the machine at the end of 5 years is Rs 5120.

Question 32:

150 workers were engaged to finish a job in a certain number of days. 4 workers

dropped out on second day, 4 more workers dropped out on third day and so on. It took


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8 more days to finish the work. Find the number of days in which the work was

completed.

Answer

Let x be the number of days in which 150 workers finish the work.


150x = 150 + 146 + 142 + …. (x + 8) terms

The series 150 + 146 + 142 + …. (x + 8) terms is an A.P. with first term 146, common

difference –4 and number of terms as (x + 8)

However, x cannot be negative.

∴x = 17

Therefore, originally, the number of days in which the work was completed is 17.

Thus, required number of days = (17 + 8) = 25

Class XI Chapter 10 – Straight Lines Maths

Exercise 10.1

Question 1:

Draw a quadrilateral in the Cartesian plane, whose vertices are (–4, 5), (0, 7), (5, –5)

and (–4, –2). Also, find its area.

Answer

Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5), and D (–

4, –2).

Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA,

the given quadrilateral can be drawn as

To find the area of quadrilateral ABCD, we draw one diagonal, say AC.

Accordingly, area (ABCD) = area (∆ABC) + area (∆ACD)

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is

Therefore, area of ∆ABC


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Area of ∆ACD

Thus, area (ABCD)

Question 2:

The base of an equilateral triangle with side 2a lies along they y-axis such that the mid

point of the base is at the origin. Find vertices of the triangle.

Answer

Let ABC be the given equilateral triangle with side 2a.

Accordingly, AB = BC = CA = 2a

Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin.

i.e., BO = OC = a, where O is the origin.

Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B

are (0, –a).


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It is known that the line joining a vertex of an equilateral triangle with the mid-point of

its opposite side is perpendicular.

Hence, vertex A lies on the y-axis.

On applying Pythagoras theorem to ∆AOC, we obtain

(AC)2 = (OA)2 + (OC)2

⇒ (2a)2 = (OA)2 + a2

⇒ 4a2 – a2 = (OA)2

⇒ (OA)2 = 3a2

⇒ OA =

∴Coordinates of point A =

Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and or

(0, a), (0, –a), and .

Question 3:

Find the distance between and when: (i) PQ is parallel to the y-axis,

(ii) PQ is parallel to the x-axis.

Answer

The given points are and .

(i) When PQ is parallel to the y-axis, x1 = x2.


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In this case, distance between P and Q

(ii) When PQ is parallel to the x-axis, y1 = y2.

In this case, distance between P and Q

Question 4:

Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Answer

Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).


a2 – 14a + 85 = a2 – 6a + 25

⇒ –14a + 6a = 25 – 85

⇒ –8a = –60

Thus, the required point on the x-axis is .

Question 5:

Find the slope of a line, which passes through the origin, and the mid-point of

the line segment joining the points P (0, –4) and B (8, 0).

Answer


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The coordinates of the mid-point of the line segment joining the points

P (0, –4) and B (8, 0) are

It is known that the slope (m) of a non-vertical line passing through the points (x1, y1)

and (x2, y2) is given by .

Therefore, the slope of the line passing through (0, 0) and (4, –2) is

.

Hence, the required slope of the line is .

Question 6:

Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1)

are the vertices of a right angled triangle.

Answer

The vertices of the given triangle are A (4, 4), B (3, 5), and C (–1, –1).

It is known that the slope (m) of a non-vertical line passing through the points (x1, y1)

and (x2, y2) is given by .

∴Slope of AB (m1)

Slope of BC (m2)

Slope of CA (m3)

It is observed that m1m3 = –1

This shows that line segments AB and CA are perpendicular to each other

i.e., the given triangle is right-angled at A (4, 4).

Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.


Page 6 of 68

Question 7:

Find the slope of the line, which makes an angle of 30° with the positive direction of y-

axis measured anticlockwise.

Answer

If a line makes an angle of 30° with the positive direction of the y-axis measured

anticlockwise, then the angle made by the line with the positive direction of the x-axis

measured anticlockwise is 90° + 30° = 120°.

Thus, the slope of the given line is tan 120° = tan (180° – 60°) = –tan 60°

Question 8:

Find the value of x for which the points (x, –1), (2, 1) and (4, 5) are collinear.

Answer

If points A (x, –1), B (2, 1), and C (4, 5) are collinear, then

Slope of AB = Slope of BC

Thus, the required value of x is 1.


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Question 9:

Without using distance formula, show that points (–2, –1), (4, 0), (3, 3) and

(–3, 2) are vertices of a parallelogram.

Answer

Let points (–2, –1), (4, 0), (3, 3), and (–3, 2) be respectively denoted by A, B, C, and D.

Slope of AB

Slope of CD =

⇒ Slope of AB = Slope of CD

⇒ AB and CD are parallel to each other.

Now, slope of BC =

Slope of AD =

⇒ Slope of BC = Slope of AD

⇒ BC and AD are parallel to each other.

Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is

a parallelogram.

Thus, points (–2, –1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.

Question 10:

Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).

Answer

The slope of the line joining the points (3, –1) and (4, –2) is

Now, the inclination (θ ) of the line joining the points (3, –1) and (4, – 2) is given by


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tan θ= –1

⇒ θ = (90° + 45°) = 135°

Thus, the angle between the x-axis and the line joining the points (3, –1) and (4, –2) is

135°.

Question 11:

The slope of a line is double of the slope of another line. If tangent of the angle between

them is , find the slopes of he lines.

Answer

Let be the slopes of the two given lines such that .

We know that if θisthe angle between the lines l1 and l2 with slopes m1 and m2, then

.

It is given that the tangent of the angle between the two lines is .

Case I


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If m = –1, then the slopes of the lines are –1 and –2.

If m = , then the slopes of the lines are and –1.

Case II

If m = 1, then the slopes of the lines are 1 and 2.

If m = , then the slopes of the lines are .

Hence, the slopes of the lines are –1 and –2 or and –1 or 1 and 2 or .

Question 12:

A line passes through . If slope of the line is m, show that

.

Answer

The slope of the line passing through is .

It is given that the slope of the line is m.

Hence,


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Question 13:

If three point (h, 0), (a, b) and (0, k) lie on a line, show that .

Answer

If the points A (h, 0), B (a, b), and C (0, k) lie on a line, then

Slope of AB = Slope of BC

On dividing both sides by kh, we obtain

Hence,

Question 14:

Consider the given population and year graph. Find the slope of the line AB and using it,

find what will be the population in the year 2010?


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Answer

Since line AB passes through points A (1985, 92) and B (1995, 97), its slope is

Let y be the population in the year 2010. Then, according to the given graph, line AB

must pass through point C (2010, y).

∴Slope of AB = Slope of BC

Thus, the slope of line AB is , while in the year 2010, the population will be 104.5

crores.


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Exercise 10.2

Question 1:

Write the equations for the x and y-axes.

Answer

The y-coordinate of every point on the x-axis is 0.

Therefore, the equation of the x-axis is y = 0.

The x-coordinate of every point on the y-axis is 0.

Therefore, the equation of the y-axis is y = 0.

Question 2:

Find the equation of the line which passes through the point (–4, 3) with slope .

Answer

We know that the equation of the line passing through point , whose slope is m,

is .

Thus, the equation of the line passing through point (–4, 3), whose slope is , is

Question 3:

Find the equation of the line which passes though (0, 0) with slope m.

Answer


is .

Thus, the equation of the line passing through point (0, 0), whose slope is m,is


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(y – 0) = m(x – 0)

i.e., y = mx

Question 4:

Find the equation of the line which passes though and is inclined with the x-

axis at an angle of 75°.

Answer

The slope of the line that inclines with the x-axis at an angle of 75° is

m = tan 75°


is .

Thus, if a line passes though and inclines with the x-axis at an angle of 75°,

then the equation of the line is given as

Question 5:

Find the equation of the line which intersects the x-axis at a distance of 3 units to the

left of origin with slope –2.


Page 14 of 68

Answer

It is known that if a line with slope m makes x-intercept d, then the equation of the line

is given as

y = m(x – d)

For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = –

3.

The slope of the line is given as m = –2

Thus, the required equation of the given line is

y = –2 [x – (–3)]

y = –2x – 6

i.e., 2x + y + 6 = 0

Question 6:

Find the equation of the line which intersects the y-axis at a distance of 2 units above

the origin and makes an angle of 30° with the positive direction of the x-axis.

Answer

It is known that if a line with slope m makes y-intercept c, then the equation of the line

is given as

y = mx + c

Here, c = 2 and m = tan 30° .


Question 7:

Find the equation of the line which passes through the points (–1, 1) and (2, –4).

Answer


Page 15 of 68

It is known that the equation of the line passing through points (x1, y1) and (x2, y2) is

.

Therefore, the equation of the line passing through the points (–1, 1) and

(2, –4) is

Question 8:

Find the equation of the line which is at a perpendicular distance of 5 units from the

origin and the angle made by the perpendicular with the positive x-axis is 30°

Answer

If p is the length of the normal from the origin to a line and ω is the angle made by the

normal with the positive direction of the x-axis, then the equation of the line is given by

xcos ω + y sin ω = p.

Here, p = 5 units and ω = 30°


x cos 30° + y sin 30° = 5

Question 9:

The vertices of ∆PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median

through the vertex R.

Answer

It is given that the vertices of ∆PQR are P (2, 1), Q (–2, 3), and R (4, 5).

Let RL be the median through vertex R.


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Accordingly, L is the mid-point of PQ.

By mid-point formula, the coordinates of point L are given by

It is known that the equation of the line passing through points (x1, y1) and (x2, y2) is

.

Therefore, the equation of RL can be determined by substituting (x1, y1) = (4, 5) and

(x2, y2) = (0, 2).

Hence,

Thus, the required equation of the median through vertex R is .

Question 10:

Find the equation of the line passing through (–3, 5) and perpendicular to the line

through the points (2, 5) and (–3, 6).

Answer

The slope of the line joining the points (2, 5) and (–3, 6) is

We know that two non-vertical lines are perpendicular to each other if and only if their

slopes are negative reciprocals of each other.


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Therefore, slope of the line perpendicular to the line through the points (2, 5) and (–3,

6)

Now, the equation of the line passing through point (–3, 5), whose slope is 5, is

Question 11:

A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in

the ratio 1:n. Find the equation of the line.

Answer

According to the section formula, the coordinates of the point that divides the line

segment joining the points (1, 0) and (2, 3) in the ratio 1: n is given by

The slope of the line joining the points (1, 0) and (2, 3) is

We know that two non-vertical lines are perpendicular to each other if and only if their

slopes are negative reciprocals of each other.

Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and

(2, 3)

Now, the equation of the line passing through and whose slope is is

given by


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Question 12:

Find the equation of a line that cuts off equal intercepts on the coordinate axes and

passes through the point (2, 3).

Answer

The equation of a line in the intercept form is

Here, a and b are the intercepts on x and y axes respectively.

It is given that the line cuts off equal intercepts on both the axes. This means that a = b.

Accordingly, equation (i) reduces to

Since the given line passes through point (2, 3), equation (ii) reduces to

2 + 3 = a ⇒ a = 5

On substituting the value of a in equation (ii), we obtain

x + y = 5, which is the required equation of the line

Question 13:

Find equation of the line passing through the point (2, 2) and cutting off intercepts on

the axes whose sum is 9.

Answer

The equation of a line in the intercept form is

Here, a and b are the intercepts on x and y axes respectively.


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It is given thata + b = 9 ⇒ b = 9 – a … (ii)

From equations (i) and (ii), we obtain

It is given that the line passes through point (2, 2). Therefore, equation (iii) reduces to

If a = 6 and b = 9 – 6 = 3, then the equation of the line is

If a = 3 and b = 9 – 3 = 6, then the equation of the line is

Question 14:

Find equation of the line through the point (0, 2) making an angle with the positive

x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of

2 units below the origin.

Answer


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The slope of the line making an angle with the positive x-axis is

Now, the equation of the line passing through point (0, 2) and having a slope is

.

The slope of line parallel to line is .

It is given that the line parallel to line crosses the y-axis 2 units below the

origin i.e., it passes through point (0, –2).

Hence, the equation of the line passing through point (0, –2) and having a slope is

Question 15:

The perpendicular from the origin to a line meets it at the point (– 2, 9), find the

equation of the line.

Answer

The slope of the line joining the origin (0, 0) and point (–2, 9) is

Accordingly, the slope of the line perpendicular to the line joining the origin and point (–

2, 9) is

Now, the equation of the line passing through point (–2, 9) and having a slope m2 is


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Question 16:

The length L (in centimetre) of a copper rod is a linear function of its Celsius

temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C

= 110, express L in terms of C.

Answer

It is given that when C = 20, the value of L is 124.942, whereas when C = 110, the

value of L is 125.134.

Accordingly, points (20, 124.942) and (110, 125.134) satisfy the linear relation between

L and C.

Now, assuming C along the x-axis and L along the y-axis, we have two points i.e., (20,

124.942) and (110, 125.134) in the XY plane.

Therefore, the linear relation between L and C is the equation of the line passing through

points (20, 124.942) and (110, 125.134).

(L – 124.942) =

Question 17:

The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs

14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship

between selling price and demand, how many litres could he sell weekly at Rs 17/litre?

Answer

The relationship between selling price and demand is linear.


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Assuming selling price per litre along the x-axis and demand along the y-axis, we have

two points i.e., (14, 980) and (16, 1220) in the XY plane that satisfy the linear

relationship between selling price and demand.

Therefore, the linear relationship between selling price per litre and demand is the

equation of the line passing through points (14, 980) and (16, 1220).

When x = Rs 17/litre,

Thus, the owner of the milk store could sell 1340 litres of milk weekly at Rs 17/litre.

Question 18:

P (a, b) is the mid-point of a line segment between axes. Show that equation of the line

is

Answer

Let AB be the line segment between the axes and let P (a, b) be its mid-point.

Let the coordinates of A and B be (0, y) and (x, 0) respectively.

Since P (a, b) is the mid-point of AB,


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Thus, the respective coordinates of A and B are (0, 2b) and (2a, 0).

The equation of the line passing through points (0, 2b) and (2a, 0) is

On dividing both sides by ab, we obtain

Thus,the equation of the line is .

Question 19:

Point R (h, k) divides a line segment between the axes in the ratio 1:2. Find equation of

the line.

Answer

Let AB be the line segment between the axes such that point R (h, k) divides AB in the

ratio 1: 2.


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Let the respective coordinates of A and B be (x, 0) and (0, y).

Since point R (h, k) divides AB in the ratio 1: 2, according to the section formula,

Therefore, the respective coordinates of A and B are and (0, 3k).

Now, the equation of line AB passing through points and

(0, 3k) is

Thus,the required equation of the line is 2kx + hy = 3hk

Question 20:

By using the concept of equation of a line, prove that the three points (3, 0),


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(–2, –2) and (8, 2) are collinear.

Answer

In order to show that points (3, 0), (–2, –2), and (8, 2) are collinear, it suffices to show

that the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2).

The equation of the line passing through points (3, 0) and (–2, –2) is

It is observed that at x = 8 and y = 2,

L.H.S. = 2 × 8 – 5 × 2 = 16 – 10 = 6 = R.H.S.

Therefore, the line passing through points (3, 0) and (–2, –2) also passes through point

(8, 2). Hence, points (3, 0), (–2, –2), and (8, 2) are collinear.


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Exercise 10.3

Question 1:

Reduce the following equations into slope-intercept form and find their slopes and the y-

intercepts.

(i) x + 7y = 0 (ii) 6x + 3y – 5 = 0 (iii) y = 0

Answer

(i) The given equation is x + 7y = 0.

It can be written as

This equation is of the form y = mx + c, where .

Therefore, equation (1) is in the slope-intercept form, where the slope and the y-

intercept are and 0 respectively.

(ii) The given equation is 6x + 3y – 5 = 0.



intercept are–2 and respectively.

(iii) The given equation is y = 0.



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y = 0.x + 0 … (3)

This equation is of the form y = mx + c, where m = 0 and c = 0.


intercept are 0 and 0 respectively.

Question 2:

Reduce the following equations into intercept form and find their intercepts on the axes.

(i) 3x + 2y – 12 = 0 (ii) 4x – 3y = 6 (iii) 3y + 2 = 0.

Answer

(i) The given equation is 3x + 2y – 12 = 0.


This equation is of the form , where a = 4 and b = 6.

Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes

are 4 and 6 respectively.

(ii) The given equation is 4x – 3y = 6.


This equation is of the form , where a = and b = –2.


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Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes

are and –2 respectively.

(iii) The given equation is 3y + 2 = 0.


This equation is of the form , where a = 0 and b = .

Therefore, equation (3) is in the intercept form, where the intercept on the y-axis is

and it has no intercept on the x-axis.

Question 3:

Reduce the following equations into normal form. Find their perpendicular distances from

the origin and angle between perpendicular and the positive x-axis.

(i) (ii) y – 2 = 0 (iii) x – y = 4

Answer

(i) The given equation is .

It can be reduced as:

On dividing both sides by , we obtain


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Equation (1) is in the normal form.

On comparing equation (1) with the normal form of equation of line

x cos ω + y sin ω = p, we obtain ω = 120° and p = 4.

Thus, the perpendicular distance of the line from the origin is 4, while the angle between

the perpendicular and the positive x-axis is 120°.

(ii) The given equation is y – 2 = 0.

It can be reduced as 0.x + 1.y = 2

On dividing both sides by , we obtain 0.x + 1.y = 2

⇒ x cos 90° + y sin 90° = 2 … (1)



x cos ω + y sin ω = p, we obtain ω = 90° and p = 2.

Thus, the perpendicular distance of the line from the origin is 2, while the angle between

the perpendicular and the positive x-axis is 90°.

(iii) The given equation is x – y = 4.

It can be reduced as 1.x + (–1) y = 4




x cos ω + y sin ω = p, we obtain ω = 315° and .


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Thus, the perpendicular distance of the line from the origin is , while the angle

between the perpendicular and the positive x-axis is 315°.

Question 4:

Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Answer

The given equation of the line is 12(x + 6) = 5(y – 2).

⇒ 12x + 72 = 5y – 10

⇒12x – 5y + 82 = 0 … (1)

On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain A =

12, B = –5, and C = 82.

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point

(x1, y1) is given by .

The given point is (x1, y1) = (–1, 1).

Therefore, the distance of point (–1, 1) from the given line

Question 5:

Find the points on the x-axis, whose distances from the line are 4 units.

Answer

The given equation of line is

On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain A =

4, B = 3, and C = –12.

Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.


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Therefore,

Thus, the required points on the x-axis are (–2, 0) and (8, 0).

Question 6:

Find the distance between parallel lines

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

(ii) l (x + y) + p = 0 and l (x + y) – r = 0

Answer

It is known that the distance (d) between parallel lines Ax + By + C1 = 0 and Ax + By +

C2 = 0 is given by .

(i) The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.

Here, A = 15, B = 8, C1 = –34, and C2 = 31.

Therefore, the distance between the parallel lines is

(ii) The given parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0.

lx + ly + p = 0 and lx + ly – r = 0

Here, A = l, B = l, C1 = p, and C2 = –r.


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Therefore, the distance between the parallel lines is

Question 7:

Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the

point (–2, 3).

Answer

The equation of the given line is

, which is of the form y = mx + c

∴ Slope of the given line

It is known that parallel lines have the same slope.

∴ Slope of the other line =

Now, the equation of the line that has a slope of and passes through the point (–2, 3)

is

Question 8:

Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept

3.

Answer

The given equation of line is .


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∴Slope of the given line

The slope of the line perpendicular to the line having a slope of is

The equation of the line with slope –7 and x-intercept 3 is given by

y = m (x – d)

⇒ y = –7 (x – 3)

⇒ y = –7x + 21

⇒ 7x + y = 21

Question 9:

Find angles between the lines

Answer

The given lines are .

The slope of line (1) is , while the slope of line (2) is .

The acute angle i.e., θ between the two lines is given by


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Thus, the angle between the given lines is either 30° or 180° – 30° = 150°.

Question 10:

The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0. at right

angle. Find the value of h.

Answer

The slope of the line passing through points (h, 3) and (4, 1) is

The slope of line 7x – 9y – 19 = 0 or is .

It is given that the two lines are perpendicular.


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Thus, the value of h is .

Question 11:

Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A

(x –x1) + B (y – y1) = 0.

Answer

The slope of line Ax + By + C = 0 or is

It is known that parallel lines have the same slope.

∴ Slope of the other line =

The equation of the line passing through point (x1, y1) and having a slope is

Hence, the line through point (x1, y1) and parallel to line Ax + By + C = 0 is

A (x –x1) + B (y – y1) = 0

Question 12:


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Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If

slope of one line is 2, find equation of the other line.

Answer

It is given that the slope of the first line, m1 = 2.

Let the slope of the other line be m2.

The angle between the two lines is 60°.

The equation of the line passing through point (2, 3) and having a slope of is


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In this case, the equation of the other line is .

The equation of the line passing through point (2, 3) and having a slope of is

In this case, the equation of the other line is .

Thus, the required equation of the other line is or

.

Question 13:

Find the equation of the right bisector of the line segment joining the points (3, 4) and

(–1, 2).

Answer

The right bisector of a line segment bisects the line segment at 90°.

The end-points of the line segment are given as A (3, 4) and B (–1, 2).

Accordingly, mid-point of AB

Slope of AB


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∴Slope of the line perpendicular to AB =

The equation of the line passing through (1, 3) and having a slope of –2 is

(y – 3) = –2 (x – 1)

y – 3 = –2x + 2

2x + y = 5

Thus, the required equation of the line is 2x + y = 5.

Question 14:

Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x –

4y – 16 = 0.

Answer

Let (a, b) be the coordinates of the foot of the perpendicular from the point (–1, 3) to

the line 3x – 4y – 16 = 0.

Slope of the line joining (–1, 3) and (a, b), m1

Slope of the line 3x – 4y – 16 = 0 or

Since these two lines are perpendicular, m1m2 = –1


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Point (a, b) lies on line 3x – 4y = 16.

∴3a – 4b = 16 … (2)

On solving equations (1) and (2), we obtain

Thus, the required coordinates of the foot of the perpendicular are .

Question 15:

The perpendicular from the origin to the line y = mx + c meets it at the point

(–1, 2). Find the values of m and c.

Answer

The given equation of line is y = mx + c.

It is given that the perpendicular from the origin meets the given line at (–1, 2).

Therefore, the line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.

∴Slope of the line joining (0, 0) and (–1, 2)

The slope of the given line is m.

Since point (–1, 2) lies on the given line, it satisfies the equation y = mx + c.

Thus, the respective values of m and c are .

Question 16:


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If p and q are the lengths of perpendiculars from the origin to the lines x cos θ – y sin θ

= k cos 2θ and x sec θ+ y cosec θ = k, respectively, prove that p2 + 4q2 = k2

Answer

The equations of given lines are

x cos θ – y sinθ = k cos 2θ … (1)

x secθ + y cosec θ= k … (2)

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

.

On comparing equation (1) to the general equation of line i.e., Ax + By + C = 0, we

obtain A = cosθ, B = –sinθ, and C = –k cos 2θ.

It is given that p is the length of the perpendicular from (0, 0) to line (1).

On comparing equation (2) to the general equation of line i.e., Ax + By + C = 0, we

obtain A = secθ, B = cosecθ, and C = –k.

It is given that q is the length of the perpendicular from (0, 0) to line (2).

From (3) and (4), we have


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Hence, we proved that p2 + 4q2 = k2.

Question 17:

In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and

length of altitude from the vertex A.

Answer

Let AD be the altitude of triangle ABC from vertex A.

Accordingly, AD⊥BC


Page 42 of 68

The equation of the line passing through point (2, 3) and having a slope of 1 is

(y – 3) = 1(x – 2)

⇒ x – y + 1 = 0

⇒ y – x = 1

Therefore, equation of the altitude from vertex A = y – x = 1.

Length of AD = Length of the perpendicular from A (2, 3) to BC

The equation of BC is


.

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A

= 1, B = 1, and C = –3.

∴Length of AD

Thus, the equation and the length of the altitude from vertex A are y – x = 1 and

units respectively.


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Question 18:

If p is the length of perpendicular from the origin to the line whose intercepts on the

axes are a and b, then show that .

Answer

It is known that the equation of a line whose intercepts on the axes are a and b is


.


= b, B = a, and C = –ab.

Therefore, if p is the length of the perpendicular from point (x1, y1) = (0, 0) to line (1),

we obtain



Page 44 of 68

Hence, we showed that .


Question 1:

Find the values of k for which the line is

(a) Parallel to the x-axis,

(b) Parallel to the y-axis,

(c) Passing through the origin.

Answer

The given equation of line is

(k – 3) x – (4 – k2) y + k2 – 7k + 6 = 0 … (1)

(a) If the given line is parallel to the x-axis, then

Slope of the given line = Slope of the x-axis

The given line can be written as

(4 – k2) y = (k – 3) x + k2 – 7k + 6 = 0

, which is of the form y = mx + c.

∴Slope of the given line =

Slope of the x-axis = 0

Thus, if the given line is parallel to the x-axis, then the value of k is 3.

(b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be

undefined.

The slope of the given line is .


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Now, is undefined at k2 = 4

k2 = 4

⇒ k = ±2

Thus, if the given line is parallel to the y-axis, then the value of k is ±2.

(c) If the given line is passing through the origin, then point (0, 0) satisfies the

given equation of line.

Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.

Question 2:

Find the values of θand p, if the equation is the normal form of the

line .

Answer

The equation of the given line is .

This equation can be reduced as


On comparing equation (1) to , we obtain


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Since the values of sin θ and cos θ are negative,

Thus, the respective values of θand p are and 1

Question 3:

Find the equations of the lines, which cut-off intercepts on the axes whose sum and

product are 1 and –6, respectively.

Answer

Let the intercepts cut by the given lines on the axes be a and b.

It is given that

a + b = 1 … (1)

ab = –6 … (2)


a = 3 and b = –2 or a = –2 and b = 3

It is known that the equation of the line whose intercepts on the axes are a and b is

Case I: a = 3 and b = –2

In this case, the equation of the line is –2x + 3y + 6 = 0, i.e., 2x – 3y = 6.

Case II: a = –2 and b = 3

In this case, the equation of the line is 3x – 2y + 6 = 0, i.e., –3x + 2y = 6.

Thus, the required equation of the lines are 2x – 3y = 6 and –3x + 2y = 6.

Question 4:

What are the points on the y-axis whose distance from the line is 4 units.

Answer

Let (0, b) be the point on the y-axis whose distance from line is 4 units.


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The given line can be written as 4x + 3y – 12 = 0 … (1)


= 4, B = 3, and C = –12.



Therefore, if (0, b) is the point on the y-axis whose distance from line is 4

units, then:

Thus, the required points are and .

Question 5:

Find the perpendicular distance from the origin to the line joining the points

Answer

The equation of the line joining the points is given by


Page 48 of 68



Therefore, the perpendicular distance (d) of the given line from point (x1, y1) = (0, 0) is

Question 6:


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Find the equation of the line parallel to y-axis and drawn through the point of

intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

Answer

The equation of any line parallel to the y-axis is of the form

x = a … (1)

The two given lines are

x – 7y + 5 = 0 … (2)

3x + y = 0 … (3)

On solving equations (2) and (3), we obtain .

Therefore, is the point of intersection of lines (2) and (3).

Since line x = a passes through point , .

Thus, the required equation of the line is .

Question 7:

Find the equation of a line drawn perpendicular to the line through the point,

where it meets the y-axis.

Answer

The equation of the given line is .

This equation can also be written as 3x + 2y – 12 = 0


∴Slope of the given line

∴Slope of line perpendicular to the given line


Page 50 of 68

Let the given line intersect the y-axis at (0, y).

On substituting x with 0 in the equation of the given line, we obtain

∴The given line intersects the y-axis at (0, 6).

The equation of the line that has a slope of and passes through point (0, 6) is


Question 8:

Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.

Answer

The equations of the given lines are

y – x = 0 … (1)

x + y = 0 … (2)

x – k = 0 … (3)

The point of intersection of lines (1) and (2) is given by

x = 0 and y = 0


x = k and y = –k


x = k and y = k

Thus, the vertices of the triangle formed by the three given lines are (0, 0), (k, –k), and

(k, k).

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is

.

Therefore, area of the triangle formed by the three given lines


Page 51 of 68

Question 9:

Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y –

3 = 0 may intersect at one point.

Answer


3x + y – 2 = 0 … (1)

px + 2y – 3 = 0 … (2)

2x – y – 3 = 0 … (3)


x = 1 and y = –1

Since these three lines may intersect at one point, the point of intersection of lines (1)

and (3) will also satisfy line (2).

p (1) + 2 (–1) – 3 = 0

p – 2 – 3 = 0

p = 5

Thus, the required value of p is 5.

Question 10:

If three lines whose equations are

concurrent, then show that

Answer


y = m1x + c1 … (1)

y = m2x + c2 … (2)

y = m3x + c3 … (3)


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On subtracting equation (1) from (2), we obtain

On substituting this value of x in (1), we obtain

is the point of intersection of lines (1) and (2).

It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of

lines (1) and (2) will also satisfy equation (3).

Hence,

Question 11:

Find the equation of the lines through the point (3, 2) which make an angle of 45° with

the line x –2y = 3.

Answer

Let the slope of the required line be m1.


Page 53 of 68

The given line can be written as , which is of the form y = mx + c

∴Slope of the given line =

It is given that the angle between the required line and line x – 2y = 3 is 45°.

We know that if θisthe acute angle between lines l1 and l2 with slopes m1 and m2

respectively, then .

Case I: m1 = 3

The equation of the line passing through (3, 2) and having a slope of 3 is:

y – 2 = 3 (x – 3)

y – 2 = 3x – 9

3x – y = 7


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Case II: m1 =

The equation of the line passing through (3, 2) and having a slope of is:

Thus, the equations of the lines are 3x – y = 7 and x + 3y = 9.

Question 12:

Find the equation of the line passing through the point of intersection of the lines 4x +

7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

Answer

Let the equation of the line having equal intercepts on the axes be

On solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0, we obtain .

is the point of intersection of the two given lines.

Since equation (1) passes through point ,

∴ Equation (1) becomes



Page 55 of 68

Question 13:

Show that the equation of the line passing through the origin and making an angle θwith

the line .

Answer

Let the equation of the line passing through the origin be y = m1x.

If this line makes an angle of θ with line y = mx + c, then angle θ is given by

Case I:


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Case II:

Therefore, the required line is given by .

Question 14:

In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line

x + y = 4?

Answer

The equation of the line joining the points (–1, 1) and (5, 7) is given by


x + y – 4 = 0 … (2)


x = 1 and y = 3

Let point (1, 3) divide the line segment joining (–1, 1) and (5, 7) in the ratio 1:k.

Accordingly, by section formula,


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Thus, the line joining the points (–1, 1) and (5, 7) is divided by line

x + y = 4 in the ratio 1:2.

Question 15:

Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y

= 0.

Answer

The given lines are

2x – y = 0 … (1)

4x + 7y + 5 = 0 … (2)

A (1, 2) is a point on line (1).

Let B be the point of intersection of lines (1) and (2).


∴Coordinates of point B are .


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By using distance formula, the distance between points A and B can be obtained as

Thus, the required distance is .

Question 16:

Find the direction in which a straight line must be drawn through the point (–1, 2) so

that its point of intersection with the line x + y = 4 may be at a distance of 3 units from

this point.

Answer

Let y = mx + c be the line through point (–1, 2).

Accordingly, 2 = m (–1) + c.

⇒ 2 = –m + c

⇒ c = m + 2

∴ y = mx + m + 2 … (1)

The given line is


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x + y = 4 … (2)


is the point of intersection of lines (1) and (2).

Since this point is at a distance of 3 units from point (– 1, 2), according to distance

formula,

Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-

axis.

Question 18:

Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line

to be a plane mirror.

Answer


x + 3y = 7 … (1)

Let point B (a, b) be the image of point A (3, 8).

Accordingly, line (1) is the perpendicular bisector of AB.


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Since line (1) is perpendicular to AB,

The mid-point of line segment AB will also satisfy line (1).

Hence, from equation (1), we have

On solving equations (2) and (3), we obtain a = –1 and b = –4.

Thus, the image of the given point with respect to the given line is (–1, –4).

Question 19:

If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find

the value of m.

Answer


y = 3x + 1 … (1)


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2y = x + 3 … (2)

y = mx + 4 … (3)

Slope of line (1), m1 = 3

Slope of line (2),

Slope of line (3), m3 = m

It is given that lines (1) and (2) are equally inclined to line (3). This means that

the angle between lines (1) and (3) equals the angle between lines (2) and (3).


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Thus, the required value of m is .

Question 20:

If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5

= 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.

Answer


x + y – 5 = 0 … (1)

3x – 2y + 7 = 0 … (2)

The perpendicular distances of P (x, y) from lines (1) and (2) are respectively given by

It is given that .


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, which is the equation of a

line.

Similarly, we can obtain the equation of line for any signs of .

Thus, point P must move on a line.

Question 21:

Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x

+ 2y + 6 = 0.

Answer


9x + 6y – 7 = 0 … (1)

3x + 2y + 6 = 0 … (2)

Let P (h, k) be the arbitrary point that is equidistant from lines (1) and (2). The

perpendicular distance of P (h, k) from line (1) is given by

The perpendicular distance of P (h, k) from line (2) is given by

Since P (h, k) is equidistant from lines (1) and (2),


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∴

9h + 6k – 7 = – 9h – 6k – 18

⇒ 18h + 12k + 11 = 0

Thus, the required equation of the line is 18x + 12y + 11 = 0.

Question 22:

A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the

reflected ray passes through the point (5, 3). Find the coordinates of A.

Answer

Let the coordinates of point A be (a, 0).

Draw a line (AL) perpendicular to the x-axis.

We know that angle of incidence is equal to angle of reflection. Hence, let

∠BAL = ∠CAL = Φ

Let ∠CAX = θ


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∴∠OAB = 180° – (θ + 2Φ) = 180° – [θ + 2(90° – θ)]

= 180° – θ – 180° + 2θ

= θ

∴∠BAX = 180° – θ

From equations (1) and (2), we obtain

Thus, the coordinates of point A are .

Question 23:

Prove that the product of the lengths of the perpendiculars drawn from the points

Answer



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Length of the perpendicular from point to line (1) is

Length of the perpendicular from point to line (2) is

On multiplying equations (2) and (3), we obtain


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Hence, proved.

Question 24:

A person standing at the junction (crossing) of two straight paths represented by the

equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation

is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

Answer


2x – 3y + 4 = 0 … (1)

3x + 4y – 5 = 0 … (2)


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6x – 7y + 8 = 0 … (3)

The person is standing at the junction of the paths represented by lines (1) and (2).


Thus, the person is standing at point .

The person can reach path (3) in the least time if he walks along the perpendicular line

to (3) from point .

∴Slope of the line perpendicular to line (3)

The equation of the line passing through and having a slope of is given

by

Hence, the path that the person should follow is .

Class XI Chapter 11 – Conic Sections Maths

Exercise 11.1

Question 1:

Find the equation of the circle with centre (0, 2) and radius 2

Answer

The equation of a circle with centre (h, k) and radius r is given as

(x – h)2 + (y – k)2 = r2

It is given that centre (h, k) = (0, 2) and radius (r) = 2.

Therefore, the equation of the circle is

(x – 0)2 + (y – 2)2 = 22

x2 + y2 + 4 – 4 y = 4

x2 + y2 – 4y = 0

Question 2:

Find the equation of the circle with centre (–2, 3) and radius 4

Answer


(x – h)2 + (y – k)2 = r2

It is given that centre (h, k) = (–2, 3) and radius (r) = 4.


(x + 2)2 + (y – 3)2 = (4)2

x2 + 4x + 4 + y2 – 6y + 9 = 16

x2 + y2 + 4x – 6y – 3 = 0

Question 3:

Find the equation of the circle with centre and radius

Answer


(x – h)2 + (y – k)2 = r2

It is given that centre (h, k) = and radius (r) = .



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Question 4:

Find the equation of the circle with centre (1, 1) and radius

Answer


(x – h)2 + (y – k)2 = r2

It is given that centre (h, k) = (1, 1) and radius (r) = .


Question 5:

Find the equation of the circle with centre (–a, –b) and radius

Answer


(x – h)2 + (y – k)2 = r2

It is given that centre (h, k) = (–a, –b) and radius (r) = .



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Question 6:

Find the centre and radius of the circle (x + 5)2 + (y – 3)2 = 36

Answer

The equation of the given circle is (x + 5)2 + (y – 3)2 = 36.

(x + 5)2 + (y – 3)2 = 36

⇒ {x – (–5)}2 + (y – 3)2 = 62, which is of the form (x – h)2 + (y – k)2 = r2, where h = –

5, k = 3, and r = 6.

Thus, the centre of the given circle is (–5, 3), while its radius is 6.

Question 7:

Find the centre and radius of the circle x2 + y2 – 4x – 8y – 45 = 0

Answer

The equation of the given circle is x2 + y2 – 4x – 8y – 45 = 0.

x2 + y2 – 4x – 8y – 45 = 0

⇒ (x2 – 4x) + (y2 – 8y) = 45

⇒ {x2 – 2(x)(2) + 22} + {y2 – 2(y)(4)+ 42} – 4 –16 = 45

⇒ (x – 2)2 + (y –4)2 = 65

⇒ (x – 2)2 + (y –4)2 = , which is of the form (x – h)2 + (y – k)2 = r2, where h =

2, k = 4, and .

Thus, the centre of the given circle is (2, 4), while its radius is .

Question 8:

Find the centre and radius of the circle x2 + y2 – 8x + 10y – 12 = 0

Answer

The equation of the given circle is x2 + y2 – 8x + 10y – 12 = 0.

x2 + y2 – 8x + 10y – 12 = 0


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⇒ (x2 – 8x) + (y2 + 10y) = 12

⇒ {x2 – 2(x)(4) + 42} + {y2 + 2(y)(5) + 52}– 16 – 25 = 12

⇒ (x – 4)2 + (y + 5)2 = 53

, which is of the form (x – h)2 + (y – k)2 = r2, where h

= 4, k = –5, and .

Thus, the centre of the given circle is (4, –5), while its radius is .

Question 9:

Find the centre and radius of the circle 2x2 + 2y2 – x = 0

Answer

The equation of the given circle is 2x2 + 2y2 – x = 0.

, which is of the form (x – h)2 + (y – k)2 = r2, where h =

, k = 0, and .

Thus, the centre of the given circle is , while its radius is .

Question 10:

Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose

centre is on the line 4x + y = 16.

Answer

Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.


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Since the circle passes through points (4, 1) and (6, 5),

(4 – h)2 + (1 – k)2 = r2 … (1)

(6 – h)2 + (5 – k)2 = r2 … (2)

Since the centre (h, k) of the circle lies on line 4x + y = 16,

4h + k = 16 … (3)


(4 – h)2 + (1 – k)2 = (6 – h)2 + (5 – k)2

⇒ 16 – 8h + h2 + 1 – 2k + k2 = 36 – 12h + h2 + 25 – 10k + k2

⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k

⇒ 4h + 8k = 44

⇒ h + 2k = 11 … (4)

On solving equations (3) and (4), we obtain h = 3 and k = 4.

On substituting the values of h and k in equation (1), we obtain

(4 – 3)2 + (1 – 4)2 = r2

⇒ (1)2 + (– 3)2 = r2

⇒ 1 + 9 = r2

⇒ r2 = 10

⇒

Thus, the equation of the required circle is

(x – 3)2 + (y – 4)2 =

x2 – 6x + 9 + y2 – 8y + 16 = 10

x2 + y2 – 6x – 8y + 15 = 0

Question 11:

Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose

centre is on the line x – 3y – 11 = 0.

Answer


Since the circle passes through points (2, 3) and (–1, 1),

(2 – h)2 + (3 – k)2 = r2 … (1)

(–1 – h)2 + (1 – k)2 = r2 … (2)

Since the centre (h, k) of the circle lies on line x – 3y – 11 = 0,


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h – 3k = 11 … (3)


(2 – h)2 + (3 – k)2 = (–1 – h)2 + (1 – k)2

⇒ 4 – 4h + h2 + 9 – 6k + k2 = 1 + 2h + h2 + 1 – 2k + k2

⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k

⇒ 6h + 4k = 11 … (4)


On substituting the values of h and k in equation (1), we obtain


Question 12:

Find the equation of the circle with radius 5 whose centre lies on x-axis and passes

through the point (2, 3).


Page 7 of 49

Answer


Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

Now, the equation of the circle becomes (x – h)2 + y2 = 25.

It is given that the circle passes through point (2, 3).

When h = –2, the equation of the circle becomes

(x + 2)2 + y2 = 25

x2 + 4x + 4 + y2 = 25

x2 + y2 + 4x – 21 = 0

When h = 6, the equation of the circle becomes

(x – 6)2 + y2 = 25

x2 – 12x +36 + y2 = 25

x2 + y2 – 12x + 11 = 0

Question 13:

Find the equation of the circle passing through (0, 0) and making intercepts a and b on

the coordinate axes.

Answer


Since the centre of the circle passes through (0, 0),

(0 – h)2 + (0 – k)2 = r2

⇒ h2 + k2 = r2

The equation of the circle now becomes (x – h)2 + (y – k)2 = h2 + k2.

It is given that the circle makes intercepts a and b on the coordinate axes. This means

that the circle passes through points (a, 0) and (0, b). Therefore,

(a – h)2 + (0 – k)2 = h2 + k2 … (1)


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(0 – h)2 + (b – k)2 = h2 + k2 … (2)

From equation (1), we obtain

a2 – 2ah + h2 + k2 = h2 + k2

⇒ a2 – 2ah = 0

⇒ a(a – 2h) = 0

⇒ a = 0 or (a – 2h) = 0

However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h = .

From equation (2), we obtain

h2 + b2 – 2bk + k2 = h2 + k2

⇒ b2 – 2bk = 0

⇒ b(b – 2k) = 0

⇒ b = 0 or(b – 2k) = 0

However, b ≠ 0; hence, (b – 2k) = 0 ⇒ k = .


Question 14:

Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).

Answer

The centre of the circle is given as (h, k) = (2, 2).

Since the circle passes through point (4, 5), the radius (r) of the circle is the distance

between the points (2, 2) and (4, 5).


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Thus, the equation of the circle is

Question 15:

Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?

Answer

The equation of the given circle is x2 + y2 = 25.

x2 + y2 = 25

⇒ (x – 0)2 + (y – 0)2 = 52, which is of the form (x – h)2 + (y – k)2 = r2, where h = 0, k

= 0, and r = 5.

∴Centre = (0, 0) and radius = 5

Distance between point (–2.5, 3.5) and centre (0, 0)

Since the distance between point (–2.5, 3.5) and centre (0, 0) of the circle is less than

the radius of the circle, point (–2.5, 3.5) lies inside the circle.


Page 10 of 49

Exercise 11.2

Question 1:

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the

length of the latus rectum for y2 = 12x

Answer

The given equation is y2 = 12x.

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

On comparing this equation with y2 = 4ax, we obtain

4a = 12 ⇒ a = 3

∴Coordinates of the focus = (a, 0) = (3, 0)

Since the given equation involves y2, the axis of the parabola is the x-axis.

Equation of direcctrix, x = –a i.e., x = – 3 i.e., x + 3 = 0

Length of latus rectum = 4a = 4 × 3 = 12

Question 2:


length of the latus rectum for x2 = 6y

Answer

The given equation is x2 = 6y.

Here, the coefficient of y is positive. Hence, the parabola opens upwards.

On comparing this equation with x2 = 4ay, we obtain

∴Coordinates of the focus = (0, a) =

Since the given equation involves x2, the axis of the parabola is the y-axis.

Equation of directrix,

Length of latus rectum = 4a = 6

Question 3:


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length of the latus rectum for y2 = – 8x

Answer

The given equation is y2 = –8x.

Here, the coefficient of x is negative. Hence, the parabola opens towards the left.

On comparing this equation with y2 = –4ax, we obtain

–4a = –8 ⇒ a = 2

∴Coordinates of the focus = (–a, 0) = (–2, 0)


Equation of directrix, x = a i.e., x = 2


Question 4:


length of the latus rectum for x2 = – 16y

Answer

The given equation is x2 = –16y.

Here, the coefficient of y is negative. Hence, the parabola opens downwards.

On comparing this equation with x2 = – 4ay, we obtain

–4a = –16 ⇒ a = 4

∴Coordinates of the focus = (0, –a) = (0, –4)


Equation of directrix, y = a i.e., y = 4


Question 5:


length of the latus rectum for y2 = 10x

Answer

The given equation is y2 = 10x.

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

On comparing this equation with y2 = 4ax, we obtain


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∴Coordinates of the focus = (a, 0)




Question 6:


length of the latus rectum for x2 = –9y

Answer

The given equation is x2 = –9y.

Here, the coefficient of y is negative. Hence, the parabola opens downwards.

On comparing this equation with x2 = –4ay, we obtain

∴Coordinates of the focus =




Question 7:

Find the equation of the parabola that satisfies the following conditions: Focus (6, 0);

directrix x = –6

Answer

Focus (6, 0); directrix, x = –6

Since the focus lies on the x-axis, the x-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form y2 = 4ax or


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y2 = – 4ax.

It is also seen that the directrix, x = –6 is to the left of the y-axis, while the focus (6, 0)

is to the right of the y-axis. Hence, the parabola is of the form y2 = 4ax.

Here, a = 6

Thus, the equation of the parabola is y2 = 24x.

Question 8:

Find the equation of the parabola that satisfies the following conditions: Focus (0, –3);

directrix y = 3

Answer

Focus = (0, –3); directrix y = 3

Since the focus lies on the y-axis, the y-axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form x2 = 4ay or

x2 = – 4ay.

It is also seen that the directrix, y = 3 is above the x-axis, while the focus

(0, –3) is below the x-axis. Hence, the parabola is of the form x2 = –4ay.

Here, a = 3

Thus, the equation of the parabola is x2 = –12y.

Question 9:

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0);

focus (3, 0)

Answer

Vertex (0, 0); focus (3, 0)

Since the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, x-axis

is the axis of the parabola, while the equation of the parabola is of the form y2 = 4ax.

Since the focus is (3, 0), a = 3.

Thus, the equation of the parabola is y2 = 4 × 3 × x, i.e., y2 = 12x

Question 10:

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0)

focus (–2, 0)

Answer


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Vertex (0, 0) focus (–2, 0)

Since the vertex of the parabola is (0, 0) and the focus lies on the negative x-axis, x-

axis is the axis of the parabola, while the equation of the parabola is of the form y2 = –

4ax.

Since the focus is (–2, 0), a = 2.

Thus, the equation of the parabola is y2 = –4(2)x, i.e., y2 = –8x

Question 11:

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0)

passing through (2, 3) and axis is along x-axis

Answer

Since the vertex is (0, 0) and the axis of the parabola is the x-axis, the equation of the

parabola is either of the form y2 = 4ax or y2 = –4ax.

The parabola passes through point (2, 3), which lies in the first quadrant.

Therefore, the equation of the parabola is of the form y2 = 4ax, while point

(2, 3) must satisfy the equation y2 = 4ax.

Thus, the equation of the parabola is

Question 12:

Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0),

passing through (5, 2) and symmetric with respect to y-axis

Answer

Since the vertex is (0, 0) and the parabola is symmetric about the y-axis, the equation

of the parabola is either of the form x2 = 4ay or x2 = –4ay.

The parabola passes through point (5, 2), which lies in the first quadrant.

Therefore, the equation of the parabola is of the form x2 = 4ay, while point


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(5, 2) must satisfy the equation x2 = 4ay.

Thus, the equation of the parabola is


Page 16 of 49

Exercise 11.3

Question 1:

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis,

the eccentricity and the length of the latus rectum of the ellipse

Answer

The given equation is .

Here, the denominator of is greater than the denominator of .

Therefore, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with , we obtain a = 6 and b = 4.

Therefore,

The coordinates of the foci are .

The coordinates of the vertices are (6, 0) and (–6, 0).

Length of major axis = 2a = 12

Length of minor axis = 2b = 8

Length of latus rectum

Question 2:



Answer


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Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with , we obtain b = 2 and a = 5.

Therefore,


The coordinates of the vertices are (0, 5) and (0, –5)




Question 3:



Answer






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Therefore,


The coordinates of the vertices are .




Question 4:



Answer





Therefore,


The coordinates of the vertices are (0, ±10).




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Question 5:



Answer





Therefore,


The coordinates of the vertices are (± 7, 0).





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Question 6:



Answer





Therefore,


The coordinates of the vertices are (0, ±20)




Question 7:


the eccentricity and the length of the latus rectum of the ellipse 36x2 + 4y2 = 144

Answer

The given equation is 36x2 + 4y2 = 144.



Page 21 of 49



On comparing equation (1) with , we obtain b = 2 and a = 6.

Therefore,






Question 8:


the eccentricity and the length of the latus rectum of the ellipse 16x2 + y2 = 16

Answer

The given equation is 16x2 + y2 = 16.



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On comparing equation (1) with , we obtain b = 1 and a = 4.

Therefore,






Question 9:


the eccentricity and the length of the latus rectum of the ellipse 4x2 + 9y2 = 36

Answer

The given equation is 4x2 + 9y2 = 36.



Page 23 of 49




Therefore,


The coordinates of the vertices are (±3, 0).




Question 10:

Find the equation for the ellipse that satisfies the given conditions: Vertices (±5, 0), foci

(±4, 0)

Answer

Vertices (±5, 0), foci (±4, 0)

Here, the vertices are on the x-axis.

Therefore, the equation of the ellipse will be of the form , where a is the

semi-major axis.

Accordingly, a = 5 and c = 4.


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It is known that .

Thus, the equation of the ellipse is .

Question 11:

Find the equation for the ellipse that satisfies the given conditions: Vertices (0, ±13),

foci (0, ±5)

Answer

Vertices (0, ±13), foci (0, ±5)

Here, the vertices are on the y-axis.


semi-major axis.

Accordingly, a = 13 and c = 5.

It is known that .


Question 12:

Find the equation for the ellipse that satisfies the given conditions: Vertices (±6, 0), foci

(±4, 0)

Answer



Page 25 of 49



semi-major axis.

Accordingly, a = 6, c = 4.

It is known that .


Question 13:

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis

(±3, 0), ends of minor axis (0, ±2)

Answer

Ends of major axis (±3, 0), ends of minor axis (0, ±2)

Here, the major axis is along the x-axis.


semi-major axis.

Accordingly, a = 3 and b = 2.


Question 14:

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis

, ends of minor axis (±1, 0)

Answer


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Ends of major axis , ends of minor axis (±1, 0)

Here, the major axis is along the y-axis.


semi-major axis.

Accordingly, a = and b = 1.


Question 15:

Find the equation for the ellipse that satisfies the given conditions: Length of major axis

26, foci (±5, 0)

Answer

Length of major axis = 26; foci = (±5, 0).

Since the foci are on the x-axis, the major axis is along the x-axis.


semi-major axis.

Accordingly, 2a = 26 ⇒ a = 13 and c = 5.

It is known that .


Question 16:


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Find the equation for the ellipse that satisfies the given conditions: Length of minor axis

16, foci (0, ±6)

Answer

Length of minor axis = 16; foci = (0, ±6).

Since the foci are on the y-axis, the major axis is along the y-axis.


semi-major axis.

Accordingly, 2b = 16 ⇒ b = 8 and c = 6.

It is known that .


Question 17:

Find the equation for the ellipse that satisfies the given conditions: Foci (±3, 0), a = 4

Answer

Foci (±3, 0), a = 4



semi-major axis.

Accordingly, c = 3 and a = 4.

It is known that .



Page 28 of 49

Question 18:

Find the equation for the ellipse that satisfies the given conditions: b = 3, c = 4, centre

at the origin; foci on the x axis.

Answer

It is given that b = 3, c = 4, centre at the origin; foci on the x axis.



semi-major axis.

Accordingly, b = 3, c = 4.

It is known that .


Question 19:

Find the equation for the ellipse that satisfies the given conditions: Centre at (0, 0),

major axis on the y-axis and passes through the points (3, 2) and (1, 6).

Answer

Since the centre is at (0, 0) and the major axis is on the y-axis, the equation of the

ellipse will be of the form

The ellipse passes through points (3, 2) and (1, 6). Hence,

On solving equations (2) and (3), we obtain b2 = 10 and a2 = 40.


Page 29 of 49


Question 20:

Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-

axis and passes through the points (4, 3) and (6, 2).

Answer

Since the major axis is on the x-axis, the equation of the ellipse will be of the form

The ellipse passes through points (4, 3) and (6, 2). Hence,

On solving equations (2) and (3), we obtain a2 = 52 and b2 = 13.



Page 30 of 49

Exercise 11.4

Question 1:

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the

latus rectum of the hyperbola

Answer


On comparing this equation with the standard equation of hyperbola i.e., , we

obtain a = 4 and b = 3.

We know that a2 + b2 = c2.

Therefore,

The coordinates of the foci are (±5, 0).



Question 2:


latus rectum of the hyperbola

Answer


trainer

Text Box

Hyperbola 11.4


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On comparing this equation with the standard equation of hyperbola i.e., , we

obtain a = 3 and .


Therefore,

The coordinates of the foci are (0, ±6).



Question 3:


latus rectum of the hyperbola 9y2 – 4x2 = 36

Answer

The given equation is 9y2 – 4x2 = 36.


9y2 – 4x2 = 36

On comparing equation (1) with the standard equation of hyperbola i.e., , we




Page 32 of 49

Therefore,




Question 4:


latus rectum of the hyperbola 16x2 – 9y2 = 576

Answer

The given equation is 16x2 – 9y2 = 576.


16x2 – 9y2 = 576




Therefore,

The coordinates of the foci are (±10, 0).



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Question 5:



Answer



obtain a = and b = 2.


Therefore, the coordinates of the foci are .



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Question 6:



Answer



49y2 – 16x2 = 784




Therefore,





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Question 7:

Find the equation of the hyperbola satisfying the give conditions: Vertices (±2, 0), foci

(±3, 0)

Answer



Therefore, the equation of the hyperbola is of the form .

Since the vertices are (±2, 0), a = 2.

Since the foci are (±3, 0), c = 3.


Thus, the equation of the hyperbola is .

Question 8:

Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±5), foci

(0, ±8)

Answer

Vertices (0, ±5), foci (0, ±8)



Since the vertices are (0, ±5), a = 5.

Since the foci are (0, ±8), c = 8.




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Question 9:

Find the equation of the hyperbola satisfying the give conditions: Vertices (0, ±3), foci

(0, ±5)

Answer

Vertices (0, ±3), foci (0, ±5)



Since the vertices are (0, ±3), a = 3.



∴32 + b2 = 52

⇒ b2 = 25 – 9 = 16


Question 10:

Find the equation of the hyperbola satisfying the give conditions: Foci (±5, 0), the

transverse axis is of length 8.

Answer

Foci (±5, 0), the transverse axis is of length 8.

Here, the foci are on the x-axis.



Since the length of the transverse axis is 8, 2a = 8 ⇒ a = 4.


∴42 + b2 = 52

⇒ b2 = 25 – 16 = 9



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Question 11:

Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±13), the

conjugate axis is of length 24.

Answer

Foci (0, ±13), the conjugate axis is of length 24.

Here, the foci are on the y-axis.



Since the length of the conjugate axis is 24, 2b = 24 ⇒ b = 12.


∴a2 + 122 = 132

⇒ a2 = 169 – 144 = 25


Question 12:

Find the equation of the hyperbola satisfying the give conditions: Foci , the

latus rectum is of length 8.

Answer

Foci , the latus rectum is of length 8.



Since the foci are , c = .

Length of latus rectum = 8


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∴a2 + 4a = 45

⇒ a2 + 4a – 45 = 0

⇒ a2 + 9a – 5a – 45 = 0

⇒ (a + 9) (a – 5) = 0

⇒ a = –9, 5

Since a is non-negative, a = 5.

∴b2 = 4a = 4 × 5 = 20


Question 13:

Find the equation of the hyperbola satisfying the give conditions: Foci (±4, 0), the latus

rectum is of length 12

Answer

Foci (±4, 0), the latus rectum is of length 12.




Length of latus rectum = 12


∴a2 + 6a = 16

⇒ a2 + 6a – 16 = 0

⇒ a2 + 8a – 2a – 16 = 0

⇒ (a + 8) (a – 2) = 0

⇒ a = –8, 2

Since a is non-negative, a = 2.

∴b2 = 6a = 6 × 2 = 12


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Question 14:

Find the equation of the hyperbola satisfying the give conditions: Vertices (±7, 0),

Answer

Vertices (±7, 0),



Since the vertices are (±7, 0), a = 7.

It is given that




Page 40 of 49

Question 15:

Find the equation of the hyperbola satisfying the give conditions: Foci ,

passing through (2, 3)

Answer

Foci , passing through (2, 3)

Here, the foci are on the y-axis.


Since the foci are , c = .


∴ a2 + b2 = 10

⇒ b2 = 10 – a2 … (1)

Since the hyperbola passes through point (2, 3),


In hyperbola, c > a, i.e., c2 > a2

∴ a2 = 5

⇒ b2 = 10 – a2 = 10 – 5 = 5

Thus, the equation of the hyperbola is

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Question 1:

If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Answer

The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such

a way that the axis of the reflector is along the positive x-axis.

This can be diagrammatically represented as

The equation of the parabola is of the form y2 = 4ax (as it is opening to the right).

Since the parabola passes through point A (10, 5), 102 = 4a(5)

⇒ 100 = 20a

Therefore, the focus of the parabola is (a, 0) = (5, 0), which is the mid-point of the

diameter.

Hence, the focus of the reflector is at the mid-point of the diameter.

Question 2:

An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m

wide at the base. How wide is it 2 m from the vertex of the parabola?

Answer


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The origin of the coordinate plane is taken at the vertex of the arch in such a way that

its vertical axis is along the positive y-axis.


The equation of the parabola is of the form x2 = 4ay (as it is opening upwards).

It can be clearly seen that the parabola passes through point .

Therefore, the arch is in the form of a parabola whose equation is .

When y = 2 m,

Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately

2.23 m.


Page 43 of 49

Question 3:

The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The

roadway which is horizontal and 100 m long is supported by vertical wires attached to

the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a

supporting wire attached to the roadway 18 m from the middle.

Answer

The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken

as the vertex of the parabola, while its vertical axis is taken along the positive y-axis.


Here, AB and OC are the longest and the shortest wires, respectively, attached to the

cable.

DF is the supporting wire attached to the roadway, 18 m from the middle.

Here, AB = 30 m, OC = 6 m, and .

The equation of the parabola is of the form x2 = 4ay (as it is opening upwards).

The coordinates of point A are (50, 30 – 6) = (50, 24).

Since A (50, 24) is a point on the parabola,

∴Equation of the parabola, or 6x2 = 625y

The x-coordinate of point D is 18.

Hence, at x = 18,


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∴DE = 3.11 m

DF = DE + EF = 3.11 m + 6 m = 9.11 m

Thus, the length of the supporting wire attached to the roadway 18 m from the middle is

approximately 9.11 m.

Question 4:

An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find

the height of the arch at a point 1.5 m from one end.

Answer

Since the height and width of the arc from the centre is 2 m and 8 m respectively, it is

clear that the length of the major axis is 8 m, while the length of the semi-minor axis is

2 m.

The origin of the coordinate plane is taken as the centre of the ellipse, while the major

axis is taken along the x-axis. Hence, the semi-ellipse can be diagrammatically

represented as

The equation of the semi-ellipse will be of the form , where a is the

semi-major axis

Accordingly, 2a = 8 ⇒ a = 4


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b = 2

Therefore, the equation of the semi-ellipse is

Let A be a point on the major axis such that AB = 1.5 m.

Draw AC⊥ OB.

OA = (4 – 1.5) m = 2.5 m

The x-coordinate of point C is 2.5.

On substituting the value of x with 2.5 in equation (1), we obtain

∴AC = 1.56 m

Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56 m.

Question 5:

A rod of length 12 cm moves with its ends always touching the coordinate axes.

Determine the equation of the locus of a point P on the rod, which is 3 cm from the end

in contact with the x-axis.

Answer

Let AB be the rod making an angle θ with OX and P (x, y) be the point on it such that AP

= 3 cm.

Then, PB = AB – AP = (12 – 3) cm = 9 cm [AB = 12 cm]

From P, draw PQ⊥OY and PR⊥OX.


Page 46 of 49

In ∆PBQ,

In ∆PRA,

Thus, the equation of the locus of point P on the rod is .

Question 6:

Find the area of the triangle formed by the lines joining the vertex of the parabola x2 =

12y to the ends of its latus rectum.

Answer

The given parabola is x2 = 12y.

On comparing this equation with x2 = 4ay, we obtain 4a = 12 ⇒ a = 3

∴The coordinates of foci are S (0, a) = S (0, 3)

Let AB be the latus rectum of the given parabola.

The given parabola can be roughly drawn as


Page 47 of 49

At y = 3, x2 = 12 (3) ⇒ x2 = 36 ⇒ x = ±6

∴The coordinates of A are (–6, 3), while the coordinates of B are (6, 3).

Therefore, the vertices of ∆OAB are O (0, 0), A (–6, 3), and B (6, 3).

Thus, the required area of the triangle is 18 unit2.

Question 7:

A man running a racecourse notes that the sum of the distances from the two flag posts

form him is always 10 m and the distance between the flag posts is 8 m. find the

equation of the posts traced by the man.

Answer

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man.

Accordingly, PA + PB = 10.


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We know that if a point moves in a plane in such a way that the sum of its distances

from two fixed points is constant, then the path is an ellipse and this constant value is

equal to the length of the major axis of the ellipse.

Therefore, the path described by the man is an ellipse where the length of the major axis

is 10 m, while points A and B are the foci.

Taking the origin of the coordinate plane as the centre of the ellipse, while taking the

major axis along the x-axis, the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form , where a is the semi-major

axis

Accordingly, 2a = 10 ⇒ a = 5

Distance between the foci (2c) = 8

⇒ c = 4

On using the relation , we obtain

Thus, the equation of the path traced by the man is .

Question 8:


Page 49 of 49

An equilateral triangle is inscribed in the parabola y2 = 4 ax, where one vertex is at the

vertex of the parabola. Find the length of the side of the triangle.

Answer

Let OAB be the equilateral triangle inscribed in parabola y2 = 4ax.

Let AB intersect the x-axis at point C.

Let OC = k

From the equation of the given parabola, we have

∴The respective coordinates of points A and B are

AB = CA + CB =

Since OAB is an equilateral triangle, OA2 = AB2.

Thus, the side of the equilateral triangle inscribed in parabola y2 = 4 ax is .

Class XI Chapter 12 – Introduction to Three Dimensional Geometry Maths

Exercise 12.1

Question 1:

A point is on the x-axis. What are its y-coordinates and z-coordinates?

Answer

If a point is on the x-axis, then its y-coordinates and z-coordinates are zero.

Question 2:

A point is in the XZ-plane. What can you say about its y-coordinate?

Answer

If a point is in the XZ plane, then its y-coordinate is zero.

Question 3:

Name the octants in which the following points lie:

(1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (–4, 2, –5), (–4, 2, 5),

(–3, –1, 6), (2, –4, –7)

Answer

The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive.

Therefore, this point lies in octant I.

The x-coordinate, y-coordinate, and z-coordinate of point (4, –2, 3) are positive,

negative, and positive respectively. Therefore, this point lies in octant IV.

The x-coordinate, y-coordinate, and z-coordinate of point (4, –2, –5) are positive,

negative, and negative respectively. Therefore, this point lies in octant VIII.

The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, –5) are positive,

positive, and negative respectively. Therefore, this point lies in octant V.

The x-coordinate, y-coordinate, and z-coordinate of point (–4, 2, –5) are negative,

positive, and negative respectively. Therefore, this point lies in octant VI.

The x-coordinate, y-coordinate, and z-coordinate of point (–4, 2, 5) are negative,

positive, and positive respectively. Therefore, this point lies in octant II.

The x-coordinate, y-coordinate, and z-coordinate of point (–3, –1, 6) are negative,

negative, and positive respectively. Therefore, this point lies in octant III.

The x-coordinate, y-coordinate, and z-coordinate of point (2, –4, –7) are positive,

negative, and negative respectively. Therefore, this point lies in octant VIII.


Page 1 of 17


Page 2 of 17

Question 4:

Fill in the blanks:

Answer

(i) The x-axis and y-axis taken together determine a plane known as .

(ii) The coordinates of points in the XY-plane are of the form .

(iii) Coordinate planes divide the space into octants.


Page 3 of 17

Exercise 12.2

Question 1:

Find the distance between the following pairs of points:

(i) (2, 3, 5) and (4, 3, 1) (ii) (–3, 7, 2) and (2, 4, –1)

(iii) (–1, 3, –4) and (1, –3, 4) (iv) (2, –1, 3) and (–2, 1, 3)

Answer

The distance between points P(x1, y1, z1) and P(x2, y2, z2) is given by

(i) Distance between points (2, 3, 5) and (4, 3, 1)

(ii) Distance between points (–3, 7, 2) and (2, 4, –1)

(iii) Distance between points (–1, 3, –4) and (1, –3, 4)

(iv) Distance between points (2, –1, 3) and (–2, 1, 3)


Page 4 of 17

Question 2:

Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Answer

Let points (–2, 3, 5), (1, 2, 3), and (7, 0, –1) be denoted by P, Q, and R respectively.

Points P, Q, and R are collinear if they lie on a line.


Page 5 of 17

Here, PQ + QR = PR

Hence, points P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1) are collinear.

Question 3:

Verify the following:

(i) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are the vertices of an isosceles triangle.

(ii) (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right angled triangle.

(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Answer

(i) Let points (0, 7, –10), (1, 6, –6), and (4, 9, –6) be denoted by A, B, and C

respectively.

Here, AB = BC ≠ CA

Thus, the given points are the vertices of an isosceles triangle.

(ii) Let (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) be denoted by A, B, and C respectively.


Page 6 of 17

Therefore, by Pythagoras theorem, ABC is a right triangle.

Hence, the given points are the vertices of a right-angled triangle.

(iii) Let (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) be denoted by A, B, C, and D

respectively.


Page 7 of 17

Here, AB = CD = 6, BC = AD =

Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are

equal.

Therefore, ABCD is a parallelogram.

Hence, the given points are the vertices of a parallelogram.

Question 4:

Find the equation of the set of points which are equidistant from the points (1, 2, 3) and

(3, 2, –1).

Answer

Let P (x, y, z) be the point that is equidistant from points A(1, 2, 3) and B(3, 2, –1).

Accordingly, PA = PB

⇒ x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 + 2z + 1


Page 8 of 17

⇒ –2x –4y – 6z + 14 = –6x – 4y + 2z + 14

⇒ – 2x – 6z + 6x – 2z = 0

⇒ 4x – 8z = 0

⇒ x – 2z = 0

Thus, the required equation is x – 2z = 0.

Question 5:

Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and

B (–4, 0, 0) is equal to 10.

Answer

Let the coordinates of P be (x, y, z).

The coordinates of points A and B are (4, 0, 0) and (–4, 0, 0) respectively.

It is given that PA + PB = 10.


On squaring both sides again, we obtain

25 (x2 + 8x + 16 + y2 + z2) = 625 + 16x2 + 200x

⇒ 25x2 + 200x + 400 + 25y2 + 25z2 = 625 + 16x2 + 200x

⇒ 9x2 + 25y2 + 25z2 – 225 = 0

Thus, the required equation is 9x2 + 25y2 + 25z2 – 225 = 0.


Page 9 of 17

Exercise 12.3

Question 1:

Find the coordinates of the point which divides the line segment joining the points (–2,

3, 5) and (1, –4, 6) in the ratio (i) 2:3 internally, (ii) 2:3 externally.

Answer

(i) The coordinates of point R that divides the line segment joining points P (x1, y1, z1)

and Q (x2, y2, z2) internally in the ratio m: n are

.

Let R (x, y, z) be the point that divides the line segment joining points(–2, 3, 5) and (1,

–4, 6) internally in the ratio 2:3

Thus, the coordinates of the required point are .

(ii) The coordinates of point R that divides the line segment joining points P (x1, y1, z1)

and Q (x2, y2, z2) externally in the ratio m: n are

.

Let R (x, y, z) be the point that divides the line segment joining points(–2, 3, 5) and (1,

–4, 6) externally in the ratio 2:3

Thus, the coordinates of the required point are (–8, 17, 3).

Question 2:


Page 10 of 17

Given that P (3, 2, –4), Q (5, 4, –6) and R (9, 8, –10) are collinear. Find the ratio in

which Q divides PR.

Answer

Let point Q (5, 4, –6) divide the line segment joining points P (3, 2, –4) and R (9, 8, –

10) in the ratio k:1.

Therefore, by section formula,

Thus, point Q divides PR in the ratio 1:2.

Question 3:

Find the ratio in which the YZ-plane divides the line segment formed by joining the

points (–2, 4, 7) and (3, –5, 8).

Answer

Let the YZ planedivide the line segment joining points (–2, 4, 7) and (3, –5, 8) in the

ratio k:1.

Hence, by section formula, the coordinates of point of intersection are given by

On the YZ plane, the x-coordinate of any point is zero.

Thus, the YZ plane divides the line segment formed by joining the given points in the

ratio 2:3.


Page 11 of 17

Question 4:

Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and are

collinear.

Answer

The given points are A (2, –3, 4), B (–1, 2, 1), and .

Let P be a point that divides AB in the ratio k:1.

Hence, by section formula, the coordinates of P are given by

Now, we find the value of k at which point P coincides with point C.

By taking , we obtain k = 2.

For k = 2, the coordinates of point P are .

i.e., is a point that divides AB externally in the ratio 2:1 and is the same as

point P.

Hence, points A, B, and C are collinear.

Question 5:

Find the coordinates of the points which trisect the line segment joining the points P (4,

2, –6) and Q (10, –16, 6).

Answer

Let A and B be the points that trisect the line segment joining points P (4, 2, –6) and Q

(10, –16, 6)


Page 12 of 17

Point A divides PQ in the ratio 1:2. Therefore, by section formula, the coordinates of

point A are given by

Point B divides PQ in the ratio 2:1. Therefore, by section formula, the coordinates of

point B are given by

Thus, (6, –4, –2) and (8, –10, 2) are the points that trisect the line segment joining

points P (4, 2, –6) and Q (10, –16, 6).


Page 13 of 17


Question 1:

Three vertices of a parallelogram ABCD are A (3, –1, 2), B (1, 2, –4) andC (–1, 1, 2).

Find the coordinates of the fourth vertex.

Answer

The three vertices of a parallelogram ABCD are given as A (3, –1, 2), B (1, 2, –4), and C

(–1, 1, 2). Let the coordinates of the fourth vertex be D (x, y, z).

We know that the diagonals of a parallelogram bisect each other.

Therefore, in parallelogram ABCD, AC and BD bisect each other.

∴Mid-point of AC = Mid-point of BD

⇒ x = 1, y = –2, and z = 8

Thus, the coordinates of the fourth vertex are (1, –2, 8).

Question 2:

Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and

(6, 0, 0).


Page 14 of 17

Answer

Let AD, BE, and CF be the medians of the given triangle ABC.

Since AD is the median, D is the mid-point of BC.

∴Coordinates of point D = = (3, 2, 0)

Thus, the lengths of the medians of ∆ABC are .

Question 3:

If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10)

and R (8, 14, 2c), then find the values of a, b and c.

Answer


Page 15 of 17

It is known that the coordinates of the centroid of the triangle, whose vertices are (x1,

y1, z1), (x2, y2, z2) and (x3, y3, z3), are .

Therefore, coordinates of the centroid of ∆PQR

It is given that origin is the centroid of ∆PQR.

Thus, the respective values of a, b, and c are

Question 4:

Find the coordinates of a point on y-axis which are at a distance of from the point P

(3, –2, 5).

Answer

If a point is on the y-axis, then x-coordinate and the z-coordinate of the point are zero.

Let A (0, b, 0) be the point on the y-axis at a distance of from point P (3, –2, 5).

Accordingly,


Page 16 of 17

Thus, the coordinates of the required points are (0, 2, 0) and (0, –6, 0).

Question 5:

A point R with x-coordinate 4 lies on the line segment joining the pointsP (2, –3, 4) and

Q (8, 0, 10). Find the coordinates of the point R.

[Hint suppose R divides PQ in the ratio k: 1. The coordinates of the point R are given by

Answer

The coordinates of points P and Q are given as P (2, –3, 4) and Q (8, 0, 10).

Let R divide line segment PQ in the ratio k:1.

Hence, by section formula, the coordinates of point R are given by

It is given that the x-coordinate of point R is 4.

Therefore, the coordinates of point R are


Page 17 of 17

Question 6:

If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the

set of points P such that PA2 + PB2 = k2, where k is a constant.

Answer

The coordinates of points A and B are given as (3, 4, 5) and (–1, 3, –7) respectively.

Let the coordinates of point P be (x, y, z).

On using distance formula, we obtain

Now, if PA2 + PB2 = k2, then

Thus, the required equation is .

Miscellaneous of chapter 13

Question 1:

Find the derivative of the following functions from first principle:

(i) –x (ii) (–x)–1 (iii) sin (x + 1)

(iv)

Answer :

(i) Let f(x) = –x. Accordingly,

By first principle,

(ii) Let . Accordingly,

By first principle,

Class XI______________________________

Chapter 13- Limits and Derivatives______________________________

Mathematics______________________________

______________________________________________________________________________________Page 1 of 58

(iii) Let f(x) = sin (x + 1). Accordingly,

By first principle,

(iv) Let . Accordingly,

By first principle,

Class XI______________________________


Mathematics______________________________

______________________________________________________________________________________Page 2 of 58

Question 2:

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are

fixed non-zero constants and m and n are integers): (x + a)

Answer :

Let f(x) = x + a. Accordingly,

By first principle,

Question 3:


fixed non-zero constants and m and n are integers):

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Mathematics______________________________

______________________________________________________________________________________Page 3 of 58

Answer :

By Leibnitz product rule,

Question 4:


fixed non-zero constants and m and n are integers): (ax + b) (cx + d)2

Answer :

Let


Question 5:



Answer :

Let

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By quotient rule,

Question 6:



Answer :

By quotient rule,

Question 7:



Answer :

Let

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By quotient rule,

Question 8:



Answer :

By quotient rule,

Question 9:



Answer :

By quotient rule,

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Question 10:



Answer :

Question 11:



Answer :

Question 12:Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): (ax + b)n

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Answer :

By first principle,

Question 13:


fixed non-zero constants and m and n are integers): (ax + b)n (cx + d)m

Answer :

Let


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Therefore, from (1), (2), and (3), we obtain

Question 14:


fixed non-zero constants and m and n are integers): sin (x + a)

Answer :

Let

By first principle,

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Question 15:


fixed non-zero constants and m and n are integers): cosec x cot x

Answer :

Let


By first principle,

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Now, let f2(x) = cosec x. Accordingly,

By first principle,

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Question 16:



Answer :

Let

By quotient rule,

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Question 17:



Answer :

Let

By quotient rule,

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Question 18:



Answer :

Let

By quotient rule,

Question 19:


fixed non-zero constants and m and n are integers): sinn x

Answer :

Let y = sinn x.

Accordingly, for n = 1, y = sin x.

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For n = 2, y = sin2 x.

For n = 3, y = sin3 x.

We assert that

Let our assertion be true for n = k.

i.e.,

Thus, our assertion is true for n = k + 1.

Hence, by mathematical induction,Question 20:



Answer :

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By quotient rule,

Question 21:



Answer :

Let

By quotient rule,

By first principle,

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From (i) and (ii), we obtain

Question 22:


fixed non-zero constants and m and n are integers): x4 (5 sin x – 3 cos x)

Answer :

Let

By product rule,

Question 23:

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fixed non-zero constants and m and n are integers): (x2 + 1) cos x

Answer :

Let

By product rule,

Question 24:


fixed non-zero constants and m and n are integers): (ax2 + sin x) (p + q cos x)

Answer :

Let

By product rule,

Question 25:



Answer :

Let

By product rule,

Let . Accordingly,

By first principle,

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Therefore, from (i) and (ii), we obtain

Question 26:



Answer :

Let

By quotient rule,

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Question 27:



Answer :

Let

By quotient rule,

Question 28:



Answer :

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Let

By first principle,

From (i) and (ii), we obtain

Question 29:


fixed non-zero constants and m and n are integers): (x + sec x) (x – tan x)

Answer :

Let

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By product rule,

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From (i), (ii), and (iii), we obtain

Question 30:



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Answer :

Let

By quotient rule,

It can be easily shown that

Therefore,

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EXCERSIE 13.1 SOULTIONS

Question 1:

Evaluate the Given limit:

Answer :

Question 2:


Answer :

Question 3:


Answer :

Question 4:


Answer :

Question 5:


Answer :

Question 6:


Answer :

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Put x + 1 = y so that y → 1 as x → 0.

Question 7:


Answer :

At x = 2, the value of the given rational function takes the form .

Question 8:


Answer :

At x = 2, the value of the given rational function takes the form .

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Question 9:


Answer :

Question 10:


Answer :

At z = 1, the value of the given function takes the form .

Put so that z →1 as x → 1.

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Question 11:


Answer :

Question 12:


Answer :

At x = –2, the value of the given function takes the form .

Question 13:

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Answer :

At x = 0, the value of the given function takes the form .

Question 14:


Answer :


Question 15:

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Answer :

It is seen that x → π ⇒ (π – x) → 0

Question 16:

Evaluate the given limit:

Answer :

Question 17:


Answer :


Now,

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Question 18:


Answer :


Now,

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Question 19:


Answer :

Question 20:


Answer :


Now,

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Question 21:


Answer :


Now,

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Question 22:

Answer :

At , the value of the given function takes the form .

Now, put so that .

Question 23:

Find f(x) and f(x), where f(x) =

Answer :

The given function is

f(x) =

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Question 24:

Find f(x), where f(x) =

Answer :


Question 25:

Evaluate f(x), where f(x) =

Answer :


f(x) =

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Question 26:


Answer :


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Question 27:


Answer :

The given function is f(x) = .

Question 28:

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Suppose f(x) = and if f(x) = f(1) what are possible values of a and b?

Answer :


Thus, the respective possible values of a and b are 0 and 4.Question 29:

Let be fixed real numbers and define a function

What is f(x)? For some compute f(x).

Answer :


Question 30:

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If f(x) = .

For what value (s) of a does f(x) exists?

Answer :


When a < 0,

When a > 0

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Thus, exists for all a ≠ 0.Question 31:

If the function f(x) satisfies , evaluate .

Answer :

Question 32:

If . For what integers m and n does and

exist?

Answer :


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Thus, exists if m = n.

Thus, exists for any integral value of m and n.

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EXCERISE 13.2 SOULTIONSQuestion 1:

Find the derivative of x2 – 2 at x = 10.

Answer :

Let f(x) = x2 – 2. Accordingly,

Thus, the derivative of x2 – 2 at x = 10 is 20.Question 2:

Find the derivative of 99x at x = 100.

Answer :

Let f(x) = 99x. Accordingly,

Thus, the derivative of 99x at x = 100 is 99.Question 3:

Find the derivative of x at x = 1.

Answer :

Let f(x) = x. Accordingly,

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Thus, the derivative of x at x = 1 is 1.Question 4:

Find the derivative of the following functions from first principle.

(i) x3 – 27 (ii) (x – 1) (x – 2)

(ii) (iv)

Answer :

(i) Let f(x) = x3 – 27. Accordingly, from the first principle,

(ii) Let f(x) = (x – 1) (x – 2). Accordingly, from the first principle,

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(iii) Let . Accordingly, from the first principle,

(iv) Let . Accordingly, from the first principle,

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Question 5:

For the function

Prove that

Answer :


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Thus,Question 6:

Find the derivative of for some fixed real number a.

Answer :

Let

Question 7:

For some constants a and b, find the derivative of

(i) (x – a) (x – b) (ii) (ax2 + b)2 (iii)

Answer :

(i) Let f (x) = (x – a) (x – b)

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(ii) Let

(iii)

By quotient rule,

Question 8:

Find the derivative of for some constant a.

Answer :

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By quotient rule,

Question 9:

Find the derivative of

(i) (ii) (5x3 + 3x – 1) (x – 1)

(iii) x–3 (5 + 3x) (iv) x5 (3 – 6x–9)

(v) x–4 (3 – 4x–5) (vi)

Answer :

(i) Let

(ii) Let f (x) = (5x3 + 3x – 1) (x – 1)


(iii) Let f (x) = x– 3 (5 + 3x)


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(iv) Let f (x) = x5 (3 – 6x–9)


(v) Let f (x) = x–4 (3 – 4x–5)


(vi) Let f (x) =

By quotient rule,

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Question 10:

Find the derivative of cos x from first principle.

Answer :

Let f (x) = cos x. Accordingly, from the first principle,

Question 11:

Find the derivative of the following functions:

(i) sin x cos x (ii) sec x (iii) 5 sec x + 4 cos x

(iv) cosec x (v) 3cot x + 5cosec x

(vi) 5sin x – 6cos x + 7 (vii) 2tan x – 7sec x

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Answer :

(i) Let f (x) = sin x cos x. Accordingly, from the first principle,

(ii) Let f (x) = sec x. Accordingly, from the first principle,

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(iii) Let f (x) = 5 sec x + 4 cos x. Accordingly, from the first principle,

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(iv) Let f (x) = cosec x. Accordingly, from the first principle,

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(v) Let f (x) = 3cot x + 5cosec x. Accordingly, from the first principle,

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(vi) Let f (x) = 5sin x – 6cos x + 7. Accordingly, from the first principle,

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(vii) Let f (x) = 2 tan x – 7 sec x. Accordingly, from the first principle,

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Exercise 14.1 solutions Question 1:

Which of the following sentences are statements? Give reasons for your answer.

(i) There are 35 days in a month.

(ii) Mathematics is difficult.

(iii) The sum of 5 and 7 is greater than 10.

(iv) The square of a number is an even number.

(v) The sides of a quadrilateral have equal length.

(vi) Answer this question.

(vii) The product of (–1) and 8 is 8.

(viii) The sum of all interior angles of a triangle is 180°.

(ix) Today is a windy day.

(x) All real numbers are complex numbers.

Answer :

(i) This sentence is incorrect because the maximum number of days in a month is 31. Hence, it is

a statement.

(ii) This sentence is subjective in the sense that for some people, mathematics can be easy and

for some others, it can be difficult. Hence, it is not a statement.

(iii) The sum of 5 and 7 is 12, which is greater than 10. Therefore, this sentence is always

correct. Hence, it is a statement.

(iv) This sentence is sometimes correct and sometimes incorrect. For example, the square of 2 is

an even number. However, the square of 3 is an odd number. Hence, it is not a statement.

(v) This sentence is sometimes correct and sometimes incorrect. For example, squares and

rhombus have sides of equal lengths. However, trapezium and rectangles have sides of unequal

lengths. Hence, it is not a statement.

(vi) It is an order. Therefore, it is not a statement.

(vii) The product of (–1) and 8 is (–8). Therefore, the given sentence is incorrect. Hence, it is a

statement.

(viii) This sentence is correct and hence, it is a statement.

(ix) The day that is being referred to is not evident from the sentence. Hence, it is not a

statement.

(x) All real numbers can be written as a × 1 + 0 × i. Therefore, the given sentence is always

correct. Hence, it is a statement.Question 2:

Give three examples of sentences which are not statements. Give reasons for the answers.

Answer :

The three examples of sentences, which are not statements, are as follows.

(i) He is a doctor.

It is not evident from the sentence as to whom ‘he’ is referred to. Therefore, it is not a statement.

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(ii) Geometry is difficult.

This is not a statement because for some people, geometry can be easy and for some others, it

can be difficult.

(iii) Where is she going?

This is a question, which also contains ‘she’, and it is not evident as to who ‘she’ is. Hence, it is

not a statement.

Exercise 14.2 solutions

Question 1:

Write the negation of the following statements:

(i) Chennai is the capital of Tamil Nadu.

(ii) is not a complex number.

(iii) All triangles are not equilateral triangle.

(iv) The number 2 is greater than 7.

(v) Every natural number is an integer.

Answer :

(i) Chennai is not the capital of Tamil Nadu.

(ii) is a complex number.

(iii) All triangles are equilateral triangles.

(iv) The number 2 is not greater than 7.

(v) Every natural number is not an integer.Question 2:

Are the following pairs of statements negations of each other?

(i) The number x is not a rational number.

The number x is not an irrational number.

(ii) The number x is a rational number.

The number x is an irrational number.

Answer :

(i) The negation of the first statement is “the number x is a rational number”.

This is same as the second statement. This is because if a number is not an irrational number,

then it is a rational number.

Therefore, the given statements are negations of each other.

(ii) The negation of the first statement is “the number x is not a rational number”. This means that

the number x is an irrational number, which is the same as the second statement.

Therefore, the given statements are negations of each other.Question 3:

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Find the component statements of the following compound statements and check whether they

are true or false.

(i) Number 3 is prime or it is odd.

(ii) All integers are positive or negative.

(iii) 100 is divisible by 3, 11 and 5.

Answer :

(i) The component statements are as follows.

p: Number 3 is prime.

q: Number 3 is odd.

Both the statements are true.

(ii) The component statements are as follows.

p: All integers are positive.

q: All integers are negative.

Both the statements are false.

(iii) The component statements are as follows.

p: 100 is divisible by 3.

q: 100 is divisible by 11.

r: 100 is divisible by 5.

Here, the statements, p and q, are false and statement r is true.


For each of the following compound statements first identify the connecting words and then

break it into component statements.

(i) All rational numbers are real and all real numbers are not complex.

(ii) Square of an integer is positive or negative.

(iii) The sand heats up quickly in the Sun and does not cool down fast at night.

(iv) x = 2 and x = 3 are the roots of the equation 3x2 – x – 10 = 0.

Answer :

(i) Here, the connecting word is ‘and’.

The component statements are as follows.

p: All rational numbers are real.

q: All real numbers are not complex.

(ii) Here, the connecting word is ‘or’.


p: Square of an integer is positive.

q: Square of an integer is negative.

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(iii) Here, the connecting word is ‘and’.


p: The sand heats up quickly in the sun.

q: The sand does not cool down fast at night.

(iv) Here, the connecting word is ‘and’.


p: x = 2 is a root of the equation 3x2 – x – 10 = 0

q: x = 3 is a root of the equation 3x2 – x – 10 = 0

Question 2:

Identify the quantifier in the following statements and write the negation of the statements.

(i) There exists a number which is equal to its square.

(ii) For every real number x, x is less than x + 1.

(iii) There exists a capital for every state in India.

Answer :

(i) The quantifier is “There exists”.

The negation of this statement is as follows.

There does not exist a number which is equal to its square.

(ii) The quantifier is “For every”.


There exist a real number x such that x is not less than x + 1.

(iii) The quantifier is “There exists”.


There exists a state in India which does not have a capital.Question 3:

Check whether the following pair of statements is negation of each other. Give reasons for the

answer.

(i) x + y = y + x is true for every real numbers x and y.

(ii) There exists real number x and y for which x + y = y + x.

Answer :

The negation of statement (i) is as follows.

There exists real number x and y for which x + y ≠ y + x. This is not the same as statement (ii).

Thus, the given statements are not the negation of each other.Question 4:

State whether the “Or” used in the following statements is “exclusive “or” inclusive. Give reasons

for your answer.

(i) Sun rises or Moon sets.

(ii) To apply for a driving licence, you should have a ration card or a passport.

(iii) All integers are positive or negative.

Answer :

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(i) Here, “or” is exclusive because it is not possible for the Sun to rise and the moon to set

together.

(ii) Here, “or” is inclusive since a person can have both a ration card and a passport to apply for a

driving licence.

(iii) Here, “or” is exclusive because all integers cannot be both positive and negative.


Rewrite the following statement with “if-then” in five different ways conveying the same meaning.

If a natural number is odd, then its square is also odd.

Answer :

The given statement can be written in five different ways as follows.

(i) A natural number is odd implies that its square is odd.

(ii) A natural number is odd only if its square is odd.

(iii) For a natural number to be odd, it is necessary that its square is odd.

(iv) For the square of a natural number to be odd, it is sufficient that the number is odd.

(v) If the square of a natural number is not odd, then the natural number is not odd.Question 2:

Write the contrapositive and converse of the following statements.

(i) If x is a prime number, then x is odd.

(ii) It the two lines are parallel, then they do not intersect in the same plane.

(iii) Something is cold implies that it has low temperature.

(iv) You cannot comprehend geometry if you do not know how to reason deductively.

(v) x is an even number implies that x is divisible by 4

Answer :

(i) The contrapositive is as follows.

If a number x is not odd, then x is not a prime number.

The converse is as follows.

If a number x is odd, then it is a prime number.

(ii) The contrapositive is as follows.

If two lines intersect in the same plane, then they are not parallel.


If two lines do not intersect in the same plane, then they are parallel.

(iii) The contrapositive is as follows.

If something does not have low temperature, then it is not cold.


If something is at low temperature, then it is cold.

(iv) The contrapositive is as follows.

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If you know how to reason deductively, then you can comprehend geometry.


If you do not know how to reason deductively, then you cannot comprehend geometry.

(v) The given statement can be written as follows.

If x is an even number, then x is divisible by 4.

The contrapositive is as follows.

If x is not divisible by 4, then x is not an even number.


If x is divisible by 4, then x is an even number.Question 3:

Write each of the following statement in the form “if-then”.

(i) You get a job implies that your credentials are good.

(ii) The Banana trees will bloom if it stays warm for a month.

(iii) A quadrilateral is a parallelogram if its diagonals bisect each other.

(iv) To get A+ in the class, it is necessary that you do the exercises of the book.

Answer :

(i) If you get a job, then your credentials are good.

(ii) If the Banana tree stays warm for a month, then it will bloom.

(iii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

(iv) If you want to get an A+ in the class, then you do all the exercises of the book.Question 4:

Given statements in (a) and (b). Identify the statements given below as contrapositive or

converse of each other.

(a) If you live in Delhi, then you have winter clothes.

(i) If you do not have winter clothes, then you do not live in Delhi.

(ii) If you have winter clothes, then you live in Delhi.

(b) If a quadrilateral is a parallelogram, then its diagonals bisect each other.

(i) If the diagonals of a quadrilateral do not bisect each other, then the quadrilateral is not a

parallelogram.

(ii) If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Answer :

(a) (i) This is the contrapositive of the given statement (a).

(ii) This is the converse of the given statement (a).

(b) (i) This is the contrapositive of the given statement (b).

(ii) This is the converse of the given statement (b).


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Show that the statement

p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by

(i) direct method

(ii) method of contradiction

(iii) method of contrapositive

Answer :

p: “If x is a real number such that x3 + 4x = 0, then x is 0”.

Let q: x is a real number such that x3 + 4x = 0

r: x is 0.

(i) To show that statement p is true, we assume that q is true and then show that r is true.

Therefore, let statement q be true.

∴ x3 + 4x = 0

x (x2 + 4) = 0

⇒ x = 0 or x2 + 4 = 0

However, since x is real, it is 0.

Thus, statement r is true.

Therefore, the given statement is true.

(ii) To show statement p to be true by contradiction, we assume that p is not true.

Let x be a real number such that x3 + 4x = 0 and let x is not 0.

Therefore, x3 + 4x = 0

x (x2 + 4) = 0

x = 0 or x2 + 4 = 0

x = 0 or x2 = – 4

However, x is real. Therefore, x = 0, which is a contradiction since we have assumed that x is not

0.

Thus, the given statement p is true.

(iii) To prove statement p to be true by contrapositive method, we assume that r is false and

prove that q must be false.

Here, r is false implies that it is required to consider the negation of statement r. This obtains the

following statement.

∼r: x is not 0.

It can be seen that (x2 + 4) will always be positive.

x ≠ 0 implies that the product of any positive real number with x is not zero.

Let us consider the product of x with (x2 + 4).

∴ x (x2 + 4) ≠ 0

⇒ x3 + 4x ≠ 0

This shows that statement q is not true.

Thus, it has been proved that

∼r ⇒∼q

Therefore, the given statement p is true.Question 2:

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Show that the statement “For any real numbers a and b, a2 = b2 implies that a = b” is not true by

giving a counter-example.

Answer :

The given statement can be written in the form of “if-then” as follows.

If a and b are real numbers such that a2 = b2, then a = b.

Let p: a and b are real numbers such that a2 = b2.

q: a = b

The given statement has to be proved false. For this purpose, it has to be proved that if p, then

∼q. To show this, two real numbers, a and b, with a2 = b2 are required such that a ≠ b.

Let a = 1 and b = –1

a2 = (1)2 = 1 and b2 = (– 1)2 = 1

∴ a2 = b2

However, a ≠ b

Thus, it can be concluded that the given statement is false.Question 3:

Show that the following statement is true by the method of contrapositive.

p: If x is an integer and x2 is even, then x is also even.

Answer :

p: If x is an integer and x2 is even, then x is also even.

Let q: x is an integer and x2 is even.

r: x is even.

To prove that p is true by contrapositive method, we assume that r is false, and prove that q is

also false.

Let x is not even.

To prove that q is false, it has to be proved that x is not an integer or x2 is not even.

x is not even implies that x2 is also not even.

Therefore, statement q is false.

Thus, the given statement p is true.Question 4:

By giving a counter example, show that the following statements are not true.

(i) p: If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.

(ii) q: The equation x2 – 1 = 0 does not have a root lying between 0 and 2.

Answer :

(i) The given statement is of the form “if q then r”.

q: All the angles of a triangle are equal.

r: The triangle is an obtuse-angled triangle.

The given statement p has to be proved false. For this purpose, it has to be proved that if q, then

∼r.

To show this, angles of a triangle are required such that none of them is an obtuse angle.

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It is known that the sum of all angles of a triangle is 180°. Therefore, if all the three angles are

equal, then each of them is of measure 60°, which is not an obtuse angle.

In an equilateral triangle, the measure of all angles is equal. However, the triangle is not an

obtuse-angled triangle.

Thus, it can be concluded that the given statement p is false.

(ii) The given statement is as follows.

q: The equation x2 – 1 = 0 does not have a root lying between 0 and 2.

This statement has to be proved false. To show this, a counter example is required.

Consider x2 – 1 = 0

x2 = 1

x = ± 1

One root of the equation x2 – 1 = 0, i.e. the root x = 1, lies between 0 and 2.

Thus, the given statement is false.Question 5:

Which of the following statements are true and which are false? In each case give a valid reason

for saying so.

(i) p: Each radius of a circle is a chord of the circle.

(ii) q: The centre of a circle bisects each chord of the circle.

(iii) r: Circle is a particular case of an ellipse.

(iv) s: If x and y are integers such that x > y, then –x < –y.

(v) t: is a rational number.

Answer :

(i) The given statement p is false.

According to the definition of chord, it should intersect the circle at two distinct points.

(ii) The given statement q is false.

If the chord is not the diameter of the circle, then the centre will not bisect that chord.

In other words, the centre of a circle only bisects the diameter, which is the chord of the circle.

(iii) The equation of an ellipse is,

If we put a = b = 1, then we obtain

x2 + y2 = 1, which is an equation of a circle

Therefore, circle is a particular case of an ellipse.

Thus, statement r is true.

(iv) x > y

⇒ –x < –y (By a rule of inequality)

Thus, the given statement s is true.

(v) 11 is a prime number and we know that the square root of any prime number is an irrational

number. Therefore, is an irrational number.

Thus, the given statement t is false.

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MISCELLANEOUS of chapter 14

Question 1:

Write the negation of the following statements:

(i) p: For every positive real number x, the number x – 1 is also positive.

(ii) q: All cats scratch.

(iii) r: For every real number x, either x > 1 or x < 1.

(iv) s: There exists a number x such that 0 < x < 1.

Answer :

(i) The negation of statement p is as follows.

There exists a positive real number x, such that x – 1 is not positive.

(ii) The negation of statement q is as follows.

There exists a cat that does not scratch.

(iii) The negation of statement r is as follows.

There exists a real number x, such that neither x > 1 nor x < 1.

(iv) The negation of statement s is as follows.

There does not exist a number x, such that 0 < x < 1.Question 2:

State the converse and contrapositive of each of the following statements:

(i) p: A positive integer is prime only if it has no divisors other than 1 and itself.

(ii) q: I go to a beach whenever it is a sunny day.

(iii) r: If it is hot outside, then you feel thirsty.

Answer :

(i) Statement p can be written as follows.

If a positive integer is prime, then it has no divisors other than 1 and itself.

The converse of the statement is as follows.

If a positive integer has no divisors other than 1 and itself, then it is prime.

The contrapositive of the statement is as follows.

If positive integer has divisors other than 1 and itself, then it is not prime.

(ii) The given statement can be written as follows.

If it is a sunny day, then I go to a beach.

The converse of the statement is as follows.

If I go to a beach, then it is a sunny day.

The contrapositive of the statement is as follows.

If I do not go to a beach, then it is not a sunny day.

(iii) The converse of statement r is as follows.

If you feel thirsty, then it is hot outside.

The contrapositive of statement r is as follows.

If you do not feel thirsty, then it is not hot outside.Question 3:

Write each of the statements in the form “if p, then q”.

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(i) p: It is necessary to have a password to log on to the server.

(ii) q: There is traffic jam whenever it rains.

(iii) r: You can access the website only if you pay a subscription fee.

Answer :

(i) Statement p can be written as follows.

If you log on to the server, then you have a password.

(ii) Statement q can be written as follows.

If it rains, then there is a traffic jam.

(iii) Statement r can be written as follows.

If you can access the website, then you pay a subscription fee.Question 4:

Re write each of the following statements in the form “p if and only if q”.

(i) p: If you watch television, then your mind is free and if your mind is free, then you watch

television.

(ii) q: For you to get an A grade, it is necessary and sufficient that you do all the homework

regularly.

(iii) r: If a quadrilateral is equiangular, then it is a rectangle and if a quadrilateral is a rectangle,

then it is equiangular.

Answer :

(i) You watch television if and only if your mind is free.

(ii) You get an A grade if and only if you do all the homework regularly.

(iii) A quadrilateral is equiangular if and only if it is a rectangle.Question 5:

Given below are two statements

p: 25 is a multiple of 5.

q: 25 is a multiple of 8.

Write the compound statements connecting these two statements with “And” and “Or”. In both

cases check the validity of the compound statement.

Answer :

The compound statement with ‘And’ is “25 is a multiple of 5 and 8”.

This is a false statement, since 25 is not a multiple of 8.

The compound statement with ‘Or’ is “25 is a multiple of 5 or 8”.

This is a true statement, since 25 is not a multiple of 8 but it is a multiple of 5.Question 6:

Check the validity of the statements given below by the method given against it.

(i) p: The sum of an irrational number and a rational number is irrational (by contradiction

method).

(ii) q: If n is a real number with n > 3, then n2 > 9 (by contradiction method).

Answer :

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(i) The given statement is as follows.

p: the sum of an irrational number and a rational number is irrational.

Let us assume that the given statement, p, is false. That is, we assume that the sum of an

irrational number and a rational number is rational.

Therefore, , where is irrational and b, c, d, e are integers.

is a rational number and is an irrational number.

This is a contradiction. Therefore, our assumption is wrong.

Therefore, the sum of an irrational number and a rational number is rational.

Thus, the given statement is true.

(ii) The given statement, q, is as follows.

If n is a real number with n > 3, then n2 > 9.

Let us assume that n is a real number with n > 3, but n2 > 9 is not true.

That is, n2 < 9

Then, n > 3 and n is a real number.

Squaring both the sides, we obtain

n2 > (3)2

⇒ n2 > 9, which is a contradiction, since we have assumed that n2 < 9.

Thus, the given statement is true. That is, if n is a real number with n > 3, then n2 > 9.Question 7:

Write the following statement in five different ways, conveying the same meaning.

p: If triangle is equiangular, then it is an obtuse angled triangle.

Answer :

The given statement can be written in five different ways as follows.

(i) A triangle is equiangular implies that it is an obtuse-angled triangle.

(ii) A triangle is equiangular only if it is an obtuse-angled triangle.

(iii) For a triangle to be equiangular, it is necessary that the triangle is an obtuse-angled triangle.

(iv) For a triangle to be an obtuse-angled triangle, it is sufficient that the triangle is equiangular.

(v) If a triangle is not an obtuse-angled triangle, then the triangle is not equiangular.

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Exercise 15.1 SOULTIONS

Question 1:

Find the mean deviation about the mean for the data

4, 7, 8, 9, 10, 12, 13, 17

Answer :

The given data is

4, 7, 8, 9, 10, 12, 13, 17

Mean of the data,

The deviations of the respective observations from the mean are

–6, – 3, –2, –1, 0, 2, 3, 7

The absolute values of the deviations, i.e. , are

6, 3, 2, 1, 0, 2, 3, 7

The required mean deviation about the mean is

Question 2:


38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Answer :

The given data is

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean of the given data,

The deviations of the respective observations from the mean are

–12, 20, –2, –10, –8, 5, 13, –4, 4, –6

The absolute values of the deviations, i.e. , are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

The required mean deviation about the mean is

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Question 3:

Find the mean deviation about the median for the data.

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Answer :

The given data is

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Here, the numbers of observations are 12, which is even.

Arranging the data in ascending order, we obtain

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

The deviations of the respective observations from the median, i.e. are

–3.5, –2.5, –2.5, –1.5, –0.5, –0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The absolute values of the deviations, , are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The required mean deviation about the median is

Question 4:

Find the mean deviation about the median for the data

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Answer :

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The given data is

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Here, the number of observations is 10, which is even.

Arranging the data in ascending order, we obtain

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

The deviations of the respective observations from the median, i.e. are

–11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The absolute values of the deviations, , are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Thus, the required mean deviation about the median is

Question 5:

Find the mean deviation about the mean for the data.

xi 5 10 15 20 25

fi 7 4 6 3 5

Answer :

xi fi fi xi

5 7 35 9 63

10 4 40 4 16

15 6 90 1 6

20 3 60 6 18

25 5 125 11 55

25 350 158

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Question 6:


xi 10 30 50 70 90

fi 4 24 28 16 8

Answer :

xi fi fi xi

10 4 40 40 160

30 24 720 20 480

50 28 1400 0 0

70 16 1120 20 320

90 8 720 40 320

80 4000 1280

Question 7:

Find the mean deviation about the median for the data.

xi 5 7 9 10 12 15

fi 8 6 2 2 2 6

Answer :

The given observations are already in ascending order.

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Adding a column corresponding to cumulative frequencies of the given data, we obtain the

following table.

xi fi c.f.

5 8 8

7 6 14

9 2 16

10 2 18

12 2 20

15 6 26

Here, N = 26, which is even.

Median is the mean of 13th and 14th observations. Both of these observations lie in the

cumulative frequency 14, for which the corresponding observation is 7.

The absolute values of the deviations from median, i.e. are

|xi – M| 2 0 2 3 5 8

fi 8 6 2 2 2 6

fi |xi – M|16 0 4 6 10 48

and

Question 8:

Find the mean deviation about the median for the data

xi 15 21 27 30 35

fi 3 5 6 7 8

Answer :

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the

following table.

xi fi c.f.

15 3 3

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21 5 8

27 6 14

30 7 21

35 8 29

Here, N = 29, which is odd.

observation = 15th observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is

30.

∴ Median = 30

The absolute values of the deviations from median, i.e. are

|xi – M| 15 9 3 0 5

fi 3 5 6 7 8

fi |xi – M| 45 45 18 0 40

∴Question 9:

Find the mean deviation about the mean for the data.

Income per day Number of persons

0-100 4

100-200 8

200-300 9

300-400 10

400-500 7

500-600 5

600-700 4

700-800 3

Answer :

The following table is formed.

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Income per day Number of persons fi Mid-point xi fi xi

0 – 100 4 50 200 308 1232

100 – 200 8 150 1200 208 1664

200 – 300 9 250 2250 108 972

300 – 400 10 350 3500 8 80

400 – 500 7 450 3150 92 644

500 – 600 5 550 2750 192 960

600 – 700 4 650 2600 292 1168

700 – 800 3 750 2250 392 1176

50 17900 7896

Here,

Question 10:


Height in cms Number of boys

95-105 9

105-115 13

115-125 26

125-135 30

135-145 12

145-155 10

Answer :


Height in cms Number of boys fi Mid-point xi fi xi

95-105 9 100 900 25.3 227.7

105-115 13 110 1430 15.3 198.9

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115-125 26 120 3120 5.3 137.8

125-135 30 130 3900 4.7 141

135-145 12 140 1680 14.7 176.4

145-155 10 150 1500 24.7 247

Here,

Question 11:

Find the mean deviation about median for the following data:

Marks Number of girls

0-10 6

10-20 8

20-30 14

30-40 16

40-50 4

50-60 2

Answer :


Marks Number ofgirlsfi

Cumulative frequency(c.f.)

Mid-pointxi

|xi –Med.|

fi |xi –Med.|

0-10 6 6 5 22.85 137.1

10-20 8 14 15 12.85 102.8

20-30 14 28 25 2.85 39.9

30-40 16 44 35 7.15 114.4

40-50 4 48 45 17.15 68.6

50-60 2 50 55 27.15 54.3

50 517.1

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The class interval containing the or 25th item is 20 – 30.

Therefore, 20 – 30 is the median class.

It is known that,

Here, l = 20, C = 14, f = 14, h = 10, and N = 50

∴ Median =

Thus, mean deviation about the median is given by,

Question 12:

Calculate the mean deviation about median age for the age distribution of 100 persons given

below:

Age Number

16-20 5

21-25 6

26-30 12

31-35 14

36-40 26

41-45 12

46-50 16

51-55 9

Answer :

The given data is not continuous. Therefore, it has to be converted into continuous frequency

distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class

interval.

The table is formed as follows.

Age Numberfi Cumulative frequency (c.f.) Mid-pointxi |xi –Med.| fi |xi –Med.|

15.5-20.5 5 5 18 20 100

20.5-25.5 6 11 23 15 90

25.5-30.5 12 23 28 10 120

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30.5-35.5 14 37 33 5 70

35.5-40.5 26 63 38 0 0

40.5-45.5 12 75 43 5 60

45.5-50.5 16 91 48 10 160

50.5-55.5 9 100 53 15 135

100 735

The class interval containing the or 50th item is 35.5 – 40.5.

Therefore, 35.5 – 40.5 is the median class.

It is known that,

Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100

Thus, mean deviation about the median is given by,


Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12

Answer :

6, 7, 10, 12, 13, 4, 8, 12

Mean,

The following table is obtained.

xi

6 –3 9

7 –2 4

10 –1 1

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12 3 9

13 4 16

4 –5 25

8 –1 1

12 3 9

74

Question 2:

Find the mean and variance for the first n natural numbers

Answer :

The mean of first n natural numbers is calculated as follows.

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Question 3:

Find the mean and variance for the first 10 multiples of 3

Answer :

The first 10 multiples of 3 are

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Here, number of observations, n = 10

The following table is obtained.

xi

3 –13.5 182.25

6 –10.5 110.25

9 –7.5 56.25

12 –4.5 20.25

15 –1.5 2.25

18 1.5 2.25

21 4.5 20.25

24 7.5 56.25

27 10.5 110.25

30 13.5 182.25

742.5

Question 4:

Find the mean and variance for the data

Answer :

The data is obtained in tabular form as follows.

xi f i fixi

6 2 12 –13 169 338

xi 6 10 14 18 24 28 30

f i 2 4 7 12 8 4 3

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10 4 40 –9 81 324

14 7 98 –5 25 175

18 12 216 –1 1 12

24 8 192 5 25 200

28 4 112 9 81 324

30 3 90 11 121 363

40 760 1736

Here, N = 40,

Question 5:

Find the mean and variance for the data

Answer :


xi f i fixi

92 3 276 –8 64 192

93 2 186 –7 49 98

97 3 291 –3 9 27

98 2 196 –2 4 8

102 6 612 2 4 24

104 3 312 4 16 48

109 3 327 9 81 243

22 2200 640

Here, N = 22,

xi 92 93 97 98 102 104 109

f i 3 2 3 2 6 3 3

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Question 6:

Find the mean and standard deviation using short-cut method.

Answer :


xi fi yi2 fiyi fiyi

2

60 2 –4 16 –8 32

61 1 –3 9 –3 9

62 12 –2 4 –24 48

63 29 –1 1 –29 29

64 25 0 0 0 0

65 12 1 1 12 12

66 10 2 4 20 40

67 4 3 9 12 36

68 5 4 16 20 80

100 220 0 286

Mean,

Question 7:

Find the mean and variance for the following frequency distribution.

xi 60 61 62 63 64 65 66 67 68

fi 2 1 12 29 25 12 10 4 5

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Answer :

Class Frequency fi Mid-point xi yi2 fiyi fiyi

2

0-30 2 15 –3 9 –6 18

30-60 3 45 –2 4 –6 12

60-90 5 75 –1 1 –5 5

90-120 10 105 0 0 0 0

120-150 3 135 1 1 3 3

150-180 5 165 2 4 10 20

180-210 2 195 3 9 6 18

30 2 76

Mean,

Question 8:

Find the mean and variance for the following frequency distribution.

Answer :

ClassFrequency

fiMid-point xi yi

2 fiyi fiyi2

0-10 5 5 –2 4 –10 20

10-20 8 15 –1 1 –8 8

20-30 15 25 0 0 0 0

Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210

Frequencies 2 3 5 10 3 5 2

Classes 0-10 10-20 20-30 30-40 40-50

Frequencies 5 8 15 16 6

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30-40 16 35 1 1 16 16

40-50 6 45 2 4 12 24

50 10 68

Mean,

Question 9:

Find the mean, variance and standard deviation using short-cut method

Answer :

Class Interval Frequencyfi Mid-pointxi yi2 fiyi fiyi

2

70-75 3 72.5 –4 16 –12 48

75-80 4 77.5 –3 9 –12 36

80-85 7 82.5 –2 4 –14 28

Heightin cms

No. of children

70-75 3

75-80 4

80-85 7

85-90 7

90-95 15

95-100 9

100-105 6

105-110 6

110-115 3

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85-90 7 87.5 –1 1 –7 7

90-95 15 92.5 0 0 0 0

95-100 9 97.5 1 1 9 9

100-105 6 102.5 2 4 12 24

105-110 6 107.5 3 9 18 54

110-115 3 112.5 4 16 12 48

60 6 254

Mean,

Question 10:

The diameters of circles (in mm) drawn in a design are given below:

Answer :

Class Interval Frequencyfi Mid-pointxi fi2 fiyi fiyi

2

32.5-36.5 15 34.5 –2 4 –30 60

36.5-40.5 17 38.5 –1 1 –17 17

40.5-44.5 21 42.5 0 0 0 0

Diameters No. of children

33-36 15

37-40 17

41-44 21

45-48 22

49-52 25

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44.5-48.5 22 46.5 1 1 22 22

48.5-52.5 25 50.5 2 4 50 100

100 25 199

Here, N = 100, h = 4

Let the assumed mean, A, be 42.5.

Mean,


From the data given below state which group is more variable, A or B?

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80

Group A 9 17 32 33 40 10 9

Group B 10 20 30 25 43 15 7

Answer :

Firstly, the standard deviation of group A is calculated as follows.

Marks Group Afi

Mid-pointxi

yi2 fiyi fiyi

2

10-20 9 15 –3 9 –27 81

20-30 17 25 –2 4 –34 68

30-40 32 35 –1 1 –32 32

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PARING THEM FOR BETTER TOMMOROW

40-50 33 45

50-60 40 55

60-70 10 65

70-80 9 75

Σ 150

Here, h = 10, N = 150, A = 45

The standard deviation of group B is calculated as follows.

Marks Group Bfi

Mid-pointxi

10-20 10 15

20-30 20 25

30-40 30 35

40-50 25 45

50-60 43 55

60-70 15 65

70-80 7 75

Σ 150

https://www.fac

0 0 0 0

1 1 40 40

2 4 20 40

3 9 27 81

–6 342


point yi2 fiyi fiyi

2

15 –3 9 –30 90

25 –2 4 –40 80

35 –1 1 –30 30

45 0 0 0 0

55 1 1 43 43

65 2 4 30 60

75 3 9 21 63

–6 366

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Since the mean of both the groups is same, the group with greater standard deviation will be

more variable.

Thus, group B has more variability in the marks.Question 1:

From the data given below state which group is more variable, A or B?

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80

Group A 9 17 32 33 40 10 9

Group B 10 20 30 25 43 15 7

Answer :

Firstly, the standard deviation of group A is calculated as follows.

Marks Group A fi Mid-point xi yi2 fiyi fiyi

2

10-20 9 15 –3 9 –27 81

20-30 17 25 –2 4 –34 68

30-40 32 35 –1 1 –32 32

40-50 33 45 0 0 0 0

50-60 40 55 1 1 40 40

60-70 10 65 2 4 20 40

70-80 9 75 3 9 27 81

150 –6 342

Here, h = 10, N = 150, A = 45

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Marks Group Bfi

Mid-pointxi

yi2 fiyi fiyi

2

10-20 10 15 –3 9 –30 90

20-30 20 25 –2 4 –40 80

30-40 30 35 –1 1 –30 30

40-50 25 45 0 0 0 0

50-60 43 55 1 1 43 43

60-70 15 65 2 4 30 60

70-80 7 75 3 9 21 63

150 –6 366

Since the mean of both the groups is same, the group with greater standard deviation will be

more variable.

Thus, group B has more variability in the marks.

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Question 2:

From the prices of shares X and Y below, find out which is more stable in value:

X 35 54 52 53 56 58 52 50 51 49

Y 108 107 105 105 106 107 104 103 104 101

Answer :

The prices of the shares X are

35, 54, 52, 53, 56, 58, 52, 50, 51, 49

Here, the number of observations, N = 10

The following table is obtained corresponding to shares X.

xi

35 –16 256

54 3 9

52 1 1

53 2 4

56 5 25

58 7 49

52 1 1

50 –1 1

51 0 0

49 –2 4

350

The prices of share Y are

108, 107, 105, 105, 106, 107, 104, 103, 104, 101

The following table is obtained corresponding to shares Y.

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yi

108 3 9

107 2 4

105 0 0

105 0 0

106 1 1

107 2 4

104 –1 1

103 –2 4

104 –1 1

101 –4 16

40

C.V. of prices of shares X is greater than the C.V. of prices of shares Y.

Thus, the prices of shares Y are more stable than the prices of shares X.Question 3:

An analysis of monthly wages paid to workers in two firms A and B, belonging to the same

industry, gives the following results:

Firm A Firm B

No. of wage earners 586 648

Mean of monthly wages Rs 5253 Rs 5253

Variance of the distribution of wages 100 121

(i) Which firm A or B pays larger amount as monthly wages?

(ii) Which firm, A or B, shows greater variability in individual wages?

Answer :

(i) Monthly wages of firm A = Rs 5253

Number of wage earners in firm A = 586

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∴Total amount paid = Rs 5253 × 586

Monthly wages of firm B = Rs 5253

Number of wage earners in firm B = 648

∴Total amount paid = Rs 5253 × 648

Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B

are more than the number of wage earners in firm A.

(ii) Variance of the distribution of wages in firm A = 100

∴ Standard deviation of the distribution of wages in firm

A ((σ1) =

Variance of the distribution of wages in firm = 121

∴ Standard deviation of the distribution of wages in firm

The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater

standard deviation will have more variability.

Thus, firm B has greater variability in the individual wages.Question 4:

The following is the record of goals scored by team A in a football session:

No. of goals scored 0 1 2 3 4

No. of matches 1 9 7 5 3

For the team B, mean number of goals scored per match was 2 with a standard

deviation 1.25 goals. Find which team may be considered more consistent?

Answer :

The mean and the standard deviation of goals scored by team A are calculated as follows.

No. of goals scored No. of matches fixi xi2 fixi

2

0 1 0 0 0

1 9 9 1 9

2 7 14 4 28

3 5 15 9 45

4 3 12 16 48

25 50 130

Thus, the mean of both the teams is same.

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The standard deviation of team B is 1.25 goals.

The average number of goals scored by both the teams is same i.e., 2. Therefore, the team with

lower standard deviation will be more consistent.

Thus, team A is more consistent than team B.Question 5:

The sum and sum of squares corresponding to length x (in cm) and weight y

(ingm) of 50 plant products are given below:

Which is more varying, the length or weight?

Answer :

Here, N = 50

∴ Mean,

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Mean,

Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the

lengths.

Miscellaneous of chapter 15 Question 1:

The mean and variance of eight observations are 9 and 9.25, respectively. If six of the

observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Answer :

Let the remaining two observations be x and y.

Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.

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From (1), we obtain

x2 + y2 + 2xy = 144 …(3)


2xy = 64 … (4)


x2 + y2 – 2xy = 80 – 64 = 16

⇒ x – y = ± 4 … (5)


x = 8 and y = 4, when x – y = 4

x = 4 and y = 8, when x – y = –4

Thus, the remaining observations are 4 and 8.

Question 2:

The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations

are 2, 4, 10, 12 and 14. Find the remaining two observations.

Answer :

Let the remaining two observations be x and y.

The observations are 2, 4, 10, 12, 14, x, y.

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From (1), we obtain

x2 + y2 + 2xy = 196 … (3)


2xy = 196 – 100

⇒ 2xy = 96 … (4)


x2 + y2 – 2xy = 100 – 96

⇒ (x – y)2 = 4

⇒ x – y = ± 2 … (5)


x = 8 and y = 6 when x – y = 2

x = 6 and y = 8 when x – y = – 2

Thus, the remaining observations are 6 and 8.

Question 3:

The mean and standard deviation of six observations are 8 and 4, respectively. If each

observation is multiplied by 3, find the new mean and new standard deviation of the resulting

observations.

Answer :

Let the observations be x1, x2, x3, x4, x5, and x6.

It is given that mean is 8 and standard deviation is 4.

If each observation is multiplied by 3 and the resulting observations are yi, then

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From (1) and (2), it can be observed that,

Substituting the values of xi and in (2), we obtain

Therefore, variance of new observations =

Hence, the standard deviation of new observations isQuestion 4:

Given that is the mean and σ2 is the variance of n observations x1, x2 … xn. Prove that the

mean and variance of the observations ax1, ax2, ax3 …axn are and a2 σ2, respectively (a ≠ 0).

Answer :

The given n observations are x1, x2 … xn.

Mean =

Variance = σ2

If each observation is multiplied by a and the new observations are yi, then

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Therefore, mean of the observations, ax1, ax2 … axn, is .

Substituting the values of xiand in (1), we obtain

Thus, the variance of the observations, ax1, ax2 … axn, is a2 σ2.

Question 5:

The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On

rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and

standard deviation in each of the following cases:

(i) If wrong item is omitted.

(ii) If it is replaced by 12.

Answer :

(i) Number of observations (n) = 20

Incorrect mean = 10

Incorrect standard deviation = 2

That is, incorrect sum of observations = 200

Correct sum of observations = 200 – 8 = 192

∴ Correct mean

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(ii) When 8 is replaced by 12,

Incorrect sum of observations = 200

∴ Correct sum of observations = 200 – 8 + 12 = 204

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Question 6:

The mean and standard deviation of marks obtained by 50 students of a class in three subjects,

Mathematics, Physics and Chemistry are given below:

Subject Mathematics Physics Chemistry

Mean 42 32 40.9

Standard deviation 12 15 20

Which of the three subjects shows the highest variability in marks and which shows the lowest?

Answer :

Standard deviation of Mathematics = 12

Standard deviation of Physics = 15

Standard deviation of Chemistry = 20

The coefficient of variation (C.V.) is given by .

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The subject with greater C.V. is more variable than others.

Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in

Mathematics.

Question 7:

The mean and standard deviation of a group of 100 observations were found to be 20 and 3,

respectively. Later on it was found that three observations were incorrect, which were recorded

as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Answer :

Number of observations (n) = 100

Incorrect mean

Incorrect standard deviation

∴ Incorrect sum of observations = 2000

⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940

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Question 1:

Describe the sample space for the indicated experiment: A coin is tossed three times.

Answer :

A coin has two faces: head (H) and tail (T).

When a coin is tossed three times, the total number of possible outcomes is 23 = 8

Thus, when a coin is tossed three times, the sample space is given by:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}Question 2:

Describe the sample space for the indicated experiment: A die is thrown two times.

Answer :

When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.

When a die is thrown two times, the sample space is given by S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}

The number of elements in this sample space is 6 × 6 = 36, while the sample space is given by:

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2),

(3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5,

5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}Question 3:

Describe the sample space for the indicated experiment: A coin is tossed four times.

Answer :

When a coin is tossed once, there are two possible outcomes: head (H) and tail (T).

When a coin is tossed four times, the total number of possible outcomes is 24 = 16

Thus, when a coin is tossed four times, the sample space is given by:

S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT,

TTHH, TTHT, TTTH, TTTT}Question 4:

Describe the sample space for the indicated experiment: A coin is tossed and a die is thrown.

Answer :


A die has six faces that are numbered from 1 to 6, with one number on each face.

Thus, when a coin is tossed and a die is thrown, the sample space is given by:

S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}Question 5:

Describe the sample space for the indicated experiment: A coin is tossed and then a die is rolled

only in case a head is shown on the coin.

Answer :

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Thus, when a coin is tossed and then a die is rolled only in case a head is shown on the coin, the

sample space is given by:

S = {H1, H2, H3, H4, H5, H6, T}Question 6:

2 boys and 2 girls are in Room X, and 1 boy and 3 girls in Room Y. Specify the sample space for

the experiment in which a room is selected and then a person.

Answer :

Let us denote 2 boys and 2 girls in room X as B1, B2 and G1, G2 respectively. Let us denote 1 boy

and 3 girls in room Y as B3, and G3, G4, G5 respectively.

Accordingly, the required sample space is given by S = {XB1, XB2, XG1, XG2, YB3, YG3, YG4,

YG5}Question 7:

One die of red colour, one of white colour and one of blue colour are placed in a bag. One die is

selected at random and rolled, its colour and the number on its uppermost face is noted.

Describe the sample space.

Answer :


Let us denote the red, white, and blue dices as R, W, and B respectively.

Accordingly, when a die is selected and then rolled, the sample space is given by

S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}Question 8:

An experiment consists of recording boy-girl composition of families with 2 children.

(i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order

of their births?

(ii) What is the sample space if we are interested in the number of girls in the family?

Answer :

(i) When the order of the birth of a girl or a boy is considered, the sample space is given by S =

{GG, GB, BG, BB}

(ii) Since the maximum number of children in each family is 2, a family can either have 2 girls or

1 girl or no girl. Hence, the required sample space is S = {0, 1, 2}Question 9:

A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession

without replacement. Write the sample space for this experiment.

Answer :

It is given that the box contains 1 red ball and 3 identical white balls. Let us denote the red ball

with R and a white ball with W.

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When two balls are drawn at random in succession without replacement, the sample space is

given by

S = {RW, WR, WW}Question 10:

An experiment consists of tossing a coin and then throwing it second time if a head occurs. If a

tail occurs on the first toss, then a die is rolled once. Find the sample space.

Answer :



Thus, in the given experiment, the sample space is given by

S = {HH, HT, T1, T2, T3, T4, T5, T6}Question 11:

Suppose 3 bulbs are selected at random from a lot. Each bulb is tested and classified as

defective (D) or non-defective (N). Write the sample space of this experiment?

Answer :

3 bulbs are to be selected at random from the lot. Each bulb in the lot is tested and classified as

defective (D) or non-defective (N).

The sample space of this experiment is given by

S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}Question 12:

A coin is tossed. If the out come is a head, a die is thrown. If the die shows up an even number,

the die is thrown again. What is the sample space for the experiment?

Answer :

When a coin is tossed, the possible outcomes are head (H) and tail (T).

When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.

Thus, the sample space of this experiment is given by:

S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62,

H63, H64, H65, H66}Question 13:

The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are put in a box

and mixed thoroughly. A person draws two slips from the box, one after the other, without

replacement. Describe the sample space for the experiment.

Answer :

If 1 appears on the first drawn slip, then the possibilities that the number appears on the second

drawn slip are 2, 3, or 4. Similarly, if 2 appears on the first drawn slip, then the possibilities that

the number appears on the second drawn slip are 1, 3, or 4. The same holds true for the

remaining numbers too.

Thus, the sample space of this experiment is given by S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2,

4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

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Question 14:

An experiment consists of rolling a die and then tossing a coin once if the number on the die is

even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this

experiment.

Answer :

A die has six faces that are numbered from 1 to 6, with one number on each face. Among these

numbers, 2, 4, and 6 are even numbers, while 1, 3, and 5 are odd numbers.


Hence, the sample space of this experiment is given by:

S = {2H, 2T, 4H, 4T, 6H, 6T, 1HH, 1HT, 1TH, 1TT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT}Question 15:

A coin is tossed. If it shows a tail, we draw a ball from a box which contains 2 red and 3 black

balls. If it shows head, we throw a die. Find the sample space for this experiment.

Answer :

The box contains 2 red balls and 3 black balls. Let us denote the 2 red balls as R1, R2 and the 3

black balls as B1, B2, and B3.

The sample space of this experiment is given by

S = {TR1, TR2, TB1, TB2, TB3, H1, H2, H3, H4, H5, H6}Question 16:

A die is thrown repeatedly until a six comes up. What is the sample space for this experiment?

Answer :

In this experiment, six may come up on the first throw, the second throw, the third throw and so

on till six is obtained.

Hence, the sample space of this experiment is given by

S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), … , (1, 5, 6), (2, 1, 6), (2, 2, 6), … , (2,

5, 6), … ,(5, 1, 6), (5, 2, 6), …}


Question 1:

A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”.

Are E and F mutually exclusive?

Answer :

When a die is rolled, the sample space is given by

S = {1, 2, 3, 4, 5, 6}

Accordingly, E = {4} and F = {2, 4, 6}

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It is observed that E ∩ F = {4} ≠ Φ

Therefore, E and F are not mutually exclusive events.Question 2:

A die is thrown. Describe the following events:

(i) A: a number less than 7 (ii) B: a number greater than 7 (iii) C: a multiple of 3

(iv) D: a number less than 4 (v) E: an even number greater than 4 (vi) F: a number not less than

3

Also find

Answer :

When a die is thrown, the sample space is given by S = {1, 2, 3, 4, 5, 6}.

Accordingly:

(i) A = {1, 2, 3, 4, 5, 6}

(ii) B = Φ

(iii) C = {3, 6}

(iv) D = {1, 2, 3}

(v) E = {6}

(vi) F = {3, 4, 5, 6}

A ∪ B = {1, 2, 3, 4, 5, 6}, A ∩ B = Φ

B ∪ C = {3, 6}, E ∩ F = {6}

D ∩ E =Φ, A – C = {1, 2, 4, 5}

D – E = {1, 2, 3},Question 3:

An experiment involves rolling a pair of dice and recording the numbers that come up. Describe

the following events:

A: the sum is greater than 8, B: 2 occurs on either die

C: The sum is at least 7 and a multiple of 3.

Which pairs of these events are mutually exclusive?

Answer :

When a pair of dice is rolled, the sample space is given by

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It is observed that

A ∩ B =Φ

B ∩ C =Φ

Hence, events A and B and events B and C are mutually exclusive.Question 4:

Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event

“two heads and one tail show”. C denote the event “three tails show” and D denote the event ‘a

head shows on the first coin”. Which events are

(i) mutually exclusive? (ii) simple? (iii) compound?

Answer :

When three coins are tossed, the sample space is given by

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Accordingly,

A = {HHH}

B = {HHT, HTH, THH}

C = {TTT}

D = {HHH, HHT, HTH, HTT}

We now observe that

A ∩ B =Φ, A ∩ C =Φ, A ∩ D = {HHH} ≠ Φ

B ∩ C =Φ, B ∩ D = {HHT, {HTH} ≠ Φ

C ∩ D = Φ

(i) Event A and B; event A and C; event B and C; and event C and D are all mutually exclusive.

(ii) If an event has only one sample point of a sample space, it is called a simple event. Thus, A

and C are simple events.

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(iii) If an event has more than one sample point of a sample space, it is called a compound event.

Thus, B and D are compound events.

Question 5:

Three coins are tossed. Describe

(i) Two events which are mutually exclusive.

(ii) Three events which are mutually exclusive and exhaustive.

(iii) Two events, which are not mutually exclusive.

(iv) Two events which are mutually exclusive but not exhaustive.

(v) Three events which are mutually exclusive but not exhaustive.

Answer :

When three coins are tossed, the sample space is given by

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

(i) Two events that are mutually exclusive can be

A: getting no heads and B: getting no tails

This is because sets A = {TTT} and B = {HHH} are disjoint.

(ii) Three events that are mutually exclusive and exhaustive can be

A: getting no heads

B: getting exactly one head

C: getting at least two heads

i.e.,

A = {TTT}

B = {HTT, THT, TTH}

C = {HHH, HHT, HTH, THH}

This is because A ∩ B = B ∩ C = C ∩ A = Φand A ∪ B ∪ C = S

(iii) Two events that are not mutually exclusive can be

A: getting three heads

B: getting at least 2 heads

i.e.,

A = {HHH}

B = {HHH, HHT, HTH, THH}

This is because A ∩ B = {HHH} ≠ Φ

(iv) Two events which are mutually exclusive but not exhaustive can be

A: getting exactly one head

B: getting exactly one tail

That is

A = {HTT, THT, TTH}

B = {HHT, HTH, THH}

It is because, A ∩ B =Φ, but A ∪ B ≠ S

(v) Three events that are mutually exclusive but not exhaustive can be

A: getting exactly three heads

B: getting one head and two tails

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C: getting one tail and two heads

i.e.,

A = {HHH}

B = {HTT, THT, TTH}

C = {HHT, HTH, THH}

This is because A ∩ B = B ∩ C = C ∩ A = Φ, but A ∪ B ∪ C ≠ SQuestion 6:

Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice ≤ 5

Describe the events

(i) (ii) not B (iii) A or B

(iv) A and B (v) A but not C (vi) B or C

(vii) B and C (viii)

Answer :

When two dice are thrown, the sample space is given by

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(vii) B and C = B ∩ C

Question 7:

Two dice are thrown. The events A, B and C are as follows:

A: getting an even number on the first die.

B: getting an odd number on the first die.

C: getting the sum of the numbers on the dice ≤ 5

State true or false: (give reason for your answer)

(i) A and B are mutually exclusive

(ii) A and B are mutually exclusive and exhaustive

(iii)

(iv) A and C are mutually exclusive

(v) A and are mutually exclusive

(vi) are mutually exclusive and exhaustive.

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Answer :

(i) It is observed that A ∩ B = Φ

∴ A and B are mutually exclusive.


(ii) It is observed that A ∩ B = Φ and A ∪ B = S

∴ A and B are mutually exclusive and exhaustive.


(iii) It is observed that


(iv) It is observed that A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠Φ

∴A and C are not mutually exclusive.

Thus, the given statement is false.

(v)

∴A and are not mutually exclusive.


(vi) It is observed that ;

However,

Therefore, events are not mutually exclusive and exhaustive.



Question 1:

Which of the following can not be valid assignment of probabilities for outcomes of sample space

S =

Assignment ω1 ω2 ω3 ω4 ω5 ω6 ω7

(a) 0.1 0.01 0.05 0.03 0.01 0.2 0.6

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(b)

(c) 0.1 0.2 0.3 0.4 0.5 0.6 0.7

(d) –0.1 0.2 0.3 0.4 –0.2 0.1 0.3

(e)

Answer :

(a)

ω1 ω2 ω3 ω4 ω5 ω6 ω7

0.1 0.01 0.05 0.03 0.01 0.2 0.6

Here, each of the numbers p(ωi) is positive and less than 1.

Sum of probabilities

Thus, the assignment is valid.

(b)

ω1 ω2 ω3 ω4 ω5 ω6 ω7



Thus, the assignment is valid.

(c)

ω1 ω2 ω3 ω4 ω5 ω6 ω7

0.1 0.2 0.3 0.4 0.5 0.6 0.7



Thus, the assignment is not valid.

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(d)

ω1 ω2 ω3 ω4 ω5 ω6 ω7

–0.1 0.2 0.3 0.4 –0.2 0.1 0.3

Here, p(ω1) and p(ω5) are negative.

Hence, the assignment is not valid.

(e)

ω1 ω2 ω3 ω4 ω5 ω6 ω7

Here,

Hence, the assignment is not valid.Question 2:

A coin is tossed twice, what is the probability that at least one tail occurs?

Answer :

When a coin is tossed twice, the sample space is given by

S = {HH, HT, TH, TT}

Let A be the event of the occurrence of at least one tail.

Accordingly, A = {HT, TH, TT}

Question 3:

A die is thrown, find the probability of following events:

(i) A prime number will appear,

(ii) A number greater than or equal to 3 will appear,

(iii) A number less than or equal to one will appear,

(iv) A number more than 6 will appear,

(v) A number less than 6 will appear.

Answer :

The sample space of the given experiment is given by

S = {1, 2, 3, 4, 5, 6}

(i) Let A be the event of the occurrence of a prime number.

Accordingly, A = {2, 3, 5}

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(ii) Let B be the event of the occurrence of a number greater than or equal to 3. Accordingly, B =

{3, 4, 5, 6}

(iii) Let C be the event of the occurrence of a number less than or equal to one. Accordingly, C =

{1}

(iv) Let D be the event of the occurrence of a number greater than 6.

Accordingly, D = Φ

(v) Let E be the event of the occurrence of a number less than 6.

Accordingly, E = {1, 2, 3, 4, 5}

Question 4:

A card is selected from a pack of 52 cards.

(a) How many points are there in the sample space?

(b) Calculate the probability that the card is an ace of spades.

(c) Calculate the probability that the card is (i) an ace (ii) black card.

Answer :

(a) When a card is selected from a pack of 52 cards, the number of possible outcomes is 52 i.e.,

the sample space contains 52 elements.

Therefore, there are 52 points in the sample space.

(b) Let A be the event in which the card drawn is an ace of spades.

Accordingly, n(A) = 1

(c) (i)Let E be the event in which the card drawn is an ace.

Since there are 4 aces in a pack of 52 cards, n(E) = 4

(ii)Let F be the event in which the card drawn is black.

Since there are 26 black cards in a pack of 52 cards, n(F) = 26

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Question 5:

A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the

probability that the sum of numbers that turn up is (i) 3 (ii) 12

Answer :

Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that

are numbered 1, 2, 3, 4, 5, and 6, the sample space is given by

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Accordingly, n(S) = 12

(i) Let A be the event in which the sum of numbers that turn up is 3.

Accordingly, A = {(1, 2)}

(ii) Let B be the event in which the sum of numbers that turn up is 12.

Accordingly, B = {(6, 6)}

Question 6:

There are four men and six women on the city council. If one council member is selected for a

committee at random, how likely is it that it is a woman?

Answer :

There are four men and six women on the city council.

As one council member is to be selected for a committee at random, the sample space contains

10 (4 + 6) elements.

Let A be the event in which the selected council member is a woman.

Accordingly, n(A) = 6

Question 7:

A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each

tail that turns up. From the sample space calculate how many different amounts of money you

can have after four tosses and the probability of having each of these amounts.

Answer :

Since the coin is tossed four times, there can be a maximum of 4 heads or tails.

When 4 heads turns up, is the gain.

When 3 heads and 1 tail turn up, Re 1 + Re 1 + Re 1 – Rs 1.50 = Rs 3 – Rs 1.50 = Rs 1.50 is the

gain.

When 2 heads and 2 tails turns up, Re 1 + Re 1 – Rs 1.50 – Rs 1.50 = – Re 1, i.e., Re 1 is the

loss.

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When 1 head and 3 tails turn up, Re 1 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 3.50, i.e., Rs 3.50 is

the loss.

When 4 tails turn up, – Rs 1.50 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 6.00, i.e., Rs 6.00 is the

loss.

There are 24 = 16 elements in the sample space S, which is given by:

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT,

THTT, TTHT, TTTH, TTTT}

∴n(S) = 16

The person wins Rs 4.00 when 4 heads turn up, i.e., when the event {HHHH} occurs.

∴Probability (of winning Rs 4.00) =

The person wins Rs 1.50 when 3 heads and one tail turn up, i.e., when the event {HHHT, HHTH,

HTHH, THHH} occurs.

∴Probability (of winning Rs 1.50) =

The person loses Re 1.00 when 2 heads and 2 tails turn up, i.e., when the event {HHTT, HTTH,

TTHH, HTHT, THTH, THHT} occurs.

∴Probability (of losing Re 1.00)

The person loses Rs 3.50 when 1 head and 3 tails turn up, i.e., when the event {HTTT, THTT,

TTHT, TTTH} occurs.

Probability (of losing Rs 3.50) =

The person loses Rs 6.00 when 4 tails turn up, i.e., when the event {TTTT} occurs.

Probability (of losing Rs 6.00) =Question 8:

Three coins are tossed once. Find the probability of getting

(i) 3 heads (ii) 2 heads (iii) at least 2 heads

(iv) at most 2 heads (v) no head (vi) 3 tails

(vii) exactly two tails (viii) no tail (ix) at most two tails.

Answer :

When three coins are tossed once, the sample space is given by

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

∴Accordingly, n(S) = 8

It is known that the probability of an event A is given by

(i) Let B be the event of the occurrence of 3 heads. Accordingly, B = {HHH}

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∴P(B) =

(ii) Let C be the event of the occurrence of 2 heads. Accordingly, C = {HHT, HTH, THH}

∴P(C) =

(iii) Let D be the event of the occurrence of at least 2 heads.

Accordingly, D = {HHH, HHT, HTH, THH}

∴P(D) =

(iv) LetE be the event of the occurrence of at most 2 heads.

Accordingly, E = {HHT, HTH, THH, HTT, THT, TTH, TTT}

∴P(E) =

(v) Let F be the event of the occurrence of no head.

Accordingly, F = {TTT}

∴P(F) =

(vi) Let G be the event of the occurrence of 3 tails.

Accordingly, G = {TTT}

∴P(G) =

(vii) Let H be the event of the occurrence of exactly 2 tails.

Accordingly, H = {HTT, THT, TTH}

∴P(H) =

(viii) Let I be the event of the occurrence of no tail.

Accordingly, I = {HHH}

∴P(I) =

(ix) Let J be the event of the occurrence of at most 2 tails.

Accordingly, I = {HHH, HHT, HTH, THH, HTT, THT, TTH}

∴P(J) =Question 9:

If is the probability of an event, what is the probability of the event ‘not A’.

Answer :

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It is given that P(A) = .

Accordingly, P(not A) = 1 – P(A)Question 10:

A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is

(i) a vowel (ii) an consonant

Answer :

There are 13 letters in the word ASSASSINATION.

∴Hence, n(S) = 13

(i) There are 6 vowels in the given word.

∴Probability (vowel) =

(ii) There are 7 consonants in the given word.

∴Probability (consonant) =Question 11:

In a lottery, person choses six different natural numbers at random from 1 to 20, and if these six

numbers match with the six numbers already fixed by the lottery committee, he wins the prize.

What is the probability of winning the prize in the game? [Hint: order of the numbers is not

important.]

Answer :

Total number of ways in which one can choose six different numbers from 1 to 20

Hence, there are 38760 combinations of 6 numbers.

Out of these combinations, one combination is already fixed by the lottery committee.

∴Required probability of winning the prize in the game =Question 12:

Check whether the following probabilities P(A) and P(B) are consistently defined

(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6

(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

Answer :

(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6

It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F).

However, here, P(A ∩ B) > P(A).

Hence, P(A) and P(B) are not consistently defined.

(ii)P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

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It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F).

Here, it is seen that P(A ∪ B) > P(A) and P(A ∪ B) > P(B).

Hence, P(A) and P(B) are consistently defined.Question 13:

Fill in the blanks in following table:

P(A) P(B) P(A ∩ B) P(A ∪ B)

(i) …

(ii) 0.35 … 0.25 0.6

(iii) 0.5 0.35 … 0.7

Answer :

(i) Here,

We know that

(ii) Here, P(A) = 0.35, P(A ∩ B) = 0.25, P(A ∪ B) = 0.6

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴0.6 = 0.35 + P(B) – 0.25

⇒P(B) = 0.6 – 0.35 + 0.25

⇒P(B) = 0.5

(iii)Here, P(A) = 0.5, P(B) = 0.35, P(A ∪ B) = 0.7

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

∴0.7 = 0.5 + 0.35 – P(A ∩ B)

⇒P(A ∩ B) = 0.5 + 0.35 – 0.7

⇒P(A ∩ B) = 0.15Question 14:

Given P(A) = and P(B) = . Find P(A or B), if A and B are mutually exclusive events.

Answer :

Here, P(A) = , P(B) =

For mutually exclusive events A and B,

P(A or B) = P(A) + P(B)

∴P(A or B)Question 15:

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If E and F are events such that P(E) = , P(F) = and P(E and F) = , find:(i) P(E or F), (ii)

P(not E and not F).

Answer :

Here, P(E) = , P(F) = , and P(E and F) =

(i) We know that P(E or F) = P(E) + P(F) – P(E and F)

∴P(E or F) =

(ii) From (i), P(E or F) = P (E ∪ F) =

Question 16:

Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually

exclusive.

Answer :

It is given that P (not E or not F) = 0.25

Thus, E and F are not mutually exclusive.Question 17:

A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i)

P(not A), (ii) P (not B) and (iii) P(A or B).

Answer :

It is given that P(A) = 0.42, P(B) = 0.48, P(A and B) = 0.16

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(i) P(not A) = 1 – P(A) = 1 – 0.42 = 0.58

(ii) P(not B) = 1 – P(B) = 1 – 0.48 = 0.52

(iii) We know that P(A or B) = P(A) + P(B) – P(A and B)

∴ P(A or B) = 0.42 + 0.48 – 0.16 = 0.74Question 18:

In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of

the class study both Mathematics and Biology. If a student is selected at random from the class,

find the probability that he will be studying Mathematics or Biology.

Answer :

Let A be the event in which the selected student studies Mathematics and B be the event in

which the selected student studies Biology.

Accordingly, P(A) = 40% = =

P(B) = 30%

P(A and B) = 10%

We know that P(A or B) = P(A) + P(B) – P(A and B)

Thus, the probability that the selected student will be studying Mathematics or Biology is 0.6.Question 19:

In an entrance test that is graded on the basis of two examinations, the probability of a randomly

chosen student passing the first examination is 0.8 and the probability of passing the second

examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability

of passing both?

Answer :

Let A and B be the events of passing first and second examinations respectively.

Accordingly, P(A) = 0.8, P(B) = 0.7 and P(A or B) = 0.95


∴0.95 = 0.8 + 0.7 – P(A and B)

⇒ P(A and B) = 0.8 + 0.7 – 0.95 = 0.55

Thus, the probability of passing both the examinations is 0.55.Question 20:

The probability that a student will pass the final examination in both English and Hindi is 0.5 and

the probability of passing neither is 0.1. If the probability of passing the English examination is

0.75, what is the probability of passing the Hindi examination?

Answer :

Let A and B be the events of passing English and Hindi examinations respectively.

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Accordingly, P(A and B) = 0.5, P(not A and not B) = 0.1, i.e.,

P(A) = 0.75


∴0.9 = 0.75 + P(B) – 0.5

⇒ P(B) = 0.9 – 0.75 + 0.5

⇒ P(B) = 0.65

Thus, the probability of passing the Hindi examination is 0.65.Question 21:

In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and

NSS. If one of these students is selected at random, find the probability that

(i) The student opted for NCC or NSS.

(ii) The student has opted neither NCC nor NSS.

(iii) The student has opted NSS but not NCC.

Answer :

Let A be the event in which the selected student has opted for NCC and B be the event in which

the selected student has opted for NSS.

Total number of students = 60

Number of students who have opted for NCC = 30

∴P(A) =

Number of students who have opted for NSS = 32

Number of students who have opted for both NCC and NSS = 24

(i) We know that P(A or B) = P(A) + P(B) – P(A and B)

Thus, the probability that the selected student has opted for NCC or NSS is .

(ii)

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Thus, the probability that the selected students has neither opted for NCC nor NSS is .

(iii) The given information can be represented by a Venn diagram as

It is clear that

Number of students who have opted for NSS but not NCC

= n(B – A) = n(B) – n(A ∩ B) = 32 – 24 = 8

Thus, the probability that the selected student has opted for NSS but not for NCC =

Miscellaneous of chapter 16 Question 1:

A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn

from the box, what is the probability that

(i) all will be blue? (ii) atleast one will be green?

Answer :

Total number of marbles = 10 + 20 + 30 = 60

Number of ways of drawing 5 marbles from 60 marbles =

(i) All the drawn marbles will be blue if we draw 5 marbles out of 20 blue marbles.

5 blue marbles can be drawn from 20 blue marbles in ways.

∴Probability that all marbles will be blue =

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(ii) Number of ways in which the drawn marble is not green =

∴Probability that no marble is green =

∴Probability that at least one marble is green =Question 2:

4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3

diamonds and one spade?

Answer :

Number of ways of drawing 4 cards from 52 cards =

In a deck of 52 cards, there are 13 diamonds and 13 spades.

∴Number of ways of drawing 3 diamonds and one spade =

Thus, the probability of obtaining 3 diamonds and one spade = .Question 2:

4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3

diamonds and one spade?

Answer :

Number of ways of drawing 4 cards from 52 cards =

In a deck of 52 cards, there are 13 diamonds and 13 spades.

∴Number of ways of drawing 3 diamonds and one spade =

Thus, the probability of obtaining 3 diamonds and one spade =Question 3:

A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with

number ‘3’. If die is rolled once, determine

(i) P(2) (ii) P(1 or 3) (iii) P(not 3)

Answer :

Total number of faces = 6

(i) Number faces with number ‘2’ = 3

(ii) P (1 or 3) = P (not 2) = 1 − P (2)

(iii) Number of faces with number ‘3’ = 1

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Question 4:

In a certain lottery, 10,000 tickets are sold and ten equal prizes are awarded. What is the

probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets?

Answer :

Total number of tickets sold = 10,000

Number prizes awarded = 10

(i) If we buy one ticket, then

P (getting a prize) =

∴P (not getting a prize) =

(ii) If we buy two tickets, then

Number of tickets not awarded = 10,000 − 10 = 9990

P (not getting a prize) =

(iii) If we buy 10 tickets, then

P (not getting a prize) =Question 5:

Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the

100 students, what is the probability that

(a) you both enter the same sections?

(b) you both enter the different sections?

Answer :

My friend and I are among the 100 students.

Total number of ways of selecting 2 students out of 100 students =

(a) The two of us will enter the same section if both of us are among 40 students or among 60

students.

∴ Number of ways in which both of us enter the same section =

∴ Probability that both of us enter the same section

(b) P(we enter different sections)

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= 1 − P(we enter the same section)

=Question 6:

Three letters are dictated to three persons and an envelope is addressed to each of them, the

letters are inserted into the envelopes at random so that each envelope contains exactly one

letter. Find the probability that at least one letter is in its proper envelope.

Answer :

Let L1, L2, L3 be three letters and E1, E2, and E3 be their corresponding envelops respectively.

There are 6 ways of inserting 3 letters in 3 envelops. These are as follows:

There are 4 ways in which at least one letter is inserted in a proper envelope.

Thus, the required probability is .Question 7:

A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35.

Find (i) P(A ∩ B) (ii) P(A′ ∩ B′) (iii) P(A ∩ B′) (iv) P(B ∩ A′)

Answer :

It is given that P(A) = 0.54, P(B) = 0.69, P(A ∩ B) = 0.35

(i) We know that P (A ∪ B) = P(A) + P(B) − P(A ∩ B)

∴P (A ∪ B) = 0.54 + 0.69 − 0.35 = 0.88

(ii) A′ ∩ B′ = (A ∪ B)′ [by De Morgan’s law]

∴P(A′ ∩ B′) = P(A ∪ B)′ = 1 − P(A ∪ B) = 1 − 0.88 = 0.12

(iii) P(A ∩ B′) = P(A) − P(A ∩ B)

= 0.54 − 0.35

= 0.19

(iv) We know that

Question 8:

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From the employees of a company, 5 persons are selected to represent them in the managing

committee of the company. Particulars of five persons are as follows:

S. No. Name Sex Age in years

1. Harish M 30

2. Rohan M 33

3. Sheetal F 46

4. Alis F 28

5. Salim M 41

A person is selected at random from this group to act as a spokesperson. What is the probability

that the spokesperson will be either male or over 35 years?

Answer :

Let E be the event in which the spokesperson will be a male and F be the event in which the

spokesperson will be over 35 years of age.

Accordingly, P(E) = and P(F) =

Since there is only one male who is over 35 years of age,

We know that

Thus, the probability that the spokesperson will either be a male or over 35 years of age is .Question 9:

If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what

is the probability of forming a number divisible by 5 when, (i) the digits are repeated? (ii) the

repetition of digits is not allowed?

Answer :

(i)When the digits are repeated

Since four-digit numbers greater than 5000 are formed, the leftmost digit is either 7 or 5.

The remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 as repetition of digits is

allowed.

∴Total number of 4-digit numbers greater than 5000 = 2 × 5 × 5 × 5 − 1

= 250 − 1 = 249

[In this case, 5000 can not be counted; so 1 is subtracted]

A number is divisible by 5 if the digit at its units place is either 0 or 5.

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∴Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 2 × 5 × 5 × 2 − 1 =

100 − 1 = 99

Thus, the probability of forming a number divisible by 5 when the digits are repeated is

.

(ii)When repetition of digits is not allowed

The thousands place can be filled with either of the two digits 5 or 7.

The remaining 3 places can be filled with any of the remaining 4 digits.

∴Total number of 4-digit numbers greater than 5000 = 2 × 4 × 3 × 2

= 48

When the digit at the thousands place is 5, the units place can be filled only with 0 and the tens

and hundreds places can be filled with any two of the remaining 3 digits.

∴Here, number of 4-digit numbers starting with 5 and divisible by 5

= 3 × 2 = 6

When the digit at the thousands place is 7, the units place can be filled in two ways (0 or 5) and

the tens and hundreds places can be filled with any two of the remaining 3 digits.

∴Here, number of 4-digit numbers starting with 7 and divisible by 5

= 1 × 2 × 3 × 2 = 12

∴Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 6 + 12 = 18

Thus, the probability of forming a number divisible by 5 when the repetition of digits is not

allowed is .Question 10:

The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The

lock opens with a sequence of four digits with no repeats. What is the probability of a person

getting the right sequence to open the suitcase?

Answer :

The number lock has 4 wheels, each labelled with ten digits i.e., from 0 to 9.

Number of ways of selecting 4 different digits out of the 10 digits =

Now, each combination of 4 different digits can be arranged in ways.

∴Number of four digits with no repetitions =

There is only one number that can open the suitcase.

Thus, the required probability is .

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