06_Redacted.pdf - EduGorilla Study Material

jee main

1. Let S be the sum of the first 2015 terms of thesequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ....... where n occurs as n terms. The sum of the digits of S is(a) 21 (b) 22 (c) 23 (d) 24

2. Two circles S1 and S2 pass through the points(0, a) and (0, –a). The line y = mx+c is a tangent to the two circles. If S1 and S2 are orthogonal, then(a) a2(m2 + 1) = c2 (b) a2(m2 + 2) = c2

(c) c2(m2 + 1) = a2 (d) c2(m2 + 2) = a2

3. OABC is a tetrahedron with all the edges are of unitlength. The shortest distance between the edges OA and BC is

(a) 23 (b)

12

(c) 12

(d) 1

4. If A =1 1 01 1 1

0 1 1

−−

, then A–1 + A2 =

(a) I + 3A (b) I – 3A(c) 3A – I (d) 2I5. An ellipse has its axes along the coordinate axes

and passes through the points (1, 1) and 12

52

,

. Its

eccentricity is

(a) 12 (b)

12 (c) 2

3(d) 2

3

jee advanced

6. ii =

(a) eπ/2 (b) e3 2π/ (c) e−π/2 (d) e−3 2π/

Maths Musing was started in January 2003 issue of with the suggestion of Shri Mahabir Singh. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material.

During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India.Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced.

comprehension

If I = sin/

−∫ 1

0

2π(c cos x) dx, c < 1 =

cn

n

n

2 1

21 2 1

−

=

∞

−∑

( ),

then for c = 1

7. 12

1 nn=

∞

∑ =

(a) π2

12(b) π2

8(c) π2

6(d) π2

16

8. ( )− −

=

∞

∑ 1 1

21

n

n n

(a) π2

12(b) π2

8(c)

π2

6(d) π2

16integer match

9. A fair coin is tossed 10 times. Let mn

, in the lowest

terms, be the probability that heads never occur on consecutive tosses. Then m is

matching list

10. Let Cn be the coefficient of xn in the expansion of(1+x+x2)n.

Column-I Column-IIP. If n = 3, the last digit of Cn is 1. 1Q. If n = 4, the last digit of Cn is 2. 3R. If n = 5, the last digit of Cn is 3. 7S. If n = 7, the last digit of Cn is 4. 9

P Q R S(a) 1 2 3 4(b) 2 3 4 1(c) 3 1 2 4(d) 3 4 1 2

See Solution set of Maths Musing 149 on page no. 81

Set 150

| June ‘158

section-isingle correct option

this section contains 8 multiple choice questions. each question has four choices (a), (b), (c) and (d) out of which onlY one is correct. [3 marks for correct answer and –1 for wrong answer]

1. On a railway there are 10 stations. The number oftypes of tickets required in order that it may bepossible to book a passenger from every station toevery other is

(a) 108

!!

(b) 10! 2! (c) 102

!!

(d) 108 2

!! !

2. If the roots of ax2 + bx + c = 0 are of the formm

mm

m−+

11 and then the value of (a + b + c)2 is

(a) b2 – 2ac (b) 2b2 – ac(c) b2 – 4ac (d) 2(b2 – 2ac)

3. If A and B are two square matrices of order 3 × 3which satisfy AB = A and BA = B, then (A + B)7 is(a) 7(A + B) (b) 7·I3 × 3 (c) 64(A + B) (d) 128 I3 × 3

4. If a1, a2, a3, a4, a5 are in H.P., thena1a2 + a2a3 + a3a4 + a4a5 is equal to(a) 2a1a5 (b) 3a1a5 (c) 8a1a5 (d) 4a1a5

5. The maximum value of (sinα1)(sinα2) ... (sinαn)under the restrictions 0 ≤ α1, α2, ... αn < p/2 and(tanα1)(tanα2) .... (tanαn) = 1 is

(a) 12n (b) 1

2n(c) 1

2 2n/(d) 1

6. In a conference 10 speakers are to give theirspeaches one after another. Find the probability ofthe event if S1 speaks before S2 and S2 speaks beforeS3 and the remaining 7 speakers have no objectionto speak at any number?(a) 1/6 (b) 1/3 (c) 1/45 (d) 3/10

7. ii is a(a) complex number(b) purely imaginary number(c) real number (d) none of these

8. If k and K are minimum and maximum values ofsin–1x + cos–1x + tan–1x respectively, then

(a) k K= =π π4

34

, (b) k = 0, K = p

(c) k K= =π π2

, (d) not defined

section-iiMultiple correct option

this section contains 4 multiple choice questions. each question has 4 choices (a), (b), (c) and (d) for its answer, out of which one or More is/are correct. [4 marks for correct answer and −1 for wrong answer]

9. If f (x) and g(x) are functions such thatf (x + y) = f(x) · g(y) + g(x) · f(y) then

f g ff g ff g f

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

α α α θβ β β θγ γ γ θ

+++

is independent of

(a) α (b) b (c) g (d) q10. Sum of the roots of the equation

(x + 1) = 2log2(2x + 3) – 2log4(1980 – 2–x)is greater than(a) 2 (b) 3 (c) 4 (d) 5

11. Let 0 < P(A) < 1, 0 < P(B) < 1 andP(A ∪ B) = P(A) + P(B) – P(A) · P(B), then

(a) P B A P BP A

( / ) ( )( )

=

(b) P(AC ∪ BC) = P(AC) + P(BC)(c) P((A ∪ B)C) = P(AC) P(BC)(d) P(A/B) = P(A)

PaPer-1

| June ‘1510

12. The origin and roots of the equation z2 + pz + q = 0form an equilateral triangle if(a) p2 = q (b) p2 = 3q(c) q2 = 3p (d) q2 = p

section-iiilinked coMprehension tYpe

this section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. each question has 4 choices (a), (b), (c) and (d), out of which one or more than one can be correct. [4 marks for correct answer and −1 for wrong answer]

Paragraph for Question Nos. 13 to 15Let p be a prime number and n be a positive integer, then exponent of p is n! is denoted by Ep(n!) and is given by

E n np

np

np

np

P k( !) ...=

+

+

+ +

2 3

where pk < n < pk + 1 and [x] denotes the integral part of x.If we isolate the power of each prime contained in any number N, then N can be written as

N = p1α1 · p2

α2 · p3α3 .... where p1, p2, p3 ... prime

numbers and α1, α2, α3, ... are natural numbers.

13. The numbers of zeroes at the end of 108! is(a) 24 (b) 25 (c) 26 (d) 21

14. The number of prime numbers among thenumbers101! + 2, 101! + 3, 101! + 4, ..., 101! + 101 is(a) 31 (b) 29(c) 53 (d) none of these

15. The last non-zero digit in 20! must be equal to(a) 2 (b) 3 (c) 4 (d) 9

Paragraph for Question Nos. 16 to 18The numbers 1, 3, 6, 10, 15, 21, 28, …… are called triangular numbers. Let tn denotes the nth triangular number then it can be observed that t1 = 1, t2 = 3, tn = tn − 1 + n.

16. t100 must be equal to(a) 5050 (b) 5151(c) 5252 (d) none of these

17. If m is nth triangular number then

(a) n m= + +1 8 12

(b) n m= + −1 8 12

(c) n m= + −1 4 12

(d) none of these

18. The number of positive integers lying between t50and t51 must be

(a) 50 (b) 51 (c) 52 (d) none of these

section-iV

Matrix-Match tYpe

this section contains 2 questions. each question contains

statements given in two columns which have to be matched.

statements (a, B, c, d) in column i have to be matched with

statements (p, q, r, s) in column ii. the answers to these

questions h a v e t o b e a p p r o p r i a t e l y bubbled.

[8 marks for correct answer and no negative marking for

wrong answer and each row 2 marks]

19. Match the following:

Column I Column II

(A) If n be the number of ways in which 12 different books can be distributed equally among 3 persons, then

( !)!

412

4n is divisible by

p. 4

(B) If l be the number of integral solutions of x + y + z = 15 such that x ≥ 1, y ≥ 2 and z ≥ 3 then l is divisible by

q. 6

(C) If l be maximum number of points of intersection of 8 straight lines then l is divisible by

r. 12

(D) A car will hold 2 persons in the front seat and 1 in the near seat. If among 6 persons only 2 can drive, the number of ways in which car can be filled is l then l is divisible by

s. 5

20. If in a triangle 2a2 + 4b2 + c2 = 4ab + 2ac, thenmatch the following:

Column I Column II

(A) cosA p. 1/4

(B) cosB q. 7/8

(C) cosC r. 0

(D) sin(A – C) s. 1

| June ‘1512

PaPer-2

section-isingle correct option

this section contains multiple choice questions. each question has 4 choices (a), (b), (c) and (d), out of which onlY one is correct. [3 marks for correct answer and −1 for wrong answer].

1. The remainder when 597 is divided by 52 is(a) 2 (b) 3 (c) 5 (d) 10

2. The number of real solutions of (x, y), wherey = |sinx|, y = cos–1(cosx), –2p ≤ x ≤ 2p, is(a) 2 (b) 1 (c) 3 (d) 4

3. In DABC, ∠A = ∠C = p/6 and radius of incircle isr, if radius of circumscribing circle R = 4 then r isequal to(a) 4 3 6− (b) 4 3 6+(c) 2 3 2− (d) 2 3 2+

4. If 2 1 3x i= − + , then the value of (1 – x2 + x)6 – (1 – x + x2)6 = (a) 32 (b) –64 (c) 64 (d) 0

section-iiMultiple correct option

this section contains multiple choice questions. each question has 4 choices (a), (b), (c) and (d) for its answer, out of which one or More is/are correct. [4 marks for correct answer and −1 for wrong answer]

5. Let {D1, D2, D3, ..., Dk} be the set of third-orderdeterminants that can be made with the distinctnon-zero real numbers a1, a2, a3, ...., a9. Then

(a) k = 9! (b) ∆ii

k=∑

=0

1

(c) at least one Di = 0 (d) none of these

6. The quadratic equation x2 – 2x – l = 0, l ≠ 0,(a) cannot have a real roots if l < –1(b) can have a rational root if l is a perfect square

of an integer(c) cannot have an integral root if n2 – 1 < l < n2 + 2n

where n = 0, 1, 2, 3, ....(d) none of these

7. Given the set of four real numbers. First threenumbers are in G.P. and last three numbers arein A.P. The common difference of A.P. is 6. Firstand fourth terms are same. According to the givendata,(a) fourth term = 4 (b) fourth term = 8(c) sum of 4 nos. = 14 (d) sum of 4 nos. = 16

8. If x + y + z = 5 and xy + yz + zx = 3 ; x, y, z ∈ R thenfor x,

(a) largest value = 133

(b) largest value = 103

(c) least value = –1 (d) least value = 5

9. In the expansion of 4 16

34

20

+

,

(a) the number of irrational terms = 19(b) middle term is irrational(c) the number of irrational terms = 15(d) 9th term is rational

section-iiiMatrix-Match tYpe

this section contains 2 questions. each question contains

statements given in two columns which have to be matched.

statements (a, B, c, d) in column i have to be matched with

statements (p, q, r, s) in column ii. the answers to these

questions h a v e t o b e a p p r o p r i a t e l y bubbled.

[8 marks for correct answer and no negative marking for

wrong answer and each row 2 marks]


Column I Column II(A) Any two small squares on a chess

board are chosen at random. Probability that they have a common side is

p. 1/10

(B) One of ten keys opens the door. If we try the keys one after another, then the probability that the door is opened on the tenth attempt is

q. 0

(C) The probability of obtaining no head in an infinite sequence of independent tosses of a coin is

r. 7144

(D) Two squares are chosen at random from the small squares drawn on a chess board. The chance that the two squares chosen have exactly one corner in common is

s. 1/18

| June ‘1514


Column I Column II(A) Mr is defined as :

Mr

r

r

r =−

−

1 1

1 11 2( )

and |Mr| is the corresponding determinant value of Mr, then lim (| | | | ... | |)

n nM M M→∞

+ + + =2 3

p. 0

(B)

If 1

11

00

0

cos coscos coscos cos


α βα γβ γ

α βα γβ γ

=

11

1

00

0



α βα γβ γ

α βα γβ γ

=

then sin2α + sin2b + sin2g =

q. 4

(C)If A =

−

3 11 1

and a matrix C is

defined as C = (BAB–1)(B–1ATB), where |B| ≠ 0, then |C| = k2, (k ∈ N) where k =

r. 2

(D)If A =

−

1 11 1

and A4 = –lI,

then l equals

s. 1

section-iVinteger answer tYpe

this section contains 8 questions. the answer to each of the questions is a single−digit integer, ranging from 0 to 9. [4 marks for correct answer and −1 for wrong answer]

12. Find the number of roots of the equation z10 = 1satisfying |arg z| < (p/2).

13. Find the number of quadratic equations whichremain unchanged by squaring their roots.

14. A(0, 0), B(4, 2) and C(6, 0) are the vertices of atriangle ABC and BD is its altitude. The line throughD parallel to the side AB intersects the side BC ata point E. Find the product of areas of DABC andDBDE.

15. A coin is tossed twice and the four possibleoutcomes are assumed to be equally likely. If A isthe event, ‘both head and tail have appeared’, andB be the event, ‘at most one tail is observed’, then5[P(B/A)] is

16. The sum of the roots of the equation

42

4 1 03 2sin cos cos( )π π+

− − + − =x x x in the

interval [0, 4p] is pp. Find p.

17. In DABC, ∠A = 30° and ∠C = 105°. Find k such that

k c ba

= −

2 2

2

2.

18. Find the number of integral values of l if (l, 2)is an interior point of DABC formed by x + y = 4,3x − 7y = 8, 4x − y = 31.

19. The equation ax2 + bx + 6 = 0 does not have twodistinct real roots, then the least value of 3a + b + 3is

answers

paper-1

1. (a) 2. (c) 3. (c) 4. (d) 5. (c)6. (a) 7. (c) 8. (a) 9. (a, b, c, d)10. (a, b) 11. (c, d) 12. (b) 13. (b) 14. (d)15. (c) 16. (a) 17. (b) 18. (a)19. (A) → (p, q, r); (B) → (s); (C) → (p); (D) → (p, s)20. (A) → (p) ; (B) → (q) ; (C) → (p) ; (D) → (r)

paper-2

1. (c) 2. (c) 3. (a) 4. (d) 5. (a, b)6. (a, c) 7. (b, c) 8. (a, c) 9. (a, b, d)10. (A) → (s) ; (B) → (p) ; (C) → (q) ; (D) → (r)11. (A) → (s) ; (B) → (r) ; (C) → (q) ; (D) → (q)12. (5) 13. (4) 14. (8) 15. (5) 16. (6)17. (3) 18. (1) 19. (1)

nn

| June ‘1516

CATEGORY – I (Q1 to Q60)

Each question has one correct option and carries 1 mark, for wrong answer 1/4 mark will

be deducted

1. In a certain town, 60% of the families own a car,30% own a house and 20% own both a car and a house. If a family is randomnly chosen, what is the probability that this family owns a car or a house but not both?(a) 0.5 (b) 0.7 (c) 0.1 (d) 0.9

2. The letters of the word COCHIN are permutedand all the permutations are arranged in alphabetical order as in English dictionary. The number of words that appear before the word COCHIN is (a) 360 (b) 192 (c) 96 (d) 48

3. Let f : R → R be a continuous function which

satisfies f x f t dtx

( ) ( )= ∫0

. Then the value of f (loge5) is

(a) 0 (b) 2 (c) 5 (d) 3

4. The value of limx

x tx

dt→ −∫

2

2

2

32

is

(a) 10 (b) 12 (c) 8 (d) 16

5. If cot tan ,23 3 3x x kx+ = cosec then the value of k is

(a) 1 (b) 2 (c) 3 (d) –1

6. If θ π π∈

2

32

, , then the value of

4 2 2 24 2

4 2 2cos sin cot cosθ θ θ π θ+ + −

is

(a) –2cotq (b) 2cotq(c) 2cosq (d) 2sinq

7. The number of real solutions of the equation(sinx – x)(cosx – x2) = 0 is (a) 1 (b) 2 (c) 3 (d) 4

8. The value of l, such that the following system ofequations has no solution, is

x – 2y + z = – 4 2x – y – 2z = 2

x + y + lz = 4(a) 3 (b) 1 (c) 0 (d) –3

9. If f xx x

x x x x xx x x x x x x x

( ) ( ) ( )( ) ( )( ) ( ) ( )

=+

− +− − − + −

1 12 1 1

3 1 1 2 1 1

Then f (100) is equal to (a) 0 (b) 1 (c) 100 (d) 1010. Which of the following is not always true?(a) | | | | | |

a b a b+ = +2 2 2 if

a b and are perpendicular to each other.

(b) | | | | a b a+ ≥λ for a l l l ∈ R i f

a b and are perpendicular to each other

(c) | | | | | | | | a b a b a b+ + − = +( )2 2 2 22

(d) | | | | a b a+ ≥λ for all l ∈ R if a is parallel to b

11. If the four points with position vectors− + + + + −2 i j k i j k j k^ ^ ^ ^ ^ ^ ^ ^, , and λ j k^ ^+ are coplanarthen l = (a) 1 (b) 2 (c) –1 (d) 0

12. If sin ...− − + − +

=1

2 3 4

2 4 8 6x x x x π

where |x| < 2

then the value of x is (a) 2/3 (b) 3/2 (c) –2/3 (d) –3/213. The area of the region bounded by the curve y = x3,its tangent at (1, 1) and x-axis is

(a) 1

12 (b) 1

6(c) 2

17(d)

215

14. If log0.2(x – 1) > log0.04(x + 5) then(a) –1 < x < 4 (b) 2 < x < 3(c) 1 < x < 4 (d) 1 < x < 3

| June ‘15 17

15. The number of real roots of equation logex + ex = 0 is(a) 0 (b) 1 (c) 2 (d) 3

16. For all real values of a0, a1, a2, a3 satisfying

aa a a

01 2 32 3 4

0+ + + = , the equation a0 + a1x + a2x2 +

a3x3 = 0 has a real root in the interval(a) [0, 1] (b) [–1, 0] (c) [1, 2] (d) [–2, –1]

17. Let f : R → R be defined as

f xx

x x( )

,sin | |,

=

0 is irrational is rational

. Then which of the

following is true?(a) f is discontinuous for all x(b) f is continuous for all x(c) f is discontinuous at x = kp , where k is an

integer(d) f is continuous at x = kp, where k is an integer.

18. If the vertex of the conic y2 – 4y = 4x – 4a alwayslies between the straight lines x + y = 3 and 2x + 2y – 1 = 0 then

(a) 2 < a < 4 (b) − < <12

2a

(c) 0 < a < 2 (d) − < <12

32

a

19. Number of intersecting points of the conic4x2 + 9y2 = 1 and 4x2 + y2 = 4 is (a) 1 (b) 2 (c) 3 (d) 0

20. The value of l for which the straight linex y z− =

−+

= −−

λλ31

23

1 may lie on the plane x – 2y = 0 is

(a) 2 (b) 0 (c) –1/2 (d) there is no such l

21. If f : [0, p/2) → R is defined as

f ( )tan

tan tantan

θθ

θ θθ

= −− −

1 11

1 1

Then the range of f is (a) (2, ∞) (b) (–∞, –2](c) [2, ∞) (d) (–∞, 2]

22. If A and B are two matrices such that AB = B andBA = A, then A2 + B2 equals(a) 2AB (b) 2BA (c) A + B (d) AB

23. If w is an imaginary cube root of unity, then the

value of the determinant

1

1

2

2 2

2 2

+ −

+ −

+ −

ω ω ω

ω ω ω

ω ω ω ω

is

(a) –2w (b) –3w2 (c) –1 (d) 0

24. The value of 2 12

43

1 1cot cot− −− is

(a) − π8

(b) 32π

(c) π4

(d) π2

25. If the point (2cosq, 2sinq), for q ∈ (0, 2p) lies inthe region between the lines x + y = 2 and x – y = 2 containing the origin, then q lies in

(a) 02

32

2, ,π π π

∪

(b) [0, p]

(c) π π2

32

,

(d) π π4 2

,

26. Number of points having distance 5 from the

straight line x – 2y + 1 = 0 and a distance 13 from the line 2x + 3y – 1 = 0 is (a) 1 (b) 2 (c) 4 (d) 5

27. Let a, b, c, d be any four real numbers. Thenan + bn = cn + dn holds for any natural number n if (a) a + b = c + d (b) a – b = c – d(c) a + b = c + d, a2 + b2 = c2 + d2

(d) a – b = c – d, a2 – b2 = c2 – d2

28. If a, b are the roots of x2 – px + 1 = 0 and g is a rootof x2 + px + 1 = 0, then (a + g)(b + g) is (a) 0 (b) 1 (c) –1 (d) p

29. Number of irrational terms in the binomialexpansion of (31/5 + 71/3)100 is (a) 90 (b) 88 (c) 93 (d) 95

30. The quadratic expression (2x + 1)2 – px + q ≠ 0 forany real x if (a) p2 – 16p – 8q < 0 (b) p2 – 8p + 16q < 0(c) p2 – 8p – 16q < 0 (d) p2 – 16p + 8q < 0

31. Let f : R → R be defined as f x x xx x

( ) = − ++ +

2

244

. Then

the range of the function f (x) is

(a) 35

53

,

(b) 35

53

,

(c) −∞

∪ ∞

, ,35

53

(d) − −

53

35

,

| June ‘1518

32. The least value of 2x2 + y2 + 2xy + 2x – 3y + 8 forreal numbers x and y is (a) 2 (b) 8 (c) 3 (d) 1

33. Let f : [–2, 2] → R be a continuous function suchthat f (x) assumes only irrational values. If f ( ) ,2 2=then

(a) f (0) = 0 (b) f ( )2 1 2 1− = −

(c) f ( )2 1 2 1− = + (d) f ( )2 1 2− =

34. The minimum value of cos sinsin

θ θθ

+ + 22

for

q ∈ (0, p/2) is (a) 2 2+ (b) 2(c) 1 2+ (d) 2 2

35. The value of 1 31 3

1 31 3

64 64+−

+ −

+

ii

ii

is

(a) 0 (b) –1 (c) 1 (d) i

36. Find the maximum value of |z| when zz

− =3 2,

z being a complex number.(a) 1 3+ (b) 3(c) 1 2+ (d) 1

37. Given that x is a real number satisfying

5 26 53 10 3

02

2x xx x

− +− +

< , then

(a) x < 15

(b) 15

3< <x

(c) x > 5 (d) 15

13

3 5< < < <x xor

38. Let xn nn = −

−

−

−+

1 1

31 1

61 1

101 1

12

2 2 2....

( )

2

n ≥ 2. Then the value of limn nx

→∞ is

(a) 1/3 (b) 1/9 (c) 1/81 (d) 0

39. The variance of first 20 natural numbers is(a) 133/4 (b) 279/12 (c) 133/2 (d) 399/4

40. A fair coin is tossed a fixed number of times.If the probability of getting exactly 3 heads equals the probability of getting exactly 5 heads, then the probability of getting exactly one head is (a) 1/64 (b) 1/32 (c) 1/16 (d) 1/8

41. If the letters of the word PROBABILITY are writ-ten down at random in a row, the probability that two B's are together is (a) 2/11 (b) 10/11(c) 3/11 (d) 6/11

42. The least positive value of t so that the linesx = t + a, y + 16 = 0 and y = ax are concurrent is (a) 2 (b) 4 (c) 16 (d) 8

43. If in a triangle ABC, a2cos2A – b2 – c2 = 0, then

(a) π π4 2

< <A (b) π π2

< <A

(c) A = π2

(d) A < π4

44. { :| cos | sin } ,x R x x∈ ≥ ∩

=0 32π

(a) 04

34

32

, ,π π π

∪

(b) 04 2

32

, ,π π π

∪

(c) 04

54

32

, ,π π π

∪

(d) 0 32

, π

45. The number of distinct real roots ofsin cos coscos sin coscos cos sin

x x xx x xx x x

= 0 in the interval − ≤ ≤π π4 4

x isis

(a) 0 (b) 2 (c) 1 (d) > 2

46. Let x1, x2, ..., x15 be 15 distinct numbers chosenfrom 1, 2, 3, ..., 15. Then the value of (x1 – 1)(x2 – 1) (x3 – 1) ... (x15 – 1) is (a) always ≤ 0 (b) 0(c) always even (d) always odd

47. Let [x] denote the greatest integer less than orequal to x. Then the value of a for which the function

f xx

xx

x( )

sin[ ][ ]

,

,=

−−

≠

=

2

2 0

0αis continuous at x = 0 is

(a) a = 0 (b) a = sin(–1)(c) a = sin(1) (d) a = 1

| June ‘15 19

48. A particle starts moving from rest from a fixedpoint in a fixed direction. The distance s from the fixed point at a time t is given by s = t2 + at – b + 17, where a, b are real numbers. If the particle comes to rest after 5 sec at a distance of s = 25 units from the fixed point, then values of a and b are respectively(a) 10, –33 (b) –10, –33(c) –8, 33 (d) –10, 33

49. lim ...n

nn n→∞

+ + + − =1 2 1

(a) 1/2 (b) 1/3 (c) 2/3 (d) 0

50. If limlog( )

x

xaxe b x

x→

− +=

0 21

3 , then the values of

a and b are respectively(a) 2, 2 (b) 1, 2 (c) 2, 1 (d) 2, 0

51. Area of the region bounded by y = |x| andy = –|x| + 2 is (a) 4 sq. units (b) 3 sq. units(c) 2 sq. units (d) 1 sq. unit

52. Let d(n) denote the number of divisors of nincluding 1 and itself. Then d(225), d(1125) and d(640) are (a) in A.P. (b) in H.P.(c) in G.P. (d) consecutive integers

53. The trigonometric equation sin–1x = 2sin–12a has areal solution if

(a) | |a > 12

(b) 1

2 212

< <| |a

(c) | |a > 12 2

(d) | |a ≤ 12 2

54. If 2 + i and 5 2− i are the roots of the equation(x2 + ax + b)(x2 + cx + d) = 0, where a, b, c, d are real constants, then product of all roots of the equation is (a) 40 (b) 9 5 (c) 45 (d) 35

55. Let P(x) be a polynomial, which when divided byx – 3 and x – 5 leaves remainders 10 and 6 respectively. If the polynomial is divided by (x – 3)(x – 5) then the remainder is (a) –2x + 16 (b) 16(c) 2x – 16 (d) 60

56. The integrating factor of the differential equationdydx

x y x y+ − + =−( tan )( )3 1 02 1 3 2 is

(a) ex2 (b) ex3 (c) e3x2 (d) e3x3

57. If y = e–x cos2x, then which of the followingdifferential equation is satisfied?

(a) d y

dx

dydx

y2

2 2 5 0+ + = (b) d y

dx

dydx

y2

2 5 2 0+ + =

(c) d y

dx

dydx

y2

2 5 2 0− − = (d) d y

dx

dydx

y2

2 2 5 0+ − =

58. Let f (x) denote the fractional part of a real number x.

Then the value of f x dx( )2

0

3 is ∫

(a) 2 3 2 1− − (b) 0(c) 2 3 1− + (d) 3 2 1− +

59. Let S = {(a, b, c) ∈ N × N × N : a + b + c = 21,a ≤ b ≤ c} and T = {(a, b, c) ∈ N × N × N : a, b, c are in A.P.}, where N is the set of all natural numbers. Then the number of elements in the set S ∩ T is (a) 6 (b) 7 (c) 13 (d) 14

60. Let y = ex2 and y = ex2sinx be two given curves.Then the angle between the tangents to the curves at any point of their intersection is (a) 0 (b) p (c) p/2 (d) p/4

CATEGORY–II (Q61 to Q70)

Each question has one correct option and carries 2 mark, for each wrong answer 1/2

mark will be deducted.

61. A person goes to office by car or scooter or busor train, probability of which are 1/7, 3/7, 2/7 and 1/7 respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is 2/9, 1/9, 4/9 and 1/9 respectively. Given that he reached office in time, the probability that he travelled by a car is (a) 1/7 (b) 2/7 (c) 3/7 (d) 4/7

62. The value of ( −− +

∫x dx

x x2

2 32 7 1 3)

{( ) ( ) } / is

(a) 320

23

4 3xx

c−+

+/

(b) 3

2023

3 4xx

c−+

+/

(c) 5

1223

4 3xx

c−+

+/

(d) 3

2023

5 3xx

c−+

+/

63. In a triangle ABC, ∠C = 90°, r and R are the inradiusand circumradius of the triangle ABC respectively, then 2(r + R) is equal to (a) b + c (b) c + a(c) a + b (d) a + b + c

| June ‘1520

64. Let a, b be two distinct roots of acosq + bsinq = c,where a, b and c are three real constants and q ∈ [0, 2p]. Then a + b is also a root of the same equation if (a) a + b = c (b) b + c = a(c) c + a = b (d) c = a

65. For a matrix, A =

1 0 02 1 03 2 1

, if U1, U2 and U3 are

3 × 1 column matrices satisfying

AU AU AU1 2 3

100

230

231

=

=

=

, , and U is a 3 × 3

matrix whose columns are U1, U2 and U3. Then sum of the elements of U–1 is (a) 6 (b) 0 (c) 1 (d) 2/3

66. Let f : R → R be differentiable at x = 0. If f (0) = 0and f ′(0) = 2, then the value of

lim [ ( ) ( ) ( ) ... ( )]x x

f x f x f x f x→

+ + + +0

1 2 3 2015 is

(a) 2015 (b) 0(c) 2015 × 2016 (d) 2015 × 2014

67. If x and y are digits such that17! = 3556xy428096000, then x + y equals

(a) 15 (b) 6 (c) 12 (d) 13

68. Let f : N → R be such that f (1) = 1 andf (1) + 2f (2) + 3f (3) + .... + nf (n) = n(n + 1)f (n), for all n ∈ N, n ≥ 2, where N is the set of natural numbers and R is the set of real numbers. Then the value of f (500) is (a) 1000 (b) 500 (c) 1/500 (d) 1/1000

69. If 5 distinct balls are placed at random into 5 cells,then the probability that exactly one cell remains empty is (a) 48/125 (b) 12/125 (c) 8/125 (d) 1/125

70. A survey of people in a given region showed that20% were smokers. The probability of death due to lung cancer, given that a person smoked, was 10 times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is 0.006, what is the probability of death due to lung cancer given that a person is a smoker?(a) 1/140 (b) 1/70 (c) 3/140 (d) 1/10

CATEGORY–III (Q71 to Q80)

Each question has one or more correct options, choosing which will fetch maximum 2 marks

on pro rata basis. however, choice of any wrong option(s) will fetch zero mark for the

question.

71. If cosx and sinx are solutions of the differential

equation ad y

dxa

dydx

a y0

2

2 1 2 0+ + = , where a0, a1, a2 are

real constants then which of the following is/are always true?(a) Acosx + Bsinx is a solution, where A and B are

real constants.

(b) A xcos +

π4

is a solution, where A is a a real

constant.(c) Acosx s inx is a solution, where A is real

constant.

(d) A x B xcos sin+

+ −

π π4 4

is a solution, where

A and B are real constants.

72. Which of the following statements is/are correct

for 02

< <θ π ?

(a) (cos ) cos/θ θ1 22

≤ (b) (cos ) cos/θ θ3 4 34

≥

(c) cos (cos ) /56

5 6θ θ≥ (d) cos (cos ) /78

7 8θ θ≤

73. Which of the following is/are always false?(a) A quadratic equation with rational coefficients has

zero or two irrational roots.(b) A quadratic equation with real coefficients has

zero or two non-real roots.(c) A quadratic equation with irrational coefficients

has zero or two rational roots.(d) A quadratic equation with integer coefficients has

zero or two irrational roots.

74. If the straight line (a – 1)x – by + 4 = 0 is normal tothe hyperbola xy = 1 then which of the following does not hold?(a) a > 1, b > 0 (b) a > 1, b < 0(c) a < 1, b < 0 (d) a < 1, b > 0

75. Let f be any continuously differentiable functionon [a, b] and twice differentiable on (a, b) such that

| June ‘15 21

f(a) = f ′(a) = 0 and f (b) = 0. Then(a) f ′′(a) = 0(b) f ′(x) = 0 for some x ∈ (a, b)(c) f ′′(x) = 0 for some x ∈ (a, b)(d) f ′′′(x) = 0 for some x ∈ (a, b)

76. A relation r on the set of real number R is definedas follows: "x r y if and only if xy > 0". Then which of the following is/are true?(a) r is reflexive and symmetric.(b) r is symmetric but not reflexive(c) r is symmetric and transitive(d) r is an equivalence relation

77. Suppose a machine produces metal parts thatcontains some defective parts with probability 0.05. How many parts should be produced in order that the probability of at least one part being defective is 1/2 or more? (Given log1095 = 1.977 and log102 = 0.3)(a) 11 (b) 12 (c) 15 (d) 14

78. Let f : R → R be such that f (2x – 1) = f (x) for allx ∈ R. If f is continuous at x = 1 and f (1) = 1, then(a) f (2) = 1 (b) f (2) = 2(c) f is continuous only at x = 1(d) f is continuous at all points

79. Let 16x2 – 3y2 – 32x – 12y = 44 represents ahyperbola. Then(a) length of the transverse axis is 2 3(b) length of each latus rectum is 32 3/(c) eccentricity is 19 3/

(d) equation of a directrix is x = 193

80. For the function f xx

( )[ ]

=

1, where [x] denotes

the greatest integer less than or equal to x, which of the following statements are true?(a) The domain is (–∞, ∞).(b) The range is {0} ∪ {–1} ∪ {1}.(c) The domain is (–∞, 0) ∪ [1, ∞).(d) The range is {0} ∪ {1}

SOluTIOnS

1. (a) : 40% 10%20%

C H

\ Only car or house = 40% + 10% = 50%\ Probability = 1/2

2. (c) : Alphabetically : C, C, H, I, N, OPermutation : CC * * * * → 4

CH * * * * → 4

CI * * * * → 4

CN * * * * → 4Next → COCHIN\ No. of words before COCHIN = 4 4 96× =

3. (c) : f x f t dt f x f xx

( ) ( ) ( ) ( )= ∫ ⇒ ′ =0

⇒ f (x) = ex

∴ = =f eee(log ) log5 55

4. (b) : limx

xt dt

x→

∫−

2

2

23

200

form

=−

′→

∫lim using Hospital rulex

xddx

t dt

ddx

xL

2

2

23

2( )[ ]

= =→

limx

x2

231

12

5. (b) :cos cos sin sin

sin cos

23 3

23 3

23 3

3

x x x x

x xkx+

= cosec

⇒ cos

cos

23 3

23

33

2

x x x

xkx k

−

⋅

= ⇒ =cosec

cosec

6. (b) :

4 4 2 24 2

4 2 2 2cos sin cos cot cosθ θ θ θ π θ+ + ⋅ −

= + + + −

4 2 12

2 2 2cos (cos sin ) cot cosθ θ θ θ π θ

= 2|cosq| + 2cotq (1 + sinq) = –2cosq + 2cotq + 2cosq

= ∈

∴ <2

232

0cot [ , cos ]θ θ π π θ∵

7. (c) :

\ Total 3 solutions exist.

| June ‘1522

8. (d) : Augmented matrix is

1 2 1 42 1 2 21 1 4

− −− −

λ

Applying R2 → R2 – 2R1, R3 → R3 – R1, we get1 2 1 40 3 4 100 3 1 8

− −−−

λ

Applying R3 → R3 – R2,we get1 2 1 40 3 4 100 0 3 2

− −−+ −

λ

\ System has no solution if l + 3 = 0 ⇒ l = –39. (a) :

f xx

x x xx x x x x

( ) ( )( ) ( )( )

= −− − −

1 02 1 0

3 1 1 2 0

Applying C3 → C3 – C1 – C2, we getf(x) = 0 ⇒ f(100) = 0

10. (b) : ∵ | | ( ) ( )a b a b a b+ = + ⋅ +λ λ λ2

= + ⋅ + ⋅ +| | | | a b a a b b2 2 2

λ λ λ

= + + + ⊥| | | | [ ] a b a b2 2 20 0 λ

= + ≥| | | | | | , a b a2 2 2 2λ

∴ + ≥ ⊥| | | | a b a a bλ if

11. (a) : Let

a b c d, , and be the given vectors. a b i b c i k c d j k− = − − = + − = − −3 2 1 2, , ( )λ

For coplanarity, we have −

− −= ⇒ −

− −=

3 0 01 0 20 1 2

0 30 2

1 20

λλ

⇒ l = 1

12. (a) : sin−

− −

= ⇒+

= ⇒ =1

12

62

212

23

xx

xx

xπ

13. (a) : Let the equation of the curve be y = x3... (1)

∴

=dydx ( , )1 1

3

PQ : y – 1 = 3(x – 1) ⇒ y = 3x – 2 ... (2)

Required area OQPO = x dx x dx3

0

1

2 3

13 2 1

12∫ − −∫ =( )/

14. (c) : log0.2(x – 1) > log0.04(x + 5)

⇒ − > +log ( ) log ( ). .0 2 0 21 12

5x x

⇒ (x – 1)2 < x + 5 [Q base < 1]⇒ x2 – 3x – 4 < 0⇒ (x – 4)(x + 1) < 0 ⇒ –1 < x < 4But log0.2(x – 1) is defined if x – 1 > 0 i.e. x > 1\ 1 < x < 4

15. (b) :

y = logex ... (1)y = –ex ... (2)

From graph it is clear that (1) and (2) meet at one point.⇒ No. of real root of logex + ex = 0 is 1.

16. (a) : Let f x a x a x a x a x( ) = + + +0 1

2

2

3

3

4

2 3 4

′ = + + +f x a a x a x a x( ) 0 1 22

33

∵ f (0) = 0 = f (1), ∵ f aa a a

( )12 3 4

001 2 3= + + + =

\ f (x) satisfies all conditions of Rolle's theorem. f ′(c) = 0 for some c ∈ (0, 1)\ f ′(x) = 0 has a real root in [0, 1]17. (d) : f(x) is continuous at x = a if lim ( ) ( )

hf a h f a

→− =

0

we consider the continuity at x = kp (k ∈ I)

\ lim ( ) ( )h

f k h f k→

+ =0

π π for continuity at x = kp (k∈I) ...(1)

| June ‘15 23

Now, lim ( )h

f k h→

+0

π

=+

+ +→

→

lim0 if is rational

lim if is irrationalh

h

k h

k h k h0

0

π

π πsin | |,

= 0 [... sin|kp| = 0 as k∈I]And f(kp) = 0 [... kp is irrational as k ∈I] \ (1) is true18. (b) : y2 – 4y = 4x – 4a⇒ (y – 2)2 = 4(x – a + 1)⇒ vertex is (a –1, 2)∵ (a –1, 2) lies between parallel lines x + y = 3 and 2x + 2y – 1 = 0\ {(a – 1) + 2 – 3} and {2(a – 1) + 2(2) – 1} will be of opposite sign.i.e. (a – 2) and (2a + 1) will be of opposite sign.

⇒ (a – 2)(2a + 1) < 0 ⇒ − < <12

2a

19. (d) : 4 9 112

13

12 22

2

2

2x y x y+ = ⇒

+

= ... (i)

and 4 41 2

12 22

2

2

2x y x y+ = ⇒ + = ...(ii)

From graph, it is seen that there is no solution of (i) and (ii).

20. (c) : If sayx y z

K−

=−+

=−−

=λ

λ31

23

1( ) ...(i)

lie on plane x – 2y = 0 ...(ii) then 3·(1) + (2 + l)·(–2) + (–1).0 = 0

⇒ 3 – 4 – 2l = 0 ⇒ = −λ 12

21. (c) :

f

R R

( )tan

tan tantan

tan

tan tan

θθ

θ θθ

θ

θ θ

= −− −

= +→ +

1 11

1 11 1

0 1 20 0 2

2 2 2 RRR R R

1

3 3 1

tanθ→ +

= = ∈ ∞ ≤ <21

02 2 0

222tan

secsec [ , )

θ

θθ θ πas

22. (c) : ∵ AB = B ...(i) and BA = A ...(ii)\ A2 + B2 = A·A + B·B

= A(BA) + B(AB) [using (i) and (ii)]= (AB)A + (BA)B [Associative law]= BA + AB [using (i) and (ii)]= A + B

23. (b) :

− −

− −

− −

+ + =

ω ω ω

ω ω ω

ω ω

ω ω

2 2

2

2

2

1

1 0[ ]∵

=

−

−

− + −

→ +

0

0

1

2

2

21 1 2

ω ω

ω ω

ω ω ω( )

( )C C C

= − +−

−= − − +

= − − +

( ) ( )( )

( )( )

1 1

1

2

24 2

2

ωω ω

ω ωω ω ω

ω ω ω

= – w2 + w + w3 – w2 = –2w2 + w + 1= – 3w2 + (w2 + w + 1) = – 3w2

24. (d) : 2 2 34

2 21 2

34

1 1 12

1tan tan tan ( ) tan− − − −− = +−

−π

= + −

−− −π tan tan1 14

334

= − − = − +

− − − −π πtan tan tan cot1 1 1 143

34

43

43

= + =

− −πθ θ π

2 21 1∵ tan cot

25. (c) : ∵(2cosq, 2sinq) is a point on circle x2 + y2 = 4and the region between x + y = 2 and x – y = 2 containing origin is ABCDA.\ All pts will be scattered along BCD on circle for

which q will belong to π π2

32

,

| June ‘1524

26. (c) : We must have| |

( )

| |x y x y− +

+ −=

+ −

+=

2 1

1 25

2 3 1

2 313

2 2 2 2and

⇒ x – 2y + 1 = ± 5 and 2x + 3y – 1 = ± 13On solving, we will get exactly 4 points27. (d) : an + bn = cn + dn ... (1)Putting n = 1, a + b = c + d ... (2)Putting n = 2, a2 + b2 = c2 + d2 ... (3)From (2), a2 + b2 + 2ab = c2 + d2 + 2cd⇒ ab = cd ... (4) [From (3)]Now, from (2),

a3+ b3 + 3ab(a + b) = c3 + d3 + 3cd(c + d)⇒ a3 + b3 + 3ab(a + b) = c3 + d3 + 3ab(a + b)

[Using (2) and (4)]\ a3 + b3 = c3 + d3 ⇒ (1) is true for n = 3From (3), a2 – d2 = c2 – b2 ⇒ (a + d)(a – d) = (c + b)(c – b)⇒ a + d = c + b [Using (2)]⇒ a – b = c – d ... (5)On multiplying (2) and (5), a2 – b2 = c2 – d2

28. (a) : Q x2 – px + 1 = 0 changes to x2 + px + 1 = 0if we put x = –x\ If a, b be the roots of x2 – px + 1 = 0 then –a, –b will be the roots of x2 – px + 1 = 0 ⇒ g = –a or –b⇒ a + g = 0 or b + g = 0⇒ (a + g)(b + g) = 029. (None of the options is correct) :

Tr + 1 = 100Cr(31/5)100 – r (71/3)r

= ⋅ ⋅−

100100

5 33 7Cr

r r

No. of rational terms will be obtained if r = 0, 15, 30, 45, 60, 75 or 90\ No. of irrational terms = total terms – 7

= 101 – 7 = 94 30. (c) : 4x2 + (4 – p)x + (q + 1) ≠ 0⇒ no real root of 4x2 + (4 – p)x + (q + 1) = 0⇒ D < 0 ⇒ (4 – p)2 – 4 · 4(q + 1) < 0⇒ p2 – 8p – 16q < 0

31. (a) : If x > 0, xx

+ ≥4 4 [using A.M. ≥ G.M]

If x < 0, xx

+ ≤ −4 4

Now, f x x xx x

xx

xx

( ) = − ++ +

=+

−

+

+

2

244

4 1

4 1

Putting = 4, =

Putting = 4, =

x f x

x f x

( )

( )

3553

−

∴ ∈

f x( ) ,35

53

32. (None of the options is correct) :Given expression

= + + + − +12

4 2 4 4 6 162 2[ ]x y xy x y

= + + ⋅ + ⋅ + ⋅ + − +12

2 2 2 1 1 2 8 162 2 2[( ) ( ) ( )]x y x y y x y y

= + + − + − ≥ −12

2 1 1 4 12

2 2[( ) ( ) ]x y y

Equality sign will hold only if 2x + y + 1 = 0 & y – 4 = 0

i.e. if y = 4, x = −52

\ Least value = − 12

33. (d) : ... f : [–2, 2] → R is continuous function andassumes only irrational values\ f(x) = k (some irrational no.)However, f k( )2 2 2= ⇒ =(given)

∴ = ⇒ − =f x f( ) ( )2 2 1 234. (a) : The minimum value will be obtained when

sinq = cosq i.e. θ π π=

4

02

in ,

\ Min value = + + = +12

12

21

2 2

35. (b) : Let andω ω=− +

=− −1 3

21 3

22i i

∴ = −−

+ −

−

Given expression 22

22

2 64

2

64ωω

ωω

= + = + =ωω

ωω

ω6464

31 1 1[ ]∵

= w + w2 = –1

36. (b) : 2 3 3 2 3= − ≥ − ⇒ ≥ −zz

zz

zz

| | | || |

⇒ |z|2 – 2|z| – 3 ≤ 0 ⇒ (|z| – 1)2 ≤ 4⇒ ||z| – 1| ≤ 2 ⇒ –2 ≤ |z| – 1 ≤ 2

| June ‘15 25

⇒ –1 ≤ |z| ≤ 3 ⇒ |z| ≤ 3\ Maximum value of |z| = 3

37. (d) : ( )( )( )( )

( )

( )

x xx x

x x

x x

− −− −

< ⇒−

−

−

−

<5 5 1

3 1 30

15

5

13

30

∴ < < < <15

13

3 5x xor

38. (d)39. (a) : Variance of first 20 natural nos. is given by

622 2

= −

∑ ∑nn

nn

=+ +

−+

n n nn

n nn

( )( ) ( )1 2 16

12

2

=+ +

−+

=+

+ − −( )( ) ( ) ( )

( )n n n n

n n1 2 1

61

41

124 2 3 3

2

=+ −

=−

=−

=( )( )

[ ]n n n

n1 112

112

20 112

202 2

∵

= =39912

1334

40. (b ): Here, p = getting a head = ∴ = − =12

1 12

q p... P(x = 3) = P(x = 5) (given)

∴ = ⇒ = = =

− −n n n n n nC q p C q p C C p q33 3

55 5

3 512

∵

⇒ n = 3 + 5 = 8

∴ = = = ⋅

=−P x C q p( )1 8 1

21

328

18 1 1

8

41. (a) : Without any restriction total number of ways

of arranging the letters = 112 2

= n S( )

Considering 2 B's as a single letter then 102

= n E( )

∴ = = × =P E n En S

( ) ( )( )

102

2 211

211

42. (d) : We have t = x – a, y = – 16, –16 = ax

∴ = − − = − +

t 16 16

αα α

α...(i)

= + −zz

z16 [ ]Let =α

=+

≥ ⋅2

16

22 16z

z zz

\ t ≥ 8

43. (b) : cos22 2

2Ab c

a=

+

But, cos2A < 1 as in a triangle ABCcosA ≠ 1

∴+

< ⇒ + − < ⇒ <b c

ab c a A

2 2

22 2 21 0 0cos

⇒ < <π π2

A

44. (a) : ∵ |cosx| ≥ sinx is true in 0 32

, π

only

if andx ∈

04

34

32

, ,π π π

∴ =

∪

Required answer 04

34

32

, ,π π π

45. (c) :sin cos cossin cos sin cos cossin cos cos sin s

x x xx x x x xx x x x

++ −+ −

2 022 iin x

= 0

Applying C1 → C1 + (C2 + C3) ; C2 → C2 – C3

⇒ (sinx + 2cosx)(sinx – cosx)1 01 11 1

0coscossin

xxx−

=

⇒ (sinx + 2cosx)(sinx – cosx)1 00 1 00 1

0cos

sin cos

x

x x− −=

Applying R2 → R2 – R1 ; R3 → R3 – R1, we get⇒ sinx + 2cosx = 0 or sinx – cosx = 0⇒ tanx = –2 or tanx = 1

⇒ x has only one solution in −

π π4 4

,

46. (b) : ∵ Exactly one of x1, x2, ..., x15 will be 1 andso exactly any one factor of (x1 – 1), (x2 – 1), (x3 – 1), ... or (x15 – 1) will be zero.\ (x1 – 1)(x2 – 1)(x3 – 1) ... (x15 – 1) = 0

47. (c) : lim ( ) lim sin[ ][ ]x x

f x xx→ →

= −−0 0

2

2

= −−

=→

lim sin( )( )

sinx 0

11

1 and f (0) = a

\ f (x) will be continuous at x = 0 if a = sin1.48. (b) : s = t2 + at – b + 17 ...(i)

dsdt

v t a= = +2 ...(ii)

| June ‘1526

∵ Particle comes to rest after 5 sec. at a distance of 25 units\ When t = 5, s = 25, v = 0\ From (ii) 0 = 10 + a ⇒ a = –10From (i), 25 = 25 + (–10)5 – b + 17 ⇒ b = –3349. (c) :

lim limn n r

n

n n nn

n nrn→∞ →∞ =

−+ + +

−

= ∑1 1 2 1 1

1

1....

= =

=∫ x dx x23

23

3 2

0

1

0

1/

50. (a) :

lim..... ....

x

ax x x b x x x

x→

+ + +

− − + −

=0

2 2 3

2

11 2 2 3 3

As limit exists, we must have,

a b a b− = + =01 2

3and

⇒ a = b = 2

51. (c) :

Here, OABC is the bounded region.

Area of right angled DOAB = × × =12

2 2 1sq. unit

\ Area of OABC = 2 × 1 = 2 sq. units52. (c) : 225 = 32·52 ⇒ d(225) = (2 + 1) (2 + 1) = 9 = 11125 = 32·53 ⇒ d(1125) = (2 + 1) (3 + 1) = 12640 = 27·51 ⇒ d(640) = (7 + 1) (1 + 1) = 16∵ 122 = 9 × 16 ⇒ d(225), d(1125) and d(640) are in G.P.

53. (d) : − ≤ ≤ ∈ −−π π2 2

1 11sin [ , ]x xfor all

⇒ − ≤ ≤ ⇒ − ≤ ≤−π π4

24

12

2 12

1sin a a

⇒ − ≤ ≤ ⇒ ≤12 2

12 2

12 2

a a| |

54. (d) : ... a, b, c, d are real.\ If 2 + i and 5 2− i be roots then other roots will be their conjugate.

\ Product of all roots = (2 + i)(2 – i) ( )( )5 2 5 2− +i i= (4 – i2) (5 – 4i2) = 4555. (a) : ... P(x) be such a polynomial which whendivided by x – 3 and x – 5 (both linear) leaves the remainder 10 and 6 respectively. (both constant)\ When P(x) will be divided by (x – 3)(x – 5), the remainder will be linear, let ax + b\ Accordingly, P(3) = 10 and P(5) = 6 gives 3a + b = 10 and 5a + b = 6On solving, a = –2, b = 16⇒ Remainder is –2x + 16

56. (b) : We can write, 11

322 1 3

+⋅ + =−

ydydx

x y xtan

On putting tan–1y = t, we get 11 2+

⋅ =y

dydx

dtdx

\ Given equation becomes dtdx

x t x+ =3 2 3

Here, P = 3x2, Q = x3

∴ = ∫ =I F. . .e eP dx x3

57. (a) : y = e–xcos2x ...(i)\ y1 = e–x(–2 sin2x) – e–x cos2x

= – 2e–xsin2x – y ...(ii)⇒ = − − −− −y e x e x yx x

22 2 2 2 1{ ( cos ) sin }

= –4y + (–y1 – y) – y1 [from (i) and (ii)]\ y2 + 2y1 + 5y = 0

58. (c) :

f x dx f x dx f x dx f x dx( ) ( ) ( ) ( )2

0

32

0

12

1

22

2

3

∫ ∫ ∫ ∫= + +

= + − + −∫∫∫ x dx x dx x dx2 2 2

2

3

1

2

0

11 2( ) ( )

=

+ −

+ −

x x x x x3

0

1 3

1

2 3

2

3

3 3 32

= + −

− −

+ − − −

13

2 23

2 13

1 3 2 3 2 23

2 2( )

= − +2 3 1

59. (b) : For S, a + b + c = 21 ...(i) For T, a + c = 2b ...(ii)

If considering S ∩ T, then on solving (i) and (ii), we get b = 7 and a + c = 14∵ a b c a≤ ≤ ∴ = 1 2 3 4 5 6 7, , , , , ,

| June ‘15 27

a b c1 7 132 7 123 7 114 7 105 7 96 7 87 7 7

No. of elements in S ∩ T = 7

60. (a) : y ex=2

...(i) y e xx=2

sin ...(ii)On solving,

e e x x x ex x x2 2 21

20= ⇒ = ⇒ = ≠sin sin [ ]π

∵

From (i), dydx

e xx= ⋅2

2 ...(iii)

From (ii), dydx

e x xe x xx x= + =2 2

0cos sin [ cos ]∵

For sinx = 1,dydx

xe xx= =2 02

[ cos ]∵ ...(iv)

\ At point of intersection (where sinx = 1), both slopes (of tangents) are same ⇒ Required angle = 061. (a) : Using Baye's theorem,

P C T P T C P CP T C P C

( / ) ( / ) ( )( / ) ( )

=∑

=−

−

+ −

+ −

+ −

1 24

17

1 29

17

1 19

37

1 49

27

1 19

=17

17

62. (a) : xx x

dx−+

⋅+

∫23

13

1 3

2

/

( )

Put xx

zx

dx dz−+

= ⇒+

=

23

53 2( )

= ⋅∫ = ⋅ ⋅ +15

15

34

1 3 4 3z dz z c/ / = −+

+320

23

4 3xx

c/

63. (c) : ∵ r s c C= −( ) tan2

\ 2r = 2(s – c)tan45° = 2s – 2c= a + b + c – 2c = a + b – c

= a + b – 2R.\ 2(r + R) = a + b

and cC

R c R c Rsin sin

.= ⇒°

= ∴ =290

2 2

64. (d) : ∵ a, b are roots of acosq + bsinq = c ... (i)\ acosa + bsina = c ... (ii) acosb + bsinb = c ... (iii)(iii) – (ii) :

a b⋅ +

−

− ⋅ +

−

22 2

22 2

sin sin cos sinα β α β α β α β = 0

⇒ +

=tan α β2

ba

... (iv)

If a + b is also a root of (i) then acos(a + b) + bsin(a + b) = c

⇒

−

+

+⋅

+

=

a ba

ba

b ba

ba

1

1

2

1

2

2 2 cc [ ( )]using iv

⇒ − ++

= ⇒ − ++

=a a b b aa b

c a a b ba b

c( ) ( )2 2 2

2 2

2 2 2

2 22 2

⇒ a = c

65. (b) : Let Uabc

Upqr

Uxyz

1 2 3=

=

=

, and

∵ AUa

a ba b c

1

100

23 2

100

=

⇒ ++ +

=

⇒ a = 1, b = –2, c = 1

∵ AUp

p qp q r

2

230

23 2

230

=

⇒ ++ +

=

⇒ p = 2, q = –1, r = –4

∵ AUx

x yx y z

3

231

23 2

231

=

⇒ ++ +

=

⇒ x = 2, y = –1, z = –3

∴ = − − −− −

∴ =− −− − −

U U1 2 22 1 1

1 4 3

1 2 07 5 3

9 6 3. Adj

∴ =−UU

U1 1| |

( )adj

\ Sum of all elements in U–1

= − − + − − − + + + =1 1 2 0 7 5 3 9 6 3 0| |

( )U

| June ‘1528

66. (c) :

lim [ ( ) ( ) ( ) ... ( )]x x

f x f x f x f x→

+ + + +

0

1 2 3 2015 00

form

=′ + ′ ⋅ + ′ ⋅ + + ′ ⋅

→lim

[ ( ) ( ) ( ) ... ( ) ]x

f x f x f x f x0

2 2 3 3 2015 20151

= f ′(0) + 2f ′(0) + 3f ′(0) + ... + 2015f ′(0)= (1 + 2 + 3 + ... + 2015) f ′(0)

= ⋅ ⋅ ′2015 2016

2( ) 2 [ (0) = 2]∵ f

= 2015 × 2016

67. (a) : ∵ 17 1 2 3 9 10 11 12 17= ⋅ ⋅( ) ( ) ( ) ... ( )( )( )( ) .... ( )

\ It must be divisible by 9 and 11.For divisibility by 9, sum of all digits must be a multiple of 9.\ 3 + 5 + 5 + 6 + x + y + 4 + 2 + 8 + 9 + 6 = 48 + x + y must be a multiple of 9 for which x + y = 6 or 15 ... (1)For divisibility by 11, we know that difference of sum of odd places digits and sum of even places digits must be either zero or a multiple of 11.Let a = 3 + 5 + x + 4 + 8 + 9 = 29 + xand b = 5 + 6 + y + 2 + 0 + 6 = 19 + yQ a – b = 10 + x – y\ x – y = 1 is possible only ... (2)But, x + y = 6 and x – y = 1 can't give integral value.Hence, x + y = 15 only.68. (d) : f (1) + 2f (2) + 3f (3) + ... + nf (n)

= n(n + 1) f(n) ... (i) for n ≥ 2Putting n = n + 1, we getf (1) + 2f (2) + 3f (3) + ... + nf (n) + (n + 1) f (n + 1)

= (n + 1)(n + 2) f (n + 1) ... (ii)(ii) – (i) : (n +1) f (n + 1) = (n + 1)(n + 2) f (n + 1)

– n(n + 1) f (n)⇒ f (n + 1) = (n + 2) f (n + 1) – n f (n)⇒ (n + 1) f (n + 1) = n f (n)⇒ 2f (2) = 3f (3) = 4 f (4) ..... = nf (n)From (i), f nf n nf n nf n nf n

n( ) ( ) ( ) ( ) .... ( )

( )1

1+ + + + +

− times = +n n f n( ) ( )1

⇒ 1 + (n – 1) n f (n) = n(n + 1) f (n)⇒ 2n f (n) = 1

∴ = ⇒ =f nn

f( ) ( )12

500 11000

69. (a) : Without any restriction : 55

Exactly one empty cell can be selected in 5C1 ways.\ Required probability

= 5

15 4

15 4

25 4

35

54 3 2 1

548

125C C C C( )− ⋅ + ⋅ − ⋅

=

70. (c) : S = smoker, S′ = Non-smoker, D = death bycancerUsing conditional probability, we can write

P(D) = P(S) P(D/S) or P(S′) P(D/S′)

0 006 20100

80100

. ( / ) ( / )= ⋅ + ⋅ ′P D S P D S

= ⋅ + ⋅1

545 10

x x

[Let P(D/S) = x and given P(D/S) = 10·P(D/S′)]

⇒ =x 3140

71. (a, b, d) : ∵ cosx and sinx are solutions of

ad y

dxa

dydx

a y0

2

2 1 2 0+ + = ,

which is a differential equation of order 2 and degree 1.\ y = Acosx + Bsinx will be a general solution.Further,

A x A x B xcos cos sin+

+

+ −

π π π4 4 4

and

both can also be expressed in Acosx + Bsinx form.

72. (a, c) : If 0 < n < 1 ... (i) and 02

< <θ π ... (ii)

then 02

< <nθ π

\ If n is a fraction, nq will be away from p/2 and

so cosnq will be away from cos π2

. 0i.e

whereas cosn q = (cosq)n may be very close to 0.

∴ < < ≤In if 0 < < 102

θ π θ θ, cos cos ,n n n

Hence, (cos ) cos cos/θ θ θ θ1 2 12

56

≤ ≤ and (cos )5/6

73. (c) : Only option (c) is always false.74. (b, d) : ∵ xy = 1

∴ = ⇒ = −yx

dydx x

1 12

\ Slope of normal = x12 > 0 at any point P(x1, y1)

| June ‘15 29

If (a – 1)x – by + 4 = 0 is a normal

then its slope = a

b− >1 0

\ a – 1 and b must be of opposite sign.⇒ a > 1, b < 0 or a < 1, b > 075. (b, c) : ∵ f (x) is continuous and differentiable on[a, b] and f (a) = 0 = f (b)\ From Rolle's theorem, f ′(c) = 0 for some c ∈ (a, b)i.e. f ′(x) = 0 for some x ∈ (a, b)Further, f ′(a) = f ′(c) [Q each = 0]f ′(x) is continuous and differentiable in [a, b]\ From Rolle's theorem again,

f ′′(x) = 0 for some x ∈ (a, c) i.e. x ∈ (a, b)76. (b, c) : ∵ 0 ∈ R but, 0 · 0 > 0 is not truei.e. (0, 0) ∉ r\ r is not reflexive∵ xy > 0 ⇒ yx > 0 ⇒ r is symmetricLet xy > 0 and yz > 0\ (xy)(yz) > 0 ⇒ (xz)y2 > 0 ⇒ xz > 0 [Q y2 > 0]\ x r y and y r z ⇒ x r z\ r is transitive.77. (c, d) : P (at least one) = 1 – P (none)Let n be number of parts produced

∴ − ≥ ≤1 0 95 12

12

( . )n n will imply (0.95)

⇒ ≤n log . log10 100 95 12

⇒ n(1.977 – 2) ≤ –0.3

⇒ ≥ ∴ =n n30023

14 or 15

78. (a, d) : ∵ f (2x + 1) = f (x)

Putting n n f x f x= + ⋅ + −

= +

12

2 12

1 12

, we get

⇒ = +

f x f x( ) 12

∴ = +

=

+ +

= + +

f x f x f

x

f x( ) 12

12

1

21 2

22

=

+ + +

= + + +

=f

x

f x1 2

21

21 2 2

2

2 2

3 ...

= + + + + +

−f x n

n1 2 2 2

2

2 1...

= + −−

= + −

f x f xn

n

n n n22 1

2 2 1 21 1

2( )

( )

\ When n → ∞, x f x fn n2

0 12

0 1 1→ → ⇒ = =, ( ) ( )

\ f (x) is constant for all x ⇒ f (2) = 1 also

79. (a, b, c) : We have( ) ( )x y− −

+=1

32

161

2 2

Here, a b= =3 4,

\ Length of transverse axis = 2a = 2 3

Length of latus rectum = 2 2 16

332

3

2ba

= =( )

e ba

= + =1 193

2

2

Equation of directrices are given by

x ae

i e x− = ± = ±1 1 319

. .

80. (b, c) : ∵ f xx

( )[ ]

=

1

\ f (x) is not defined when [x] = 0i.e. when 0 ≤ x < 1\ Df = R – [0, 1) = (–∞, 0) ∪ [1, ∞)When x = 1, f (x) = 1

When x > 1, 0 1 1 1 0< < ⇒

=[ ] [ ]x x

When –1 ≤ x < 0, [x] = –1

⇒

= −1 1[ ]x

When –∞ < x < –1, [x] ≤ –2

⇒ − < <1 1 0[ ]x

⇒

= −1 1[ ]x

Hence, Rf = {0} ∪ {–1} ∪ {1}nn

| June ‘1530

| june ‘15 31

1. Let a1, a2,...., a100 be an ordered set of numbers. At each move it is allowed to choose any two numbers an, am and change them to the numbers

aa

nm

aa

aaa

mn

aa

an

m

m

nm

m

n

n

mn

2 2 2 2− −

− −

and

respectively. Determine if it is possible, starting with

the set with ai = 15

for i = 20, 40, 60, 80, 100 and ai = 1

otherwise, to obtain a set consisting of integers only.2. Does there exist a bijection f of f of f(a) a plane with itself (b) three-dimensional space with itself such that

for any distinct points A, B , l ine AB and line f (f (f A) f (f (f B) are perpendicular?

3. Find all triplets (x, y ,z) of natural numbers such that y is a prime number, y is a prime number, y y and 3 do not divide y and 3 do not divide y z, and x3 – y3 = z2. 4. Prove that the equation x3 + y3 + z3 + t3 = 1999 has infinitely many integral solutions.

5. Determine the maximum value of l such that if f(f(f x) = x3 + ax2 + bx + c is a cubic polynomial with all its roots non negative, then f(f(f x) ≥ l (x – a)3 for all x ≥ 0. Find the equality condition.6. Determine all functions f : R –{0, 1} → R satisfying

the functional relation f x fx

xx x

( ) ( )( )

+−

= −−

11

2 1 21

.

7. Show that 1993 – 1399 is a positive integer divisible by 162.

8. A sergeant appoints arbitrarily chosen 9 or 10 soldiers every night, out of a 33 member guard team, for duty. Determine the least number of days within which all guards can be on duty the same number of times. How can the guard duty be arranged to achieve this?

9. Find all primes p for which the quotient ( )2 11p

p

− −

is a square.

10. Prove that

1 11001

11002

11003

13001

43

< + + + + <...

11. Three congruent circles have a common point Oand lie inside a triangle such that each circle touches a pair of sides of the triangle. Prove that the incentre and the circumcentre of the triangle and the point O arecollinear.

12. Find the number of permutations, (P1, P2, ..., P6), of (1, 2, ..., 6) such that for any k, 1 ≤ k ≤ 5, (P1, P2, ..., Pk) does not form a permutation of 1, 2, ..., k.[That is, P1 ≠ 1; (P1, P2) is not a permutation of1, 2, 3, etc.]

13. Let f be a bijective (one-one and onto) function f be a bijective (one-one and onto) function ffrom the set A = {1, 2, 3, ...,n} to itself. Show that there is a positive integer M > 1 such that M > 1 such that M f M(i) = f(f(f i) for eachi ∈ A .[f[f[ M denotes the composite function fo fo ............repeated M times.]M times.]M

solutions1. Suppose we change

a aaa

nm

aa

an nn

m

m

nmto ′ = − −

2 2and

a to aaa

mn

aa

am mm

n

n

mn′ = − −

2 2

Observe that′

+′a

nam

n m equals

1 1 1 12 2 2 2

naa m

aa

am m

aa n

aa

n

m

m

n

m m

n

n

m. . . .−

+

+ −

+

ann

aspirants

*

Problems from

| june ‘1532

which equals an

am

n m+ . Thus, the quantity aii

i=∑

1

100 is

invariant under the given operation. Originally, this

sum equals Iaii

i1

1

99 1500

= +=∑ .

When each of the numbers a a a1 2 991 2 99

, ,..., is written as

a fraction in lowest terms, none of their denominators are divisible by 125. On the other hand, 125 does

divide the denominator of1

500. Thus when written as

a fraction in lowest terms, I1 must have a denominator divisible by 125.Now suppose, by way of contradiction, that we could make all the numbers equal to integers b1, b2, ...,b100

in that order. Then in Ibii

i2

1

100=

=∑ , the denominator of

each of the fraction bii is not divisible by 125. Thus,

when I2 is written as a fraction in lowest terms, its denominator is not divisible by 125 either. Then I2 cannot possibly equal I1, a contradiction. Therefore, we can never obtain a set consisting of integers only.

2. (a) Yes: simply rotate the plane 90° about some axisperpendicular to it. For example, in the xy-plane, we could map each point (x, y) to the point (y, – x). (b) Suppose such a bijection existed. Label the three-dimensional space with x-, y-, and z-axes. Given any point P = (x0, y0, z0), we also view it as the vector p from (0, 0, 0) to (x0, y0, z0). The given condition says that (a – b)·(f(a) – f(b)) = 0 for any vectors a, b.Assume without loss of generality that f maps the origin to itself otherwise, g (p) = f(p) – f(0) is still a bijection and still satisfies the above equation. Plugging b = (0, 0, 0) into the equation above, we have a·f(a) = 0 for all a. Then the above equation reduces to a·f(b) +b ·f(a) = 0.Given any vectors a,b,c and any reals m, n, we then have

m(a·f(b) + b·f(a)) = 0 n(a·f(c) + c·f(a)) = 0a · f(mb + nc) + (mb + nc)· f(a) = 0. Adding the first two equations and subtracting the third gives a· (m f(b) + n f(c) – f(mb + nc)) = 0. Because this must be true for any vector a, we must have f(mb + nc) = m f(b) + n f(c).Therefore f is linear, and it is determined by how it transforms the unit vectors i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1). If f(i) = (a1, a2, a3), f(j) = (b1, b2, b3)

and f(k) = (c1, c2, c3), then for a vector x, we have

f xa b ca b ca b c

x( ) =

1 1 1

2 2 2

3 3 3Applying f(a)·a = 0 with a = i, j, k, we have a1 = b2 = c3 = 0Then applying a·f(b) + b·f(a) with (a, b) = (i, j), (j, k), (k, i) we have b1 = –a2, c2 = –b3, c1 = –a3. Setting k1 = c2, k2 = – c1, and k3 = b1, we find that f(k1i + k2j + k3k) = k1 f(i) + k2 f(j) + k3 f(k) = 0Because f is injective and f(0) = 0, this implies that k1 = k2 = k3 = 0. Then f(x) = 0 for all x, contradicting the assumption that f was surjective. Therefore our original assumption was false, and no such bijection exists.

3. We rewrite the equation in the form(x – y)(x2 + xy + y2) = z2

Any common divisor of x – y and x2 + xy + y2 also divides both z2 and (x2 + xy + y2) – (x + 2y)(x – y) = 3y2. Because z2 and 3y2 are relatively prime by assumption, x – y and x2 + xy + y2 must be relatively prime as well. Therefore, both x – y and x2 + xy + y2 are perfect squares. Writing a x y= − , we have x2 + xy + y2 = (a2 + y)2 + (a2 + y)y + y2

= a4 + 3a2y + 3y2

and 4(x2 + xy + y2) = (2a2 + 3y)2 + 3y2

Writing m x xy y n a y= + + = +2 2 32 2 2and , we have m2 = n2 + 3y2 or (m – n)(m + n) = 3y2,so (m – n, m + n) = (1, 3y2), (y, 3y) or (3, y2).In the first case, 2n = 3y2 – 1 and 4a2 = 2n – 6y = 3y2 – 6y – 1Hence, a2 ≡ 2(mod 3), which is impossible.In the s econd cas e , n = y <2a 2 + 3y = n , a contradiction.In the third case, we have 4a2 = 2n – 6y = y2 – 6y – 3 < (y – 3)2. When y ≥ 10 we have y2 – 6y – 3 > (y – 4)2. Hence, we must actually have y = 2, 3, 5, or 7. In this

case, we have ay y

=− −2 6 32

, which is real only when

y = 7, a = 1, x = y + a2 = 8 and z = 13. This yields the unique solution (x, y, z) = (8, 7, 13).4. Observe that (m – n)3 + (m + n)3 = 2m3 + 6mn2.Now suppose we want a general solution of the form

x y z t a b a b c d c d, , , , , ,( ) = − + − +

2 2 2 2

for integers a, b and odd integers c, d. One simple solution to the given equation is

| june ‘15 33

(x, y, z, t) = (10, 10, – 1, 0), so we try setting a = 10 and c = – 1. Then

x y z t b b d d, , , , , ,( ) = − + − − − +

10 10 12 2

12 2

is a solution exactly when

( )2000 60 1 34

1999 80 122

2 2+ − + = ⇔ − =b d d b

The second equation is a Pell’s equation with solution (d1, b1) = (9, 1)We can generate infinitely many more solutions by setting (dn+1, bn+1) = (9dn + 80bn, 9bn + dn) for n = 1, 2, 3, .....This can be proven by induction, and it follows from a general recursion (pn+1, qn+1) = (p1pn + q1qnD, p1qn + q1pn)for generating solutions to p2 – Dq2 = 1 given a non trivial solution (p1, q1).A quick check also shows that each dn is odd. Thus because there are infinitely many solutions (bn, dn) to the Pell’s equation (and with each dn odd), there are infinitely many integral solutions

( , , , ) , , ,x y z t b bd d

n n n n n nn n= − + − − − +

10 10 12 2

12 2

to the original equation.

5. Let a, b, g be the three roots. Without loss ofgenerality, suppose that 0 ≤ a ≤ b ≤ g. We havex – a = x + a + b + g ≥ 0 and f(x) = (x – a)(x – b)(x – g). If 0 ≤ x ≤ a, then (applying the AM-GM inequality to obtain the first inequality below)

− = − − − ≤ + + −f x x x x x( ) ( )( )( ) ( )a b g a b g127

3 3

≤ + + +( ) = −127

127

3 3x x aa b g ( )

so that f x x a( ) ( )≥ − −127

3 .

Equality holds exactly when a – x = b – x = g – x in the first inequality and a + b + g – 3x = x + a + b + g in the second; that is, when x = 0 and a = b = g. If b ≤ x ≤ g, then (again applying the AM-GM inequality to obtain the first inequality below)

− = − − − ≤ + − −f x x x x x( ) ( )( )( ) ( )a b g g a b127

3

≤ + + + = −127

127

3 3( ) ( )x x aa b g

so that again f x x a( ) ( )≥ − −127

3 .

Equality holds exactly when x – a = x – b = g – x in the first inequality and x + g – a – b = x + a + b + g; that is, when a = b = 0 and g = 2x.

Finally, when a < x < b or x > g then

f x x a( ) ( )> ≥ − −0 127

3

Thus, l = − 127

works. From the above reasoning we can

find that l must be at most − 127

or else the inequality

fai ls for the polynomial f x x x x( ) ( ) .= − =2 1 12

at

Equality occurs either when a = b = g and x = 0; or when

a = b = 0, g is any non negative real, and x = g2

.

6. The given function relation can be written as

f x fx x x

( ) +−

= +−

11

2 21 ...(1)

Let t be any real number other than 0 and 1. Put

xt

=−1

1in (1); then 1

11 1

−= −

x t and we have

ft

ft

tt

11

1 1 2 2−

+ −

= − + ...(2)

When x = 1 1 11

−−

=t x

t, and now (1) yields

ft

f t tt

t1 1 21

2−

+ =−

−( ) ...(3)

Also replacing x by t in (1), we get

f t ft t t

( ) +−

= +−

11

2 21

..(4)

Now adding (3) and (4) and subtracting (2), we obtain

f t tt

( ) = +−

11

7. In the binomial expansion of 1 613

15+

, the third

term 15 141 2

613

132⋅

⋅

> .

Therefore 1 613

1315

+

> ⇒ +

>1 613

1331

2

⇒ 1931 > 1333 ⇒ 1993 > 1399

In the binomial expansion of (1 + 18)93 the first term is 1; the second term is 93. 18 ≡ 12·18(mod 81); all other terms are divisible by 81. In the expansion of (1 + 12)99 the first term is 1; the second term is 99. 12 ≡ 18. 12 (mod 81); the next two terms 99 98

1 212 99 98 97

1 2 3122 3⋅

⋅⋅ ⋅ ⋅

⋅ ⋅, are divisible by 81; the other

terms being multiples of 124 are also divisible by 81. Hence we have 1993 ≡ 1399(mod 81). \ 162 |1993 – 1399 ( 1993 – 1399 is even)

| june ‘1534

8. Let x (resp. y) be the number of days on which9 (resp. 10) soldiers do guard duty. Let k be the number of days on which every soldier gets duty. Then9x + 10y = 33k ...(1)So we have to minimize x + y subject to (1). Obviously when k = 1, (1) has no non - negative integral solutions for x, y. So k ≥ 2, then 10(x + y) ≥ 9x + 10y = 33k ≥ 66 ⇒ x + y ≥ 7x + y attains this minimum when k = 2; in that case, y = 3, and x = 4.Here is one way of allocating the duty to the soldiers.Let D1 = {1, 2, 3, 4, 5, 6, 7, 8, 9} D2 = {10, 11, 12, 13, 14, 15, 16, 17, 18}, D3 = {19, 20, 21, 22, 23, 24, 25, 26, 27}, D4 = {28, 29, 30, 31, 32, 33, 1, 2, 3}, D5 = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, D6 = {14, 15, 16, 17, 18, 19, 20, 21, 22, 23}and D7 = {24, 25, 26, 27, 28, 29, 30, 31, 32, 33}For k = 1,..., 7, and j = 1,...,33, the soldier Sj assigned duty on kth day if j ∈ Dk.

9. For p = 2, the given quotient is not an even integerand so we can assume p is an odd prime. Then by Fermat’s Little theorem 2p–1 – 1 is divisible by p. Suppose that for some integer a, 2p–1 – 1 = pa2. Since p is odd we have that

2 1 2 11

21

2 2p p

pa−( ) −( )

−

+

=

Both the factors on the L.H.S. are odd and hence they are relatively prime to each other which implies that p divides exactly one of the two factors. Therefore either

2 1 2 11

2 21

2 2p p

px y−( ) −( )

− = + =,

or 2 1 2 11

2 21

2 2p p

x py−( ) −( )

− = + =,

Case – I: Suppose 2 1 2 11

2 21

2 2p p

px y−( ) −( )

− = + =, .

But then 2 1 11

2( )

( )( )p

y y−

= − + for which the only solution is y = 3 i.e., p = 7. When p = 7 observe that the given quotient is indeed a square.

Case – II: Suppose 2 1 2 11

2 21

2 2( ) ( )

, .p p

x py− −

− = + =

Then if p > 3 we get that x2 leaves remainder 3 when divided by 4 which is impossible. Thus p can only be equal to 3; when p = 3 also we observe that the given quotient is indeed a square.

Thus the primes p for which the quotient ( )2 11p

p

− −

is a square are p = 3, 7.

10. Consider 2001 numbers 1 1001 3001k

k, ≤ ≤

Using AM-HM inequality, we get

kkk k= =

∑ ∑

> ( )1001

3001

1001

300121 2001

But kk =∑ = ( )1001

300122001

Hence we get the inequality1 1

1001

3001

kk =∑ >

On the other hand, grouping 500 terms at a time, we also have

Skk

==∑ 1

1001

3001

< + + + +5001000

5001500

5002000

5002500

13001

< + + + +12

13

14

15

13000

= <38513000

43

Note: We can sharpen the above inequality. Consider the sum

Skk n

n=

= +

+

∑ 1

1

3 1

There are 2n + 1 terms in the sum and the middle term

is 1

2 1n+. We can write the sum in the form

Sn n k n kk

n=

++

+ ++

+ −

=∑1

2 11

2 11

2 11

=+

++( ) +

−+

=∑1

2 12

2 11

12 1

21n n k

nk

n

For we have0 12

< <a ,

1 11

1 2+ <−

< +aa

a

Thus we get the bounds

| june ‘15 35

12 1

22 1

12 1

2

1n nk

nS

k

n

++

++

+

<=∑

and Sn n

knk

n<

++

++

+

=

∑12 1

22 1

1 22 1

2

1

This on simplification gives

1 22 1

1 42 13

23

1

2

1+

+< < +

+= =∑ ∑

( ) ( )nk S

nk

k

n

k

n

Now using the identity kn n n

k

n2

1

1 2 16=

∑ =+ +( )( )

, the

inequality simplifies to

11

3 2 11 2

31

2 12 2++

+< < +

+

+

n n

nS

n n

n

( )

( )

( )

( )

But for n ≥ 1 , we also have 29

1

2 1142≤

+

+≤

n n

n

( )

( )

This leads to 2927

76

< <S

11. Let K, L, M be the centres of the three circles ofequal radii, meeting in a common point O, and pairwise touching some side of the triangle ABC in which they lie. Since the three circles have equal radii, we see that O is equidistant from the points K, L, M and so is the circumcentre of the triangle KLM. Also for same reason, the sides of the triangle KLM are parallel to the corresponding sides of the triangle ABC. (For instance, KL is parallel to AB, as K and L are equidistant from AB.)

Further AK, BL, CM are the bisectors of angles A, B, C of triangle ABC. So not only they meet in I, the incentre of the triangle ABC but also KI, LI, MI are the bisectors of the angles of triangle KLM, implying that I is also the incentre of triangle KLM.

It follows, from the relation IKIA

ILIB

IMIC

= = that triangle

KLM is homothetic to triangle ABC with respect to I, the centre of homothety. (This simply means that triangle ABC is a dilation (or an enlargement) of triangle KLM as seen from I).A property of homothety is that the centre of homothety and any two corresponding points of the homothetic figures are collinear. Here, in particular, I and the circumcentres of triangles KLM and ABC have to be collinear. This is precisely what was to be proved.

12. We shall look at the problem from a generalviewpoint. For any positive integer n, let Tn denote the number of permutations of (P1, P2, P3, ....., Pn) of 1, 2, 3, ..., n such that for each k, 1 ≤ k ≤ n, (P1, P2, P3, .... Pn) is not a permutation of 1, 2, ..., k. We shall obtain a formula for Tn which expresses Tn in terms of T1, T2, ... , Tn–1 (n > 1). (Such a relation is called a recurrence relation for Tn.)Consider any permutation (P1, P2, P3, ....., Pn) of 1, 2, ..., n. There is always a least positive integer k such that (P1, P2, P3, ..... Pk) is a permutation of 1, 2, ..., k. In fact k may be any integer in the set {1, 2, ..., n}; and those permutations for which k = n are (1, 2, ..., n) for all of which k is the least positive integer satisfying the above property is Tk. (n – k)!, by our definition of Tn. The second factor corresponds to the permutations of k + 1, k + 2, ..., n which fill up the remaining (n – k) places. Since there are n! permutations in all, we obtain

n T n kkk

n! .( )!= −

=∑

1

= T1.(n – 1)! + T2.(n – 2)! + ...... + Tn–1.1! + Tn.0! HenceTn = n! – T1.(n – 1)! – T2.(n – 2)! .... – Tn–1.1!Clearly T1 = 1T2 = 2! – T1 . 1! = 2 – 1 = 1T3 = 3! – T1 . 1! – T2 . 1! = 6 – 2 – 1 = 3T4 = 4! – T1 . 1! – T2 . 2! – T3 . 1! = 24 – 6 – 2 – 3 = 13.T4 = 5! – T1 . 4! – T2 . 3! – T3 . 2! – T4 . 1!

= 120 – 24 – 6 – 6 – 13 = 71T6 = 6! – T1 . 5! – T2 . 4! – T3 . 3! – T4 . 2! – T5 . 1! = 720 – 120 – 24 – 18 – 26 – 71 = 461.Thus the required number is 461.

13. 1st Solution:i) If f : A → A is a bijective function then thereis a unique bijective function g : A → A such that

| june ‘1536

fog = gof = IA, then identity function on A. The function g is called the inverse of f and is denoted by f –1. Thus, fof –1 = IA = f –1ofii) fo IA = f = IA ofiii) If f and g are bijections from A to A, then so aregof and fog.iv) If f, g, h are bijective functions from A to A andfog = foh, then g = h.Apply f –1 at left to both sides to obtain g = h.Coming to the problem, since A has n elements, we see that there are only finitely many (in fact, n!) bijective functions from A to A as each bijective function f gives a permutation of {1, 2, 3, ..., n} by taking {f(1), f(2), ..., f(n)}. Since f is a bijective function from A to A, so is each of the function in the sequence:

fof = f 2, fofof = f 3, ........f n, ...... All these cannot be distinct, Since there are only finitely many bijective functions from A to A. Hence for some two distinct positive integers m and n, m > n say, we must have f m = f n If n = 1, we take M = m, to obtain the result. If n > 1, multiply both sides by (f –1)n–1, to get f m – n + 1 = f. We take M = m – n + 1 to get the relation f M = f, (M > 1)Note : this means f M(i) = f(i) for all i ∈ A. 2nd Solution: Take any element r in the set A and consider the sequence of elements

r, f(r), (fof)(r), (fofof)(r), ......obtained by applying f successively. Since A has only n elements there must be repetitions in the above sequence. But when the first repetition occurs, this must be r itself; for, if the above sequence looks (for instance) like r, a, b, c, d, e, c, ...where the first repetition is an element c other than r, this would imply f(b) = c and f(e) = c,contradicting the fact that f is a bijection. Thus for some positive integer lr ≥ 1, we have f lr(r) = rThis is true for each r in the set A = 1, 2, ..., n. By taking M to be the l.c.m. of l1, l2, ..., lr we get f M(r) = r for each r ∈ A.[Note: If f itself is the identity function the above proof fails because each lr = 1. But in this case we may take M to be any integer greater than or equal to 2].

| june ‘15 37

CATEGORY – IFor each correct one mark will be awarded, whereas, for each wrong answer, 25% of total marks (1/4) will be deducted. If candidates mark more than one answer, negative marking will be done.

1. Let x1 and y1 be real numbers. If z1 and z2 arecomplex numbers such that |z1| = |z2| = 4, then |x1z1 – y1z2|2 + |y1z1 + x1z2|2 = (a) 32 1

212( )x y+ (b) 16 1

212( )x y+

(c) 4 12

12( )x y+ (d) 32

2. Let z1 and z2 be complex numbers such that

z i z1 2 0+ =( ) and arg ( ) .z z1 2 3= p Then arg ( )z1 is

(a) p3

(b) p (c) p2

(d) 512p

3. Let the complex numbers z1, z2, z3 and z4 denotethe vertices of a square taken in order. If z1 = 3 + 4i and z3 = 5 + 6i, then the other two vertices z2 and z4 are respectively(a) 5 + 4i, 5 +6i (b) 5 + 4i, 3 + 6i(c) 5 + 6i, 3 + 5i (d) 3 + 6i, 5 + 3i

4. Let a and b are the roots of the equationpx2 + qx + r = 0. If p, q, r are in A.P. and a + b = 4, then ab =(a) –9 (b) 9 (c) 5 (d) –5

5. If a is a root of the equation x2 – 3x + 1 = 0 then the

value of aa

3

6 1+=

(a) 114

(b) 115

(c) 116

(d) 118

6. The quadratic equation (x – a) (x – b) + (x – b)(x – c) + (x – c) (x – a) = 0 has equal roots, if (a) a ≠ b, b = c (b) a = b, b ≠ c(c) a ≠ b, b ≠ c (d) a = b = c

7. If a1 = 4 and an+1 = an + 4n for n ≥ 1, then the valueof a100 is (a) 19804 (b) 18904(c) 18894 (d) 19904

8. If the first term of a G.P. is 729 and its 7th term is64, then the sum of first seven terms is (a) 2187 (b) 2059 (c) 1458 (d) 2123

9. Let a1, a2, a3, a4 be in A.P. If a1 + a4 = 10 anda2a3 = 24, then the least term of them is(a) 1 (b) 2 (c) 3 (d) 4

10. An A.P. has the property that the sum of first tenterms is half the sum of the next ten terms. If the second term is 13, then the common difference is (a) 3 (b) 2 (c) 5 (d) 411. If nCr – 1 = 36 and nCr = 84 then(a) 13r – 3n – 3 = 0 (b) 10r – 3n – 30 = 0(c) 10r + 3n – 3 = 0 (d) 10r – 3n – 3 = 0

12. If 8 13

54

9!! !

,+

= Pr

then the value of r is

(a) 4 (b) 5 (c) 3 (d) 213. The number of onto functions from the set{1, 2, ...., 11} to the set {1, 2, ...., 10} is (a) 5 × 11! (b) 10!

(c) 112

! (d) 10 × 11!

PRACTICE PAPERThe entire syllabus of Mathematics of WB-JEE is being divided in to six units, on each unit there will be a Mock Test Paper (MTP) which will be published in the subsequent issue.

UNIT-I : algebra

| june ‘1538

14. Let n be a positive even integer. The ratio of thelargest coefficient and the 2nd largest coefficient in the expression of (1 + x)n is 11 : 10. Then the number of terms in the expansion of (1 + x)n is(a) 20 (b) 21 (c) 10 (d) 11

15. If (1 + x + x2)n = 1 + a1x + a2x2 + ... + a2nx2n, then2a1 – 3a2 + ... –(2n + 1)a2n = (a) n (b) –n (c) n + 1 (d) –n + 1

16. The value of n if C0 + 2C1 + 3C2 + ... + (n + 1)Cn= 576, is(a) 7 (b) 5 (c) 6 (d) 917. Let A and B are square matrices of order 'n' suchthat A2 – B2 = (A – B) (A + B), then which of the following will be true?(a) Either A or B is zero matrix(b) A = B(c) AB = BA(d) Either A or B is an identity matrix.

18. If the matrix2 35 1−

= +A B, where A is symmetric

and B is skew symmetric, then B =

(a) 2 44 1−

(b)

0 22 0

−

(c) 0 11 0−

(d)

0 11 0

−

19. The value of the determinant

1 2 2

2 1 2

2 2 1

2 2

2 2

2 2

+ − −

− +

− − −

a b ab b

ab a b a

b a a b

is equal to

(a) 0 (b) 1 + a2 + b2

(c) (1 + a + b2)3 (d) (1 + a2 + b2)3

20. If the determinant of the adjoint of a (real) matrixof order 3 is 25, then the determinant of the inverse of the matrix is

(a) 0.2 (b) ± 5 (c) 16255 (d) ± 0.2

21. If n(A) = 43, n(B) = 51 and n(A ∪ B) = 75, thenn((A – B) ∪ (B – A)) = (a) 53 (b) 45 (c) 56 (d) 66

22. If A and B are non-empty sets such that A ⊃ B,then(a) B′ – A′ = A – B (b) B′ – A′ = B – A(c) A′ – B′ = A – B (d) A′ ∩ B′ = B – A

23. For any two real numbers A and B, we define aRbif and only if sin2a + cos2b = 1. The relation R is(a) reflexive but not symmetric(b) symmetric but not transitive(c) transitive but not reflexive(d) an equivalence relation24. The identity mapping Ic : S → S is defined asIs(x) = x for x ∈ S. Suppose f : A → B is a bijection. Then which one of the following is true?(a) f–1of ≠ IA but fof–1 = IB(b) f–1of = IA and fof–1 = IB(c) f–1of = IA but fof–1 ≠ IB(d) f–1of ≠ IA and fof–1 ≠ IB

25. The value of x for which log3 (5·3x – 1 + 1),log9 (31 – x + 1) and 1 are in A.P. is

(a) log353

(b) log335

(c) log332

(d) log325

26. log(sin1°) × log(sin2°) × log(sin3°) .....log(sin 179°)(a) is positive(b) is negative(c) lies between 1 and 180(d) zero27. If n = 2015! then the value of

1 1 1 1

2 3 4 2015log log log...

logn n n n+ + + + is

(a) n! (b) log n!(c) 1 (d) none of these28. If log0.3 (x + 1) ≤ log0.09 (x – 1), then(a) x < 2 (b) x > 2(c) x ≥ 2 (d) none of these

29. If 41

22

n

nn

n+< ( )!

( !), then P(n) is true for

(a) n ≥ 1 (b) n > 0 (c) n < 0 (d) n ≥ 230. If P(n) is a statement (n ∈ N) such that, if P(K) istrue, P(K + 1) is true for k ∈ N, then P(n) is true(a) for all n (b) for all n ≥ 3(c) for all n > 4 (d) none of these

CATEGORY – IIFor each correct answer 2 marks will be allotted, whereas, for each wrong answer, 25% of total marks (1/2) will be deducted.31. If the sum of the n terms of an A.P. is cn(n – 1)where c ≠ 0, then the sum of the squares of these terms is

| june ‘15 39

(a) c2n2(n + 1)2 (b) 23

1 2 12c n n n( )( )− −

(c) 23

1 2 12c n n n( )( )+ + (d) 13

2 2c n

32. Let c be a complex number such that |c| = 1. If theequation cz2 + z + 1 = 0 has purely imaginary root, then tan (arg c) =

(a) 5 12− (b) 5 1

2+

(c) 5 12− (d) 5 1

2+

33. If A =

3 71 2

then the value of the determinant

|A2015 – 3A2014| = (a) 8 (b) –8 (c) 7 (d) –7

34. If A =−

sin coscos sin

a aa a

00

0 0 1then A–1 is equal to

(a) A (b) I(c) AT (d) none of these

35. A five digit number divisible by 3 is to be formedusing the numbers 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is (a) 216 (b) 240 (c) 600 (d) 3125

CATEGORY–IIIIn this section more than 1 answer can be correct. Candidates will have to mark all the correct answers, for which 2 marks will be awarded. If candidates marks one correct and one incorrect answer then no marks will be awarded. But if, candidates marks only correct, without making any incorrect, formula below will be used to allot marks. 2 × (no. of correct response/total no. of correct options).

36. Let A = {1, 2, 3}, we define R = {(1, 1), (2, 2), (3, 3)}then it is (a) reflexive(b) symmetric(c) ordered relation on A.(d) none of these

37. If f(x) = 2x – sinx and g(x) = x3 , then(a) range of gof is R(b) gof is one-one(c) both f and g are one-one(d) both f and g are onto.

38. If A =0 11 0

−

and B

ii

=

00

, then

(a) A2 = I (b) B2 = I(c) A2 = –I (d) B2 = –I

39. Let pl4 + ql3 + rl2 + sl + t =l l l l

l l ll l l

2 3 1 31 2 43 3 3

+ − −− − −− +

be an identity in l, where p, q, r, s and t are constants. Then the value of t is (a) –3 (b) 3 (c) –6 (d) 6

40. If 10! = 2p · 3q · 5r · 7s, then(a) p = 2q (b) pqrs = 64(c) number of divisors of 10! is 280(d) number of ways of putting 10! as a product of two

number is 135

solutions

1. (a) : |x1z1 – y1z2|2 + |y1z1 + x1z2|2

= |x1z1|2 + |y1z2|2 – 2Re (x1y1z1z2) + |y1z1|2 + |x1 z2|2

+ 2Re (x1y1z1z2)

= |x1z1|2 + |y1z2|2 + |y1z1|2 + |x1z2|2

= x12 |z1|2 + y1

2 |z2|2 + y12 |z1|2 + x1

2 |z2|2

= x1242 + y1

242 + y1242 + x1

242 [

|z1| = |z2| = 4]

= 2(x12 + y1

2)42

= 32(x12 + y1

2)

2. (d) : Given that z iz1 2 0+ =⇒ + =( )z iz1 2 0⇒ − = ⇒ + = ⇒ = −z iz iz z z iz1 2 1 2 2 10 0

Further, arg ( )z z1 2 3= p

⇒ − = = −arg ( ( )) [ ]z iz z iz1 1 2 13p

⇒ + − + =arg ( ) arg ( ) arg ( )z i z1 1 3p

⇒ − =22 31arg ( )z p p

⇒ =arg z1512p

3. (b) : Co-ordinates of the points A(z1) = (3, 4)and C(z3) = (5, 6) respectively.\ Length of diagonal = + =4 4 8 units

| june ‘1540

The length of another diagonal must be 8That is possible only when we take the other two vertices as 5 + 4i and 3 + 6i.

4. (a) : Given that a, b are the roots of the equationpx2 + qx + r = 0

\ + = − =a b abqp

rp

and

a b+ = \ − = ⇒ = −4 4 4qp

q p

Also, we have 2q p r p q r= + ∈[ , , ] A.P.

\ 2(–4p) = p + r ⇒ r = –9p

Thus ab = = − = −rp

pp

99

5. (d) : Given that a is a root of x2 – 3x + 1 = 0\ a2 – 3a + 1 = 0 ⇒ a2 = 3a – 1and a6 + 1 = (a2 + 1)(a4 – a2 + 1)

= (3a) (9a2 – 6a + 1 – a2 + 1)= 3a (8a2 – 6a + 2)= 6a (4a2 – 3a + 1)

Now aa

aa a a

aa a

3

6

3

2

2

21 6 4 3 1 6 4 3 1+=

− +=

− +( ) ( )

=−

= −aa a

a a2

2 22

6 43 1

( )[ ]

=×a

a

2

26 3 = 118

6. (d) : The given equation is(x – a)(x – b) + (x – b)(x – c) + (x – c)(x – a) = 0At a = b = c = m (say), the above equation becomes (x – m)2 = 0This is a quadratic equation of equal roots.7. (a) : a1 = 4

a2 = a1 + 4 × 1a3 = a2 + 4 × 2a4 = a3 + 4 × 3:.a100 = a99 + 4 × 99

Adding all the above equations, we get(a1 + a2 + a3 + … + a100) = (4 + a1 + a2 + a3 + … + a99)

+ 4(1 + 2 + 3 + … + 99)

⇒ = + ××

=a100 4 499 100

219804

8. (b) : Let the 1st term = a and common ratio = rThen a1 = 729 and a7 = 64Q a 7 = a1r6 \ 64 = 729 r6

⇒ = =

\ =r r6

664729

23

23

Now, S arr7 1

77

11

7291 2

3

1 23

2059=−−

=

−

−

=

9. (b) : Let a1 = a – 3d, a2 = a – d, a3 = a + d anda4 = a + 3dGiven that a1 + a4 = 10 ⇒ 2a = 10 \ a = 5and a2a3 = 24 ⇒ a2 – d2 = 24 \ d2 = 1⇒ d = 1The A.P. is 2, 4, 6, 8.10. (b) : Sum of first 10 terms

= 12

(sum of next 10 terms)

⇒ = −S S S10 20 1012

( )

⇒ 3S10 = S20

⇒ +

= +3 102

2 9 202

2 19( ) [ ]a d a d

⇒ 2a = 11d⇒ 2(13 – d) = 11d [ second term is 13 \ a = 13 – d]⇒ 13d = 26 ⇒ d = 2

11. (d) : nrC n

n r r− = ⇒− + −

=1 361 1

36!( )!( )!

⇒− + − −

=nn r n r r

!( )( )!( )!1 1

36 ...(i)

Again nrC n

r n r= ⇒

−=84 84!

!( )!

⇒− −

=nr r n r

!( )!( )!1

84 ...(ii)

Dividing (i) by (ii), we get r

n r− +=

13684

⇒ 7r = 3n – 3r + 3 ⇒ 10r – 3n – 3 = 0

12. (b) : 8 13

54

9!! !+

= Pr

⇒+

=84 5

49!

!Pr

⇒×

=−

8 94

99

!!

!( )!r

⇒ (9 – r)! = 4! ⇒ r = 5

| june ‘15 41

13. (d) : Let A = {1, 2, 3, ....., 11} and B = {1, 2, 3, ....., 10}\ n(A) = 11, n(B) = 10For onto function, 9 elements in B has one pre-image in A and 1 element in B will have 2 pre-image in A.(i) When each element in B has only one pre-image

in A.For 1st element in B we have 11 options in AFor second element in B we have 10 options in A............................................................................For 10th element in B we have 2 options in A

\ Total number = 11·10·9. ........ ·2 = 11(ii) When one element in B has 2 pre-images in A.

Any one element in B can be chosen to have a second pre-image in A i.e. in 10 ways\ Required number of onto functions = 10 × 11

14. (b) : n is a positive even integer\ n = 2K (say), where K ∈ NIn (1 + x)n = (1 + x)2K, the largest and second largest co-efficients are 2KCK and 2KCK–1 respectively.\ 2KCK : 2KCK – 1 = 11 : 10 (given)

⇒ ⋅− +

=2 1 1

21110

KK K

K K

K

⇒+

= ⇒ = =K

KK n

1 1110

10 20and

\ Number of terms = 2115. (b) : (1 + x + x2)n = 1 + a1x + a2x

2 + ... + a2nx2n ...(i)At x = –1, 1 = 1 – a1 + a2 – a3 + ......... + a2n ....(ii)Differentiating (i) with respect to x, we getn(1 + x + x2)n – 1 (1 + 2x) = a1 + 2a2x + 3a3x2 + ......+

2na2nx2n–1

At x = –1–n = a1 – 2a2 + 3a3 – ..... – 2n a2n ...(iii)Subtracting (ii) from (iii), we get

–n – 1 = –1 + 2a1 – 3a2 + 4a3 – ....... –(2n + 1)a2n⇒ –n = 2a1 – 3a2 + 4a3 – ..... – (2n + 1)a2n

16. (a) : C0 + 2C1 + 3C2 + ..... + (n + 1)Cn= (C0 + C1 + C2 + ....... + Cn) + (C1 + 2C2 + 3C3 + ......

+ nCn)

= + + ⋅−

+ ⋅− −

+ +

2 2

12

31 23

n nn n n n n

n( )

!( )( )

!.....

= + + − +− −

+ +

2 1 1

1 22

1n n nn n( )( )

!....

= 2n + n(1 + 1)n–1 = 2n + n·2n–1

= (n + 2)2n–1

\ (n + 2)2n–1 = 576 = 26 × 32 = (2 + 7)27–1

\ n = 7

17. (c) : A2 – B2 = A2 – BA + AB – B2

⇒ 0 = –BA + AB ⇒ AB = BAOption (c) is the only necessary condition.

18. (d) : B A A= − ′12

( )

=−

−−

12

2 35 1

2 53 1

=−

12

0 22 0

=−

0 11 0

19. (d) :

1 2 2

2 1 2

2 2 1

2 2

2 2

2 2

+ − −

− +

− − −

a b ab b

ab a b a

b a a b

Applying C ′1 → C1 – bC3 and C′2 = aC3 + C2

= + +−

− − −

( )11 0 20 1 2

1

2 2 2

2 2

a bb

a

b a a b

= (1 + a2 + b2)2 [1(1 – a2 – b2 + 2a2) – 2b (–b)]= (1 + a2 + b2)2 (1 + a2 + b2)= (1 + a2 + b2)3

20. (d) : |adj A| = 25, n = 3We have, |adj A| = |A|n–1

\ 25 = |A|2 ⇒ |A| = ±5

\ = = ± = ±−| || |

.AA

1 1 15

0 2

21. (c) : Let n(A ∩ B) = xNow n(A – B) + n(B – A) + n(A ∩ B) = n(A ∪ B)⇒ 43 – x + 51 – x + x = 75⇒ x = 19\ n((A – B) ∪ (B – A)) = n(A – B) + n(B – A)

= 43 – x + 51 – x = 94 – 38 = 56

| june ‘1542

22. (a) : A ⊃ B\ B′ – A′ = A – B

23. (d) : aRb if sin2a + cos2b = 1 sin2a + cos2a = 1 " a ∈ R\ aRa ⇒ R is reflexive. sin2a + cos2b = 1 ⇒ 1 – cos2a + 1 – sin2b = 1⇒ sin2b + cos2a = 1 ⇒ bRa\ R is symmetric.Further, let aRb and bRci.e. sin2a + cos2b = 1 and sin2b + cos2c = 1⇒ sin2a + 1 + cos2c = 2⇒ sin2a + cos2c = 1 ⇒ aRc\ R is transitive.\ R is an equivalence relation.24. (b) : See inverse mapping25. (b) : log3(5·3x–1 + 1), log9(31–x + 1), 1 are in A.P.\ 2 log9 (31–x + 1) = log3(5·3x–1 + 1) + 1

⇒ + = ⋅ + +− −22

3 1 5 3 1 131

31log ( ) log ( )x x

⇒+

⋅ +

=

−

−log log3

1

1 33 1

5 3 13

x

x

⇒+

⋅ +=

−

−3 1

5 3 13

1

1

x

x

⇒ 5⋅3x + 3 = 3·3–x + 1⇒ 5(3x)2 + 3·3x = 3 + 3x

⇒ 5(3x)2 + 2·3x – 3 = 0⇒ 5(3x)2 + 5·3x – 3·3x – 3 = 0⇒ 5·3x (3x + 1) –3(3x + 1) = 0⇒ (5·3x – 3)(3x + 1) = 0

⇒ 5·3x – 3 = 0 ⇒ = \ =

3 3

5353

x x log

Also 3x + 1 = 0 \ 3x = – 1 (not possible)26. (d) : log(sin1°) × log(sin2°) × log(sin3°) × .....log (sin179°)= log (sin1°)·log (sin2°) · log (sin3°) ......... log (sin90°)..... log (sin 179°)= log (sin1°) log (sin2°) .... log 1 ..... log (sin179°) = 0

27. (c) : 1 1 1 1

2 3 4 2015log log log......

logn n n n+ + + +

= logn 2 + logn 3 + logn 4 + ...... + logn 2015= logn (2⋅3⋅4⋅ .......... 2015)= log2015! (2015!) = 1

28. (c) : log0.3(x – 1) ≤ log0.09(x – 1)

⇒−

≤−log( )

log .log( )log .

x x10 3

10 09

⇒−

≤−log( )

log .log( )

log .x x10 3

12 0 3

⇒ log0.3 (x – 1)2 – log0.3(x – 1) ≤ 0⇒ log0.3 (x – 1) ≤ 0

⇒−

≤log( )

log .x 10 3

0

⇒ log(x – 1) ≥ 0 [ log100.3 = log103 – log1010 < 0]⇒ x – 1 ≥ 100 ⇒ x – 1 ≥ 1 ⇒ x ≥ 2

29. (d) : P nn

nn

n( ) : ( )!

( !)4

12

2+<

For n = 1, P(n) is not true.

For n = 2, P(2) : 42 1

42

163

244

2

2+< ⇒ <!

( )is true.

Let for n = m > 2, P(n) is true i.e 41

22

m

mm

m+< ( )!

( !)

Now 42

41

4 12

1m m

m mm

m

+

+=

+⋅ +

+( ) < ⋅ +

+( )!( !)

( )2 4 122

mm

mm

= + + + ++ + + +

( )!( )( ) ( )( )( )( )( !) ( ) ( )2 2 1 2 2 4 1 12 1 2 2 1 2

2

2 2m m m m mm m m m m

= ++

⋅ ++ +

[ ( )]![( )!]

( )( )( )

2 11

2 12 1 22

2mm

mm m

< ++

++ +

< " >

[ ( )]![( )!]

( )( )( )

,2 11

2 12 1 2

1 02

2mm

mm m

m

Hence, for n ≥ 2, P(n) is true.30. (d) : Unless we prove P(1) is true, nothing can besaid.31. (b) : Given that Stn = cn(n – 1) = Sn (say)\ tn = Sn – Sn – 1

= cn(n – 1) – c(n – 1)(n – 2)= c(n – 1) (n – n + 2)

tn = 2c(n – 1)Now tn

2 = 4c2(n – 1)2

\ Stn2 = 4c2S(n2 – 2n + 1)

= 4c2[Sn2 – 2Sn + n]

= + + − + +

4 1 2 16

2 12

2c n n n n n n( )( ) ( )

= + + − − +

4 2 3 1 6 6 66

22

c n n n n

| june ‘15 43

= − +4 2 3 16

2 2c n n n[ ]

= − −23

1 2 12c n n n( )( )

32. (d) : Let z = iy, y ∈ R and c = cosa + isina\ cz2 + z + 1 = 0⇒ –(cosa + isina)y2 + iy + 1 = 0⇒ (cosa + i sina)y2 – iy – 1 = 0⇒ (cosa · y2 – 1) + (sina · y2 – y)i = 0⇒ y2cosa – 1 = 0 and y2sina – y = 0⇒ y2cosa = 1 and ysina = 1, y ≠ 0⇒ y2cosa – y2sin2a = 0⇒ y2(sin2a – cosa) = 0\ 1 – cos2a – cosa = 0 ⇒ cos2a + cosa – 1 = 0

⇒ = − ± − ⋅ −⋅

= − ±cos ( )a 1 1 4 1 12 1

1 52

2

⇒ = − \ = +

cos seca a5 1

25 12

22

\ = +

− = +tan2

25 12

1 5 12

a

\ = = +tan tan(arg )a c 5 12

33. (d) : |A2015 – 3A2014|=|A2014| |A – 3I|=|A|2014 |A – 3I|

=−

= − = − =

10 71 1

1 7 73 71 2

( ) A

34. (d) : Here | |sin coscos sinA =

−=

a aa a

00

0 0 11

By definition AA

A A A− = = =1 1 1| |

( | | )adj adj

\ A–1 = adj A35. (a) : Using 1, 2, 3, 4, 5, we can form 5 digits numberdivisible by 3 in 5 ways. Now we fix '0' at unit place, tenth place, hundredth place and thousand place then number of ways = 4 × 4!\ Total number of ways = 120 + 96 = 21636. (a, b, c)

37. (a, b, c, d) : Here f(x) = 2x – sinx and g x x( ) = 3

Now gof(x) = g(f(x))gof x x= −23 sin

Clearly, the range of gof is R

Now ddx

gof xx x

( ) ( cos )( sin ) /= −

−13

22 2 3 > 0 " x ∈ R

⇒ gof(x) is strictly increasing function and hence itis one-one function.⇒ Both f(x) and g(f) are one-one functions.Since f(x) → ± ∞ and g(x) → ± ∞ as x → ± ∞Hence they are also onto.Thus both f and g are one-one and onto.

38. (c,d) : A A=−

\ =

−

−

0 11 0

0 11 0

0 11 0

2

=−

−

= −

1 00 1

I

\ A2 = –I

Bi

iB

ii

ii

=

⇒ =

00

00

00

2

=−

−

= −

1 00 1

I

\ B2 = –I39. (a) : Here pl4 + ql3 + rl2 + sl + t

=+ − −− − −− +

l l l ll l ll l l

2 3 1 31 2 43 3 3

Putting l = 0 in the above identity, we get

t =− −

− −−

0 1 31 0 43 3 0

= 1(0 – 12) –3(–3) = –12 + 9 = –3

40. (a, b, c, d) : 10! = 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10= (2 × 4 × 6 × 8 × 10) × 3 × 5 × 7 × 9

= 25(1 × 2 × 3 × 4 × 5) × 3 × 5 × 7 × 9 = 28 × 34 × 52 × 71

\ p = 8, q = 4, r = 2, s = 1\ p = 8 = 2 × 4 = 2q, Thus option (a) is correct.Now pqrs = 8 × 4 × 2 × 1 = 64, option (b) is correct.Again 10! = 28 × 34 × 52 × 7

= 25 (22 × 7 × 2 × 5) × 34 × 5 = 25 × 280 × 34 × 5

\ 10! is divisible by 280. Thus option (c) is correct.The product of two integers in 10! is 135 (15 × 9 = 135).Thus option (d) is correct.

SetSA set is a well-defined collection of objects. Sets are usually denoted by capital letters A, B, C, etc. and their elements by small letters a, b, c, etc.Let A be any set of objects and let 'a' be a member of A, then we write a ∈ A and read it as 'a belongs to A' or 'a is an element of A' or 'a is a member of A'. If a is not an object of A, then we write a ∉ A and read it as 'a does not belong to A' or 'a is not an element of A' or a' is not a member of A'.

RepReSentation of SetSThere are two ways of expressing a set. These are:

SetS

Tabular form or Roaster Method Set-builder form or Rule Method

types Definition example

tabular form or Roaster Method

In this method we list all the members of the set separating them by commas and enclosing them in curly brackets {}.

Let A be the set consisting of the numbers 1, 3, 4 and 5, then we write A = {1, 3, 4, 5}.

Set-builder form or Rule Method

In this method, instead of listing all elements of a set, we write the set by some special property or properties satisfied by all its elements and write it as A = {x | x has the property P(x)} and read it as 'A is the set of all elements x such that x has the property P'. The symbol ' : ' or ' | ' stands for 'such that'.

Let A be the set consisting of the elements 2, 3, 4, 5, 6, 7, 8, 9, 10. Then the set A can be written as : A = {x : 2 ≤ x ≤ 10 and x ∈ N}

typeS of Set

Finite & Infinite set Equal & equivalent set

types of set

Super set Subset

Power set

Proper subset

Universal set

Null or empty or void set

| june ‘1544

Finite and Infinite Set : • A set whose elements can be counted i.e. a set which has only finite number of elements, is called a finite set and which has an infinite number of elements is called an infinite set.equal and equivalent set : • Two sets A and B arecalled equal if A and B have identical elements. Iftwo sets A and B have equal number of elementsthen A and B are called equivalent sets.Note: Equal sets are always equivalent but equivalentsets may or may not be equal.Null or empty or void set :• A set having no elementin it is called a null set. We denote it by φ or { }.Subset :• A set A is said to be subset of set B ifevery member of set A is also the member of setB. It is denoted as A ⊆ B. If A is not a subset ofB, we write A ⊄ B.Note :(i) Empty set is the subset of every set.(ii) Every set is the subset of itself.(iii) Total number of subsets of a set having nelements is 2n.Superset :• If A ⊆ B, then B is called superset of A,and we write, B ⊇ A.Proper Subset :• If A ⊂ B and A ≠ B, then A iscalled proper subset of B and we write A⊂ B.Power Set :• Set of all the subsets of a set is calledpower set of a set which is denoted by P(A).i.e., P(A) = {S : S ⊂ A}, where S is subset of setA.Universal Set :• If all sets under consideration aresubsets of a larger set, then this larger set is calleduniversal set, denoted by U.

complement of a Set

The complement of a given set is a set which•contains all those members of the universal setthat do not belong to the given set.The complement of the set • A is denoted by A′ orby Ac = {x : x ∈ U, x ∉ A}.

opeRationS on SetSUnion of Sets

Let • A and B be two given sets. Then the union of Aand B is the set of all those elements which belongto either A or B or both.The union of • A and B is denoted by A ∪ B and

is read as 'A union B'. The symbol ∪ stands for union. Symbolically, A ∪ B = {x : either x ∈ A or x ∈ B}

Intersection of Sets Let • A and B be two given sets. Then the intersectionof A and B is the set of elements which belong toboth A and B. In other words, the intersectionof A and B is the set of common members of Aand B.The intersection of • A and B is denoted by A ∩ Band is read as 'A intersection B'. The symbol∩ stands for intersection. Symbolically, A ∩ B= {x : x ∈ A and x ∈ B}

Disjoint SetsIf • A ∩ B = φ, then A and B are said to be disjointsets.

Difference of two SetsLet • A and B be two given sets. The difference ofsets A and B is the set of elements which are in Abut not in B. It is written as A – B and read as 'Adifference B'. Symbolically,

A – B = {x : x ∈ A and x ∉ B} Similarly, B – A = {x : x ∈ B and x ∉ A}Symmetric Difference of two sets

Let • A and B be two given sets. The symmetricdifference of sets A and B is the set (A – B) ∪(B – A) and is denoted by A D B. Symbolically,A D B = {x : x ∉ A ∩ B}

impoRtant pRopeRtieS

Properties of Union•(i) A ∪ B = B ∪ A (Commutative law)(ii) (A ∪ B) ∪ C = A ∪ (B ∪ C) (Associative

law)(iii) A ∪ φ = A (Law of identity element)(iv) A ∪ A = A (Idempotent law)(v) U ∪ A = U (Law of U)

Properties of Intersection•(i) A ∩ B = B ∩ A (Commutative law)(ii) (A ∩ B) ∩ C = A ∩ (B ∩ C) (Associative

law)(iii) φ ∩ A = φ, U ∩ A = A (Law of φ and U)(iv) A ∩ A = A (Idempotent law)(v) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) (Distributive

law) i.e., ∩ distributes over ∪.Complement laws•(i) A ∪ A′ = U

| june ‘15 45

(ii) A ∩ A′ = φ De Morgan’s law•(i) (A ∪ B)′ = A′ ∩ B′(ii) (A ∩ B)′ = A′ ∪ B′

Law of double complementation•(A′)′ = A

Laws of empty set and universal set•φ′ = U and U′ = φ

application of SetS

Let • A and B be finite sets and A ∩ B = φ then n(A∪ B) = n(A) + n(B)

If • A and B are finite sets such that A ∩ B ≠ φ thenn(A ∪ B) = n(A) + n(B) – n(A ∩ B)If • A, B and C are three finite sets, thenn(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(A∩ C) – n(B ∩ C) + n(A ∩ B ∩ C)n• (A – B) = n(A) – n(A ∩ B)n• (ADB) = n(A) + n(B) – 2n (A ∩ B)

Venn DiagRamS

In Venn diagrams, the universal set • U is representedby using rectangle and its subsets are representedby using circle (a closed curve).

A & B are two sets contained in universal set U. (a) Shaded area represents A ∪ B (b) Shaded area represents A ∩ B(c) Shaded area represents A′ (d) No shaded area, i.e., A & B are disjoint sets(e) Shaded area represents A – B (f) Shaded area represents B – A(g) Shaded area represents A – B; Shaded area represents B – A; Shaded area represents A ∩ B(h) This venn diagram shows A ⊆ B

Venn Diagrams

U

A'

A

A B

U

A B

U

A B

U

U

BA

A B

A B∪A B– B A–

U

(a) (b)

(c)

(d)(e)

(f)

(g)

(h)

Very short answer type

1. If P is the set of all prime numbers and E is the setof all even numbers; find P ∩ E.

2. If S and T are two sets such that S has 21 elements,T as 32 elements, and S ∩ T has 11 elements, howmany elements does S ∪ T have?

3. A is the set of letters in word moon. Find the powerset of A.

4. Show by an example, A ∩ B = A ∩ C need not implyB = C.

5. Prove that the shaded region in the given figure isB – (A ∪ C).

| june ‘1546

short answer type

6. Let A = {x : x ∈R, |x| < 2}, B = {x : x ∈ R,|x – 2| ≥ 2} and A ∪ B = R – C, then find the set C.

7. If U = { x : x ∈N and 2 ≤ x ≤ 12},A = {x : x is an even prime}, B ={x : x is a factorof 24}, then prove that A – B ≠ B ∩ A′.

8. In a class of 175 students, the following datashows the number of students opting one or moresubjects.Mathematics 100; Physics 70; Chemistry 40;Mathematics and Physics 30; Mathematicsand Chemistry 28; Physics and Chemistry 23;Mathematics, Physics and Chemistry 18.Then find the number of students who optedMathematics alone.

9. Show that for any sets A and B,A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = A ∪ B.

10. Let B be a subset of a set A andlet P(A : B) = {X ∈ P(A) | X ⊃ B}(i) Show that P(A : φ) = P(A).(ii) If A = {a, b, c, d} and B = {a, b}. List all the

members of the set P(A : B).

Long answer type

11. In a certain town, 25% families own a phone and15% own a car, 65% families own neither a phonenor a car, 2000 families own both a car and aphone.Consider the following statements in this regard.I. 10% families own both a car and a phone.II. 35% families own either a car or a phone.III. 40000 families live in the town.Then prove that only II and III are correct.

12. Let A and B be sets, if A ∩ X = B ∩ X = φ andA ∪ X = B ∪ X for some set X, prove that A = B.Also, prove that (A – B) ∩ (B – A) = φ.

13. If U = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, A = {2, 4, 7},B = {3, 5, 7, 9, 11} and C = {7, 8, 9, 10, 11},compute(i) (A ∩ U) ∩ (B ∪ C);(ii) C – B;(iii) B – C;(iv) (B – C)′(v) U′(vi) φ′

14. If X = {8n – 7n – 1| n ∈N} and Y = {49n – 49 |n∈N}, then prove that X = Y.

15. On its annual sports day, School awarded 35 medalsin athletics, 15 Judo and 18 in swimming. If thesemedals goes to a total of 58 students and only threeof them got medals in all the three sports. Find thenumber of students who received medals in exactlytwo of the three sports.

soLutions

1. P = Set of all prime numbers= {2, 3, 5, 7, 11, ...}

E = Set of all even numbers= {2, 4, 6, 8, 10, ...}

\ P ∩ E = {2, 3, 5, 7, 11, ...} ∩ {2, 4, 6, 8, 10, ...} = {2}

2. n(S) = 21, n(T) = 32, n(S ∩ T) = 11, n(S ∪ T) = ?n(S ∪ T) = n(S) + n(T) – n(S ∩ T)

= 21 + 32 – 11 = 42Hence, S ∪ T has 42 elements.

3. Power set of A = P(A) = {φ, {m}, {o}, {n}, {m, o}, {m, n}{o, n}, {m, o, n}}

4. Let A = {1, 2}, B = {1, 3} and C = {1, 4}Now, A ∩ B = {1, 2} ∩ {1, 3} = {1}and A ∩ C = {1, 2} ∩ {1, 4} = {1}\ A ∩ B = A ∩ C but B ≠ C.

5. In the venn diagram of B – (A ∪ C), whole set Bis shaded except set A ∪ C.

6. Given that A = {x : x ∈R, | x | < 2}and B = {x : x ∈R, |x – 2| ≥ 2}⇒ A = –2 < x < 2and B = {x : x ∈R, x – 2 ≤ – 2 or x – 2 ≥ 2}

= {x : x ∈R, x ≤ 0 or x ≥ 4}⇒ A = ] – 2, 2[ & B = (– ∞, 0] ∪ [4, ∞)Now A ∪ B = (– ∞, 2) ∪ [4, ∞) = R – [2, 4)

\ C = [2, 4) i.e. 2 ≤ x < 47. U = {x : x ∈N and 2 ≤ x ≤ 12}

⇒ U = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} A = {x : x is an even prime} ⇒ A = {2} B = {x : x is a factor of 24}⇒ B = {2, 3, 4, 6, 8, 12}⇒ A – B = {2} – {2, 3, 4, 6, 8, 12} = φHence, A – B is an empty set , is a true statement.Now A′ = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12},

| june ‘15 47

B ∩ A′ = {3, 4, 6, 8, 12}Hence A – B ≠ B ∩ A′

8. Given, n(M) = 100, n(P) = 70, n(C) = 40 n(M ∩ P) = 30, n(M ∩ C) = 28 n(P ∩ C) = 23and n(M ∩ P ∩ C) = 18\ n(M ∩ P′ ∩ C′) = n[M ∩ (P ∪ C)′]

(De-Morgan's Law) = n(M) – n[M ∩(P ∪ C)]

= n(M) – n[(M ∩ P) ∪ (M ∩ C)][Q n(A ∩ B′) = n(A) – n(A ∩ B)]

= n(M) – [n(M ∩ P) + n(M ∩ C)–n(M ∩ P ∩ C)]

= 100 – [30 + 28 – 18] = 100 – 40 = 60

60 35

10 518

M P

C7

12

9. (A ∩ B) ∪ (A – B) = (A ∩ B) ∪ (A ∩ B′)[Q A – B = A ∩ B′]

= A ∩ (B ∪ B′) [Distributive law] = A ∩ U, where U is universal set = AA ∪ (B – A) = A ∪ (B ∩ A′)

[Q B – A = B ∩ A′]= (A ∪ B) ∩ (A ∪ A′) [Distributive law]= (A ∪ B) ∩ U, where U = A ∪ A′ is universal set= A ∪ B (Q A ∪ B ⊂ U)

10. (i) We have,P(A : B) = {X ∈ P(A) | B ⊂ X}= Set of all those subsets of A which contain B\ P(A : φ) = Set of all those subsets of A whichcontain φ= Set of all subsets of set A= P(A)(ii) We have,A = {a, b, c, d} and B = {a, b}\ P(A : B) = Set of all those subsets of set Awhich contain B= {{a, b, c}, {a, b, d}, {a, b, c, d}}

\ Elements of P(A : B) are {a, b, c}, {a, b, d} and {a, b, c, d}

11. Let there be x families in the town. Let P and Cdenote the set of families using phone and car,respectively. Then,

n P x x( ) ,= =25100 4

n C x x( ) = =15100

320

and n P C x x( )∩ = =65100

1320

Also, n(P ∩ C) = 2000

Now, n P C x( )∩ = 1320

⇒ n P C x( )∪ = 1320

⇒ n U n P C x( ) ( )− ∪ = 1320

⇒ x n P n C n P C x− + − ∩ ={ ( ) ( ) ( )} 1320

⇒ x x x x− + −

=4

320 2000 13

20

⇒ x20 2000=

⇒ x = 40000Thus, Statement III is correct.10% of the total families in the town is 4000 and it is given that 2000 families own both a car and a phone. So, Statement I is not correct.Now, n(P ∪ C) = n(P) + n(C) – n(P ∩ C)

⇒ n P C x x( )∪ = + −4320 2000

⇒ n(P ∪ C) = 10000 + 6000 – 2000 = 14000⇒ n(P ∪ C) = 35% of 40000Thus, Statement II is correct.

12. We have for some set X,A ∪ X = B ∪ X ⇒ A ∩ (A ∪ X) = A ∩ (B ∪ X)⇒ A = (A ∩ B) ∪ (A ∩ X)

[Q A ∩ (A ∪ X) = A]⇒ A = (A ∩ B) ∪ φ

[Q A ∩ X = φ (given)]⇒ A = A ∩ B⇒ A ⊆ B ...(1)Again,A ∪ X = B ∪ X ⇒ B ∩ (A ∪ X) = B ∩ (B ∪ X)⇒ (B ∩ A) ∪ (B ∩ X) = B

[Q B ∩ (B ∪ X) = B]⇒ (B ∩ A) ∪ φ = B⇒ B ∩ A = B [Q B ∩ X = φ (given)]⇒ B ⊆ A ...(2)From (1) and (2), we get A = B.Let x ∈ A – B ⇒ x ∈ A and x ∉ B ⇒ x ∉ B – A

| june ‘1548

Further any x ∈ B – A ⇒ x ∈ B and x ∉ A ⇒ x ∉ A – BThis proves that (A – B) ∩ (B – A) = φ.

13. (i) Here A ∩ U = {2, 4, 7};B ∪ C = {3, 5, 7, 8, 9, 10, 11}.(A ∩ U) ∩ (B ∪ C) = {2, 4, 7} ∩ {3, 5, 7, 8, 9, 10,11} = {7}(ii) C – B is a set of members which belong to C,but do not belong to B.\ C – B = {8, 10}(iii) B – C is a set of members which belong to B,but do not belong to C.\ B – C = {3, 5}(iv) From (iii), B – C = {3, 5}\ (B – C)′ = {2, 4, 6, 7, 8, 9, 10, 11}(v) U′ = Set of members in U which are not in U

= φ, since there is no such member(vi) φ = {}, the empty set,φ′ = Set of members in U which are not in φ = U

14. Given, X = {8n – 7n – 1 : n ∈N} and Y = {49n – 49 : n ∈ N}Now, 8n – 7n – 1 = (7+ 1)n – 7n – 1 = 7n + nC17n – 1 + nC27n –2 + ... + nCn – 272

+ nCn–17 + nCn –7n–1[... by binomial expansion, (x + 1)n = xn + nC1xn–1

+ nC2xn–2 + ... + nCn–2x2 + nCn–1x + nCn]= nCn7n + nC17n–1 + nC27n–2 + ...+ nCn–272 + 7n + 1 – 7n – 1

(... nCn = nC0 =1, nCn–1= nC1 = n)= nCn 7

n + nC17n–1 + nC27n–2 + ... + nC272

(... nCn–2 = nC2)= nC272 + ... +nC27n–2 + nC17n–1 + nCn7n

= 72[nC2 + ...+ nCn7n–2]= 49[nC2 + ... + nCn7n–2]

\ 8n – 7n – 1 is a multiple of 49 for n ≥ 2For n = 1, 8n–7n – 1 = 8 – 8 = 0

\ 8n – 7n – 1 is a multiple of 49 for all n ∈N.X contains elements which are multiples of 49 and clearly Y contains all multiples of 49.\ X = Y.

15. Let A denote the set of students who receivedmedal in athletics, J be the set of students whogot medal in Judo and S be the set of studentswho got medal in swimming.... n(A) = 35, n(J) = 15, n(S) = 18

n(A ∪ J ∪ S) = 58 and n(A ∩ J ∩ S) = 3Now n(A ∪ J ∪ S)

= n(A) + n(J) + n(S) – n(A ∩ J) – n(A ∩ S)– n(J ∩ S) + n(A ∩ J ∩ S)

58 = 35 + 15 + 18 + 3 – n(A ∩ J) – n(A ∩ S) – n(J ∩ S)

\ n(A ∩ S) + n(A ∩ J) + n(J ∩ S) = 71 – 58 = 13\ Number of students who received medals in exactly two of the three= n(A ∩ J) + n(J ∩ S) + n(S ∩ A) – 3n(A ∩ S ∩ J)= 13 – 3 × 3 = 13 – 9 = 4.

| june ‘1550

Relations

Let A and B be two non-empty sets. The relation R between A and B is a subset of A × B. Symbolically, we write the relation between A and B as: R : A → B if and only if R ⊆ A × B

Domain and Range of a Relation

DomainIf R : A → B, then the domain of R is the set of all first elements of the ordered pairs (x, y) which belong to R.

Symbolically, Domain = {x : (x, y) ∈ R, x ∈ A for some y ∈ B}

Range The range of a relation is the set of all second elements of the ordered pairs which belong to R.

Symbolically, Range = {y : (x, y) ∈ R, y ∈ B for some x ∈ A}

Relation in a setLet R be a relation from A to B. If A = B, then we say that R is a relation in A or R is a relation on A.Thus, a relation R in a set A is a subset of the cartesian product A × A.

Total Number of RelationsLet A and B be two non-empty finite sets having n and m elements respectively,\ Total number of relations from A to B = 2nm

types of Relation

| june ‘15 51

Name Definition

Reflexive Relation A relation R on a set A is reflexive if (a, a) ∈R, ∀ a ∈A

Symmetric Relation A relation R on a set A is symmetric if (a1, a2) ∈R ⇒ (a2, a1)∈R ∀ a1, a2 ∈A

Transitive Relation A relation R on a set A is transitive if (a1, a2) ∈R and (a2, a3) ∈R ⇒ (a1, a3) ∈ R ∀a1, a2, a3 ∈ A.

Anti-symmetric Relation

A relation R on a set A is called an anti-symmetric relation, if (a, b) ∈R and (b, a) ∈ R ⇒ a = b. If a ≠ b, then a may be related to b or b may be related to a, but never both.

Equivalence Relation

A relation R is an equivalence relation if R is reflexive, symmetric and transitive.

Void Relation A relation R on a set A is called an empty or void relation, if no element of A is related to any element of A.

Universal Relation A relation R on a set A is called universal relation, if each element of A is related to every element of A.

Identity Relation The relation IA on set A is identity relation, if every element of A is related to itself only.

Inverse Relation If R is a relation on set A, then R–1 on A is defined as R–1 = {(b, a) : (a, b) ∈ R}

Composition of Relations

Let R ⊆ A × B, S ⊆ B × C be two relations, then composite relation of R and S is SoR ⊆ A × C or SoR = {(a, c); (a, b) ∈R, (b, c) ∈ S}

functionsLet A and B be two non-empty sets. If there exists a rule of correspondence by which each element x ∈ A is related to a unique element y ∈ B, then such correspondence is called the function from A to B. it is written as:f : A → B such that x ∈ A and y ∈ B, where y = f(x)The element y ∈ B is called the image of x under f and x is called the pre-image of y under f.

The number of functions from a finite set z A into set B = [n(B)][n(A)]

There may exist some elements in set z B which are not the images of any element in set A.The element y ∈ B that is associated to x under f is denoted by f(x) and is called the value of f at x.

Domain, Co-domain, Range of Function f(x)

Domain The set A is called the domain of function f.

Co-domain The set B is called the co-domain of function f.

Range The set {f(x) : x ∈ A and f(x) ∈ B} for all values taken by f is called the range of f. Obviously, it is a subset of set B.

types of functions

| june ‘1552

One-one function (Injective function) : z A function f : A → B is defined to be one-one if the images of distinct elements of A under f are distinct in B.

fA B

x1

x2

x3

y1

y2

y3

Onto function (Surjective function) : z

Let f : A → B. If every element in B has at least one pre-image in A, then f is said to be an onto function or surjective function.

If the range of the function equals to the codomain z

of the function, the function is onto.

fA B

y1

y2

x1

x2

x3x4

Many one function : z Let f : A → B. If two or more than two elements of set A have the same image in B, then f is said to be many-one.If z x1 ≠ x2 ⇒ f(x1) ≠ f(x2), for every x1, x2 ∈ domain, then f is one-one or else many one.If z f(x1) = f(x2) ⇒ x1 = x2, for every x1, x2 ∈ domain, then f is one-one or many one.

fA B

y1

y2

x1

x2

x3x4 y3

Into function : z Let f : A → B. If there exists a single element in B having no pre-image in A, then f is said to be an into function.

fA By1x1

x2

x3x4 y4

y2

y3

y5

Bijective function : z If a function is one-one and onto, then it is bijective.

orA function f : A → B is bijective if (a) f is one-one, i.e., f(x) = f(y) ⇒ x = y ∀ x, y ∈A

(b) f is onto i.e., ∀ y ∈B, $ x ∈A such that f(x) = y

Identity function : z Identity function IA on a non-empty set A defined by IA : A → A, IA(x) = x, ∀ x ∈A. This is clearly a one-one function with domain A and range A.

composition of functions and inveRtible function

Inverse function : z Inverse of a function f : A → B defined by f –1 : B → A

\ f –1(y) = x ⇔ f(x) = yA function z f is invertible if and only if f is one-one onto.fog z ≠ gof, fo(goh) = (fog)oh, If f : A → B, then foIA = IB of = f

x y f x= ( ) z g f x= { ( )}

gof

A f B g C

Composition of functions : z Let f : A → B and g : B → C be two functions of non empty sets A, B, C, then gof : A → C is called composition of f and g defined as gof (x) = g{f(x)} ∀ x ∈A

binaRy opeRationDefinition : Let A be non-empty set, then the function from A × A into A is called a binary operation. Symbolically, a function '*' which is

* : A × A → Ais called a binary operation. The image of any (a, b) ∈ A × A under '*' is denoted by a * b.(i) A binary operation '*' over a set A is said to be

commutative if a * b = b * a ∀ a, b ∈ A(ii) A binary operation '*' over a set A is said to be

associative if (a * b) * c = a * (b * c) ∀ a, b, c ∈ A(iii) Let * be a binary operation on a set A. An element

e ∈ A is said to be an identity element for the binary operation * if a * e = a = e * a ∀ a ∈ A i.e., e when combined with any element a ∈ A produces a itself.

(iv) Let * be a binary operation on a set A and let e be the identity element of the set A for this operation *. An element b ∈ A is said to be the inverse of an element a ∈ A if a * b = e = b * a.

| june ‘15 53

Very short answer type1. The relation R is defined on the set of naturalnumbers as {(a, b) : a = 2b}. Then, find R–1.2. Let L denote the set of all straight lines in a plane. Leta relation R be defined by aRb ⇔ a ^ b, a, b ∈L. Then, prove that R is symmetric.3. Find the domain for which the functionsf(x) = 2x2 – 1 and g(x) = 1 – 3x are equal.4. Let f : A → B and g : B → C be two functions. Ifboth f and g are one-to-one, show that gof is also one-to-one function.5. Prove that the function f : N → N given by f(x) = 2xis injective but not surjective.

short answer type

6. Consider the set N × N, the set of ordered pairs ofnatural numbers. Let R be the relation in N × N which is defined by (a, b) R(c, d) if and only if a + d = b + c.Prove that R is an equivalence relation.

7. Show that f : N → N defined by f(x) =

n n

n n

+

12

2

,

,

if is odd

if is evenis many-one onto function.

8. In the set N of natural numbers define the binaryoperation * by m * n = L.C.M (m, n); m, n ∈ NIs the operation commutative and associative?9. If f(x) = (ax2 – b)3, then prove that thefunction g such that f{g(x)} = g{ f(x)} is given by

g x x ba

( )/ /

= +

1 3 1 2

10. Find the range of

f x x x x( ) ( cos ) ( cos ) ( cos ) ....= − − − ∞1 1 1

Long answer type

11. Prove that the function f : R → {x ∈ R : – 1 < x < 1}defined by f(x) =

+x

x1 | |, x ∈ R is both one-one and onto.

12. Consider f : R → [–5, ∞) given by f(x) = 9x2 + 6x – 5.

Show that f is invertible with f –1(y) = y + −

6 13

.

13. Let * be a binary operation on the set R – {1} ofall real numbers defined by the rule a * b = a + b – ab for all a, b ∈ R – {1}, whereas on R.H.S. we have usual addition, subtraction and multiplication of real numbers. Show that * is commutative as well as associative. Also find the identity element. What is the inverse of an element a?

14. Show that each of the relation R in the setA = {x ∈ Z : 0 < x ≤ 12}, given by(i) R = {(a, b) : |a – b| is a multiple of 4}(ii) R = {(a, b) : a = b}is an equivalence relation. Find the set of all elements related to 1 in each case.

15. If f xx

g x xx

( ) ( ) ,= = −+

1 11

and find Dgof and Dfog.

Also, find (gof )(x) and (fog)(x).

soLUtioNs

1. R = {(2, 1), (4, 2), (6, 3)...}So, R–1 = {(1, 2), (2, 4), (3, 6), ....}

2. Given, aRb ⇔ a ^ bSince a ^ b ⇔ b ^ a ⇒ bRaHence, R is symmetric.

3. f(x) = g(x) ⇒ 2x2 – 1 = 1 – 3x

⇒ (x + 2) (2x – 1) = 0 ⇒ x = –2, 12

Thus, f(x) and g(x) are equal on the set {–2, 1/2}.

4. Let (gof) (x) = (g of) (y) for x, y ∈ A. Theng(f(x)) = g(f(y)). So, f(x) = f(y) as g is one-to-one. Therefore, x = y as f is also one-to-one. Thus, gof is one-to-one.

5. The function f is one-one, as f(x1) = f (x2)⇒ 2x1 = 2x2 ⇒ x1 = x2. Further, f is not onto, as for1 ∈ N, there does not exist any x in f(x) such that f(x) = 2x = 1.

6. (i) Reflexivity∀ (a, b) ∈ N × N, we havea + b = b + a, i.e., (a, b) R (a, b)\ R is reflexive.(ii) Symmetricity Suppose (a, b) R (c, d). Then a + d = b + c, which implies that c + b = d + a i.e., (c, d) R (a, b)Hence R is symmetric.(iii) Transitivity Now, suppose (a, b) R (c, d) and (c, d) R (e, f)Then a + d = b + c and c + f = d + eThus, (a + d) + (c + f) = (b + c) + (d + e)Subtracting c + d from both sides, we get a + f = b + e\ (a, b) R (e, f) and hence R is transitive.Since R is reflexive, symmetric and transitive, therefore, R is an equivalence relation.

| june ‘1554

7. We have, f(1) = 1 12

1+ = and f(2) = 22

1= .

Thus, 1 ≠ 2 but f(1) = f(2). So, f is a many-one function.Onto : Let n be an arbitrary element of N.If n is an odd natural number, then 2n – 1 is also an odd natural number such that

f(2n – 1) = 2 1 12

n n− + =

If n is an even natural number, then 2n is also an even number such that

f(2n) = 22n n= .

Thus, for every n ∈ N (whether even or odd), there exists its pre-image in N. So, f is onto.Hence, f is a many-one onto function.

8. (i) The operation is clearly commutative, sinceL.C.M. (m, n) = L.C.M. (n, m)

\ m * n = n * m(ii) Again for l, m, n ∈ N,

m * n = L.C.M. (m, n) and n * l = L.C.M. (n, l)

Now (m * n) * l = L.C.M. (m * n, l) = L.C.M. [L.C.M.(m, n), l]= L.C.M. [m, L.C.M. (n, l)] = L.C.M. [m, n * l] = m * (n * l)

Hence (m * n) * l = m * (n * l)Hence * is associative.

9. y = f(x) = (ax2 – b)3

⇒ ax2 – b = y1/3

⇒ xy b

a2

1 3=

+/ ⇒ x

y ba

=+

1 3 1 2/ /

Let g x x ba

( )/ /

= +

1 3 1 2

\ = +

f g x f x b

a{ ( )}

/ /1 3 1 2

= +

−

a x b

ab

1 3 3/= (x1/3)3 = x

and g{f (x)} = g{(ax2 – b)3}

= − +

= =ax b b

ax x

2 1 22 1 2

//( )

⇒ g{f(x)} = f{g(x)}

Thus, g x x ba

( )/ /

= +

1 3 1 2

10. Given

f x x x x( ) ( cos ) ( cos ) ( cos ) ...= − − − ∞1 1 1

= − − − ∞( cos ) ( cos ) ( cos ) ........1 1 112

14

18x x x

= − = − = −+ + + ∞ −( cos ) ( cos )

........1 1 1

12

14

18

1 21 1 2x x xcos

⇒ Range of f(x) is [0, 2]11. It is given that f : R → {x ∈ R : – 1 < x < 1} is

defined as f(x) = x

x1+ | |, x ∈ R.

Suppose f(x) = f(y), where x, y ∈ RIt can be observed that if x is positive and y is negative, then, we have x

x1+=

yy1−

(as x > 0 and y < 0)

⇒ 2xy = x – ySince x is positive and y is negative, then

x > y ⇒ x – y > 0But 2xy is negative. Then, 2xy ≠ x – y.Thus, the case of x being positive and y being negative can be ruled out.Under a similar argument, the case of x being negative and y being positive can also be ruled out. Therefore, x and y have to be either positive or negative. When x and y are both positive, we have

f(x) = f(y)

⇒ xx1+

= yy1+

⇒ x + xy = y + xy⇒ x = yWhen x and y are both negative, we have

f(x) = f(y)

⇒ xx1−

= y

y1−⇒ x – xy = y – yx⇒ x = yTherefore, f is one-one. Now, let y ∈ R such that – 1 < y < 1. If y is negative, then there exists

x = yy1+

∈ R such that

| june ‘15 55

f(x) = f yy1+

=

yyy

y

yyyy

1

11

1

11

+

++

=+

+ −+

= yy y1+ −

= y

If y is positive, then there exists x = yy1−

∈ R such

that f(x) = f y

y1−

=

yy

yy

yyy

y

1

11

1

11

−

+−

=−

+−

= yy y1− +

= y

Also, for y = 0, we have x = 0 ∈ RTherefore, f is onto. Hence, f is both one-one and onto.

12. Let y be an arbitrary element in range of f.Let y = 9x2 + 6x – 5 = 9x2 + 6x + 1 – 6

= (3x + 1)2 – 6⇒ y + 6 = (3x + 1)2 or 3x + 1 = y + 6

\ x = y

g y+ −

=6 1

3( )

Let g : Range f → R+ , such that g(y) = y + −6 1

3gof (x) = g(f(x)) = g (9x2 + 6x – 5)

= g [(3x + 1)2 – 6]

= ( )3 1 6 6 1

3

2x + − +( ) −= + − =3 1 1

3x x

⇒ gof (x) = x

Now, fog (y) = f(g(y)) = f y + −

6 13

= 36 1

31 6

2y

y+ −

+

− =

\ fog(y) = y⇒ gof = Ix, fog = Iy ⇒ f is invertible.

⇒ f –1 (y) = g(y) =y + −6 1

3.

13. a * b = a + b – ab ∀ a, b ∈ R – {1} ... (1) (given)Interchanging a and b in (1),

b * a = b + a – ba = a + b – ab ... (2)[Q Addition and multiplication are commutative on R – {1}]From (1) and (2), we have a * b = b * a ∀ a, b ∈ R – {1}\ * is commutative on R – {1}.Associativity : For all a, b, c ∈ R – {1}, we have(a * b) * c = (a + b – ab) * c (By (1))

= (a + b – ab) + c – (a + b – ab) c (By (1))= a + b – ab + c – ac – bc + abc= a + b + c – ab - bc – ac + abc ... (3)

Again a * (b * c) = a * (b + c – bc) (By (1))= a +(b + c – bc) – a (b + c – bc) (By (1))= a + b + c – bc – ab – ac + abc

= a + b + c – ab – bc – ac + abc ... (4)From (3) and (4), we have (a * b) * c = a * (b * c) ∀ a, b, c ∈ R – {1}\ Binary operation * is associative in R – {1}. Identity element : Let e ∈ R – {1} be the identity for binary operation *.\ a * e = a ∀ a ∈ R – {1}⇒ a + e – ae = a [Putting b = e in (1)]or e – ae = 0 or e(1 – a) = 0 ∀ a ∈ R – {1}But 1 – a ≠ 0 [Q a ∈ R – {1} ⇒ a ≠ 1 ⇒ 1 – a ≠ 0] \ e = 0 is the identity element of R – {1}. ... (5)Inverse of an element : Let b ∈ R – {1} be the inverse of an element a ∈ R – {1}.Then a * b = e\ a + b – ab = 0 [By (5)]

or b (1 – a) = –a \ b = −−

=−

aa

aa1 1

Since a ∈ R – {1} ⇒ a ≠ 1 ⇒ a – 1 ≠ 0

Again b ≠ 1 [Q if b = 1 then aa −

=1

1 or

a = a – 1 or 0 = – 1 (Impossible)]

\ b = aa −1

∈ R – {1} is the inverse of a ∈ R – {1}.

14. The set A = {x ∈ Z : 0 < x < 12} = {0, 1,2, ..., 12}(i) R = {(a, b) : |a – b| is a multiple of 4} |a – b| = 4k or b = a + 4k\ R = {(1, 5), (1, 9), (2, 6), (2, 10), (3, 7), (3, 11),

(4, 8), (4, 12), (5, 9), (6, 10), (7, 11), (8, 12), (0, 0), (1, 1), (2, 2), ..., (12, 12)}

(a) a – a = 0 = 4k, where k = 0 ⇒ (a, a) ∈ R, \ R is reflexive.(b) If |a – b| = 4k, then |b – a| = 4k i.e. (a, b) and (b, a)

both belong to R \ R is symmetric.

| june ‘1556

(c) a – c = a – b + b – cWhen a – b and b – c both are multiples of 4 then a – c is also a multiple of 4. This shows if (a, b), (b, c) ∈ R, then a – c also ∈ R. \ R is transitive.

\ R is an equivalence relation. Then set related to 1 are {(1, 5), (1, 9)}(ii) R = {(a, b) : a = b} = {(0, 0), (1, 1), (2, 2),

..., (12, 12)}(a) a = a ⇒ (a, a) ∈ R \ R is reflexive.(b) Again if (a, b) ∈ R, then (b, a) also ∈ R

⇒ R is symmetric.(c) If (a, b) ∈ R and (b, c) ∈ R ⇒ a = b = c \ a = c ⇒ (a, c) ∈ R. Hence R is transitive.

Hence, R is an equivalence relation.Set related to 1 is {1}.

15. Given f xx

g x xx

( ) ( )= = −+

1 11

and

Here, Df = {x ∈ R : x ≠ 0} = R – {0}

and R y R yx

y R xy

Rf = ∈ ={ } = ∈ = ∈ −

: : { }1 1 0

= {y ∈ R : y ≠ 0} = R – {0}Also, Dg = {x ∈ R : 1 + x ≠ 0} = R – {–1}

and R y R y xx

y R xyy

Rg = ∈ = −+{ } = ∈ =

−+

∈ − −

: : { }11

11

1

= {y ∈ R : y ≠ –1} = R – {–1}gof : Here Rf ⊄ Dg, but Rf ∩ Dg ≠ f\ gof is defined and Dgof = {x ∈ Df : f(x) ∈ Dg}

= ∈ − ∈ − −{ }= ∈ ≠ ≠ −{ } = − −

x Rx

R

x R xx

R

{ } : { }

: , { , }

0 1 1

0 1 1 0 1

Also, for all x ∈ R – {0, –1},

(gof)(x) = g(f(x)) =

=−

+= −

+g

xx

x

xx

1 1 1

1 111

.

fog : Here, Rg ⊄ Df, but Rg ∩ Df ≠ f\ fog is defined and Dfog = {x ∈ Dg : g(x) ∈ Df}

= ∈ − − −+

∈ −

x R xx

R{ } : { }1 11

0

= ∈ ≠ − −+

≠

= − −x R x xx

R: , { , }1 11

0 1 1

Also, for all x ∈ R – {–1, 1}

( )( ) ( ( )) .fog x f g x f xx x

x

xx

= = −+

=−+

= +−

11

111

11

| june ‘15 57

1. The angle between the two vectors

i j k i j k^ ^ ^ ^ ^ ^+ + − +and 2 2 2 is equal to

(a) cos−

1 23 (b) cos−

1 16

(c) cos−

1 56

(d) cos−

1 118

(e) cos−

1 13

2. If

a i j k b i j k= + + = + +^ ^ ^ ^ ^ ^, 4 3 4

and c i j k= + +^ ^ â b are coplanar and

c = 3, then

(a) a b= =2 1, (b) a = 1, b = ± 1 (c) a = ± 1, b =1 (d) a = ± 1, b = – 1(e) a = –1, b = ± 1

3. Let P(1, 2, 3) and Q(–1, –2, –3) be the two pointsand let O be the origin. Then | |PQ OP

+ =(a) 13 (b) 14 (c) 24 (d) 12(e) 84. Let ABCD be a parallelogram. If

AB i j k AD i j k

= + + = + −^ ^ ^ ^ ^ ^,3 7 2 3 5 and

p is a unit

vector parallel to AC

, then

p is equal to

(a) 13

2 2( )^ ^ î j k+ + (b) 13

2 2( )^ ^ î j k− +

(c) 17

3 6 2( )^ ^ î j k+ + (d) 17

6 2 3( )^ ^ î j k+ +

(e) 17

6 2 3( )^ ^ î j k+ −

5. Let OB i j k OA i j k

= + + = + +^ ^ ^ ^ ^ ^.2 2 4 2 2andThe distance of the point B from straight line passing

through A and parallel to the vector 2 3 6i j k^ ^ ^+ + is

(a) 7 59

(b) 5 7

9(c) 3 5

7(d) 9 5

7

(e) 9 7

56. If

a i j k b i j k= + + = + +λ λ^ ^ ^ ^ ^ ^2 2 2 2and are atright angle, then the value of | | | |

a b a b+ − − is equal to(a) 2 (b) 1 (c) 0 (d) –1(e) –2

7. Let the position vectors of the points A, B andC be

a b c, and respectively. Let Q be the point ofintersection of the medians of the triangle ABC. Then QA QB QC

+ + =

(a)

a b c+ +2

(b) 2

a b c+ +

(c)

a b c+ + (d)

a b c+ +3

(e)

0

8. The angle between the lines 2x = 3y = – z and6x = – y = – 4z is

(a) p6

(b) p4

(c) p3

(d) p2

(e) 23p

9. The projection of the line segment joining(2, 0, –3) and (5, –1, 2) on a straight line whose direction ratios are 2, 4, 4 is equal to

(a) 116

(b) 103

(c) 133

(d) 136

(e) 113

solved paper 2015

Kerala PET

| june ‘1558

10. The angle between the straight line

r i j k s i j k= + + + − +( ) ( )^ ^ ^ ^ ^ ^2 and the planer i j k⋅ − + =( )^ ^ ^2 4 is

(a) sin−

1 2 23

(b) sin−

1 26

(c) sin−

1 23

(d) sin−

1 23

(e) sin−

1 23

11. If a straight line makes angles a, b, g with the

coordinate axes, then 1

112

22

22−

++ − =

tan

tan secsin

a

a bg

(a) –1 (b) 1 (c) –2 (d) 2(e) 0

12. The equation of the plane which bisects the linesegment joining the points (3, 2, 6) and (5, 4, 8) and is perpendicular to the same line segment, is(a) x + y + z = 16 (b) x + y + z = 10(c) x + y + z = 12 (d) x + y + z = 14(e) x + y + z = 15

13. The foot of the perpendicular from the point

(1, 6, 3) to the line x y z1

12

23

=−

=−

is

(a) (1, 3, 5) (b) (–1, –1, –1)(c) (2, 5, 8) (d) (–2, –3, –4)

(e) 12

2 12

, ,

14. The plane x + 3y + 13 = 0 passes through the lineof intersection of the planes 2x – 8y + 4z + p = 0 and 3x – 5y + 4z – 10 = 0. If the plane is perpendicular to the plane 3x – y – 2z – 4 = 0, then the value of p is equal to(a) 2 (b) 5 (c) 9 (d) 3(e) –1

15. If a straight line makes the angles 60°, 45° and awith x, y and z axes respectively, then sin2a =

(a) 34

(b) 32

(c) 12

(d) 1

(e) 14

16. Let A and B be two events. Then 1 + P(A ∩ B) –P(B) – P(A) is equal to(a) P A B( )∪ (b) P A B( )∩(c) P A B( )∩ (d) P A B( )∪(e) P A B( )∩

17. The mean of five observations is 4 and theirvariance is 5.2. If three of these observations are 2, 4, and 6, then the other two observations are(a) 3 and 5 (b) 2 and 6(c) 4 and 4 (d) 8 and 10(e) 1 and 7

18. The first term of an A.P. is 148 and the commondifference is –2. If the A.M. of first n terms of the A.P. is 125, then the value of n is(a) 18 (b) 24 (c) 30 (d) 36(e) 48

19. If the combined mean of two groups is 403

and if

the mean of one group with 10 observations is 15, then the mean of the other group with 8 observations is equal to

(a) 463

(b) 354

(c) 454

(d) 414

(e) 434

20. A function f satisfies the relation f(n2) = f(n) + 6 forn ≥ 2 and f(2) = 8. Then the value of f(256) is(a) 24 (b) 26 (c) 22 (d) 28(e) 32

21. lim sinsin

sinsin

sinsinx

xx

xx

xx→

0

109

910

87

78

65

56

=

43

34 2

sinsin

sinsin

xx

xx

(a) 63256

(b) 16

(c) 65

(d) 12

(e) 25663

22. The number of points at which the function

f xxe

( )log | |

= 1 is discontinuous, is

(a) 1 (b) 2 (c) 3 (d) 4(e) infinitely many

23. The value of lim siny

yy y→∞

−

1 1 is equal to

(a) 1 (b) ∞ (c) –1 (d) 0(e) –∞

24. limx

xx

x→∞ −

−

=

2

3 2 3

(a) 13

(b) 23

(c) −23

(d) −29

(e) 29

| june ‘15 59

25. The functions f, g and h satisfy the relationsf ′(x) = g(x + 1) and g′(x) = h(x – 1). Then f ′′(2x) is equal to(a) h(2x) (b) 4h(2x)(c) h(2x – 1) (d) h(2x + 1)(e) 2h′(2x)

26. If f xx

x( ) sin

cossin

,=−

−1 1 22

then |f ′(x)| is equal to(a) |sinx| (b) x(c) 0 (d) |cosx|(e) 127. If y2 = 100 tan–1x + 45 sec–1x + 100 cot–1x +45 cosec–1x, then dy

dx=

(a) x

x

2

21

1

−

+(b)

x

x

2

21

1

+

−

(c) 1 (d) 0

(e) 1

12x x −

28. If f(x) = 3x2 – 7x + 5, then lim( ) ( )

x

f x fx→

−0

0 is equal

to(a) 6 (b) –7 (c) 7 (d) –6(e) 5

29. If y = sec(tan–1x), then dydx

is equal to

(a) x

x1 2+(b) x x1 2+

(c) 1 2+ x (d) 1

1 2+ x(e) x

x1+30. If then| | , sin , tan ,t x t

ty t

tdydx

< =+

=−

=1 21

212 2

(a) 1x

(b) 12

(c) − 12

(d) − 1x

(e) 1

31. If y xx

xx

d ydx

=+

++

11 2

2, then at x = 1 is equal to

(a) 74

(b) 78

(c) 14

(d) −78

(e) −74

32. Let f(x) = (3sin2(10x + 11) – 7)2 for x ∈ R. Then themaximum value of the function f, is(a) 9 (b) 16 (c) 49 (d) 100(e) 121

33. If a circular plate is heated uniformly, its areaexpands 3c times as fast as its radius, then the value of c when the radius is 6 units, is(a) 4p (b) 2p (c) 6p (d) 3p(e) 8p34. The slope of the tangent to the curve y = 3x2 – 5x + 6at (1, 4) is(a) –2 (b) 1 (c) 0 (d) –1(e) 2

35. The chord joining the points (5, 5) and (11, 227)on the curve y = 3x2 – 11x – 15 is parallel to tangent at a point on the curve. Then the abscissa of the point is(a) –4 (b) 4 (c) –8 (d) 8(e) 6

36. The function f(x) = sinx – kx – c, where k and c areconstants, decreases always when(a) k > 1 (b) k ≥ 1 (c) k < 1 (d) k ≤ 1(e) k < –1

37. If y = mlogx + nx2 + x has its extreme values atx = 2 and x = 1, then 2m + 10n =(a) –1 (b) –4 (c) –2 (d) 1(e) –3

38. If s = 2t3 – 6t2 + at + 5 is the distance travelledby a particle at time t and if the velocity is –3 when its acceleration is zero, then the value of a is(a) –3 (b) 3 (c) 4 (d) –4(e) 2

39. ( )

cot( )

1+∫

x e

xedx

x

x is equal to

(a) log |cos(xex)| + C (b) log|cot(xex)| + C

(c) log|sec(xe–x)| + C (d) log |cos(xe–x)| + C

(e) log |sec(xex)| + C

40. x

xdx

5

31+=∫

(a) 29

1 92 3+ − +x x C( ) (b) 29

9 13 2x x C− + +( )

(c) 29

1 3+ +x C (d) 29

1 23 3+ − +x x C( )

(e) 29

1 92 3+ + +x x C( )

| june ‘1560

41. 42 5

ee e

dxx

x x−=−∫

(a) 4log|ex – 5| + C (b) 14

52log | |e Cx − +

(c) log|2ex – 5e–x| + C (d) 4log|2ex – 5| + C(e) log|2e2x – 5| + C

42. xx

dx+

=∫1 2

(a) x x x C2

22+ + +log | |

(b) x x C2

22+ + +log | |

(c) x x x C2

2+ + +log | |

(d) x x x C2

22 2+ + +log | |

(e) x x x C2

22− + +log | |

43. xx a

dxn

n

−

+=∫

1

2 2

(a) 1 1na

xa

Cn

tan−

+ (b) n

axa

Cn

tan−

+

1

(c) na

xa

Cn

sin−

+

1 (d) na

xa

Cn

cos−

+

1

(e) 1 1na

xa

Cn

cot−

+

44. ( )

( )

x

x xdx

+

+=∫

1

1

2

2

(a) log |x(x2 + 1)| + C(b) log|x| + C(c) log|x| + 2 tan–1x + C

(d) log 11 2+

+

xC

(e) 2 log|x| + tan–1x + C

45. 1

2x xdx

log=∫

(a) 12

2log | log |x C+ (b) log|logx2| + C

(c) 2log|logx2| + C (d) 4log|logx2| + C

(e) 14

2log | log |x C+

46. 116 252 2

0

1

( )( )x xdx

+ +=∫

(a) 15

14

14

15

15

1 1tan tan− −

−

(b) 19

14

14

15

15

1 1tan tan− −

−

(c) 14

14

14

15

15

1 1tan tan− −

−

(d) 19

15

14

14

15

1 1tan tan− −

−

(e) 19

34

14

45

15

1 1tan tan− −

−

47. x x x dx( )( )1 11

1− + =

−∫

(a) 13

(b) 23

(c) 1 (d) –1

(e) 0

48. The area bounded by y = x2 + 3 and y = 2x + 3 is (insq. units)

(a) 127

(b) 43

(c) 34

(d) 83

(e) 38

49. The value ofsin2

1 7x dxx+−

∫p

p is equal to

(a) 7p (b) p (c) p2

(d) 2p

(e) 7p

50. 2 3

0

22x x dx

p /sin( )∫ =

(a) 12

14

+

p(b)

12

14

−

p

(c) 12 2

1p −

(d) 1

21

2−

p

(e) 12 4

1p −

51. If x2 + y2 = 1, then(a) yy′′ + (y′)2 + 1 = 0 (b) yy′′ + 2(y′)2 + 1 = 0(c) yy′′ – 2(y′)2 + 1 = 0 (d) yy′′ + (y′)2 – 1 = 0(e) yy′′ – (2y′)2 – 1 = 0

| june ‘15 61

52. The solution of the differential equation y′(y2 – x) = yis(a) y3 – 3xy = C (b) y3 + 3xy = C(c) x3 – 3xy = C (d) y3 – xy = C(e) x3 – xy = C53. The order and degree of the differential equation

22

2

2 3 2 3

3d ydx

dydx

d ydx

+

=

/

are respectively

(a) 2 and 2 (b) 2 and 1(c) 3 and 2 (d) 3 and 3(e) 2 and 454. The slope of a curve at any point (x, y) other thanthe origin, is y y

x+ . Then the equation of the curve is

(a) y = Cxex (b) y = x(ex + C)(c) xy = Cex (d) y + xex = C(e) (y – x)ex = C

55. The domain of the function f x x xe( ) log= − +7 3is(a) 0 < x < ∞ (b)

73≤ < ∞x

(c) 0 73

< ≤x (d) –∞ < x < 0

(e) −∞ < ≤x 73

56. If f(1) = 1, f(2n) = f(n) and f(2n + 1) = (f(n))2 – 2 forn = 1, 2, 3, ......, then the value of f(1) + f(2) + .... + f(25) is equal to(a) 1 (b) –15 (c) –17 (d) –1(e) 1357. Let X and Y be two non-empty sets such thatX ∩ A = Y ∩ A = f and X ∪ A = Y ∪ A for some non-empty set A. Then(a) X is a proper subset of Y(b) Y is a proper subset of X(c) X = Y(d) X and Y are disjoint sets(e) X|A = f58. The set (A|B) ∪ (B|A) is equal to(a) [A|(A ∩ B)] ∩ [B|(A ∩ B)](b) (A ∪ B)|(A ∩ B) (c) A|(A ∩ B)(d) ( ) |( )A B A B∩ ∪ (e) ( ) |( )A B A B∪ ∪

59. The range of the function f xx

( )cos

=−

12 3

is

(a) (–2, ∞) (b) [–2, 3]

(c) 13

2,

(d) 1

21,

(e) 13

1,

60. In a class of 80 students numbered 1 to 80, all oddnumbered students opt for Cricket, students whose numbers are divisible by 5 opt for Football and those whose numbers are divisible by 7 opt for Hockey. The number of students who do not opt any of the three games, is(a) 13 (b) 24 (c) 28 (d) 52(e) 6761. If Re(1 + iy)3 = – 26, where y is a real number, thenthe value of |y| is(a) 2 (b) 3 (c) 4 (d) 6(e) 962. If z = x + iy is a complex number such that|z| = Re(iz) + 1, then the locus of z is(a) x2 + y2 = 1 (b) x2 = 2y – 1(c) y2 = 2x – 1 (d) y2 = 1 – 2x(e) x2 = 1 – 2y63. Let i2 = –1. Then

ii

ii

ii

ii

1011

1112

1213

1314

1 1 1

1

−

+ −

+ −

+ −

+ +

=i

i14

151

(a) –1 + i (b) –1 – i(c) 1 + i (d) –i(e) i

64. If f zzz

( ) ,=−−

11

3

where z = x + iy with z ≠ 1, then

Re{ ( )}f z = 0 reduces to(a) x2 + y2 + x + 1 = 0 (b) x2 – y2 + x – 1 = 0(c) x2 – y2 – x + 1 = 0(d) x2 – y2 + x + 1 = 0(e) x2 – y2 + x + 2 = 065. If z = 1 + i, then the argument of z2ez–i is

(a) p2

(b) p6

(c) p4

(d) p3

(e) 066. If the difference between the roots ofx2 + 2px + q = 0 is two times the difference between the

roots of x qx p24

0+ + = , where p ≠ q, then

(a) p – q + 1 = 0 (b) p – q – 1 = 0(c) p + q – 1 = 0 (d) p + q + 1 = 0(e) q – 4p + 1 = 067. Sum of the roots of the equation|x – 3|2 + |x – 3| – 2 = 0 is equal to(a) 2 (b) 4 (c) 6 (d) 16

| june ‘1562

(e) –2

68. The quadratic equation whose roots are three timesthe roots of the equation 2x2 + 3x+ 5 = 0, is(a) 2x2 + 9x + 45 = 0 (b) 2x2 + 9x – 45 = 0(c) 5x2 + 9x + 45 = 0 (d) 2x2 – 9x + 45 = 0(e) 2x2 + 9x + 49 = 0

69. If x is real number, thenx

x x2 5 9− + must lie

between

(a) 1

111and (b) −1 1

11and

(c) –11 and 1 (d) 1 and 11

(e) − 111

1and

70. If the roots of the equation (x – a)(x – b) + (x – b)(x – c) + (x – c)(x – a) = 0 are equal then a2 + b2 + c2 =(a) a + b + c (b) 2a + b + c(c) 3abc (d) ab + bc + ca(e) abc

71. If one root of a quadratic equation is 1

1 3+, then

the quadratic equation is(a) 2x2 + x – 1 = 0 (b) 2x2 – 2x – 1 = 0(c) 2x2 + 2x + 1 = 0 (d) 2x2 + x + 1 = 0(e) 2x2 + 2x – 1 = 0

72. The 5th and 8th terms of a G.P. are 1458 and 54respectively. The common ratio of the G.P. is(a) 1

3 (b) 3 (c) 9 (d) 1

9

(e) 18

73. Let S(n) denote the sum of the digits of a positiveinteger n. For example, S(178) = 1 + 7 + 8 = 16. Then the value of S(1) + S(2) + S(3) + ...... + S(99) is equal to(a) 476 (b) 998 (c) 782 (d) 900

74. An A.P. consists of 23 terms. If the sum of the 3terms in the middle is 141 and the sum of the last 3 terms is 261, then the first term is(a) 6 (b) 5 (c) 4 (d) 3(e) 2

75. If 4th term of a G.P. is 32 whose common ratio ishalf of the first term, then the 15th term is(a) 212 (b) 218 (c) 214 (d) 216

(e) 210

76. Let S1 be a square of side 5 cm. Another squareS2 is drawn by joining the midpoints of the sides of S1. Square S3 is drawn by joining the midpoints of the sides of S2 and so on. Then (area of S1 + area of S2 + area of S3 + ..... + area of S10) =

(a) 25 1 1210−

(b) 50 1 1210−

(c) 2 1 1210−

(d) 1 1210−

(e) 10 1 1210−

77. If a, b, c are in A.P. and if their squares taken in thesame order form a G.P., then (a + c)4 =(a) 16a2c2 (b) 4a2c2

(c) 8a2c2 (d) 2a2c2

(e) a2c2

78. The value of x satisfying the relation 11(xC3) = 24((x + 1)C2) is(a) 8 (b) 9 (c) 11 (d) 10(e) 1279. The power of x in the term with the greatest

coefficient in the expansion of 12

10+

x is

(a) 2 (b) 3 (c) 4 (d) 5(e) 680. The number of 5-digit numbers (no digit isrepeated) that can be formed by using the digits 0, 1, 2, ...., 7 is(a) 1340 (b) 1860 (c) 2340 (d) 2160(e) 320081. If m1 and m2 satisfy the relationm

mm

mP m P++

+= −51

3112

1( )( ), then m1 + m2 is equal to

(a) 10 (b) 9 (c) 13 (d) 17(e) 1582. Sum of coefficients of the last 6 terms in theexpansion of (1 + x)11 when the expansion is in ascending powers of x, is(a) 2048 (b) 32 (c) 512 (d) 64(e) 102483. If C0, C1, C2, ......, C15 are binomial coefficients in

(1 + x)15, then CC

CC

CC

CC

1

0

2

1

3

2

15

142 3 15+ + + + =.....

(a) 60 (b) 120 (c) 64 (d) 124(e) 144

84. If and then∆ ∆= ′ =1 2 32 3 53 6 12

4 8 153 6 122 3 5

,

(a) D′ = 2D (b) D′ = –2D(c) D′ = D (d) D′ = –D(e) D′ = 3D

| june ‘15 63

85. The roots of the equation1 3 5

2 2 52 3 4

0+

++

=x

xx

are(a) 2, 1, –9 (b) 1, 1, –9(c) –1, 1, –9 (d) –2, –1, –8(e) –2, 1, 1

86. If0 3 22 0 14 1 6

b

−

is singular, then the value of b is

equal to(a) –3 (b) 3 (c) –6 (d) 6(e) –2

87. If Ax xx

=−

12 1

and if det A = –9, then the

values of x are

(a) 32

3, − (b) −23

3,

(c) 23

3, (d) −32

3,

(e) − −32

3,

88. The value of the determinant

cos cos cot

sin cot sin

cot cos co

2 2

2 2

2

54 36 135

53 135 37

135 25

° ° °

° ° °

° ° ss2 65°

is equal to

(a) –2 (b) –1 (c) 0 (d) 1(e) 2

89. If A and B are square matrices of the same orderand if A = AT, B = BT, then (ABA)T =(a) BAB (b) ABA(c) ABAB (d) ABT

(e) (AB)T

90. If |x – 3| < 2x + 9, then x lies in the interval(a) (–∞, –2) (b) (–2, 0)(c) (–2, ∞) (d) (2, ∞)(e) (–12, –2)

91. The area and perimeter of a rectangle are A and Prespectively. Then P and A satisfy the inequality(a) P + A > PA (b) P2 ≤ A(c) A – P < 2 (d) P2 ≤ 4A(e) P2 ≥ 16A

92. For any two statements p and q, the statement~(p ∨ q) ∨(~ p ∧ q) is equivalent to(a) ~p (b) p(c) q (d) ~q(e) p ∨q

93. Let p, q, r be three statements. Then ~ (p ∨(q ∧r))is equal to(a) (~p ∧~ q) ∧(~ p ∧~ r)(b) (~p ∨~ q) ∧(~ p ∨~ r)(c) (~p ∧~ q) ∨(~ p ∧~ r)(d) (~p ∨~ q) ∨(~ p ∧~ r)(e) (~p ∧~ q) ∨(~ p ∨~ r)

94. Consider the two statements P : He is intelligentand Q : He is strong. Then the symbolic form of the statement "It is not true that he is either intelligent or strong" is(a) ~ P ∨Q (b) ~ P ∨~ Q(c) ~ P ∧Q (d) P ∨ ~ Q(e) ~ (P ∨Q)

95. If tan ,θ2

12

= then the value of sinq is

(a) 45

(b) 35

(c) 12

(d) 1

(e) 25

96. If x = 5 + 2secq and y = 5 + 2tanq, then(x – 5)2 – (y – 5)2 is equal to(a) 3 (b) 1 (c) 0 (d) 4(e) 2

97. The value of tan 15° + tan 75° is equal to

(a) 2 3 (b) 2

(c) 2 3− (d) 4 3(e) 4

98. The period of the function f(x) = cos4x + tan3x is

(a) π

12(b)

π6 (c)

π2

(d) p

(e) 2p

99. If sin(q + f) = n sin(q – f), n ≠ 1, then the value oftantan

θφ

is equal to

(a) n

n − 1(b)

nn

+−

11

(c) nn1 −

(d) nn

−+

11

(e) 11

+−

nn

| june ‘1564

100. 2 13

14

1 1tan tan− −

+

=

(a) tan−

1 1613 (b) tan−

1 1723

(c) π4

(d) 0

(e) tan−

1 1112

101. If cos cos ,− −+ =1 1 27

x y πthen the value of

sin–1x + sin–1y is equal to

(a) 47π (b)

37π

(c) 27π (d)

67π

(e) 57π

102. If cos–1x > sin–1x, then x lies in the interval

(a) 12

1,

(b) (0, 1]

(c) −

1 12

, (d) [–1, 1]

(e) [0, 1]

103. If 0 ≤ x ≤ 2p, then the number of solutions of the equation sin8x + cos6x = 1 is(a) 2 (b) 3 (c) 4 (d) 5(e) 8

104. Let A(0, 0) and B(8, 0) be two vertices of a right angled triangle whose hypotenuse is BC. If the circumcentre is (4, 2), then the point C is(a) (2, 4) (b) (0, 8) (c) (0, 4) (d) (0, 6)(e) (2, 8)

105. If coordinates of the circumcentre and the orthocentre of a triangle are respectively (5, 5) and (2, 2), then the coordinates of the centroid are(a) (1, 1) (b) (3, 1) (c) (3, 3) (d) (2, 2)(e) (4, 4)

106. If the points A(3, 4), B(x1, y1) and C(x2, y2) are such that both 3, x1, x2 and 4, y1, y2 are in A.P., then(a) A, B, C are vertices of an isosceles triangle(b) A, B, C are collinear points(c) A, B, C are vertices of a right angled triangle(d) A, B, C are vertices of a scalene triangle(e) A, B, C are vertices of an equilateral triangle

107. If p1 and p2 are respectively lengths of perpendicular from the origin to the straight lines x secq + y cosecq = a and x cosq – y sinq = a cos2q, then 4p1

2 + p22 =

(a) 1 (b) a2 (c) 12a

(d) a

(e) 1a

108. The distance between the point (1, 2) and the point of intersection of the lines 2x + y = 2 and x + 2y = 2 is

(a) 173

(b) 163

(c) 175

(d) 193

(e) 195

109. If a straight line is perpendicular to 2x + 8y = 10 and meets the x-axis at (5, 0), then it meets the y-axis at(a) (0, –2) (b) (0, –8)(c) (0, –10) (d) (0, –16)(e) (0, –20)

110. If the straight line 5x + y = k forms a triangle with the coordinate axes of area 10 sq. units, then the values of k are(a) ±15 (b) ±10 (c) ±5 (d) ±20(e) ±7

111. A circle passes through the point (6, 2). If segments of the straight lines x + y = 6 and x + 2y = 4 are two diameters of the circle, then its radius is

(a) 4 (b) 8 (c) 5 (d) 2 5

(e) 4 5

112. The parametric form of equation of the circle x2 + y2 – 6x + 2y – 28 = 0 is

(a) x y= − + = − +3 38 1 38cos , sinθ θ

(b) x y= =28 28cos , sinθ θ

(c) x y= − − = − +3 38 1 38cos , sinθ θ

(d) x y= + = − +3 38 1 38cos , sinθ θ

(e) x = 3 + 38 cosq, y = –1 + 38 sinq

113. The point on the circle (x – 1)2 + (y – 1)2 = 1 which is nearest to the circle (x – 5)2 + (y – 5)2 = 4 is

(a) (2, 2) (b) 32

32

,

(c) 2 1

22 1

2+ +

, (d) ( , )2 2

(e) ( , )1 2 1 2+ +

| june ‘15 65

114. The line segment joining (5, 0) and (10 cosq, 10 sinq) is divided internally in the ratio 2 : 3 at P. If q varies, then the locus of P is(a) a pair of straight lines(b) a circle(c) a straight line(d) a parabola(e) an ellipse

115. ABCD is a square with side a. If AB and AD are along the coordinate axes, then the equation of the circle passing through the vertices A, B and D is

(a) x y a x y2 2 2+ = +( )

(b) x y a x y2 22

+ = +( )

(c) x2 + y2 = a(x + y)(d) x2 + y2 = a2(x + y)(e) a(x2 + y2) = x + y

116. The distance between the directrices of the hyperbola x2 – y2 = 9 is

(a) 92

(b) 53

(c) 32

(d) 3 2

(e) 5 3117. If the line y = kx touches the parabola y = (x – 1)2, then the values of k are(a) 2, –2 (b) 0, 4(c) 0, –2 (d) 0, 2(e) 0, –4

118. If the semi-major axis of an ellipse is 3 and the latus

rectum is 169

, then the standard equation of the ellipse is

(a) x y2 2

9 81+ = (b)

x y2 2

8 91+ =

(c) x y2 2

93

81+ = (d) 3

8 91

2 2x y+ =

(e) x y2 2

98

31+ =

119. If a point P(x, y) moves along the ellipse

x y2 2

25 161+ = and if C is the centre of the ellipse, then,

4max{CP} + 5min{CP} =(a) 25 (b) 40 (c) 45 (d) 54(e) 16

120. The one end of the latus rectum of the parabola y2 – 4x – 2y – 3 = 0 is at(a) (0, –1) (b) (0, 1)(c) (0, –3) (d) (3, 0)(e) (0, 2)

solutions

1. (e) : Let a i j k b i j k= + + = − +, 2 2 2

If q is angle between two vectors, then

cos| | | |

θ =⋅

a b

a b

= − + =⋅

= =2 2 23 12

23 2 3

26

13

⇒ =

−θ cos 1 13

2. (c) : Since given vectors are coplanar

⇒ 1 1 14 3 41

0 1α β

β= ⇒ =

⇒ =Since | |c 3

⇒ + + = ⇒ + = ⇒ =1 3 2 12 2 2 2 2α β α β α\ a = ±1

3. (b) : PQ OQ OP PQ OP OQ

= − ⇒ + =

∴ + = = + + =PQ OP OQ

( ) ( ) ( )1 2 3 142 2 2

4. (c) : AD BC i j k = = + −2 3 5

AC AB BC i j k = + = + +3 6 2

∴ =p ACAC

| |

A B

CD

= + +17

3 6 2( )i j k

5. (d) : Perpendicular distance from a r a b2 1o isn = + λ

da a b

b=

− ×| ( ) || |

2 1

where a i j k1 4 2 2= + + a i j k2 2 2= + +

b i j k= + +2 3 6

| |b = + + =4 9 36 7

| june ‘1566

( )

a a bi j k

j k2 1 3 0 02 3 6

18 9− × = = − +

| ( ) | ( ) ( ) a a b2 1

2 218 9 405− × = + =

6. (c)

7. (e) :

Q a b c= + +3

∴ = − + + = − −QA a a b c a b c

32

3

∴ + + =QA QB QC

0

8. (d) : Given lines arex y z x y z12

13

1 16

1 14

= =−

=−

=−

and

Let <a1, b1, c1 > = < − >12

13

1, ,

and <a2, b2, c2 > = < − − >16

1 14

, ,

are direction ratios of given lines.

since 12

16

13

1 1 14

0.

+ × −

+ − × −

= ⇒ θ π=

2

9. (e) : Projection of a bon

= ⋅ = − + ⋅ + +

+ += =

a bb

i j k i j k| |

( ) ( )

( ) ( ) ( )

3 1 5 2 4 4

2 4 4

226

12 2 2

113

10. (a) : If q is the required angle, then

sinq = cosine of angle between i j k i j k − + − +and 2

= 2 + + =1 13 6

2 23

⇒ θ =

−sin 1 2 2

3

11. (c) : 11

12

22

22−

++ − =tan

tan secsinα

α βy

= cos2a + cos2b + 1 – 2sin2y – 1Since cos2a + cos2b + cos2y = –1\ Required value = –2

12. (d) : Midpoint of line joining (3, 2, 6) and (5, 4, 8)is (4, 3, 7) satisfying x + y + z = 14

13. (a) : Let Q be the foot of ^r PQ

lie on line x y z1

12

23

=−

= −P

QA B

(1, 6, 3)

x y – 1 z – 21 2 3= =

Let x y z k1

12

23

=−

= − = (say)

⇒ x = k, y = 2k + 1, z = 3k + 2⇒ Point Q = (k, 2k + 1, 3k + 2) lie on line AB.Since PQ ^AB\ (k – 1)·1 + (2k + 1 – 6)·2 +(3k + 2 – 3)·3 = 0⇒ k – 1 + 2(2k – 5) + (3k – 1)·3 = 0⇒ k – 1 + 4k – 10 + 9k – 3 = 0⇒ 14k – 14 = 0 ⇒k = 1\ Required point Q = (1, 3, 5)14. (d)15. (a) : cos260° + cos245° + cos2a = 1

⇒14

12

12+ + =cos α

⇒ 1 34

2− =cos α ⇒ sin2 34

α =

16. (e) : P(A ∪ B) = P(A) + P(B) – P(A ∩ B)⇒ P(A) + P(B) = P(A ∪ B) + P(A ∩ B)\ 1 + P(A ∩ B) – P(B) – P(A)

= 1 + P(A ∩ B) – P(A ∪ B) – P(A ∩ B)= 1 – P(A ∪ B) = P(A ∪ B)c = ∩P A B( )

17. (e) : Let the unknown data be a and b

Variance .= − =∑1 5 22 2n

x x

⇒ + + + + − =15

4 16 36 16 5 22 2( ) .a b

⇒ a2 + b2 = 50 ...(i)Also 2 + 4 + 6 + a + b = 20 ⇒ a + b = 8 ...(ii)It is clear that the two observations are 1 and 7which statisfies both (i) and (ii).18. (b) : a = 148 ; d = –2sum of first n terms = + − −n n

2296 1 2( ( ) ( ))

Since A. M. = 125

n⇒ sum of n terms = n × 125\ 298 – 2n = 250 ⇒ n = 2419. (c) : There are a total of 18 observations

\ combined mean =× + ×( ) ( )a a1 210 8

18where a1 and a2 denote the means of two groups respectively.

But a1 = 15 ⇒ =+

⇒ =403

150 818

454

22

aa

| june ‘15 67

20. (b) : f(256) = f(162) = f(16) + 6f(16) = f(42) = f(4) + 6, f(4) = f(22) = f(2) + 6But f(2) = 8\ f(4) = 14 ; f(16) = 20 and f(256) = 26

21. (d) : 109

910

87

78

65

56

43

34

12

12

. . . . . . . . =

Using limx

xx→

=

0

1sin

22. (b) : The domain is not explicitly given; it isR′ –{0, ±1} and the function is continuous on its domain.

23. (a) : Required limit limsin

=

−

= − =→ ∞y

y

yy

1

11 1 0 1

24. (e) : lim( )

limx x

xx

x→ ∞ → ∞−

=−

=23 3 2

2

3 3 229

25. (a) : Given, f ′(x) = g(x + 1)⇒ f ′′(x) = g′(x + 1). Also, g ′(x) = h(x – 1)⇒ g′(x + 1) = h(x) ⇒ f′′(x) = h(x)⇒ f′′(2x) = h(2x)

26. (e) : f x xx

( ) sin cossin

= −

−1 1 22

= sin–1 (sin x) (using cos2x = 1 – 2sin2x)⇒ f ′(x) = ±1 ⇒ | f ′(x)| = 1

27. (d) : y2 = 100 + 45 = 145(∵ tan–1x + cot–1x = 1, sec–1x + cosec–1x = 1)

⇒ 2yy′ = 0 ⇒ y′ = 0 (∵ y ≠ 0)

28. (b) : The required limit = − = −→

lim ( )x

x0

3 7 7

29. (a) : y = sec(tan–1x)

= +

−sec sec 1 21 x

⇒ y xdydx

x

x= + ⇒ =

+1

1

22

30. (e) : If t = tan q, then

sin tantan

sin , tan tantan

tanx y=+

= =−

=21

2 21

22 2θ

θθ θ

θθ

\ x = 2q, y = 2q

⇒ y = x ⇒ =dydx

1

31. (a) : We have, yx x

= −+

+ +1 11

1 1

⇒ dydx x x

=+

−11

12 2( )

⇒ d ydx x x

2

2 3 32

12= −

++

( )

= − + =14

2 1at x =74

32. (c) : [3 sin2 (10x + 11) – 7]2 is maximumWhen sin2 (10x + 11) is minimum∵ Minimum value of 3sin2(10x + 11) = 0\ Maximum value = (–7)2 = 49. 33. (a) : If A sq. units is the area measure when theradius is r units, then A = pr2

∴ =dAdt

r drdt

2π But dAdt

c drdt

= 3

\ 2pr = 3c \ c = 4p, when r = 6

34. (b) : dydx

x= −6 5

= 6 – 5 = 1 at (1, 4)

35. (d) : 6 11 227 511 5

2226

37x − = −−

= =

⇒ 6x = 48 ⇒ x = 836. (b) : f ′(x) = cosx – kHence, f decreases if cos x ≤ k⇒ for decreasing k ≥ 1 [Q –1 ≤ cosx ≤ 1]37. (e) : We have, y = mlogx + nx2 + x

⇒ dydx

mx

nx= + +2 1

⇒ m n2

4+ + 1 = 0 at x = 2 …(i)

m + 2n + 1 = 0 at x = 1 …(ii)from (i) and (ii), we get 6n + 1 = 0

⇒ = − = −n m16

23

&

⇒ 2 10 43

53

3m n+ = − − = −

38. (b) : Acceleration = d sdt

2

2 = 12t – 12

acceleration is zero when t = 1

velocity at time t = dsdt

= 6t2 – 12t + a\ –3 = 6 –12 + a⇒ a = 3

| june ‘1568

39. (e) :( )

cot( ) cotif

1 += =∫ ∫

x e dx

x edu

uu x e

x

xx

= log |sec u| + C= log |sec (x ex)| + C

40. (d) : x dx

x

uu

du u x5

33

1

13

1 1+

= − = +∫∫( ) if

= −

∫

13

1uu

du

= −

+13

23

23 2 1 2u u C/ /

= + + − +2 19

1 33

3x x C( )

= + − +29

1 23 3x x C( )

41. (e) : 42 5

42 5

2

2e

e edx e

edx

x

x x

x

x−=

−−∫ ∫

= = −∫duu

u e xif 2 52

= log |u| + C= log |2 e2x – 5| + C

42. (a) : xx

dx xx

dx+

= + +

∫∫

1 2 12

= + + +12

22x x x Clog | |

43. (a) :x

x adx

ndu

a uu x

n

nn

−

+=

+=∫∫

1

2 2 2 21 ; if

= +−1 1 1n a

ua

C. tan =

+−1 1

naxa

Cn

tan

44. (c) : ( )( )x

x xdx

x xdx+

+= +

+

∫∫

11

1 21

2

2 2

= log|x| + 2tan–1x + C

45. (e) : dxx x

duu

u xlog

(if log )2 2= =∫∫

= + ′12

log | |u C = + ′12

log | log |x C

= +14

2log | log |x C

46. (b) : dxx x( ) ( )2 2

0

1

16 25+ +∫

=+

−+

∫19

116

1252 2

0

1

( ) ( )x xdx

= −

− −19

14 4

15 5

1 1

0

1. tan tanx x

= −

− −19

14

14

15

15

1 1tan tan

47. (e) : x x x dx( )( )1 11

1− +

−∫ = 0

Using property f x dxa

a( ) ,=

−∫ 0 if f(x) is an odd function.

Since f(x) = x(1 – x)(1 + x) is an odd function[f(–x) = –x(1 + x)(1 – x) = –f(x)]48. (b) : Point of intersect ion of y = x2 + 3 &y = 2x + 3 are x = 0 or 2

\ Required area measure = −∫ ( )x x dx2

0

22

= − =83

4 43

sq. units

49. (c)

50. (b) : 2 3 2

0

2

0

4x x dx x x dxsin sin

/ /π π

∫ ∫=

(By Putting x2 = t)= − − −[ ( cos ) ( sin )] /x x x 0

4π

(Using Intergration by parts)

= − +π4 2

12

= −

12

14π

51. (a) : x2 + y2 = 1Differentiating w.r.t. 'x', we get 2x + 2yy′ = 0Again differentiating w.r.t. x, we get

1 + (y′)2 + yy′′ = 0 52. (a) : We have, y′(y2 – x) = y

⇒ dxdy y

x y+ =1

I.F. = e ydy1

∫ = y

| june ‘15 69

Required solution is x y y dy A⋅ = +∫ 2

or 3xy = y3 – C53. (c) : On squaring the equation, we get

2 2

2

2 3 3

3

2d y

dx

dydx

d y

dx+

=

\ Order = 3, degree = 2

54. (a) : We have, dydx

y yx

= +

⇒dyy

dxx

dx= +

⇒ log |y| = log|x| + x + a⇒ y = elog|x|·ex·ea = Cx ex

55. (c) : For function to be defined,

7 3 0 73

0− ≥ ⇒ ≤ >x x xand

⇒ < ≤0 73

x

56. (b) : f(2n + 1) = –1 for all nQ f(4) = f(8) = f(16) = 1f(1) = f(2) = 1 and f(2n) = 1f(2n) = –1 when n is odd\ sum = 5 – 20 = –15.57. (c)58. (b)59. (e) : Since, –1 ≤cos 3x ≤ 1⇒ 1 ≤ 2 – cos 3x ≤ 3⇒ 1

31

2 31≤

−≤

cos x60. (c) : n(C) = 40, n(F) = 16, n(H) = 11n(C ∩ F) = 8, n(C ∩ H) = 6n(F ∩ H) = 2, n(C ∩ F ∩ H) = 1n(C ∪ F ∪ H) = 67 –16 + 1

= 52n(C ′ ∩ F ′ ∩ H ′) = 80 – 52 = 2861. (b) : Re(1 + iy)3 =1 – 3y2 = –26⇒ y2 = 9 ⇒|y| = 3.

62. (e) : x y y2 2 1+ = − +

⇒ x2 + y2 = y2 – 2y + 1⇒ x2 = 1 – 2y.63. (a) : i10 + i11 + i12 + i13 = 0

and 1 1 1 1 011 12 13 14i i i i+ + +

=

\ Remaining expression = ii

ii

1415

23

1 1+ = + = –1 + i

64. (d) : f(z) = 1 + z + z2

Re(f(z)) = 1 + x2 – y2 + xRe ( ( ))f z x y x= ⇒ − + + =0 1 02 2

65. (a) : z2 = 2iz2 ez – i = 2i e1 + i –i

= 2i e = 2e[i]

\ Arg of z ez i22

− = π

66. (d) : 4 4 22 2p q q p− = −⇒ 4p2 – 4q = 4(q2 – p)⇒ p2 – q2 + p – q = 0⇒ p + q + 1 = 067. (c) : We have, |x – 3|2 + |x – 3|–2 = 0⇒ |x – 3| = 1⇒ x – 3 = ±1 ⇒ x = 2, 4

68. (a) : α β α β+ = − ⇒ + = −32

3 92

( )

αβ αβ= ⇒ =52

9 452

\ Required equation is, x x2 92

452

0+ + =

or 2x2 + 9x + 45 = 0.

69. (e) : Let y xx x

=− +2 5 9

or x2y –(5y + 1)x + 9y = 0For real x, D≥ 0i.e. (5y + 1)2 – 36y2 ≥ 0⇒ (5y + 1 – 6y) (5y + 1 + 6y) ≥ 0⇒ (–y + 1) (11y + 1) ≥ 0⇒ (y – 1) (11y + 1) ≤ 0

⇒ ∈ −

y 111

1,

70. (d) : 3x2 – 2(a + b + c) x + ab + bc + ca = 0⇒ 4(a + b + c)2 = 12(ab + bc + ca) (Q D = 0)⇒ (a + b + c)2 = 3(ab + bc + ca)⇒ a2 + b2 + c2 = ab + bc + ca.

71. (e) : One root =1

1 31 3

212

32+

= −−

= − +

\ Other root = − −12

32

Sum = –1, product = 14

34

12

− = −

\ Required equation is x x2 12

0+ − =

or 2x2 + 2x – 1 = 0.

| june ‘1570

72. (a) : a5 = ar4 = 1458, a8 = ar7 = 54

⇒ = ⇒ =r r3 127

13

73. (d) 74. (d)

75. (d) : a ar r a4

3 322

= = =,

⇒ a4 = 256 ⇒ a = 4⇒ r = 2⇒ a15 = 4 × 214 = 216.

76. (b) : 25 504

54 10+ + 2 + +.... area ( . )of seriesS G P

where a r= =25 12

,

⇒ =−

−= −

Sum of G.P series25 1 1

2

1 12

50 1 12

10

10

77. (a) : (a + c)4 = (2b)4 = 16a2c2 (∵b2 = ac)

78. (d) : 113

24 12 1

=+

− −( )

( )( )x

x x

Only x = 10 satisfy it.

79. (b) : 12

1 12

14

1010

110

22+

= + +x C x C x. . . .

+ + +103

3 104

418

116

C x C x. . . . ....

By inspection, 103

18

C . is the highest coefficient.

\ Power of x is 3.80. (*) : No. of ways = 7 × 7 × 6 × 5 × 4 = 588081. (c) : (m + 4) (m + 5) = 22(m – 1)

(using given relation)⇒ m2 – 13m + 42 =0⇒ m = 7, 6 ⇒ m1 + m2 = 1382. (e) : Sum of coeff. of last 6 terms

= = =22

2 102411

10

(Using relation nC0 + nC1 +..... + nCn = 2n)

83. (b) : Sincen

rn

r

C

Cn r

r−= − −

1

1( )

⇒ rC

Cn r

nr

nr −

= + −1

1

⇒ rC

Cr

nr

nrr r−= =

∑ ∑= −11

15

1

1516( )

= × − × ×16 15 12

15 16 = 240 – 120 = 120

84. (d) : D = (36 –30) –2(24 – 15) + 3(12 – 9) = –3D′ = 4(–6) –8(–9) + 15(–3)

= 3 = –D.

85. (b) : We have,

1 3 5

2 2 5

2 3 4

0

+

+

+

=

x

x

x

Apply C1 → C1 + C2 + C3

⇒++ ++ +

=9 3 59 2 59 3 4

0xx xx x

Apply R2 → R2 – R1, R3 → R3 – R1

⇒ + −−

=( )91 3 50 1 00 0 1

0x xx

⇒ (9 + x) (x – 1)2 = 0 ⇒ x = –9 or x = 1, 1

86. (c) : Let A

b

=

−

0 3 2

2 0 1

4 1 6

Since |A| = 0⇒ –3(12 – 4) + 2b(–2) = 0⇒ –24 – 4b = 0 ⇒ 4b = –24⇒ b = –687. (d) : |A| = x – 2x(x – 1) = –9⇒ x – 2x2 + 2x = –9⇒ 2x2 – 3x – 9 = 0

x = ± + = ±3 9 724

3 94

x = −3 32

or

88. (c) : Apply C1 → C1 + C2 + C3 and using cos(90– q) = sinq, cos2q + sin2q = 1, we get

0 36 135

0 135 37

0 25 65

0

2

2

2 2

cos cot

cot sin

cos cos

° °

° °

° °

=

89. (b) : (ABA)T = ATBTAT = ABA(Using A = AT, B = BT)

| june ‘15 71

90. (c) : |x – 3| < 2x + 9⇒ –(2x + 9) < (x – 3) < 2x + 9⇒ either –2x – 9 < x –3 ⇒ –3x < 6 ⇒ x > –2 …(i)or x – 3 < 2x + 9 ⇒ – x < 12 ⇒ x > –12 …(ii)From (i) and (ii), we get x > –2

91. (e) : A = xy, P = 2(x + y)Since (x + y)2 ≥ 4xy (Using A.M. ≥ G.M. for real numbers x, y)

⇒

≥P A2

42

⇒ P2 ≥ 16A.

92. (a) : ~ (p ∨ q) ∨ (~ p ∧ q)= (~p ∧ ~q) ∨ (~p ∧ q) = ~p ∧ (~q ∨ q) = ~p

93. (c) : ~[p ∨ (q ∧ r)] = ~p ∧ ~(q ∧ r)= ~p ∧ (~q ∨ ~r)= (~p ∧ ~q) ∨ (~p ∧ ~r)

94. (e)

95. (a) : sintan

tanθ

θ

θ=

+=

× 1

+=

=2

2

12

22

1 14

154

452

96. (d) : (x – 5)2 –(y – 5)2 = 4sec2q – 4tan2q = 4

97. (e) : tan15° + tan75° = − + + =2 3 2 3 4

98. (d) : f(x) = cos4x + tan3x

Period of cos4x = 24 2π π=

Period of tan3x = π3

\ period of f(x) = LCM of π π3 2

,

= p

99. (b) : sin(q + f) = n sin(q – f)

⇒ +−

=sin( )sin( )

θ φθ φ

n1

⇒Nr DrNr Dr

nn

+−

⇒ + + −+ − −

= +−

sin( ) sin( )sin( ) sin( )

θ φ θ φθ φ θ φ

11

⇒ = +−

22

11

sin coscos sin

θ φθ φ

nn

⇒ = +−

tantan

θφ

nn

11

100. (a) : 2 13

14

1 1tan tan− −+

=−

+

− −tan.

tan1 12 1

3

1 19

14

=

+

− −tan tan1 134

14

=+

−

−tan.

134

14

1 34

14

=

−tan 1 1613

101. (e) : cos cos− −+ =1 1 27

x y π

⇒ −

+ −

=− −π π π2 2

27

1 1sin sinx y

⇒ + = − =− −sin sin1 1 27

57

x y π π π

102. (c) : cos–1x > sin–1x

⇒ > −− −cos cos1 12

x xπ

⇒ >−22

1cos x π ⇒ cos− >14

x π

⇒ x x< ≥ −12

1and

103. (d) : Given equation is sin8x + cos6x = 1⇒ sin8x + (1 – sin2x)3 = 1⇒ sin8x – sin6x + 3sin4x – 3sin2x = 0⇒ (sin4x + 3) (sin4x – sin2x) = 0⇒ sin4x = sin2x⇒ either sinx = 0 ⇒ x = 0, p, 2p (3 solutions)

or sin2x = 1 ⇒ =x π π2

32

, (2 solutions)

\ 5 solutions exist.104. (c) : Let P be the circumcentre of the triangle

⇒h k+ +

=82

02

4 2, ( , )

[In a right angled triangle, circumcentre is mid-point of hypotenuse]

| june ‘1572

⇒ h = 0, k = 4\ The point C is (0, 4).

105. (e) : G C(5, 5)2 : 1

O(2, 2)

G 10 23

10 23

4 4+ +

=, ( , ) (Using section formula)

106. (b) : We have, x1 – 3 = x2 – x1 and y1 – 4 = y2 – y1[As both 3, x1, x2 and 4, y1, y2 are in A.P.]

⇒−−

=−−

yx

y yx x

1

1

2 1

2 1

43

\ slope of AB = slope of BC\ A, B, C are collinear.

107. (b) : p a a1 2 2= −

+=

sec cosecsin cos

θ θθ θ

= a2

2sin θ

p a a2 2 2

2 2= −

+=cos

cos sincosθ

θ θθ

⇒ (2p1)2 + (p2)2 = a2 ⇒ 4p12 + p2

2 = a2.108. (a) : 2x + y = 2 …(i), x + 2y = 2 …(ii)Solving (i) and (ii), we get

x y= =23

23

,

d = −

+ −

=1 23

2 23

173

2 2

109. (e) : The required line is 4x – y = k which passes through (5, 0).⇒ 20 – 0 = k ⇒ k = 20\ Required equation is 4x – y = 20.

or,x y5 20

1− = or,x y5 20

1+−

=

\ It meets the y-axis at (0, –20).

110. (b) : We have, x intercept = k5

and y intercept = k1

Area = × × = =12 5 1 10

102k k k

⇒ k = ±10111. (d) : We have x + y = 6 …(i), x + 2y = 4 …(ii)Solving (i) and (ii), we get y = –2, x = 8 ⇒ (8, –2)\ centre : (8, –2)

r = − + − − = +( ) ( )8 6 2 2 4 162 2 = 20 = 2 5

112. (d) : We have, 2g = –6 ⇒ g = –3,2f = 2 ⇒ f = 1, c = –28

r g f c= + − = + + =2 2 9 1 28 38

centre : (h, k) = (–g, –f) = (3, –1)\ x = h + r cosq, y = k + r sinq are parametric form.⇒ = + = − +x y3 38 1 38cos , sinθ θ

113. (c)

114. (b) :

x y= + = +20 155

20 05

cos , sinθ θ

⇒ 5x –15 = 20 cosq, 5y = 20 sinq

⇒ −

+

=5 320 4

12 2( )x y

⇒ (x – 3)2 + y2 = 16 is the required equation ofcircle.115. (c) : General equation of the circle isx2 + y2 + 2gx + 2fy + c = 0 …(i)It passes through (0, 0) ⇒ c = 0it passes through (a, 0) ⇒ 2g = –ait passes through (0, a) ⇒ 2f = –a\ Required equation of the circle is x2 + y2 – ax – ay = 0⇒ x2 + y2 = a(x + y)116. (d) : We have, a = 3, b = 3

so, e = 1 1 1 22

2+ = + =ba

Distance between the directrices = 2ae

= ⋅ =2 32

3 2

117. (e)

118. (c) : We have, a ba

b= = ⇒ =3 2 169

83

22,

⇒ Required equation of ellipse isx y x y2 2 2 2

9 83

19

38

1+ = ⇒ + =

119. (b) : a = 5, b = 4Now, 4 max {CP} + 5min {CP}= (4 × 5) + (5 × 4) = 40120. (a)

nn

| june ‘15 73

1. If z1, z2, z3 are complex numbers such that

| | | | | | and ,z z zz

z zz

z zz

z z1 2 312

2 3

22

3 1

32

1 21 1= = = + + = −

then find the value of |z1 + z2 + z3|–Abhijit Kr. (Bhubaneshwar)

Ans. Let z = z1 + z2 + z3. Then

z z z zz z z

= + + = + +1 2 31 2 3

1 1 1

=+ +z z z z z zz z z

1 2 2 3 3 1

1 2 3⇒ = + +z b z z z z z z1 2 2 3 3 1, where b = z1z2z3.Now,

zz z

zz z

zz z

z z z z z z12

2 3

22

3 1

32

1 213

23

33

1 2 31+ + = − ⇒ + + = −

⇒ + + − = −z z z z z z b13

23

33

1 2 33 4

⇒ (z1 + z2 + z3)[(z1 + z2 + z3)2 – 3(z1z2 + z2z3 + z3z1)]= –4b

⇒ z z z b b( )2 3 4− = −⇒ z3 – 3|z|2 b + 4b = 0⇒ z3 = (3|z|2 – 4)b⇒ |z|3 = |3|z|2– 4| (since |b| = |z1z2z3| = 1)Case 1 : Suppose that 3|z|2 ≥ 4. Then

|z|3 = 3|z|2 – 4⇒ |z|3 – 3|z|2 + 4 = 0⇒ (|z| – 2)(|z|2 – |z| – 2) = 0⇒ (|z| – 2)(|z| – 2)(|z| + 1) = 0⇒ |z| = 2Case 2 : Suppose that 3|z|2 < 4. Then

|z|3 = |3|z|2 – 4| = 4 – 3|z|2

⇒ |z|3 + 3|z|2 – 4 = 0⇒ (|z| – 1)(|z|2 + 4|z| + 4) = 0⇒ (|z| – 1)(|z| + 2)2 = 0 ⇒|z| = 1.

Do you have a question that you just can’t get answered?Use the vast expertise of our mtg team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough.The best questions and their solutions will be printed in this column each month.

2. Find the general solution of the equationsin4 2x + cos42x = sin2x cos 2x.

–Anand (W.B.)

Ans. We have,(sin2 2x + cos2 2x)2 – 2sin2 2x cos2 2x

= sin 2x cos 2x⇒ 2 sin2 2x cos2 2x + sin 2x cos 2x – 1 = 0Put sin 2x cos 2x = t. Therefore, we have

2t2 + t – 1 = 0

⇒ (2t – 1)(t + 1) = 0 ⇒ t = −12

1,

Case 1 : When t = –1, we havesin 2x cos 2x = –1⇒ sin 4x = –2, which is impossible.Therefore t = –1 is to be rejected.Case 2 : When t = 1/2, we havesin 4x = 1⇒ 4x = 2np + (p/2)

⇒ = + = +x n nπ π π2 8

4 18

( )

3. A person has to go through three successive tests.The probability of his passing first test is p. If hefails in one of the tests, then the probability of hispassing next test is p/2, otherwise it remains thesame. For selection, the person must pass atleasttwo tests. Find the probability that the person hasto be selected.

–Devika (Kochi)

Ans. Let Ei (i = 1, 2, 3) be the event of the person passing the ith test and E is the event that he is selected.ThenE E E E E E E E E= ∩ ∪ ∩ ∩ ∪ ∩ ∩( ) ( ) ( )1 2 1 2 3 1 2 3ThereforeP E P E P E E P E P E E P E E

P E P E E P E( ) ( ) ( | ) ( ) ( | ) ( | )

( ) ( | ) (= +

+1 2 1 1 2 1 3 2

1 2 1 3 || )E2

= ⋅ + − + − ⋅p p p p p p p p( ) ( )12

12

= 2p2 – p3.

4. If x a x b c d× + ⋅ =( ) , then prove that

x a a a d ca c a

= + × × ×⋅

λ ( )( ) 2

, where l is a scalar.–Rajkumar (Patna)

Ans. Here x a x b c d× + ⋅ =( )

\ { ( ) } x a x b c c d c× + ⋅ × = ×⇒ ( ) x a c d c× × = × [∵

c c× = 0 ]

Y UASKWE ANSWER

| June ‘1574

⇒ ( ) ( ) x c a a c x d c⋅ − ⋅ = ×\ a x c a a c x a d c× ⋅ − ⋅ = × ×{( ) ( ) } ( )⇒ − ⋅ × = × ×( )( ) ( ) a c a x a d c { }∵

a a× = 0

⇒ x a a d ca c

× = × ×⋅

( )

⇒ a x a a a d ca c

× × = × × ×⋅

( ) ( )

⇒ ( ) ( ) ( ) a a x a x a a a d ca c

⋅ − ⋅ = × × ×⋅

⇒ a x a x a a a d ca c

2 = ⋅ + × × ×⋅

( ) ( )

\

x a xa

a a a d ca c a

= ⋅ + × × ×⋅

( ) ( )( )2 2

= λ

a a a d ca c a

+ × × ×⋅( )

( ),2

where l =

a xa⋅2

= scalar.

5. Let PM be the perpendicular from the pointP(1, 2, 3) to the xy plane. If OP makes an angle qwith the positive direction of the z-axis and OMmakes an angle f with the positive direction of thex-axis, where O is the origin, then find q and f.

–Rahil (U.P.)

Ans. We know that if P = (x, y, z) then x = rsinq ⋅ cosf, y = rsinq ⋅ sinf, z = rcosq

\ 1 = rsinq ⋅ cosf, 2 = rsinq ⋅ sinf, 3 = rcosq⇒ 12 + 22 = r2sin2q ⋅ cos2f + r2sin2q ⋅ sin2f

= r2sin2q(cos2f + sin2f) = r2sin2q⇒ 5 = r2sin2q \ rsinq = 5

(clearly, q and f are acute)

\ rr

sincos

θθ

= 53

or tanq = 53

\ q = tan–1 53

Also, 12

= ⋅⋅

rrsin cossin sin

θ φθ φ

⇒ cotf = 12⇒ tanf = 2.

\ f = tan–12.

6. Find the value of sin cossin

x xx

dx++∫ 9 16 20

4π

–Vikas (M.P.)

Ans. Let I = d x xx

(sin cos )( sin )

−− − +∫ 9 16 1 2 160

4π

= d x xx x

(sin cos )(sin cos )

−− −

∫25 16 2

0

4π

= 116 5

4

220

4 d x x

x x

(sin cos )

(sin cos )

−

− −

∫π

Put sin x – cos x = z; thenx = 0 ⇒ z = –1; x = π

4⇒ z = 0

\ I = 1

16 54

116

1

2 54

5454

221

0

1

0

dz

z

z

z

−

= ⋅⋅

+

−

−

−

∫ log

= 140

5 45 4

140

1 191

0log log log+

−

= −{ }−

zz

= 140

9 120

3log log .=

nn

Your favourite MTG Books/Magazines available in

| June ‘15 75

1. If a, b, c, d are positive, then limx

c dx

a bx→∞

++

+

=1 1

(a) ed b/(b) ec/a

(c) ec da b

( )( )

++ (d) e

2. limcos

x

xx→

+ −( )−1

1 2 11

(a) exists and it equals 2(b) exists and it equals − 2(c) does not exist because x – 1 → 0(d) does not exist because L.H.L is not equal to

R.H.L

3. The functionlog log1 1+( ) − −( )ax bx

x is not

defined at x = 0. The value which should be assigned to f at x = 0 so that it is continuous at x = 0 is (a) a – b (b) 1 + b(c) loga + logb (d) none of these

4. If a line OP through the origin O makes angle a,45° and 60° with x, y and z axes respectively, then the direction cosines of OP are

(a) 12

12

12

, , (b) 12

12

12

, ,

(c) 12

12

12

, , (d) none of these

5. If O is the origin and the line OP of length r makesan angle a with x-axis and lies in the xz plane, then the coordinates of P are

(a) (r cosa, 0, r sina) (b) (0, 0, r sina)(c) (0, 0, r cos a) (d) (r cos a, 0, 0)

6. Evaluate : lim cos( sin )

./ /x

xx→ −π 2 2 31

7. Evaluate : lim sin cosx

x

x x→∞+

1 1

8. If f and g are differentiable at a ∈ R suchthat f(a) = g(a) = 0 and g′(a) ≠ 0, then show that

lim( )( )

( )( )

.x a

f xg x

f ag a→

=′′

9. Show that the function g, defined byg(x) = sina + cosa – 1, α = −sin { } ,1 x { . }denotes fractional part function, is an even function.

10. Evaluate : limx

xx→−∞

++

2 14 1

2

SolutionS

1. (a) : limx

c dx

a bx→∞

++

+

1 1

= ++

→∞

+++

limx

a bxc dxa bx

a bx1 1

=→∞

+

+e

x

d

b

cxax

lim

Math Archives, as the t it le itself suggests, is a col lection of various challenging problems related to the topics of JEE (Main & Advanced) Syllabus. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for JEE (Main & Advanced). In every issue of MT, challenging problems are offered with detailed solution. The readers' comments and suggestions regarding the problems and solutions offered are always welcome.

thrchives

M10 Best Problems10 Best Problems

Prof. Shyam Bhushan*10 Best Problems

| JuNE ‘1576

2. (c) : Put x – 1 = t

limcos

x

xx→

+ −( )−1

1 2 11

=→lim

cost

tt0

2

=→lim cost

tt0

2 does not exist because

t → 0 ⇒ (x – 1) → 0

3. (d) : For continuity at x = 0

limx

f x f→

( ) = ( )0

0

⇒ limlog( ) log( )

( )x

a axax

b bxbx

f→

⋅ ++

⋅ −−

=0

1 10

⇒ a + b = f(0)4. (c) : Direction cosines of OP are cosa , cos45°,cos 60° Now cos2a + cos2 45° + cos2 60° = 1

⇒ +

+

= ⇒ =cos cos2 212

14

1 14

α α

⇒ =cos α 12

5. (a) : Let the coordinates of P be (x, y, z)Since OP lies in xz plane and makes an angle a with

the x–axis, it makes angleπ α2

−

with z–axis and π2

with y–axis. So,

x r y r z r= = = −

cos , cos , cosα π π α2 2

are the required

coordinates and therefore are (r cosa, 0, r sina)

6. lim cos( sin )

lim sin( cos )/ / /x t

xx

tt→ →−

=−π 2 2 3 0 2 31 1

=→lim sin( / )cos( / )

( sin ( / )) /t

t tt0 2 2 3

2 2 22 2

=→lim cos( / )

(sin( / ))./

/t

tt0

1 31 32 2

2

Limit value is ∞ as t → 0+ and limit value is –∞ as t → 0–. Thus, limit does not exist.

7. lim sin cosx

x

x x→∞+

=1 1 e xx

x xlim sin cos→∞

+ −

1 1 1

= =→∞−

e ex

xx x x

lim sin( / )/

cos sin1 21 2

12

12

8. lim( )( )

lim

( ) ( )

( ) ( )( )( )

,x a x a

f xg x

f x f ax a

g x g ax a

f ag a→ →

=

−−−−

=′′

as f(a) = g(a) = 0 and g′(a) ≠ 0.

9. g x x x( ) sin(sin { }) cos(sin { })= + −− −1 1 1

= + − −−{ } cos(cos { })x x1 1 1

= + − −{ } { } .x x1 1

If x ∈ I, then {x} = 0 ⇒ g(x) = 0⇒ g(x) = g(–x)If x ∉ I, then {–x} = 1 – {x}

⇒ g(–x) = 1 1− + − ={ } { } ( )x x g x

Hence g is an even function.

10. lim lim ,x x

xx

x

x

x→− ∞ →−∞

++

= −+

+<2 1

4 1

2 1

4 10

2 2as

= − 12 2

.

nn

| JuNE ‘15 77

1. What least value must be given to * so that thenumber 97215 * 6 is divisible by 11 ? (a) 1 (b) 2 (c) 3 (d) 5

2. The traffic lights at three different road crossingschange after every 48 secs, 72 secs and 108 secs respectively. If they all change simultaneously at 8 : 20 : 00 hrs; then they will again change simultaneously at (a) 8 : 27 : 12 hrs (b) 8 : 27 : 24 hrs(c) 8 : 27 : 36 hrs (d) 8 : 27 : 48 hrs

3. The difference between a discount of 40% on ` 500and two successive discounts of 36% and 4% on the same amount is (a) `0 (b) `2 (c) `1.93 (d) `7.20

4. PQRS is a square. SR is aP

Q R

S OT

tangent (at point S) to the circle with centre O and TR = OS. Then, the ratio of area of the circle to the area of the square is (a) p : 3 (b) 11 : 7 (c) 3 : p (d) 7 : 11

5. A and B walk from X to Y, a distance of 27 kmat 5 km/hr and 7 km/hr respectively. B reaches Y and immediately turns back meeting A at Z. What is the distance from X to Z ?(a) 25 km (b) 22.5 km (c) 24 km (d) 20 km

6. The ratio of the number of boys and girls in aschool is 3 : 2. If 20% of the boys and 25% of the girls are scholarship holders, what percentage of the students does not get the scholarship ?(a) 56% (b) 70% (c) 78% (d) 80%

7. A can lay railway track between two given stationsin 16 days and B can do the same job in 12 days. With the help of C, they did the job in 4 days only. Then, C alone can do the job in

(a) 9 15

days (b) 9 25

days

(c) 9 35

days (d) 10 days

8. Mr. Thomas invested an amount of ̀ 13,900 dividedin two different schemes A and B at the simple interest of rate 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be `3508, what was the amount invested in Scheme B ?

(a) `6400 (b) `6500(c) `7500 (d) None of these

9. If the radius of circle is diminished by 10%, then itsarea is diminished by(a) 10% (b) 19% (c) 20% (d) 36%

10. Peter bought an item at 20% discount on itsoriginal price. He sold it with 40% increase on the price he bought it. The new sale price is by what percent more than the original price ?(a) 7.5% (b) 8% (c) 10% (d) 12%

11. In a box, there are 8 red, 7 blue and 6 green balls.One ball is picked up randomly. What is the probability that it is neither blue nor green ?

(a) 23

(b) 34

(c) 719

(d) 821

QUANTITATIVEAPTITUDE

| June ‘1578

12. 11

11

11(log ) (log ) (log )a b cbc ca ab+

++

++

is equal to

(a) 1 (b) 32

(c) 2 (d) 3

13. The income of a broker remains unchanged thoughthe rate of commission is increased from 4% to 5%. The percentage of slump in business is (a) 1% (b) 8% (c) 20% (d) 80%

14. The value of the expression 1014 × 986 is(a) 998924 (b) 999864(c) 999804 (d) 996724

15. The dimensions of an open box are 52 cms, 40 cms,and 26 cms. Its thickness is 1 cm. If 1 cubic cm. of the metal used in the box weighs 1 gm, then the weight of the box is (a) 8.48 kg (b) 6.58 kg(c) 7.28 kg (d) 658 gm

16. The unit digit in the sum (264)102 + (264)103 is(a) 0 (b) 4 (c) 6 (d) 8

17. The least number, which when increased by 1 isexactly divisible by 12, 18, 24, 32, 40 is (a) 1439 (b) 1440 (c) 1449 (d) 1459

18. One litre of water weighs 1 kg. How many cubicmillimetres of water will weigh 0.1 gram ?(a) 10 (b) 100 (c) 0.1 (d) 1

19. A man has `480 in the denominations ofone-rupee notes, five-rupee notes and ten-rupee notes. The number of notes are equal. What is the total number of notes he has ?(a) 45 (b) 60 (c) 75 (d) 90

20. 5 mangoes and 4 oranges cost as much as 3mangoes and 7 oranges. What is the ratio of the cost of one mango to that of one orange?(a) 4 : 3 (b) 1 : 3 (c) 3 : 2 (d) 5 : 2

solutions

1. (c) : For the number to be divisible by 11, thedifference between the sum of digits at odd places and that of at even places should be either 0 or divisible by 11.∴ (9 + 2 + 5 + 6) – (7 + 1 + *) = 14 – *So, * = 3

2. (a) : Interval of change= (L.C.M. of 48, 72, 108) = 432secsSo, the lights will change after every 432 seconds i.e. 7 mins 12 secs.So, the next simultaneous change will take place at 8 : 27 : 12 hrs.

3. (d) : Price after 40% discount = 60% of `500= `300.

Price after 36% discount = 64% of `500 = `320.Pr ice af ter next 4% discount = 96% of `320

= `307.20.∴ Difference in two prices = `7.20.

4. (a) : Let the radius of the circle be rThen area of the circle is pr2.Now OR = OT + TR = r + r = 2r

[ TR = OS = r (given)]Now, In DOSR by pythagoras theorem, SR = 3rSo, area of square PQRS = 3r2

∴ Required ratio = π πrr

2

23 3= or p : 3

5. (b) : Time taken by A in covering (27 – x) km issame as time taken by B in covering (27 + x) km.

X Z Y

(27– ) kmx x km

⇒ − = +( ) ( )275

277

x x

⇒ =x 276

∴ = − = × =XZ 27 276

27 56

22 5. km.

6. (c) : Let boys = 3x and girls = 2x.Number of those who do not get scholarship= (80% of 3x) + (75% of 2x)

= 80100

3 75100

2 3910

×

+ ×

=x x x .

∴ Required percentage = 3910

15

100 78xx

× ×

=% %.

7. (c) : (A + B + C)'s 1 day work = 14

,

A's 1 day work = 116

, B's 1 day work = 112

.

| June ‘15 79

∴ C's 1 day work =14

116

112

14

748

548

− +

= −

= .

So, C alone can do the work in 485

9 35

= days.

8. (a) : Let the sum invested in Scheme A be `x andthat in Scheme B be `(13900 – x).

Then, x x× ×

+ − × ×

=14 2100

13900 11 2100

3508( )

⇒ 28x – 22x = 350800 – (13900 × 22)⇒ 6x = 45000 ⇒ x = 7500.So, sum invested in Scheme B = `(13900 – 7500)

= `6400.

9. (b) : Let the original radius be R units

New radius = (90% of R) = 90100

910

×

=R R units.

Original area = pR2.∴ Diminished area

= π π πR R R22

2910

1 81100

−

= −

sq. units sq.. units

sq. units=

19100

2πR .

Decrease % = 19100

1 100 192

2π

πR

R× ×

=% %.

10. (d) : Let the original price be `100Then, C.P = `80.

S.P. = 140% of `80 = ` 140100

80×

= `112.

∴ Required percentage = (112 – 100)% = 12%.

11. (d) : Total number of balls = (8 + 7 + 6) = 21.Let E be the event that the ball drawn is neither blue nor greeni.e; E be the event that the ball drawn is red.∴ n (E) = 8

∴ P (E) = 821

.

12. (a) : Given expression can be written as;1 1 1

log log log log log loga a b b c cbc a ca b ab c++

++

+

= 1 1 1

log ( ) log ( ) log ( )a b cabc abc abc+ +

= logabc a + logabc b + logabc c = logabc (abc) = 1.

13. (c) : Suppose the business value changes from x to y.

4% of x = 5% of y x y y x⇒ = ⇒ =4100

5100

45

.

∴ Change in business = x x x−

=45 5

.

Percentage of slump = xx51 100 20× ×

=% %.

14. (c) : 1014×986 = (1000 + 14) (1000 – 14)= (1000)2 – (14)2 = (1000000 – 196) = 999804.

15. (b) : Volume of metal= (52 × 40 × 26 – 50 × 38 × 25) cu. cm.= (54080 – 47500) cu. cm. = 6580 cu. cm.∴ Weight of metal = 6580 gm = 6.58 kg.

16. (a) : (264)102 + (264)103 = (264)102 × [1 + 264]= (264)102 × 265

Unit digit in (264)4 is 6Unit digit in [(264)4]25 = Unit digit in 625 = 6Unit digit in (264)100 × (264)2 is 6Unit digit in (264)102 × 265 = unit digit in 6 × 5 = 0

17. (a) : 22 1212, 1818, 2424, 3232, 4040232

6,3,1,1,

9,9,3,3,

12,6,2,1,

16,8,8,4,

2010105

∴ L.C.M. of 12, 18, 24, 32, 40 = 1440∴ Required number = (1440 – 1) = 1439.

18. (b) : 1000 cu. cm weighs 1000 gms.i.e. 1000 gms is the weight of (1000 × 1000) cu. mm

0.1 gm is the weight of 1000 10001000

0 1× ×

. cu. mm

= 100 cu. mm.

19. (d) : Let the number of each type of notes be x.Then, x + 5x + 10x = 480 ⇒ x = 30.∴ Total number of notes = 30 + 30 + 30 = 90.

20 (c) : Let cost of each mango be x paise and that of each orange be y paise, then 5x + 4y = 3x + 7y ⇒ 2x = 3y ⇒ x : y = 3 : 2.

nn

| June ‘1580

Solution Set-149

1. (c) : Consider the blocks, 1; 2, 1; 2, 2, 1; 2, 2, 2, 1;.....while the nth block consists of (n – 1)2’s and one 1.The sum of terms in n blocks is 1 + 3 + 5 + ... .. + (2n – 1) = n2

The number of terms in n blocks is 1 + 2 + 3 + ... ........+ n = n n( )+1

2Taking n = 63 , sum of terms = 632 = 3969 and number of terms = 2016. \ The sum of 2015 terms is 39682. (d) : f(a – x) = f (a + x) …(i)f (b – x ) = f (b + x) …(ii)Let x = a in (i) and x = b in (ii)⇒ f (0) = f (2a) = f (2b) ⇒ f {2b + 2(a – b)} = f (2a).\ f (x) is periodic with period 2 (a–b)3. (b) : A (x1 , y1) , B (x2 , y2), C (x3 , y3)

The centroid Gx yi i=

∑ ∑

3 3

, .

PGx x x

xy y y

y2 1 2 32

1 2 32

3 3=

+ +−

+

+ +−

⇒ 9 PG2 = [3(PA2 + PB2 + PC2) – (a2 + b2 + c2)]= 210 – 110

\ PG = 103

. P describes a circle of radius

103

4. (a) : ( ) ( )1 1 1 1+ = +x ex x ln x

( )( ......)

( ......)

12 2

1 12 3

2 3

2

2

+ − + = − +

= ⋅ −

− +

− + −

x e ex e e ex

e e

xx x

x x

ee ex

x xx x

ex

+

= + − +

+

− +

= +

2

12 3

2 32

13

18

22 2

2 ( )

\ Limit = 13

18

1124

+ =

5. (c), (d) : The distance of A from the line x + y = 4is 1

2 = AC

⇒ ∠ BAC = 60°Slope of angle istan (45° ± 60°)= tan 105° , tan (–15)°= –2 + 3 , – 2 – 3

6. (a), (b) : Tangent y = mx + 1m

x = t2

, y = t2

2 2 lies on x2 = 2y .

Normal is y = – xt

t+ +12 2 2

2\ t = – 1

m,

t2 + 2 2 t + 2 = 0 ⇒ t = – 2

m = 12

, L is x – 2 y + 2 = 0

Distance of (0, 0) is 23

.

L is also y-axis, giving distance 0.7. (b) : A = (at2 1, 2at1) and B(at

2 2, 2at2)

⇒ t1t2 = 2 and t1 + t2 + t = 0The orthocentre of DABC is given by O(x, y)x = a(t2 – 6), y = –at.Eliminating t, the locus is y2 = a(x + 6a).L.R. = a

8. (a) : The centroid of DABC is G a t23

2 02 −( )

,

The circumcentre S divides OG in the ratio 3 : – 1.

\ S(x, y), x = a2

(t2 + 2), y = at2

.

Eliminating t, the locus is y2 = a2

(x – a), L.R. = a2

.

9. (5) : Perimeter P = OA + OB + AB

OA = 1 + 12

cotq,

OB = 12

+ tanq

AB = 12

cosecq + secq

B

y

xAO

1, 12( )

\ P = 32

12

+ (cotq + cosecq) + tanq + secq

If t = tan θ2

, then P = 12

12

21

+ +−t t

= f(t)

f ′(t) = 0 ⇒ t = 13

, Pmin = 5

10. (b) : 1 1 1 1 121

124

128

181 2 3r r r r

= + + = + + =

\ r = 8, D = r r r r1 2 3 = 336

s – a = ∆r1

33621

= = 16 ⇒ s – b = ∆r2

33624

= = 14

s – c = ∆r3

33628

= = 12. Adding, 3s – (a + b + c) = s = 42

a = 26, b = 28, c = 30.nn

y

x

A

B

B

C

O (1, 0) (4, 0)

(0, 4)

(1, 2)

2

2 12

| June ‘15 81

Directions (1 - 5) : Answer the questions on the basis of the information given below.In one of my visits to a small but beautiful town, while Iwas out on a shopping spree at the supermarket, a little statue of Buddha took my fancy. When I opened mywallet to pay for it, I realized to my horror that I had run out of money. The woman at the cash counter told me where the nearest bank was but I reached there only tofind that on that day of the week it was closed. Frustrated I drove back home and the first thing I asked my hostwas which place the supermarket, the department store, and the bank was open on which day of the week. She left me none the wiser when she started to tell me the curiosities of logic town (the town I was visiting).In logic town the supermarket, the department store and the bank are open together on one day each week.1. Each of the three places is open four days a week.2. On Sunday all three places are closed.3. None of the three places is open on three

consecutive days.4. On six consecutive days I observed that

- the department store was closed on the first day- the supermarket was closed on the second day- the bank was closed on the third day- the supermarket was closed on the fourth day- the department store was closed on the fifth day- the bank was closed on the sixth day

1. Of the six consecutive days that my host spoke of, which day is first?(a) Saturday (b) Friday(c) Thursday (d) Tuesday

2. On which day of the week was I at the supermarket?(a) Wednesday (b) Friday(c) Monday (d) Saturday

3. Which of the statement(s) is/are correct?I. Department store and bank are both open on

Tuesday.II. Bank and supermarket are both closed on

Thursday.III. Supermarket and department store are both open

on Friday.IV. Bank, department store and supermarket are all

closed on Sunday.(a) IV and I only (b) IV and II only(c) IV and III only (d) All of these

4. If you arrived at logic town on Sunday then which place you could not be able to go to the next day?(a) Bank(b) Department store(c) Supermarket(d) Both bank and Supermarket.

5. On which day of the week are all three places in logic town open?(a) Thursday (b) Friday(c) Saturday (d) Tuesday

Directions (6 & 7) : Answer the questions on the basis of the following information.

A student on holiday for d days observed that :1. It rained seven times morning or afternoon.2. When it rained in the afternoon, it was clear in the

morning.3. There were five clear afternoons.4. There were six clear mornings.

6. How many days had clear mornings but rainy afternoons?(a) 2 (b) 3 (c) 4 (d) 5

7. What is the value of d ?(a) 10 (b) 9 (c) 8 (d) 7

BrainBrainBrainBrainBrainBrainBrainBrainBrainBrainBrainBrainTeasersTeasersTeasersTeasersTeasersTeasersTeasersTeasersTeasersTeasersTeasersTeasers?

| June ‘1582

Directions (8 – 12) : Answer the questions on the basis of the information given below. The members of Gayan - a singing group - assembled on an es t ate one we ekend to p er for m in two programmes.A. Some of the members were to sing only in 'classical

songs' programme and the rest of the members were to sing only in 'contemporary songs' programme.

B. A strange thing happened that weekend. Any statement made by a member who was to sing in 'classical songs' programme was true, and any statement made by a member who was to sing in 'contemporary songs' programme was false.

Before the programmes could start, a member of Gayan was murdered. (1) Police then zeroed in on four suspects who made

the following statements :Moti : If the programme that the murderer was to sing in was, not the programme that I was to sing in, then the murderer's room is next to at least one room whose occupant was to sing in the programme that I was to sing in.neta : No two of us four who were to sing in the programme that I was to sing in, have adjacent rooms.onkar : The murderer was neither Neta nor Prem.Prem : If the murderer was to sing in the programme that I was to sing in, then I am not the murderer.(2) The four suspects had these connected rooms on

the estate.

Moti Neta Onkar Prem

8. With regard to the statements made by thesuspects what can be said?(a) Moti's hypothesis was false but conclusion was

true.(b) Moti's hypothesis was true but conclusion was

false.(c) Moti's hypothesis was false and conclusion was also

false.(d) Moti's hypothesis was true and conclusion was also

true.

9. Who among the following lied?(a) Moti (b) Onkar(c) Prem (d) Both Onkar and Prem

10. Whom among the following told the truth?(a) Onkar (b) Moti(c) Prem (d) Both Moti and Prem

11. Which of the following combinations about Premis true?(a) Prem's hypothesis was false but conclusion was

true.(b) Prem's hypothesis was true but conclusion was

false.(c) Prem's hypothesis was true and conclusion was

also true.(d) Prem's hypothesis was false and conclusion was

also false.

12. Who was the murderer?(a) Onkar (b) Moti (c) Neta (d) Prem

solutions

1. (a) 2. (c) 3. (d) 4. (a) 5. (b)1-5 : If Sunday is the first of the six consecutive days mentioned, then from (1), (2) and (4), the supermarket is closed only on Sunday, Monday and Wednesday. This is impossible, from (3).If Monday is the first of the six consecutive days mentioned, then from (2) and (4) at least one place is closed each day. This is impossible because all three places are open together on one day each week.If Tuesday is the first of the six consecutive days mentioned, then from (1), (2) and (4) the department store is closed only on Tuesday, Saturday and Sunday. This is impossible from (3).If Wednesday is the first of the six consecutive days mentioned, then from (1), (2) and (4) the bank is closed only on Sunday, Monday and Friday, and the supermarket is closed only on Sunday, Thursday and Saturday. This is impossible from (3).If Thursday is the first of the six consecutive days mentioned then from (1), (2) and (4), the bank is closed only on Tuesday, Saturday and Sunday. This is impossible, from (3).If Friday is the first of the six consecutive days mentioned, then from (1), (2) and (4) the supermarket is closed only on Monday, Saturday and Sunday. This is impossible, from (3).So, Saturday is the first of the six consecutive days mentioned; then from (1), (2) and (4) (C stands for closed, O for open).

sun Mon tue Wed Thurs Fri sat

Bank C C O O C O O

Department store C O O C O O C

supermarket C O C – – – –

From the above table, Friday must be the day all the three places are open.

| June ‘1584

To complete the table : From (1) and (3) , the supermarket cannot be closed on Wednesday or Saturday. So the supermarket must be closed on Thursday.So the table looks like


Bank C C O O C O O

Department store

C O O C O O C

supermarket C O C – C O –

Now as each of the three places is open four days a week, we have the complete table.


Bank C C O O C O O

Department store

C O O C O O C

supermarket C O C O C O O

6. (c) 7. (b)

6-7 : Make a tabular column to have a good idea.

rainy afternoon clear afternoon

Rainy morning x y

Clear morning z w

By the problem, x + y + z + w = d ... (i)As it rained seven times morning or afternoon

⇒ x + y + z = 7 ... (ii)As there were five clear afternoons

⇒ y + w = 5 ... (iii)As there were six clear mornings

⇒ z + w = 6 ... (iv)When it rained in the afternoon, it was clear in the morning. ⇒ x = 0 ... (v)With x = 0, equations (i) & (ii) reduces to

y + z = 7 ... (i)y + w = 5 ... (iii)z + w = 6 ... (iv)

Adding them y + z + w = + + =12

7 5 6 9( )

\ y = (y + z + w) – (z + w) = 9 – 6 = 3Similarly, z = 4 and w = 2. d = y + z + w = 9.

8. (d) 9. (b) 10. (d) 11. (a) 12. (c)

8-12 : The following reasoning uses A, B and (1).Suppose Moti lied : Then his hypothesis was true and his conclusion was false. Then the murderer told the truth and the murderer's room was not next to a liar's room. Then Moti was not the murderer, and from (2), Neta was not the murderer. Then either Onkar or Prem was the murderer.(i) Suppose Prem was the murderer ; then Prem told the truth and his conclusion was false; then his hypothesis was false; but, because he was the murderer, his hypothesis was true. This situation is impossible, so Onkar was the murderer. N O P

F T TThen Onkar told the truth, and Prem told the truth because his conclusion was true. Then, from (2), Neta didnot tell the truth, so Neta lied. Then from (2), because Onkar's room was next to a member who was to sing in either programme, Moti's conclusion was true. Then Moti told the truth. So if Moti lied, then he told the truth. This situation is impossible, so Moti told the

truth. M N O PT L T

Then Neta did not tell the truth, from (2); so Neta lied (Recall that Neta's room is adjacent to Moti's). Then Onkar lied, from (2), because either Onkar and Neta lied, or Onkar and Prem both lied. Then either Neta or Prem was the murderer.Suppose Prem was the murderer. Then Prem lied because his hypothesis was true and his conclusion was false. Then Moti's hypothesis was true and because it has already been established that Onkar lied, his conclusion was false. Then Moti lied. But it has already been established that Moti told the truth. This situation is impossible, so Prem was not the murderer. Then Neta was the murderer. Then Prem told the truth because his conclusion was true, Prem's hypothesis was false, while Moti's hypothesis and conclusion were both true.We summarise,(i) Moti told the truth, his hypothesis and conclusion

were both true.(ii) Neta was the murderer.(iii) Prem told the truth, his hypothesis was false but

conclusion true.(iv) Onkar lied.

nn

| June ‘15 85

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