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THE ADVANCED MATHEMATICS

(IN ENGLISH)

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Permutations and Combinations MAIN EXAM

ORIENTED QUESTIONS

1. If 1

6! +

1

7! =

𝑥

8!, then x = ?

(1) 32 (2) 48 (3) 56 (4) 64 2. If n-1P3 : nP4 = 1 : 9, then n = ?

(1) 12 (2) 11 (3) 9 (4) 10

3. 𝑛𝑃𝑛

𝑛𝑃𝑛−2

= ?

(1) 1

2 (2) 2

(3) 1

𝑛−2 (4) n (n - 1)

4. If 15Pr = 2730, then r = ? (1) 3 (2) 4

(3) 5 (4) 6 5. 7P3 = ?

(1) 105 (2) 140 (3) 210 (4) 175 6. If nP5 = 20 . nP3, then n = ?

(1) 8 (2) 9 (3) 10 (4) 11 7. If 15Pr-1 :

16Pr-2 = 3 : 4, then r = ? (1) 8 (2) 10

(3) 12 (4) 14 8. If nC10 = nC14, then n = ?

(1) 4 (2) 24 (3) 14 (4) 10 9. If nC3 = 220, then n = ?

(1) 9 (2) 10 (3) 11 (4) 12

10. If nCr + nCr+1 = n+1Cx, then x = ? (1) r - 1 (2) r

(3) r + 1 (4) n 11. 36C34 = ?

(1) 1224 (2) 612 (3) 630 (4) none of these

12. 𝑛𝐶𝑟

𝑛𝐶𝑟−1

= ?

(1) 𝑛−𝑟

𝑟 (2)

𝑛−𝑟−1

𝑟

(3) 𝑛−𝑟+1

𝑟 (4) None of these

13. If nC18 = nC12, then 32Cn = ?

(1) 248 (2) 496 (3) 992 (4) none of these

14. 60C60 = ? (1) 60! (2) 1

(3) 1

60 (4) none of these

15. In how many ways can 5 persons occupy 3 seats?

(1) 15 (2) 20 (3) 30 (4) 60

16. In how many ways can 5 children stand in a queue?

(1) 5 (2) 25 (3) 60 (4) 120

17. In how many ways can 4 different books be arranged in a shelf?

(1) 4 (2) 8 (3) 24 (4) 16

18. Ten students are participating in a race. In how many different ways can the first prize be won?

(1) 30 (2) 60 (3) 120 (4) 720

19. Three different rings are to be worn in 4 fingers. In how many ways can this be done?

(1) 12 (2) 24 (3) 64 (4) 81

20. There are 6 periods on each working day of a school. In how many ways can one arrange 5 subjects such that each subject is allowed at least one period?

(1) 360 (2) 720 (3) 3600 (4) 1800

21. How many words with or without meaning can be formed by using all the letters of the word, ‘DELHI’, using each letter exactly once?

(1) 20 (2) 60 (3) 120 (4) 5

22. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?

(1) 24 (2) 120 (3) 720 (4) 2880

23. How many 3-digit numbers are there? (1) 648 (2) 729

(3) 900 (4) 1000



24. How many 3-digit numbers are there with no digit repeated?

(1) 648 (2) 720 (3) 729 (4) none of these

25. How many 4-digit numbers can be formed with no digit repeated by using the digits 3, 4, 5, 6, 7, 8 and 0 ?

(1) 280 (2) 560 (3) 720 (4) 840

26. How many 3-digit even numbers can be formed with no digit repeated by using the digits 0, 1, 2, 3, 4 and 5?

(1) 50 (2) 52 (3) 54 (4) 56

27. The number of positive integers greater than 6000 and less than 7000 which are divisible by 5 with no digit repeated, is

(1) 28 (2) 56 (3) 84 (4) 112

28. How many 10-digit numbers can be formed by using the digits 1 and 2 ?

(1) 10P2 (2) 10C2 (3) 210 (4) 10!

29. How many words beginning with T and ending with E can be made with no letter repeated out of the letters of the word ‘TRIANGLE’?

(1) 8P6 (2) 720 (3) 722 (4) 1440

30. How many words can be formed from the letters of the word ‘DAUGHTER’ so that the vowels always come together?

(1) 720 (2) 2160 (3) 4320 (4) none of these

31. How many words can be formed form the letters of the word ‘LAUGHTER’ so that the vowels are never together?

(1) 3600 (2) 4320 (3) 36000 (4) 40320

32. In how many ways can the letters of the word ‘MACHINE’ be arranged so that the vowels may occupy only odd positions?

(1) 288 (2) 576 (3) 5040 (4) none of these

33. In how many ways can the letters of the word ‘PENCIL’ be arranged so that N is always next to E?

(1) 120 (2) 240 (3) 720 (4) 1440

34. In how many ways can the letters of the word ‘APPLE’ be arranged?

(1) 6 (2) 60 (3) 90 (4) 120

35. How many words can be formed by using all the letters of the word ‘ALLAHABAD’?

(1) 9! (2) 1890 (3) 3780 (4) 7560

36. How many words can be formed using the letter A thrice, the letter B twice and the letter C once?

(1) 6 (2) 60 (3) 90 (4) 120

37. In how many ways can 10 books be arranged in a shelf so that a particular pair of books shall be always together?

(1) 8! (2) 9!

(3) 2 8! (4) 2 9! 38. In how many ways can 10 books be arranged in a

shelf so that a particular pair of books shall be never together?

(1) 8! (2) 9!

(3) 2 9! (4) 8 9! 39. How many 4-digit numbers are there when a digit

may be repeated any number of times in each number?

(1) 5040 (2) 9000 (3) 10000 (4) 4500

40. In how many ways can 6 boys be arranged in a row?

(1) 5! (2) 6!

(3) 6 (4) 2 6! 41. In how many ways can 6 girls be seated in a

circle? (1) 6 (2) 6!

(3) 5! (4) 2 5! 42. How many diagonals are there in a polygon of n

sides?

(1) 1

2 n (n - 1) (2)

1

2 n (n - 2)

(3) 1

2 n (n - 3) (4)

1

2 n (n + 1)



43. How many diagonals are there in an octagon? (1) 28 (2) 24

(3) 20 (4) 36 44. A polygon has 54 diagonals. Number of sides in

this polygon is : (1) 9 (2) 12

(3) 15 (4) 16 45. There are 10 points in a plane, out of which 4

points are collinear. The number of line segments obtained from the pairs of these points is:

(1) 39 (2) 40 (3) 41 (4) 45

46. There are 10 points in a plane, out of which 4 points are collinear. The number of triangles formed with vertices as these points is :

(1) 20 (2) 120 (3) 116 (4) none of these

47. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

(1) 330 (2) 1050 (3) 6300 (4) 25200

48. In how many ways can a committee of 5 members be selected from 6 men and 5 ladies, consisting of 3 men and 2 ladies?

(1) 25 (2) 50 (3) 100 (4) 200

49. Out of 5 men and 2 women, a committee of 3 is to be formed. In how many ways can it be formed if at least one woman is included in each committee?

(1) 21 (2) 25 (3) 32 (4) 50

50. A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many ways can this be done when each committee may have at the most 2 ladies?

(1) 120 (2) 160 (3) 180 (4) 186

51. How many different teams of 7 players can be chosen out of 10 players?

(1) 720 (2) 70 (3) 120 (4) none of these

52. 12 persons meet in a room and each shakes hands with all the others. How many handshakes are there?

(1) 144 (2) 132 (3) 72 (4) 66

53. In how many ways can we select 9 balls out of 6 red balls, 5 white balls and 5 blue balls if 3 balls of each colour are selected?

(1) 40 (2) 200 (3) 2000 (4) 400

54. In how many ways can a cricket team be chosen out of a batch of 15 players, if a particular player is always chosen?

(1) 1364 (2) 364 (3) 1001 (4) none of these

55. In how many ways can a cricket team be chosen out of a batch of 15 players, if a particular player is never chosen?

(1) 364 (2) 1001 (3) 1364 (4) none of these

56. For the post of 5 teachers, there are 23 applicants. 2 posts are reserved for SC candidates and there are 7 SC candidates among the applicants. In how many ways can the selection be made?

(1) 5880 (2) 11760 (3) 3920 (4) none of these

57. In how many ways can 5 white balls and 3 black balls be arranged in a row so that no two black balls are together?

(1) 192 (2) 40 (3) 20 (4) 120

58. In an examination, a candidate has to pass in each of the five subjects. In how many ways can he fail?

(1) 5 (2) 10 (3) 21 (4) 31

59. An examination paper contains 12 questions consisting of two parts, A and B. Part A contains 7 questions and part B contains 5 questions. A candidate is required to attempt 8 questions, selecting at least 3 from each part. In how many ways can the candidate select the question?

(1) 210 (2) 175 (3) 420 (4) none of these



ANSWERS SHEET

1.4 2.3 3.2 4.1 5.3 6.1 7.4

8.2 9.4 10.3 11.3 12.3 13.2 14.2

15.4 16.4 17.3 18.4 19.2 20.3 21.3

22.4 23.3 24.1 25.3 26.2 27.4 28.3

29.2 30.3 31.3 32.2 33.1 34.2 35.4

36.2 37.4 38.4 39.2 40.2 41.3 42.3

43.3 44.2 45.2 46.3 47.4 48.4 49.2

50.3 51.3 52.4 53.3 54.3 55.1 56.2

57.3 58.4 59.3

SOLUTION

1. 1

6! +

1

7! =

𝑥

8!

8 ×7

8 ×7 ×(6!) +

8

8 ×(7!) =

𝑥

8!

56

8! +

8

8! =

𝑥

8!

x = 56 + 8 = 64.

2. 𝑛−1𝑃3

𝑛𝑃4

= 1

9

(𝑛−1)!

(𝑛−1−3)!

(𝑛−4)!

𝑛 ! =

1

9

1

𝑛 =

1

9 n = 9.

3. 𝑛𝑃𝑛

𝑛𝑃𝑛−2

= (n!) {𝑛−(𝑛−2)}!

𝑛 ! = 2! = 2.

4. 2730 = (15 14 13) = 15P3 r = 3.

5. 7P3 = 7!

(7−3)! =

7 ×6 ×5 × 4!

4! = 210.

6. 𝑛𝑃5

𝑛𝑃3

= 20 𝑛 !

(𝑛−5)!

(𝑛−3)!

𝑛 ! = 20

𝑛−3 𝑛−4 {(𝑛−5)!}

(𝑛−5)! =

20 (n - 3) (n - 4) = 20 n2 – 7n – 8 = 0

(n - 8) (n + 1) = 0 n = 8.

7. 15𝑃𝑟−1

16𝑃𝑟−2

= 3

4

15!

{15−(𝑟−1)}!

{16−(𝑟−2)}!

16! =

3

4

1

16

(18−𝑟)!

(16−𝑟)! =

3

4 (18 - r) (17 - r) = 12

r2 – 35r + 294 = 0 (r - 21) (r - 14) = 0

r = 14 [r 16].

8. nCP = nCq p + q = n. nC10 = nC14

n = (10 + 14) = 24.

9. nC3 = 220 𝑛 𝑛−1 (𝑛−2)

6 = 220

n (n -1)(n - 2) = 1320

n = 12 [ 12 11 10 = 1320].

10. We know that nCr + nCr+1 = n+1Cr+1. So, x = (r + 1).

11. 36C34 = 36C(36-34) = 36C2 =

36 × 35

2 = 630.

12. 𝑛𝐶𝑟

𝑛𝐶𝑟−1

= 𝑛 !

𝑟 ! ×(𝑛−𝑟)! ×

𝑟−1 !×(𝑛−𝑟+1)!

𝑛 ! =

𝑟−1 !× 𝑛−𝑟+1 × 𝑛−𝑟 !

𝑟 . 𝑟−1 ! ×(𝑛−𝑟)! =

(𝑛−𝑟+1)

𝑟

13. nC18 = nC12 n = (18 + 12) = 30.

32Cn = 32C30 = 32C2 = 32 ×31

2= 496.

14. 60C60 = 1 [nCn = 1].

15. Required number of ways = 5P3 = 5!

(5−3)!

= 5 × 4 ×3 ×2 ×1

2 ×1 = 60.

16. Required number of ways = 5P5 = 5! =

(54321) = 120.

17. Required number of ways = 4! = (4321) = 24. 18. Required number of ways = 10P3

= (10 9 8) = 720.

19. Required number of ways = (432) = 24. 20. Out of 6 periods, 5 may be arranged for 5

subjects in 6P5 ways. Remaining 1 period may be arranged for any one of the five subjects in 5P1ways.

required number of ways = (6P5 5P1)

= (654325) = 3600. 21. Required number of words = number of

arrangements of 5 letters taken all at a time

= 5P5 =5! = (54321) = 120. 22. In a row of 9 seats, the 2nd, 4th, 6th, and 8th are the

even places. These 4 places can be occupied by 4 woman in 4P4 ways = 24 ways. Remaining 5 places can be occupied by 5 men in 5P5 ways = 120 ways.

total number of seating arrangements

= (24 120) = 2880. 23. The hundreds place can be filled by any of the 9

nonzero digits. So, there are 9 ways of filling this place. The tens place can be filled by any of the 10 digits. So, there are 10 ways of filling it. The units place can be filled by any of the 10 digits. So, there are 10 ways of filling it.

total number of 3-digit number

= (91010) = 900. 24. The hundreds place can be filled by any of the 9

nonzero digits. So, there are 9 ways of filling the hundreds place. The tens digit can be filled by any of the remaining 9 digits. So, there are 9



ways of filling the tens place. The units place can now be filled by any of the remaining 8 digits. So, there are 8 ways of filling the units digit.

Required number of numbers = (998) = 648. 25. Thousands place can be filled by any of the 6

nonzero digits. So, there are 6 ways to fill this place. Hundreds place can be filled by any of the remaining 6 digits. So, there are 6 ways to fill this place. Tens place can be filled by any of the remaining 5 digits. So, there are 5 ways to fill this place. Units place can be filled by any of the remaining 4 digits. So, there are 4 ways to fill this

place. Required number of numbers = (6654) = 720.

26. Number with 0 at units place = (541) = 20.

Numbers with 2 at units place = (441) = 16.

Numbers with 4 at units place = (4 41) = 16. Total numbers = (20 + 16 + 16) = 52.

27. Clearly, thousands digit is 6. Number of numbers with units digit 0 = (1× 8 × 7 × 1) = 56. Number of numbers with units digit 5 = (1 × 8 × 7 × 1) = 56. Required number of numbers = (56 + 56) = 112.

28. Each place of the number can be filled in 2 ways.

required number of numbers = 210. 29. Fixing T at the beginning and E at the end, the

remaining 6 letters can be arranged in 6 places in 6! = 720 ways.

required number of words = 720. 30. Take all the vowels A U E together and take them

as one letter. Then, the letters to be arranged are D, G, H, T, R (A U E). These 6 letters can be arranged in 6 places in 6! ways. Now, 3 letters A, U, E among themselves can be arranged in 3! = 6 ways.

required number of words = 6! × 6 = (6×5×4×3×2×1×6) = 4320.

31. Total number of words formed by using all the 8 letters at a time =8P8 = 8! = 40320. Number of words in which vowels are never together = (total number of words) – (no. of words in which vowels are always together) = (40320 - 4320) = 36000.

32. The given word has 7 letters out of which there are 3 vowels and 4 consonants. Let us mark the

positions of these letters as (1) (2) (3) (4) (5) (6) (7). Now, the 3 vowels can be placed at any of the 3 places out of four marked 1, 3, 5, 7.

number of ways of arranging vowels = 4P3

= (4×3×2) = 24. Now, 4 consonants may be arranged at the remaining four positions in 4P4

= 4! = 24 ways. Required number of ways = (24× 24) = 576.

33. Keeping EN together and considering it as one letter, we have to arrange 5 letters at 5 places. This can be done in 5P5= 5! = 120 ways.

34. There are in all 5 letters out of which there are 2P, 1A, 1L and 1E.

required number of ways = 5!

2! 1! 1! (1!)= 60.

35. There are 9 letters in all. Out of these A is repeated 4 times, L is repeated 2 times and the rest are different.

Required number of words. = 9!

4! .(2!) = 7560.

36. There are 6 letters in all. Out of these A is repeated thrice, 8 is repeated twice and C is taken only once.

required number of words = 6!

3! 2! (1!)

= 6 ×5 ×4 ×3 ×2 ×1

3×2×2 = 60.

37. Let us keep the two particular books together and treat them as one. Now, 9 books can be arranged among themselves in 9! ways. Also, 2 books can be arranged among themselves in 2! = 2 ways.

required number of ways = 2 × 9!. 38. Number of ways in which 10 books may be

arranged = 10!. Number of ways in which 10 books may be arranged with two particular books together = (2×9!). Required number of ways in which 2 particular books are never together = (10!) – (2×9!) = (10×9!) – (2×9!) = (8 × 9!).

39. Clearly, 0 cannot be placed at the thousands place. So, this place can be filled in 9 ways. Each of the hundreds, tens and units digits can be filled in 10 ways.

required number of numbers = (9×10×10×10) = 9000.

40. 6 boys can be arranged in a row in 6! ways.



41. 6 girls can be arranged in a circle in 5! ways. 42. In a polygon of n sides, number of diagonals

= 1

2 n (n-3)

43. Putting n = 8 in 1

2 n (n-3), we get 20.

number of diagonals in an octagon = 20.

44. 1

2 n (n-3) = 54 n (n-3) = 108 n2 – 3n – 108 = 0

n2 – 12n + 9n – 108 = 0 n (n-12) + 9(n - 12)

= 0 (n -12) (n + 9) = 0 n = 12. 45. Number of line segments formed by joining pairs

of points out of 10 =10C2 = 10 ×9

2 = 45.

Number of line segments formed by joining pairs

of 4 points = 4C2 = 4 × 3

2 = 6.

But, there points being collinear give only one line.

required number of line segments = (45 – 6 + 1) = 40.

46. Number of triangles obtained form 10 points = 10C3 =

10 × 9 ×8

3 ×2 ×1 = 120.

Number of triangles obtained from 4 points =4C3 = 4C1 = 4. But, these 4 points being collinear will give no triangle.

required number of triangles = (120 - 4) = 116. 47. Number of ways of selecting 3 consonants out of

7 and 2 vowels out of 4 = (7C3 × 4C2)= 7 × 6 × 5

3 × 2 × 1 ×

4 × 3

2 ×1 = 210.

Now, 5 letters can be arranged among themselves in 5! ways = 120. Required number of words = (210 × 120) = 25200.

48. Number of ways of selecting 3 men out of 6 and 2

ladies out of 5 = (6C3× 5C2) = 6 ×5 ×4

3 ×2 ×1 ×

5 × 4

2 ×1 =200.

49. We may have. (1) 1 women and 2 men or (2) 2 women and 1 man.

required number of ways

= (2C1× 5C2) + (2C2× 5C1) = 2 ×5 × 4

2 ×1 + (1 × 5)

= (20 + 5) = 25.

50. We may have : (1) (1 lady out of 4) and (4 gents out of 6) or (2) (2 ladies out of 4) and (3gents out of 6).

required number of ways = (4C1× 6C4) + (4C2× 6C3)

= 4 ×6 × 5

2 ×1 + (6 × 20) = (60 + 120) = 180.

51. Required number of teams = 10C7 = 10C3

= 10 × 9 × 8

3 × 2 ×1 = 120.

52. Number of handshakes = 12C2 = 12 × 11

2 = 66.

53. Required number of ways = (6C3 × 5C3 × 5C3)

= (6C3 × 5C2 × 5C2) = (6 × 5 × 4

3 × 2 ×1×

5 × 4

2 ×1×

5 × 4

2 ×1) = 2000.

54. When a particular player is always chosen, then we have to select 10 players out of 14.

required number of ways = 14C10 = 14C4

= 14 × 13 × 12 ×11

4 ×3 × 2 ×1 = 1001.

55. When a particular player is never chosen, then we have to select 11 players out of 14.

required number of ways = 14C11 = 14C3

= 14 × 13 × 12

3 × 2 ×1= 364.

56. We have to select 2 posts out of 7 SC and 3 posts out of 16. Required number of ways

=( 7C2 × 16C3 ) = ( 7 × 6

2 ×

16 × 15 × 14

3 × 2 ×1) = 11760

57. Let us arrange the white balls (shown by W) and leave a space in between every pair as shown below. X W X W X W X W X W X Now, 3 black balls may be arranged in 6 places in 6C3 ways =

6 × 5 ×4

3 ×2 ×1 = 20.

58. The candidate can fail by failing in 1 or 2 or 3 or 4 or 5 subjects out of 5 in each case.

required number of ways = 5C1 + 5C2+ 5C3+ 5C4+5C5 = 5C1 + 5C2+ 5C(5-3)+ 5C(5-4)+1 = 5C1 + 5C2+ 5C2+ 5C1+1 = 2 (5C1 + 5C2) + 1 = 2 (5 +

5 × 4

2 ×1 ) + 1

= (30 + 1) = 31. 59. He must select :

(1) (3 out of 7 from A) and (5 out of 5 from B) or (2) (4 out of 7 from A) and (4 out of 5 from B) or (3) (5 out of 7 from A) and (3 out of 5 from B). The number of ways of these selections are : (1) 7C3× 5C5= (35 × 1) = 35 (2) 7C4× 5C4= (35 × 5) = 175 (3) 7C5× 5C3= (21 × 10) = 210.

required number of ways = (35 + 175 + 210) = 420.



SETS, RELATIONS & FUNCTIONS

1. Which one of the following is a null set?

(1) ,x │x R, 4x2 – 1 = 0}

(2) ,x │ x N, x is odd, x + 3 is even}

(3) ,x │ x R, x2 < 3 }

(4) ,x │ x R, x2 < 0}

2. If P = {3m : m N} and Q = {3n : n N}, then

(1) P Q (2) Q P

(3) P = Q (4) P Q = N 3. The number of all possible subsets of a set

containing n elements is: (1) n (2) 2n (3) n! (4) 2n

4. The number of non-empty subsets of {a, b c, d} is (1) 3 (2) 4 (3) 15 (4) 16

5. Which one of the following statements is correct?

(1){1} {1, 2} (2) 1 {1, 2}

(3) {1, 3} ,x│x is a + ve integer-

(4) {1, 2} {2, 1} 6. If A and B are two sets given in such a way

that n (A) = 70, n (B) = 60 and n (AB) = 110,

then n (A B) = ? (1) 240 (2) 50 (3) 40 (4) 20

7. If A = {1, 2, 3, 4}, B ={2, 4, 5, 6}and C = {1, 2,

5, 7, 8}, then (A C) B = ? (1) {1, 2, 5} (2) {2, 4, 5} (3) {1, 2, 4, 5, 7, 8} (4) {1, 2, 3, 4, 5, 7, 8}

8. If U = ,1, 2, 3, 4…..10-, A =,1, 3, 5, 7- and B =

{1, 2, 3, 4, 5}then (A' B') = ? (1){2, 4, 6, 8, 9, 10} (2){2, 6, 7, 8, 9, 10} (3){2, 4, 6, 7, 9, 10} (4){2, 4, 6, 7, 8, 9, 10 }

9. The number of all possible subsets of A {1, {2, 3}} is (1) 2 (2) 4 (3) 6 (4) 8

10. If A = {x : x is a multiple of 4} and B = {x : x is a

multiple of 6}, then A B consists of all multiples of: (1) 24 (2) 2 (3) 12 (4) 4

11. If x is a positive integer, then the solution set of the equation x + 2 = 0 is: (1){-2} (2){2}

(3) (4){} 12. If A is the set of all positive integers and B is the

set of all negative integers, then A B = ?

(1){0} (2)

(3) 0 (4){} 13. Which of the following is an infinite set ?

(1){x : x N, x < 50}

(2) {x : x I, x < 50}

(3){x : x I, x is a factor of 50} (4){x : x is a whole number , x < 50}

14. Which of the following is a finite set?

(1){x : x I, x < 1}

(2){x : x R, 0 < x < 1}

(3){x : x N, x > 5} (4) None of these

15. Which of the following is a singleton set?

(1){x : x R, x2 = x}

(2) {x : x N, 3x = 4}

(3){x : x R, x2 = -1} (4){x : x is an integer which is neither + ve nor- ve}

16. Which of the following is a correct statement?

(1) {a, b, c} (2){} {a, b, c}

(3) {a, b, c} (4) None of these 17. Which one of the following is an empty set?

(1){x : x N : 5x = 9}

(2){x I : 0 < x < 4 & x is even}

(3){x I : (x - 1) (x - 2) = 0 & x is odd} (4){0}

18. If X = {(4n – 3n -1) : n N} and Y = {9 (n -1) : n N}, then

(1) X Y (2) Y X

(3) X Y = (4) None of these 19. If A = {a, b}, then the power set of A is:

(1){aa, bb} (2){ab, ba}

(3){a2, b2} (4){,{a},{b},{a, b}}



20. If A = {1, 2,{3, 4}}, then which of the following is a correct statement?

(1) 3 A (2){1} A

(3){2} A (4) 4 A 21. The number of all possible proper subsets of

{1, 3, 5} is (1) 8 (2) 7 (3) 3! (4) 3

22. Which of the following is a true statement?

(1){x R :│x│= -1} =

(2){x R : 1 < x < 2 } is finite (3){a, b, c 1, 2, 3} is not a set

(4){a, b} {{a} b, c} 23. If A has 3 elements and B has 6 elements, then

the minimum number of elements in A B is : (1) 3 (2) 6 (3) 9 (4) 18

24. If A =,1, 2, 3 ….9-, B = ,2, 4, 6, 7, 8- and C = ,3, 4,

5, 8, 9, 10}, then (A - B) C = ? (1){1, 3, 4, 5, 8, 9, 10} (2){1, 2, 3, 4, 5, 6, 7, 8, 9} (3){2, 4, 6, 7, 8} (4){1, 3, 4, 5, 8, 9}

25. Which of the following is a true statement?

(1) (A B)'= (A'B')

(2) (AB)' = (A B')

(3) (A B)' = (A'B)

(4) (AB)' = (A'B') 26. For any three sets A, B and C :

(1) A – (BC) = (A-B) (A-C)

(2) A – (BC) = (A-B) (A-C)

(3) A – (BC) = (AB) – (AC) (4) None of these

27. In a survey, it was found that 63% Indians like apples and 76% like oranges. How many Indians like both? (1) 13% (2) 6.5% (3) 19.5% (4) 39%

28. In a group of 1000 persons, 750 can speak English and 400 can speak Hindi. How many persons can speak Hindi only? (1) 150 (2) 250 (3) 350 (4) 600

29. In a group of 52 persons, 16 drink tea but not coffee, while 33 drink tea. How many persons drink coffee but not tea? (1) 17 (2) 19 (3) 36 (4) 23

30. In a class of 45 students, 22 can speak Hindi only and 12 can speak English only. The number of students who can speak both Hindi and English is: (1) 11 (2) 23 (3) 33 (4) 34

31. Let S be the set of all straight lines in a plane. A

relation R is defined on S by a R b a b, Then R is: (1) Reflexive but neither symmetric nor transitive (2) Symmetric but neither reflexive nor transitive. (3) Transitive but neither reflexive nor symmetric. (4) An equivalence relation.

32. Let A = {1, 2, 3}. Then, the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)} on A is: (1) Reflexive and transitive but not symmetric. (2) Symmetric and transitive but not reflexive. (3) Reflexive and symmetric but not transitive. (4) Reflexive, symmetric & transitive.

33. Let R be a relation in the set I of all integers,

defined by : a R b (a - b) is divisible by 3. Then, R is (1) An equivalence relation. (2) Reflexive and symmetric but not transitive. (3) Symmetric and Transitive, but not reflexive. (4) None of these.

34. Let S be the set of all real numbers. Define a

relation R on S by : a R b │a│ b. Then, R is (1) Reflexive but neither symmetric nor transitive. (2) Symmetric but neither reflexive nor transitive. (3) Transitive but neither reflexive nor symmetric. (4) None of these.

35. Let f : N N : f (x) = 2x, Then, f is:



(1) One-one, onto (2) One-one, into (3) Many-one, onto (4) Many-one, into

36. Let f : R R : f (x) = x2, Then, f is: (1) One-one, onto (2) One-one, into (3) Many-one, onto (4) Many-one, into

37. If f (x) = (x2 - 1) and g (x) = (3x + 1) then (gof) (x) = ? (1) 9x2 + 6x (2) 3x2 - 1 (3) 2x2 - 1 (4) 3x2 - 2

38. If f (x) = *x+ and g (x) =│x│, then (go f) (-15/3) – (f o g) (-5/3) = ? (1) 0 (2) 1 (3) 2 (4) 1/2

39. Let f (x) = 1/(1 - x), then (f o f o f) (x) = ? (1) x/(1 - 3x) (2) 1/(1-3x) (3) x (4) None of these

40. If f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)} then g o f = ? (1){(1,3), (3, 1), (4, 3)} (2) {(1,5), (2,5), (5, 2)} (3){(3,1), (1,3), (3,4)} (4) {(5,1), (5,2), (2,5)}

41. Let f (x) = {9 – x2} then Dom (f) = ?

(1) ] - - 3 ] (2) [3, ]

(3) [-3, 3] (4) ]- , -3] [3, ]

42. Let f (x) = 𝑥−1

𝑥−4 . Then, Dom (f) = ?

(1) [1, ] (2) [4, [

(3) ]-, -1] ]4, [ (4) None of these

43. Let f (x) = log (1 - x) + {x2 – 1}. Then, Dom (f) = ?

(1) ]1, ] (2) ]-, -1] (3) [-1, 1[ (4) ]0, 1]

44. The domain of f (x) = sin-1 2x is (1) [-1, 1] (2) [-1/2, 1/2]

(3) [-/2, /2] (4) [-/4, /4] 45. let f (x) = x/(x2 – 3x + 2). Then, Dom (f) = ?

(1) R (2) R – {1} (3) R – {1, 2} (4) None of these

46. Let g (x) = 1/(1 – x2). Then Range (g) = ?

(1) [1, [ (2) ]-, 1] (3) [-1, 1] (4) None of these

47. The range of f (x) = x2/(1 + x2) is :

(1) [1, [ (2) [0, 1[ (3) [-1, 1] (4) ]0, 1]

48. The range of f (x) = (3x2 + 7x +10) is

(1) [70/3, [ (2) [71/12, [

(3) [0, [ (4) ]-, -3] 49. The range of f (x) = cosh x is:

(1) R (2) [1, [

(3) [0, [ (4) None of these 50. The range of f (x) = x + 1/x, is :

(1) [2, [ (2) [-2, 2]

(3) ]-, -2] (4) None of these

ANSWER SHEET

1.4 2.2 3.4 4.3 5.4 6.4 7.2

8.4 9.2 10.3 11.3 12.2 13.2 14.4

15.4 16.3 17.1 18.1 19.4 20.2 21.2

22.1 23.2 24.1 25.4 26.2 27.4 28.2

29.2 30.1 31.2 32.3 33.1 34.3 35.2

36.4 37.4 38.2 39.3 40.1 41.3 42.3

43.2 44.2 45.3 46.1 47.2 48.2 49.2

50.4

Explanation

1. since there is no real number whose square is negative, so the set given in (d) is a null set.

2. P = ,3, 6, 9, 12, 15 …..- Q = ,3, 9, 27, 81, …..- Clearly, every element of Q is in P.

Q P. 3. do yourself 4. Number of non-empty subsets of {a, b, c, d}

is (24 - 1) = 15. 5. do yourself



6. n (AB) = n (A) + n (B) – n (A B) = (70 + 60 - 110) = 20.

7. AC = {1, 2, 3, 4, 5, 7, 8}

(AC) B = {2, 4, 5} 8. A' = {2, 4, 6, 8, 9, 10}, B' = {6, 7, 8, 9, 10}

A' B' = {2, 4, 6, 7, 8, 9, 10} 9. The set A has 2 elements. So, it has 22 = 4

subsets. 10. A = ,4, 8, 12, 16, 20, 24, …..-

B = ,6, 12, 18, 24, 30, …..-

A B = ,12, 24,……- = {x : x is a multiple of 12} Note l.c.m. of 4 and 6 is 12.

11. There is no positive integer satisfying the equation,

x + 2 = 0.

its solution set is

12. A B = 13. do yourself 14. do yourself 15. do yourself

16. is a subset of every set. 17. do yourself 18. Putting n = 1, 2, 3, 4, ……, we get:

X = ,0, 9, 54, …- and Y = ,0, 9, 18, 36, 54, …..-

x y 19. Power set of A is the set of all possible

subsets of A. 20. do yourself 21. Number of all proper subsets of {1, 3, 5} is (23

-1). i.e. 7. 22. There is no real number whose modulus is -1.

23. Clearly, A B has at least 6 elements. 24. (A - B) = {1, 3, 5, 9}

(A - B) C = {1, 3, 4, 5, 8, 9, 10}. 25. By De-Morgan’s Law

(A B)' = (A' B').

26. A – (B C) = (A – B) (A - C). 27. Let A and B denote the sets of Indians who

like apples and oranges respectively. Then,

n (AB) = 100, n (A) = 63 and n (B) = 76.

n (AB) = n (A) + n (B) – n (AB) = (63 + 76 – 100) = 39.

28. n (E) = 750, n (H) = 400 and n (E H) = 1000.

N (E H) = n (E) + n (H) – n (E H) (750 + 400 - 1000) = 150.

n (E H) + n (H - E) = n (H)

n (H - E) = n (H) – n (E H) = (400 - 150) = 250.

29. Let A and B be the sets of persons who drink tea and coffee respectively. Then,

n (AB) = 52, n (A - B) = 16, n (A) = 33.

n (A) + n (B - A) = n (AB)

n (B - A) = n (AB) – n (A) = (52 - 33) = 19.

30. n (AB) = n (A – B) + n (AB) + n (B - A)

n (AB) = n (AB) – n (A - B) – n (B - A) = (45 – 22 - 12) = 11

31. a b b a.

R is symmetric. No line is perpendicular to itself. So, R is not reflexive.

a b, b c does not imply that a c. So, it is not transitive. 32. Since (1, 1), (2, 2), (3, 3) are in R, it is

reflexive.

Also, (a, b) R (b, a) R,

R is symmetric.

But, (3, 2) R, (2, 1) R, while (3, 1) R.

R is not transitive. 33. a – a = 0, is divisible by 3.

a R a for each a I.

R is reflexive.

a R b (a – b) is divisible by 3

(b - a) is divisible by 3

b R a.

R is symmetric. a R b and b R c

(a – b) is divisible by 3 and (b - c) is divisible by 3

[(a - b) + (b - c)] is divisible by 3

a R c. R is transitive. Hence, R is an equivalence relation. 34. Let a = -1 then, │a│= 1, which is not less than

or equal to -1.

R is not reflexive. -1 R 1, but 1 is not related to -1. R is not symmetric.



Let a R b and b R c. Then, │a│ b and │b│ c

│a│ b │b│ c, │a│ c, a R c.

R is transitive. Hence, R is transitive but neither reflexive not symmetric.

35. f (x1) = f (x2) 2x1 = 2x2, x1 = x2

f is one – one.

But, 1 N such that 1 has no pre-image in N.

f is into, f is one - one, into.

36. -1 R and 1 R such that f (-1) = (-1)2 = 1 and f (1) = 12 = 1.

f (-1) = f (1), f is Many – one.

-1 R having no pre image in R. i.e. there

exists no x R s.t. f (x) = -1.

f is into, Hence, f is Many – one, into. 37. (g o f) (x) = g [f (x)] = g (x2 - 1)

= 3 (x2 - 1) + 1 = (3x2 - 2) 38. (g o f) (-5/3) = g [f (-5/3)]

g = [-2 + 1/3] = g (-2) = │-2│ = 2. (f o g ) (-5/3) = f [g (-5/3)+ = f (│-5/3│) = f (5/3) = f (1 + 2/3) = [1 + 2/3] = 1.

Given exp. = (2 -1) = 1. 39. (f o f) (x) = f [f (x)]

[f o f o f] (x) = f [(f o f) (x)] = x.

40. Dom (g o f) = Dom (f): (g o f) (1) = f [f (1)] = g (2) = 3. (g o f) (3) = f [f (3)] = g (5) = 1. (g o f) (4) = f [f (4)] = g (1) = 3.

g o f = {(1, 3), (3, 1), (4, 3)}

41. f (x) is defined when (9 – x2) 0

i.e.when x2 9, i.e, when -3 x 3

Dom (f) = [-3, 3] 42. f (x) is defined when

x – 4 0 and {x-1}/{x-4} 0

i.e, x 4 and (x 4 or x 1)

i. e, when x > 4 or x 1

Dom (f) = ]- , 1] ] 4, [ 43. Let f (x) = g (x) + h (x), where

g (x) = log (1 - x) and h (x) = {x2 – 1}. g (x) is defined only when 1 – x > 0.

i.e, when x < 1, Dom (g) = ] - , 1 [.

h (x) is defined when x2 – 1 0

i.e, when x2 1

i.e, when x 1 or x - 1.

Dom (h) = ] - -1 ] [1, [.

Dom (f) = Dom (g) Dom (h)

= ] - , -1]

44. sin-12x is defined only when 2x [-1, 1]

i.e, when x [-1/2, 1/2].

Dom (f) = (-1/2, 1/2) 45. f (x) = x/{(x - 1) (x - 2)}

Clearly, f (x) is not defined when x – 1 = 0 or x – 2 = 0 i. e, when x = 1 or x = 2

Dom (f) R – {1, 2}

46. y = 1/(1 – x2) x = 1- 1

𝑦

Clearly, x is not defined when y = 0 or 1- 1/y < 0 i.e, when y = 0 or y < 1

Range (g) = [1, [

47. y = x2/(1+x2) x = 𝑦

1−𝑦

x is defined when y/(1-y) 0

and (1-y) 0.

i.e when 0 y 1.

48. y = 3x2 + 7x + 10 3x2 + 7x + (10 - y) = 0

x = -7 {12y – 71}/6 Clearly, x is defined when

12y – 71 0, i.e, y 71/12.

Range (f) = [71/12, [ 49. f (x) = cosh x = 1/2 (ex + e-x)

Clearly, f (x) 1 for all x R.

Range (f) = [1, [.

50. y = {x2 +1} /x x2 – xy + 1 = 0

x = {y y2 – 4 }/2

Clearly, x is defined when y2 – 4 0

i.e, when y2 4

i.e, when y 2 or y -2

Range (f) ] -, -2] [2, [.



Complex Number

1. If 𝑥−1

3+𝑖 +

𝑦−1

3−𝑖 = i then

(1) x = -4, y = 6 (2) x = 4, y = -6 (3) x = 4, y = 6 (4) x = -4, y = -6

2. {7 – 24i }= ?

(1) (3 – 4i) (2) (4 – 3i)

(3) (4 + 3i) (4) (3 + 4i)

3. (-3 2 + 3 2i) when expressed in polar form, is:

(1) 6 (cos 𝜋

4 + i sin

𝜋

4)

(2) – 6 (cos 𝜋

4 – i sin

𝜋

4)

(3) 6 (cos 3𝜋

4 + i sin

3𝜋

4 )

(4) None of these 4. Show that the locus of a complex number z

satisfying arg ( 𝑧−1

𝑧+1 ) =

𝜋

2 is a circle. Find its centre

and the radius. (1) (1,0), 3 (2) (1,1), 4 (3) (0,0), 1 (4) (1,0), 1

5. The value of (−1+𝑖√3

2)29 + (

−1−𝑖√3

2)29 is

(1) -2 (2) 2 (3) 1 (4) -1

6. The value of (𝑖+√3

2)100 + (

𝑖−√3

2)100 is

(1) 1 (2) -1

(3) (4) 2 7. (2 + i)3 = ?

(1) 8 + 6i (2) 2 + 11i (3) 8 – 3i (4) None of these

8. (i39 + 1

𝑖69 ) = ?

(1) 0 (2) 2i (3) – 2i (4) 1 - i

9. i109 + i114 + i119 + i124 = ? (1) i (2) 2 (3) -2 i (4) 0

10. 6i54 + 5i37 – 2i11 + 6i68 = ? (1) 0 (2) 7i (3) 13i (4) -3i

11. -16 -9 = ?

(1) 12 (2) 12 (3) -12 (4) None

12. The smallest integer n for which (1+𝑖

1−𝑖)n = 1 is

(1) 4 (2) 8 (3) 12 (4) 16

13. The smallest integer n for which (1 + i)2n = (1 - i)2n is

(1) 2 (2) 4 (3) 8 (4) 12

14. (5 - -7) (5 + -7) (1 - -1) (1 + -1) = ? (1) 25 + 7i (2) 29 – 3i (3) 32 + 5i (4) None of these

15. If a2 + b2 = 1, then (1+𝑏+𝑖𝑎

1+𝑏−𝑖𝑎) = ?

(1) a + ib (2) b + ia (3) i (a + b) (4) None of these

16. (1 – 2i)-3 = ?

(1) (- 11

125 –

2

125i)

(2) (- 11

125 +

2

125i)

(3) (11

125 –

2

125i)

(4) None of these

17. (-2 + 5i) (2 – 3i) = ?

(1) (19

13 –

4

13i) (2) (-

19

13 +

4

13i)

(3) (- 19

13–

4

13) (4) None of these

18. The multiplicative inverse of (2 + i)2 is

(1) (3

25 +

4

25i) (2) (-

3

25 +

4

25𝑖)

(3) (3

25 –

4

25i) (4) None of these

19. Arg (1+i) = ?

(1) (2) /2

(3) /3 (4) /4

20. Arg (-1+i3) = ?

(1) /3 (2) 2/3

(3) (4) None 21. (sin 1200 – i cos 1200) when expressed in polar

form is : (1) cos 600 + i sin 600 (2) cos 300 + i sin 300 (3) cos 1500 + i sin 1500 (4) None of these

22. The complex number z such that │𝑧−𝑖

𝑧+𝑖│= 1 lies on:

(1) the x-axis (2) the line y = 1 (3) a circle (4) None of these

23. The complex number z such that │𝑧−5𝑖

𝑧+5𝑖│= 1 lies

on: (1) the y-axis (2) the x-axis



(3) on a circle (4) None of these

24. The argument of (1−𝑖√3

1+𝑖√3) is:

(1) 600 (2) 1200 (3) 2100 (4) 2400

25. {5+12i} = ?

(1) (2 +3i) (2) (3+2i)

(3) (2-3i) (4) (3-2i)

26. 2i = ? (1) (1+i) (2) (1-i)

(3) -2i (4) None of these

27. The amplitude of 1+𝑖√3

√3+1 is

(1) /6 (2) /4

(3) /3 (4) None

28. (1+2𝑖

1−𝑖) lies in

(1) I quadrant (2) II quadrant (3) III quadrant (4) IV quadrant

29. (1−𝑖

1+𝑖)2 = ?

(1) 1 (2) -1/2

(3) 1/2 (4) -1

30. If i = -1 and n is a positive integer, then (in + in+1 + in+2 + in+3) is equal to

(1) 1 (2) i (3) in (4) 0

31. {1

1−2𝑖 +

3

1+𝑖} (

3+4𝑖

2−4𝑖) = ?

(1) (1

2 +

3

2 i) (2) (

1

4 +

9

4i)

(3) (3

4 –

1

2i) (4) None of these

32. If z is a complex number, then arg (z) + arg (𝑧) = ?

(1) 0 (2) /4

(3) /2 (4) 2 33. If │z + (𝑧) │= │z - (𝑧) │, then the locus of z is

(1) a pair of straight lines (2) a rectangular hyperbola (3) a line (4) a set of four lines

34. If │z + (𝑧) │+ │z - (𝑧) │= 2, the z lies on (1) a straight line (2) a set of four lines (3) a circle (4) None of these

35. If (2−√−25

1−√−16) = x + iy, then

(1) x = 3/17, y = 11/17 (2) x = 22/17, y = 3/17 (3) x = 9/17, y = 5/17

(4) None of these

36. If (a + ib) = (1+𝑖)

√(1−𝑖), then the value of (a2 + b2) is

(1) 1 (2) -1 (3) 2 (4) -2

37. If 1, , 2 be cube roots of unity, then

(1-+2)5 + (1+-2)5 = ? (1) 4 (2) 8 (3) 16 (4) 32


(1+)3 – (1+2)3 = ? (1) 2 (2) -2

(3) 0 (4) 2


(1-+2) (1-2+4) (1-4+8)……. to 2n factors is equal to :

(1) 2n (2) 22n (3) 0 (4) 1

40. (−1+√−3

2)100 + (

−1−√−3

2)100 = ?

(1) 2 (2) 0 (3) -1 (4) 1

41. If 1, , 2 be cube roots of unity, then the value

of (2+5+22)6 is (1) 576 (2) 625 (3) 729 (4) None

42. If 1, , 2 be cube roots of unity, then the value

of (2 -) (2 - 2) (2 - 10) (2 - 11) is (1) 49 (2) 36 (3) 56 (4) None

43. (−1+𝑖√3

2)17 + (

−1−𝑖√3

2))17

(1) 1 (2) -1 (3) 0 (4) None



ANSWER SHEET

1.1 2.2 3.3 4.3 5.4 6.2 7.2

8.3 9.4 10.2 11.3 12.1 13.1 14.4

15.2 16.1 17.2 18.3 19.4 20.2 21.2

22.1 23.2 24.4 25.2 26.1 27.3 28.2

29.4 30.4 31.2 32.4 33.1 34.2 35.2

36.1 37.4 38.3 39.2 40.3 41.3 42.1

43.2



Explanation

1. (x-1)/(3+i) + (y-1)/(3-i) =

(x-1) (3-i) + (y-1) (3+i) = i(3+i) (3-i)

(3x+3y +6) + i (y-x) 10i

3x 3y = 0, y-x = 10

x + y = 2, y – x = 10

x = -4, y = 6

The correct answer is (a). 2. Here a = 7, b = 24.

a2 + b2 = 49 + 576 = 625 = 25

x = (a + a2 + b2/2)1/2 = (7+25/2)1/2 = 4

y = (a2 + b2 –a/2)1/2 = (25-7/2)1/2 = 3

a-ib = (x-iy) = (44-3i). The correct answer is (b).

3. Let (-32+32i) = r (cos + i sin ). Then,

r cos = -32 and r sin = 32

r2 = (18 + 18) = 36 r = 6.

cos = -1/2, sin = 1/2

tan = -1 = -tan (/4) = tan ( - /4)

tan = tan 3/4, i.e, = 3/4

-32 + 32i = 6 (cos 3/4 + i sin 3/4) 4. Let z = (x+iy). Then,

arg (z-1/z+1) = arg (z-1) – arg (z+1) = arg (x+iy -1) – arg (x + iy +1) = arg [(x-1) + iy] – arg [(x +1) + iy] = tan-1 y/(x-1) –tan-1 y/(x+1) = tan-1 {y/(x-1) – y/(x+1)/1+y/(x-1).y/(x+1)}

= tan-1 2y/(x2-1+y2) = /2

x2 -1 + y2 = 0

x2 + y2 = 1, which is a circle with centre at (0, 0) and radius 1.

5. Let -1+i3 = . Then, -1-i3/2 = 2

Given Exp. = 29 + (2)29

= 29 + 58 = 2 + = -1

The correct answer is (d).

6. i+3/2 = i(i+3/2i) = (-1+3i/2i) = /i

i-3/2 = i(i-3/2i) = (-1-3i/2i) = 2/i

(i + 3/2)100 + (i-3/2)100

= (/i)100 + (2/i)100

= (100/i100 + 200/i200) = (100 + 200)

= ( + 2) = -1. [99 = 1&198 = 1]

The correct answer is (b). 7. (2+i)3 = 23 + i3 + 6i (2+i)

= 8 – I + 12i + 6 = (2 + 11i).

8. i39 = i36+3 = i3 = -1 i69 = i68+1 = i1 = i (i39 + 1/i69) = (-I + 1/i) = (-i + i3/i4) = (-i - i) = -2i.

9. i109 + i114 + i119 + i124 = i109 (1 + i5 + 110 + i124) = i109 (1 + i + i2 + i3)

= i108+1 (1+ i -1 -i) = i 0 = 0 10. Given Exp.

= 6i52+2 + 5i36+1 – 2i8+3 + 6i68 = 6i2 + 5i – 2i3 +6 = -6 + 5i + 2i + 6 = 7i

11. -16 -9 = (4i) (3i), = 12i2 = -12 12. (1 + i/1-i) = (1+i)2/(1-i) (1+i) = 2i/2 = i

(1+i/1-i)n = 1 in = 1. Clearly, n= 4.

13. (1+i/1-i) = (1+i)2/(1-i) (1+i) = 2i/2 = i

(1+i/1-i)2n = 1 i2n = 1

(i2)n = 1 (-1)n = 1. Clearly, n = 2.

14. Given Exp. = (5 - 7i) (5+7i) (1 - i) (1 + i)

= (25 + 7) (1+1) = (322) = 64 15. (1+b+ia)/(1+b-ia)

= (1+b+ia)2/(1+b-ia) (1+b+ia) = 1 + b2 – a2 + 2b + 2iab + 2ia/(1+b)2 + a2

= 2(1+b) (b+ia)/2(1+b) = (b = ia) [Putting (1-a2) = b2]

16. (1-2i)-3 = 1/(1-2i)3 = 1/1-8i3 – 6i (1-2i)

= 1/(-11 + 2i) (-11 -2i)/(-11-2i) = (-11-2i)/125 = (-11/125 – 2/125i)

17. Given Exp.

= (-2+5i)/(2-3i) (2+3i)/(2+3i) = -19 + 4i/13 = [(-19/13) + (4/13)i]

18. z = (2 + i)2 = (4 + i2 +4i) = (3+4i)

z-1 = 1/(3+4i) (3-4i/3-4i) = (3/25 – 4/25i)

19. Let (1+i) = r (cos + i sin ). Then,

r cos = 1 and r sin = 1



r = 2

cos = 1/2 and sin = 1/2.

= /4

z = (1+i) = 2 (cos /4 + i sin /4).

arg (z) = /4

20. z = (-1 + i3)

= 2 (-1/2 + i3/2)

= 2 (cos 2/3 + i sin 2/3)

arg (z) = 2/3. 21. (sin 1200 – i cos 1200)

= sin (900 + 300) – i cos (900 + 300) = (cos 300 + i sin 300).

22. │z-1/z+1│ = 1 │x+iy-i/x+iy+i│= 1

│x+(y-1)i/x+(y+1)i│ = 1

│x+(y-1)i│2/│x+(y+1)i│2 = 12

x2 + (y - 1)2 = x2 + (y + 1)2

4y = 0, i.e. y = 0.

z lies on x - axis. 23. do your self.

24. z = (1-i3)2/(1+i3) (1- i3)

= (-2-2i3/4)

= (-1/2 -3/2i) r (cos + i sin ).

= r cos = -1/2 and r sin = -3/2

r = 1, cos = -1/2 and sin = -3/2

r = 1, = 2400

Arg z = 2400 25. Here a = 5, b = 12

a2 + b2 = 25 + 144 = 13

a ib = (x+iy), where

x = (a + a2 + b2/2)1/2 = (5+13/2)1/2 = 3

y = (a2+b2-a/2)1/2 = (13-5/2)1/2 = 2

5+12i = (3+2i). 26. Here a = 0 and b = 2

a2 + b2 = 2

x = (a + a2 + b2/2)1/2 = (0+2/2)1/2 = 1

y = (a2+b2 –a/2)1/2 = (2-0/2)1/2 = 1.

2i = (x+iy) = (1+i).

27. z = (x+iy) = 1/(3+1) + i3/(3+1)

tan = y/x = 3

= /3 28. (1+2i/1-i) = (i+2i) (1+i)/(1-i) (1+i)

= (-1/2, 3/2), which is in II quadrant. 29. (1-i/1+i) = (1-i)2/(1+i) (1-i) = -2i/2 = -1

(1-i/1+i)2 = (-i)2 = -1

30. Given Exp. in [1+i+i2+i3] = in [1+i-1-i] = 0.

31. Given Exp. = {(1+2i)/(1-2i) (1+2i) + 3(1-i)/(1+i) (1-i)} = (3+4i) (2+4i)/(2-4i) (2+4i) = [(1/5 +2/5i) + (3/2 -3/2i)] (-1/2 + i) = (17/10 – 11/10i) (-1/2 + i) = (1/4 9/4 i)

32. Let z = r (cos + i sin ). Then,

z = r (cos - i sin )

= r [cos (2 - ) + i sin (2 - )]

arg (z) = and arg (z) = 2 -

arg (z) + arg (z) = 2 33. Let z = z + i y. Then, z = z-iy.

z + z = 2x and z – z = 2iy. │z + z│= │z - z│

│2x│ = │2iy│, i.e. │x│= │y│

x = y So, z represents a pair of straight lines.

34. Let z = x+iy. Then z = x-iy. z + z = 2x and z – z = 2iy.

│z+z│+│z-z│= 2

│2x│+│2iy│= 2

│x│+│y│ = 1

x+y = 1, x-y = 1, -x+y = 1 and –x –y = 1. Thus, it represents four straight lines.

35. L.H.S = (2-5i)/(1-4i) = (2-5i) (1+4i)/(1-4i)(1+4i) = (22/3i/17) = (22/17 + 3/17i) x = 22/17, y = 3/17.

36. (a+ib) = 1+i/1-i

(a-ib) = 1-i/1+i (a2+b2) = (a+ib) (a-ib)

= 1+i/1-i . 1-i/1+i = 1.

37. 1++2 = 0

1+ = -2 and 1+2 = -

(1-+2)5 + (1+-2)5

= (-2)5 + (-22)5

= -325 -3210

= -32 (5+10) = -32 (2+) = (-32) (-1) = 32.

38. (1+)3 – (1+2)3

= (-2)3 – (-)3 [1++2= 0]

= -6+3 = -1+1 = 0 39. Given Exp.



= (1-+2) (1-2+) (1-+2)…..to 2n factors

= (-2) (-22) (-2) (-22) …. to n pairs

= (43) (43) ……to n factors

= 4 4 ….. taken n times = 4n = 22n

40. Let (-1+-3/2) = , Then (-1--3/2) = 2

Given Exp. = 100 + (2)100

= (100+200) = (99+1) + (198+2)

= 99 +198 2

= (+2) = -1.

41. (2+5+22)6 = (2+22+5)6

= [2(1+2)+5]6

= [2(-) +5]6 = (3)6

= 36

6 = 36 = 729.

42. (2-) (2-2) (2-10) (2-11)

= (2-) (2-2) (2-) (2-2)

= [(2-) (2-2)]2

= [4-2(+2) + 3]2 = [4 – 2 (-1)]2 = 72 = 49

43. Let (-1+i3/2) =

Then, (-1-i3/2) = 2

Given Exp.

= 17 + (2)17

= 17 + 34 = 2 + = -1.

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