June 16 CBSE/JEE 2019 Questions · 2019-06-16 · 4 9. Out of 1600 students in a college, 390...

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XI GRADE CBSE/JEE Questions

June 16

2019TOPICS: SETS, FUNCTIONS, TRIGONOMETRY, COMPLEX NUMBERS, MATHEMATICAL INDUCTION, INEQUALITY AND PERMUTATIONS

PART I

1


INDEX

No Name of the topic Page number

1 Sets 2 – 5

2 JEE Questions 6 - 7

3 Functions 8 - 11

4 Trigonometry 12 - 39


6 JEE Questions on trigonometric equations 48 - 51

7 JEE Questions on triangles 52 – 57

8 Complex numbers 58 – 66

9 JEE Questions 67 – 73

10 Inequality 74 – 78

11 Mathematical Induction 79 – 84

12 Permutations and combinations 85 – 96


2


SETS

1. Given, the set of natural numbers less than or equal to 20. List the elements of the

following subsets of N.

a) A, the multiples of 3.

b) A , the complement of A

c) B, the multiples of 4.

d) A – B, the difference of sets A and B.

2. If A = {1, 2, 3}, determine the total number of subsets of A. Answer: 8

3. Write the set ,70:{ 2 xx Nx } in roster form. Answer: {1, 2, 3, …8}

4. Let A = {1, 2, 3, 4 }, B = {0}, C = {2, 4, 6, …}. Indicate by a ‘T’ or ‘F’ whether

each of the following is true or false.

a) CB b) A4 c) A4

d) B e) A4 ,3 f) B

Answer: a) T b) F c) T

d) F e) F f) F

5. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9} be the universal set.

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

B = {2, 4, 6, 8}

C = {3, 4, 5, 6}.

3


Find a) BA b) CB

c) CA d) BA

Answer:

)BA {2, 4, 5, 6, 7, 8, 9}

CB {1, 3, 4, 5, 6, 7, 9}

CA {7, 8, 9}

BA {5, 7, 9}

6. Let A be the set of all primes less than 100. Let B be the set of all numbers whose unit

digit is 7. Let C be the set of all numbers whose leading digit is a multiple of 3. List

the elements of the set CBA . Answer: 97 ,67 ,6147, ,37 ,3117, ,7 ,3

7. An investigator interviewed 100 students to determine their preferences for three

drinks: Milk, Coffee and Tea. He reported the following.

10 students had all the three drinks Milk, Coffee and Tea; 20 had Milk and Coffee; 30

had Coffee and Tea; 25 Milk and Tea; 12 had milk only; 5 had Coffee only; 8 had Tea

only. Find how many did not take any of the three drinks? Answer: 20

8. In a class of 60 students, 23 play Hockey, 15 play Basketball and 20 play Cricket. 7

play Hockey and Basketball, 5 play Cricket and Basketball, 4 play Hockey and

Cricket and 15 students do not play any of these games. Find

a) How many play Hockey, Basketball and Cricket? Answer: 3

b) How many play Hockey but not cricket? Answer: 19

c) How many play Hockey and Cricket but not Basketball? Answer: 1

4


9. Out of 1600 students in a college, 390 played cricket, 450 played hockey and 580

played basket ball, 90 played both cricket and hockey, 125 played hockey and basket

ball and 155 played cricket and basket ball; 50 played all three games.

a) How many students did not play any game? Answer: 500

b) How many played only cricket? Answer: 195

c) how many played only one game? Answer: 830

d) how many played only two games? Answer: 220

10. In a survey of 100 students it was found that 28 read magazine A, 30 read magazine

B, 42 read magazine C, 8 read magazines A and B, 10 read magazines A and C, 5 read

magazines B and C, and 3 read all these magazines.

a) Find how many read none of the magazines. Answer: 20

b) How many read magazine C only? Answer: 30

11. In a certain town 25% families own a cell phone, 15% families own a scooter and

65% families own neither a cell phone nor a scooter. If 1500 families own both a cell

phone and a scooter, find the total number of families in the town. Answer: 30,000

12. In a class 18 students offered physics, 23 students offered chemistry and 24 students

offered mathematics. Of these, 13 are in both chemistry and mathematics, 12 are in

physics and chemistry, 11 in mathematics and physics and 6 in all the three subjects.

Find

a) How many students are there in the class? Answer 35 MCPn

5


b) How many offered mathematics but not chemistry? Answer: 11CMn

c) How many are taking exactly one of the three subjects? Answer: 11

13. Ninety-two people went out for an outing. Forty-seven went for swimming, 38 for

riding, 42 for cycling, 28 both for swimming and riding, 31 for both riding and

cycling, 26 both for swimming and cycling. Twenty five went for swimming, riding

and cycling. Some preferred to walk instead of swimming, riding and cycling. How

many people preferred to walk? Answer: 25

6


JEE QUESTIONS ON SETS

1. In a certain town only two newspapers A and B are published. It is known that 25% of

the population reads A and 20% reads B while 8% reads both A and B. It is also

known that 30% of those read A but not B look into advertisements and 40% of those

read B but not A look into advertisements while 50% of those read A and B look into

advertisements. What percentage of the population reads an advertisement?

2. AIEEE 2013 (42): Let A and B be two sets containing 2 elements and 4 elements

respectively. The number of subsets of BA having 3 or more elements is:

A) 219 B) 211 C) 256 D)220

3. JEE (Advanced) 2007(II) – 64: Let 65

56)(

2

2

xx

xxxf

Match the expressions/statements in column I with the expressions/statements in

column II.

I II

A) If –1 < x < 1, then f(x) satisfies p) 0 < f(x) < 1

B) If 1 < x < 2, then f(x) satisfies q) f(x) < 0

C) If 3 < x < 5, then f(x) satisfies r) f(x) > 0

D) If x >5, then f(x) satisfies s) f(x) < 1

4. JEE (ADV) 2018(I): Let X be the set consisting of the first 2018 terms of the

arithmetic progression 1, 6, 11 … and Y be the set consisting of the first 2018 terms of

7


the arithmetic progression 9, 16, 23, . ... Then, the number of elements in the set

YX is __________.

HINTS AND ANSWERS

1. 13. 9%.

2. A). Subtract the number of subsets with 0, 1 and 2 elements from the total number of subsets of BA . Use

combinations.

3. (A, p), (B, q), (C, q), (D, r). Hint: )2)(3(

11)(

xx

xxf

4. 3748

8


FUNCTIONS

A. 1. Given A = {2, 3}, B = {4, 5}, C = {5, 6}. Find

1) )( CBA Answer: )6 ,3(),5 ,3(),4 ,3(),6 ,2(),5 ,2(),4 ,2(

2) )( CBA Answer: )5 ,3(),5 ,2(

3) )()( CBBA Answer : )6 ,5(),5 ,5(),6 ,4(),5 ,4(),5 ,3(),4 ,3(),5 ,2(),4 ,2(

B. 1) If BA {(a, 1) (a, 5), (a, 2), (b, 2), (b, 5), (b, 1)}, find A, B and AB .

Answer: A = {a, b}, B = {1, 2, 5}

AB {(1, a), (2, a), (5, a), (1, b), (2, b), (5, b)}

2) If 82 : ), ( yxyxR is a relation on N, write the range of R.

Answer: Range = {1,2,3}

3) Let R be a relation defined on the set of natural numbers N as follows:

242 and , : ), ( yxNyNxyxR

Find the domain and range of the relation R.

Answer: Domain = {1, 2, 3, ,..,11} Range = {2,4,6,…,22}

4) Let 5 than lessnumber prime a is :)3 ,( aaaR be a relation. Find its

range. Answer: Range = {8, 27}

5) Let A = {1, 2, 3, …..14}. Define a relation R from A to A by

R = {(x, y): Ayxyx , ,03 }. Depict this relationship using an

arrow diagram. Write down its domain and range.

9


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

Domain = {1, 2, 3, 4}, Range = {3, 6, 9, 12}

C. Draw the graph of the following functions.

1)

0 if 1

0 if 0

0 if 1

)(

x

x

x

xf

2)

4 if 4

41 if 3

41 if

)(

xx

xx

xx

xf

3)

2 if 5

20 if 1

0 if 1

)( 2

x

xx

x

xf

4)

20 if 20 if

)(xx

xxxf

5) 1)( xxf , 30 x , where [x] is the greatest integer not greater

than x.

6) Draw the graph of the function xxf )( .

Hint: Reflect the graph of x in x-axis.

7) Name the function 113

74)(

23

x

xxxxf and give its domain.

Answer: Rational Function , Domain = 3

11R

10


D. 1) If f and g be two real functions defined by xxf )( , Rx and ,

Rx find

1) gf 2) gf 3) gf

Answers:

i)

0for 0

0for 2)()(

x

xxxgxf

,

ii)

0for 2

0for 0)()(

xx

xxgxf

iii)

0for

0for )()(

2

2

xx

xxxgxf

E. 1) If qxxxf 5)( 2 and f (q) = –9, determine the value of f (7). Answer: 81

2) The function )(xf satisfies the equation )1()1()( xfxfxf for all values of

x. If 1)1( f and 3)2( f , what is the value of )2008(f ? Answer: )2008(f = –1

Hint: Find )0(f , )1(f , )2(f .. )5(f and so on. Observe the pattern.

3) If axx

xf 1

)( and5

28

5

1

f , find ‘a’. Answer: a = 3

F. Determine the domain of the following functions:

1) 48)( 2 xxxf Answer: [2, 8] and 2x

xxg )(

11


2)xx

y

1

Answer: x < 0

3) xx

y

1

Answer : x > 0

G. Determine the range of the following functions:

1)21 x

xy

Answer:

21

21 y

2)x

xy

sin1

sin

Answer:

210 y

3)22

3

xy

Answer: y > 1.5 or y < 0

4) x

ysin34

1

Answer: 1

7

1 y

5)5

1)(

xxf

Answer: y > 0

6) Find the domain and range of the function 29)( xxf .

Answer: Domain: [-3, 3], Range: [0,3]

H. Find the range of the following functions.

1) xxf 32)( , Rx , x > 0 Answer: Range: ( , 2)

2) 2)( 2 xxf , x is a real number. Answer : Range: (2, )

12


TRIGONOMETRY – I

θsin = y co-ordinate of the terminal side Q

θ cos = x co-ordinate of the terminal side Q

when the initial arm OR turns through an angleθ .

cosθ

sinθ

1

θ

Y

XO

(1, 0)

Q

13


1. What values between 0 and 360 may A have when

a) 2

1sin A b) 3cot A

c)2

1cos A d)

3

2sec A

e) 1tan A f) 2cos ecA

Reference angle

is the acute positive angle by the terminal side of and the x-axis.

Value of a trigonometric function of an angle

= value of trigonometric function of the reference angle.

sinθ

θ

y

x

14


sin π -θ( )θ π - θ

y

x

θ

sin π + θ( )

π + θ

y

x

15


2. Fill in all blanks in the following table:

Asin Acos )sin( A )cos( A )sin( A Atan Asec

2

3

2

1

5

3

5

4

3. Prove that:

a) 1390sin)300cos(390cos420sin

b) 0390cos330sin510sin570cos

c) 0675cot765tan405cot225tan

d) 0)270sin()180cos()270sin(cos AAAA

e) 0)360tan()90tan()180tan(cot AAAA

θ

sin 2π - θ( )

2π - θ

y

x

16


f) 01)90tan( )270tan()90sec( )270sec( AAAA

ADDITION FORMULAE FOR SINE AND COSINE

(From proof without words, Roger B. Nelson, MAA publication)

sin

= DC = EB + BF

cos = AD = AE – DE

SUBTRACTION FORMULAE FOR SINE AND COSINE

sin = BC = BF – CF

cos = AB = GE + EF

sin(β)

cos(β)

sin(α) sin(β)

cos(α) sin(β)

sin(α) cos(β)

α

cos(α) cos(β)

1

α

β

D

C F

A E

G

B

17


4. Prove that:

a) )sin()45sin()45sin()45cos()45cos( BABABA

b) )cos()45sin()45cos()45cos()45sin( BABABA

c) 0cos cos

)sin(

cos cos

)sin(

cos cos

)sin(

AC

AC

CB

CB

BA

BA

d) 45cos105cos105sin

e) 15cos105cos15sin75sin

f) AAnAnAnAn 2cos)1cos()1cos()1sin( )1sin(

g) AAnAnAnAn cos)2cos()1cos()2sin( )1sin(

α

βα

1

cos(α) cos(β)

α

sin(α) cos(β)

sin(α) sin(β)

cos(α) sin(β)

cos(β)

sin(β)

D

C F

GA

B

E

18


5. Express as a sum or difference the following:

a) 7sin 5sin2

b) 5sin 7cos2

c) 3co 11sin2 s

d) 66sin 54sin2

6. Prove that:

a)

5sin2sin2

11sin

2

3sin

2

7sin

2sin

b) 2

5sin5sin

2

9cos3cos

2cos2cos

c) 0coscos3cossinsin3sin AAAAAA

d) AAAAA 2cos)54cos()54cos()36cos()36cos(

e) AAA 2cos2

1)45sin( )45sin(

f) 14

3tan

4tan

g) 14

cot4

cot

h) AA

A sec2

tan tan1

i) AA

tA sec12

co tan

19


7. a) If 1

tan

n

nA and

12

1tan

nB , find )tan( BA 1:Answer

b) If 34

3tan

A and

34

3tan

B , prove that

8

3)tan( BA

c) If 6

5tan and

11

1tan , prove that

4

d) If 2

1tan A and

3

1tan B , find the values of

i) A2tan ii) )2tan( BA iii) )2tan( BA

Answer: 3

4, 3,

13

9

e) If 20A and 25B , find the value of BA tan1)tan1( . Answer: 2

Area of triangle ABC = θ2sin Area of triangle ABC = cos θsin 2

(From proof without words, Roger B. Nelson, MAA publication)

θ

22θ

1

sin2θ

θ

B

CD

A

θ

2cosθ

2sinθ

2

D

B C

A

20


8. a) Find the value of 2sin , when

i) 5

3cos ii)

13

12sin iii)

63

16tan

Answer: a)25

24 b)

169

120 c)

4225

2016

b) Find the value of 2cos , when

i) 17

15cos ii)

5

4sin iii)

12

5tan

Answer: a)289

161 b)

25

7 c)

169

119

9. Prove that

a) AA

Atan

2cos1

2sin

b) A

A

Acot

2cos1

2sin

c) AA

A 2tan2cos1

2cos1

d)

tan

2coscos1

2sinsin

e) 2

tancossin1

cossin1

f) AAAec cot2cot2cos

g) AAA 2cot2cottan h) A

A

A

A

2tan

8tan

14sec

18sec

i)

Aec

A

A2cos

45tan1

45tan12

2

j) AecAA 2cos2cottan

k) AAA 2sec22sec1sec2 l) AAAecA sin2cos2cot2cos

21


10. Prove that:

a) 2

cos4sinsincoscos 222

b) 2

cos4sinsincoscos 222

c) 2

sin4sinsincoscos 222

d) )cos(22sin2sin

)3sin()3sin(BA

BA

BABA

e) AAAAAA tan2tan3tantan2tan3tan

f) 110tan60tan60tan20tan20tan10tan

g) )tan(cos sincos sin

sinsin 22

BABBAA

BA

h)

2tan24

tan4

tan

i) AAA

AA

AA

AA2tan2

sincos

sincos

sincos

sincos

j) A

AAA

2sin21

2cos415tan15cot

k) 380tan60tan40tan20tan

l)

2sec24

sec4

sec

22


11. a)

15tan

20cos10cos

20sin10sin

b)

2tan

2tan

sinsin

sinsin

12. i) Given2

145cos , prove that

2

22

2

122cos

ii) Given 3

2tan and

20

, find the values of 2sin , 2cos and 4cos .

Answer:13

12,

13

5 and

169

119

13. Find the values of

a) 217sin Answer:

8

264 ,

b) 217cos Answer:

8

264

14. Find the general solution of the following equations:

i) sin9sin Answer:

4

n ,

10

)12( n

ii) 2sin3sin Answer: 5

)12( n, n2

iii) 4cos5cos Answer: 9

2 n, n2

iv) 1tan2tan Answer: 6

)12( n

23


v) cot3tan Answer: 8

)12( n

vi) 33

2tan

3tantan

Answer:

123

n

vii) xxx 2sin4sin6sin Answer: 4

n ,

6

n

viii) 4sin7sinsin Answer: 4

n ,

93

2

n

ix) 4cos7coscos Answer: 8

)12(

n,

93

2

n

x) 7sin3sinsin Answer: 3

n,

122

n

xi) sec2tancot Answer: 2

)12( n,

6)1(

nn

xii) 5sin6cos4 22 xx Answer: 4

n

xiii) xx 2cos4cos Answer: 3

n

xiv) 22sinsin2 22 xx Answer: 4

n ,

22

n

xv) xx 2tan12sec2 . Answer:2

nx ,

8

3

2

nx

xvi) 03tan31tan2 xx . Answer:4

n ,

3

n

xvii) 01cot3

13cot2

xx . Answer:

6

n ,

3

n

24


xviii) 12sectan 2 . Answer: n , 3

n

LAW OF SINES

C

c

B

b

A

a

sinsinsin

In any triangle sides are proportional to sines of opposite angles.

LAW OF COSINES

Pythagoras theorem extended to any triangle

90cos2222 bccba Abccba cos2222

c b

a

A

BC

c

a

b cb

a CC

A

B

A

B

25


15. a) In a triangle ABC, 6cos4sin3 BA and 1sin4cos3 BA . Find the measure

of C . Answer: 30C

b) If in a triangle ABC, ca

b

bc

a

c

C

b

B

a

A

cos2

coscos2 find the value of A in

degrees. Answer: 90A

c) If BbsAa cosco then prove that the triangle is either isosceles or right angled.

16. In any triangle ABC, prove that:

i) 0sinsinsin BAcACbCBa

ii) If BaAb coscos , then prove that a = b.

iii) Abc

cbBC

sincotcot

22

iv) CBA

CBA

cba

cba

cotcotcot

2cot

2cot

2cot

)(222

2

v) 2

cot2

tanA

cb

cbCB

vi) 2

sin )(2

cos A

cbCB

a

vii) 0cotcotcot 222222 CbaBacAcb

viii) 2

cot22

tan2

tan)(C

cBA

cba

ix) cbaCbaBacAcb cos)(cos)(cos)(

26


x) 0sinsinsinsinsinsin BAcACbCBa

xi) 2

sin2

cos 22222 Cba

Cbac

xii) 2

co 2

sinA

sa

cbCB

xiii) CabBcaAbccba coscoscos2222

xiv) 02cos2cos22cos2cos22cos2cos2

BAcACbCBa

xv) AbcBcCb sin22sin2sin 22

xvi) 22cos cos cbBcCba

xvii) 2

sin)(2cos cos 2 AcbCBa

xviii) 2

cos)(2cos cos 2 AcbBCa

xix) 2

22

)sin(

)sin(

a

cb

CB

CB

xx) 2

cot 2

tanBABA

ba

ba

xxi) 2

sin)(2

sinA

cbBA

a

xxii) abc

cba

BaAb

C

AcCa

B

CbBc

A

2cos cos

cos

cos cos

cos

cos cos

cos 222

27


xxiii)

02sin2sin2sin2

22

2

22

2

22

Cc

baB

b

acA

a

cb

xxiv) 2

co 2

cos)(CB

saCB

cb

17. In triangle ABC, let F denote the midpoint of side BC.

Prove that cot2cotcot .

18. In triangle ABC, AB 3 , prove that )(222 ababac .

19. If AC = 4, BC = 3, AB = 5 and 30DCB find CD. Answer: 7

24CD

θ

β α

FB C

A

30°

3

45

D

C B

A

28


20. The sides of a triangle are three consecutive natural numbers and its largest angle is

twice the smallest one. Determine the sides of the triangle. Answer: 4, 5, 6

21. Find the greatest angle of the triangle whose sides are 12 xx , 12 x and 12 x .

Answer: 120

22. a) In a ABC if 60C , prove that cbacbca

311

b) If cbacbca

311 prove that 60C .

23. i) In a triangle ABC, if2

tanA

,2

tanB

, 2

tanC

are in A.P. Show that Acos , Bcos , Ccos

are in A.P.

ii) If2a ,

2b and 2c are in A.P., prove that Acot ,

Bcot and Ccot are also in A.P.

24. With usual notation, if in a triangle ABC, 131211

baaccb

; then prove that

25

cos

19

cos

7

cos CBA

25. Let the angles A, B, C of a triangle ABC be in A.P. and let 2 : 3 : cb . Find A .

Answer: 75A

26. The sides of a triangle are sin , cos and cos sin1 for some2

0

. Find

the greatest angle of this triangle. Answer: 120

29


27. DE is a tower standing on a horizontal plane and ABCD is a straight line in the plane.

The height of the tower subtends an angle at A, 2 at B, 3 at C. If AB and BC are

respectively 50 and 20 meters, find the height of the tower and the distance CD.

Answer: 33.07 m, 17.5m

28. A vertical pole (more than 10 meters high) consists of two parts, the lower part being

rd3

1 of the whole. From a point in a horizontal plane through the foot of the pole and

4 meters from it, the upper part subtends an angle whose tangent is2

1. Find the height

20 m 50m

3θ 2θ θA

BD

E

C

30


of the pole. Answer: 12 m

β

αh

2h

4 m

A

B C

D

31


TRIGONOMETRY - II

1. What is the acute angle (measured in radians) between the hour hand and the minute

hand of a circular clock at quarter past one? Answer: 2152

2. Prove that: 10sin70sin50sin =0

3. Evaluate: 20cos50sin80cos Answer: 8

1

4. If53cos A , 2 A , find Atan and A2sin . Answer:

3

4 and

25

24

5. If 900 x and7

242tan x , determine the value of xsin Answer:

5

4

6. If13

12cos A ,

2

3 A find Asin and A2cos . Answer:

13

5 and

169

119

7. a) Find2

sinx

, 2

cosx

and2

tanx

, if3

4tan x , where x is in the II quadrant.

Answer:5

1cos

2x ,

5

2sin

2x , 2

2tan

x

b) Given 5

4sin where

2

2

3 , find

2sin

,

2cos

. Answer:

5

1,

5

2

8. If5

42sin A , find Atan . Answer: 2,

2

1

9. Prove that 80sin210sin50sin70sin820cos 2

10. Prove that: A

A

AAA

AAA

5sin

3sin

7sin5sin23sin

5sin3sin2sin

32


11. Prove that AAAAA

AAAA4tan

7cos5cos3coscos

7sin5sin3sinsin

12. Evaluate the exact value of 2122tan Answer: 12

13. Find the value of 4111tan . Answer:

12

2241

14. Prove that:

8cos1

8

3cos1

8cos1

8cos1

=

8

1

15. If 45sin15sinsin , find . Answer: 75

16. Prove that yx

yx

yx

yx

tantan

tantan

)sin(

)sin(

17. If2 and , prove that: tan2tantan

18. Find the value of: 390cos600sin150sin480cos Answer: 1

19. Prove that: 4

3200cos140cos140cos100cos100cos20cos

20. Prove that: 350tan70tan10tan

21. Prove that:

10sin

1170sin50sin30sin10sin

22. Prove that: 80sin70sin50sin40sin20sin10sin

23. a)Prove that 3cos 4

160cos60coscos

b) Prove that

3sin 3

sin3

sinsin4

33


24. Prove that: xxx 2sectan2tan1

25. Prove that: xxx

xxxtan

cos5cos

sin3sin25sin

26. Prove that: xxxxxx 3cos 2cos cos46cos4cos2cos1

27. Prove that: xecxx

x2coscot

4cos1

4sin

.

28. Prove that: BABABA 22 sinsin)sin()sin(

29. Prove that : 22cos)(2sin2sin2cos 2cos

30. Prove that:

2tan

sin4 3sincos 2cos

3cos 6sincos 8sin

31. Prove that: 16


32. Prove that: 8

180cos40cos20cos

33. Prove that: 72

1

15

7cos

15

6cos

15

5cos

15

4cos

15

3cos

15

2cos

15cos

34. Prove that: 16


35. Prove that 178tan66tan42tan6tan

36. Find the value of: 81tan63tan27tan9tan Answer: 2

37. Prove the identity:

8sin8

sin

8cos

4cos

2cos

34


38. Prove that: 2

cot4cot4cos2coscos

ececec

39. Prove that: cot16cot168tan84tan42tan2tan

40. If 25tantan BA and 30cotcot BA , find the value of )tan( BA . Answer: 150

41. Evaluate:

135tan210tan1

135tan210tan Answer: 23

42. Prove that:

tan1

tan1

2cos

2sin1

43. Prove that: AAA

AAtan

12cos2sin

12cos2sin

44. a)Evaluate: )16531998sin()2371998sin( Answer: 4

1

b)Evaluate: )16592004sin()2312004sin( Answer: 4

1

45. If 2

3sinsin BA and

2

1coscos BA , find A + B Answer: 120

46. If AAAA 2sin4cos2cos4sin ,4

0

A , find A4tan . Answer: 3

47. Prove that:

54tan

9sin9cos

9sin9cos

48. Without using trigonometric tables prove that 8

1

18

7sin

18

5sin

18sin

49. Solve the following equations:

a) 3cos5sin7cos9sin Answer:24

)12(

n or 4

n

35


b) 32tan tan32tantan xxxx Answer:93

nx

c) 01cossin2cos2sin Answer: 4

n ,

3

22

n

d)

x2sin164 = xsin62 Answer: 6

)1(

nnx ,2

)1(

nnx

e) xxx cos2sectan Answer:

2)1(

nnx ,

6)1(

nnx

f) 22 sec4

13tan72 Answer:

6

n

g) xxxx 4sin3sin2sinsin 2222 Answer: 5

n

, 2

n

50. If cot , cot and cot are in G.P, prove that)cos(

)cos(2cos

.

51. If 4

BA , prove that 2tan1)tan1( BA

52. If 4

BA , evaluate 1cot)1(cot BA Answer: 2

53. Without using trigonometric tables, prove that

44cot

16cot76cot

16cot76cot3

54. Prove that8

1

7

3cos

7

2cos

7cos

55. Prove that xxx 2cos1tan2sin

56. Prove thatx

xxxx

tan1

tansincos

4cos2

36


57. Without using trigonometric tables, prove that: 10cos5tan10cot ec

58. If and are two distinct real numbers satisfying the equation cxbxa sincos ,

prove that 22

2sin

ba

ab

.

59. Prove that:

xxx

4tan2tan2sec

60. In a right angled triangle, acute angles and satisfy

70tantantantantantan 3322 . Determine the angles of the

triangle. Hint: Let t cotant Answer:

75 ,

15

61. i) Find all values of x between 2 and such that2

1cos x .

Answer: 3

4

, 3

2

ii) Find the exact values of 195sin , 195cos and 195tan .

Answer: 22

31, 22

31 ,

13

13

iii) Find the exact values of 3

4tan

,

6

7sin

Answer: 3 ,

2

1

62. Given 5

1sin ,

5

1sin and

2

5< , <

2

7

i) Evaluate cos and cos Answer : cos = cos = 5

62

ii) Evaluate )cos( Answer: 1

37


63. If x0 and 2

1sincos xx , find .tan x Answer :

74

7312

64. If 2

0

x and 5

1sincos xx , find

2tan

x Answer: 2

65. a) Evaluate:

80sin

3

80cos

1 Answer: 4

b) Evaluate: 20sec20cos3 ec Answer: 4

66. Show that 18sin9sin148sin39sin21sin12sin 222222

67. Prove that:

1sin

1cos

89cos 88cos

1

2cos 1cos

1

1cos 0cos

12

Use:

)tan()1tan(

)1cos( )cos(

1sinxx

xx, Problem from IMO.

68. If ba

ba

yx

yx

)sin(

)sin(, find the value of

y

x

tan

tan. Answer:

b

a

69. Prove that: )tan()24cos()24cos(

)24sin()24sin(BA

ABBA

ABBA

70. Prove that:

4cos2cos4

3tan5tan

3tan5tan

71. Prove that: AAAAA

AAAA4cot

13cos9cos5coscos

13sin9sin5sinsin

72. Prove that: AAAAAAA 6cos 5cos 4cos415cos7cos5cos3cos

73. Prove that: AAsAAsAAsA

AAAAAA9tan

13co 4sin6co 3sin2co sin

13sin 4sin6sin 3sin2sin sin

38


74. Prove that:2

3

8

7cos

8

5cos

8

3cos

8cos 4444

75. Prove that: 2

3

8

7sin

8

5sin

8

3sin

8sin 4444

76. If tan , tan are the roots of 02 cbxx and cot , cot are the roots of the

equation 02 qpxx , express pq in terms of b and c . Answer: 2c

bpq

77. If , are two different values of lying between 0 and 2 which satisfy the

equation 9sin8cos6 find the value of )sin( . Answer: 25

24

78. If a sin sin and b cos cos . Show that:

a) 22

2sin

ba

ab

b) 22

22

cosab

ab

79. If ba

cossin , prove that bsba 2co 2sin

80. If )tan()tan( n , then prove that 2sin )1(2sin )1( nn .

81. If the roots of the quadratic equation 02 qpxx are 30tan and 15tan , find the

value of pq 2 . Answer: 3

82. Prove the identity

cotsin

24tan)sin1(

39


83. Prove the identity

2sin

23sin

23

sin26

cos4

84. Prove the identity cos21cos2cos

3cos2coscos12

xxx

xxx

40


TRIGONOMETRY

JEE Questions

1. JEE (Advanced) 2017(II) – 45: Let and be nonzero real numbers such that

1cos cos)coscos2 . Then which of the following is/are true?

A) 02

tan32

tan

B) 0

2tan

2tan3

C) 02

tan32

tan

D) 0

2tan

2tan3

2. JEE (Advanced) 2016(I) – 39: Let 126

. Suppose 1 , 1 are the roots of

the equation 01sec 22 xx and 2 , 2 are the roots of the equation

01tan 22 xx . If 1 > 1 and 2 > 2 then 1 + 2 equals

A) 2( )tansec B) 2 sec

C) tan2 D) 0

3. JEE (Advanced) 2016(II) – 40: The value of

64sin

6

)1(

4sin

1

13

1

k kk

is

equal to

A) 33 B) 2( 33 )

C) )13(2 D) )32(2

41


4. JEE (Advanced) 2015(II) – 56: If

11

6sin3 1 and

9

4cos3 1 , where the

inverse trigonometric functions take only the principal values, then the correct

option(s) is (are)

A) cos > 0 B) sin < 0 C) )cos( > 0 D) cos <0

5. JEE (Advanced) 2014 II)– 60(Q): The number of points in the interval ]13 ,13[-

at which 22 cossin xxf(x) attains the maximum value is _______

6. JEE (Advanced) 2012(II) – 57: Let Rf )1 ,1(: be such that

2sec2

2)4(cos

f

for

2 ,

4

4 ,0

. Then the value(s) of

3

1f is (are)

A) 1 – 2

3 B) 1 +

2

3 C) 1 –

3

2 D)1+

3

2

7. JEE (Advanced) 2010(I) – 53: The maximum value of the expression

22 cos5cossin3sin

1

is _______

8. In PQR , 90R , 2

tan P and 2

tanQ

are the roots of 0 2 cbxxa , 0a , then

A) c = a + b B) a = b+ c C) b = a + c D) b = c

9. A: 0coscoscos

B: 0sinsinsin

42


If 2

3)cos()cos()cos(

then A) A is true and B is false.

B) A is false and B is true.

C) Both A and B are true.

D) Both A and B are false.

10. ABC is a triangle such that21)2sin()sin()2sin( CBACBA . If A, B and C are

in arithmetic progression, determine the values of A, B and C.

11. In a triangle ABC, the sides opposite to the angles A, B and C are a, b and c

respectively. Then

(A) 2

sin22

cos)(CB

aA

cb

(B) 2

cos22

sin)(A

aCB

cb

(C) 2

sin2

cos)(CB

aA

cb

(D) 2

cos2

sin)(A

aCB

cb

12. IIT JEE 1999 : For a positive integer n, let

)2sec1()4sec1)(2sec1)(sec1( 2

tan)(

nnf

.Then

43


A) 116

2

f B) 1

323

f

C) 164

4

f D) 1

1285

f

13. General value of satisfying the equation 12sec2tan is ……

14. IIT JEE 98 : Thee number of values of x in the interval ]5 ,0[ satisfying the

equation 02sin72sin3 xx is

A) 0 B)5 C)6 D)10

15. 2)(

4sec2

yx

xy

true if and only if

A) 0 yx B) yx , 0x

C) yx D) 0x , 0y

16. Prove that the values of the function xx

xx

cos3sin

3cossin do not lie between

3

1and 3 for any

real x.

17. If the roots of 022 qpxx are 30tan , 15tan then q =

A) 1 +2p B) 1+ p C) 1 – p D) 1 –2p

18. If 2

1sin ,

3

1cos ,

2 ,0

, then lies in the interval ____________

44


19. k 25sin25cos , then 50cos _________

A) 21 kk B)

22 kk C) 22 kk D)

21 kk

20. 3 ,65

21sinsin ,

65

27coscos , then

2

)(cos

=_____________

A) 130

3 B)

65

6 C)

65

6 D)

130

3

21. Simplify: 5sin2

sin23sin2

3sin3 6644

A) 0 B) 1 C)3 D) 6cos4sin

22. If xxA 42 cossin , then for all real x

A) 11613 A B) 21 A

C) 16

13

43 A D) 1

43 A

23. The number of integral values of k for which the equation 12sin5cos7 kxx has a

solution is

A) 4 B) 8 C) 10 D)12

24. The angle of elevation of a cloud from a point h meters above a lake is and the

angle of depression of its reflection in the lake is ; prove that its height is

45


sin

sin h .

25. The angles of elevation of the top of a tower, standing on a horizontal plane, from two

points distant a meters and b meters from the base and in the same straight line with it

are complementary. Prove that the height of the tower is ab meters, and if be the

h

βα

46


angle subtended at the top of the tower by the line joining the two points, then

ba

ba

sin .

26. Aptitude – 2017(28): The value of

255sin 3

1

285cos

1, is :

1) 22 2) 3

24 3)

3

22 4) 23

ANSWERS AND HINTS

1) A), C). Hint: Express all functions in terms of tangent of half angles and simplify.

2) C) Hint:

2

tan2

sec = 1

3) C) Hint: Replace 1 in the numerator by

6

)1(

464sin 2

kk and use )sin( BA

4) B), C) and D) Hint: 12

6

11

6

3sin

,

8

4

9

4

3cos

5) Answer = 4. Hint:

2

4sin2 xf(x)

, 13 radians=3.6 radians = 6.206 , approximately.

θ

b

a

A CB

D

47


6) B) Hint: Let 3

14cos ,

2cos

2cos1

2sec2

2

7) Answer = 2. Hint: Denominator = )2sin(56 , where 54sin

8)A) 9) C) 10)

45A , 60B , 75C

11) C

12) B) Hint: Write the expression for n = 1, 2, 3 ,.. Simplify

13) 3

n

14) C 15) B

16) Proof Hint: The given expression =

x

x

3tan

tan, simplify to get x2tan

)3(

)13(

17) A 18) 3

2 ,

2

19) B 20) A 21) B

22) D 23) B

24) Proof

25)Proof 26) Option 2.

48


JEE QUESTIONS ON

TRIGONOMETRIC EQUATIONS

1. JEE (Advanced) 2016(I) – 40: Let

2

,0 : ) ,(

xxS . The sum of all distinct

solutions of the equation 0)cot(tan2cossec3 xxecxx in the set S is equal to

A) 9

7 B)

9

2 C) 0 D)

9

5

2. JEE (Advanced) 2015(I) – 46: The number of distinct solutions of the equation

2sincossincos2cos4

5 66442 xxxxx

in the interval ]2 ,0[ is __________

3. JEE (Advanced) 2014(II) – 45 : For ) ,0( x the equation 33sin2sin2sin xxx

has

A) infinitely many solution B) three solutions

C) one solution D) no solution

4. JEE (Advanced) 2013(II) – 58(R): If

xxxxxxxxxx 2cos4

cossec 2sincossec 2sinsin2cos 4

cos , then possible

value of xsec is _____

49


5. JEE (Advanced) 2012(I) – 53 : Let ]2 ,0[ , be such that

1cos2

cot2

tan sin)sin1(cos2 2 θθ , 0)2tan( and 2

3sin1 ,

then cannot satisfy

A) 2

0

B) 3

4

2

C) 2

3

3

4

D)

2

2

3

6. JEE (Advanced) 2010(I) – 52: The number of values of in the interval

2 ,

2

such that 5

n for n = 0, 1 , 2 and 5cottan as well as 4cos2sin is ____

7. JEE (Advanced) 2009(II) – 27: For 2

0

, the solutions of

244

cos 4

)1(cos6

1

mecmecm

, is (are)

A) 4

B)

6

C)

12

D)

12

5

8. JEE (Advanced)2009(II) – 30A: Roots of the equation 22sinsin2 22 are

_________

p) 6

q)

4

r)

3

s)

2

t)

50


9. JEE (Advanced) 2007(I) – 53: The number of solutions of the pair of equations

02cossin2 2

0sin32cos2

in the interval 2 ,0 is

A)zero B) one C) two D) four

10. JEE (Main) 2016 – E(88): If 20 x , then the number of real values of x, which

satisfy the equation, 04cos3cos2coscos xxxx , is

A) 3 B) 5 C) 7 D) 9

11. JEE (ADV) 2018(I): Let a, b c be three non-zero real numbers such that the equation

cxbxa sin2cos 3 .

2 ,

2

x .

has two distinct real roots and with3

. Then, the value of

b

a is ___.

HINTS AND ANSWERS

1. C)

2. Answer = 8

3. D) Hint: Use the formula for x3sin . Transfer xsin to the RHS and complete LHS as a perfect square

in xcos . RHS value is unbounded, whereas LHS value vary between – 6 to 3.

4. Answer = 2

5. A), C), D). Hint: Simplify LHS to get 2

cos21)sin(

, fix the domain for using the given

conditions on .

51


6. Answer = 3

7. C), D). Hint: Express the expression after the summation in terms of ines function. Write the

numerator as 4

)1(4

sin2 mm . Use )sin( BA and simplify.

8. q), s). 9. C) 10. C)

11. Substitute the roots in the given equation, subtract, use sum and difference formula. Answer = 0.5

52


JEE QUESTIONS ON

LAW OF SINES AND COSINES

1. JEE (Advanced) 2015(I) – 60(A): In a triangle XYZ, let a, band c be the lengths of

the sides opposite to the angles X, Y and Z respectively. If 2222 cba

and

Z

YX

sin

)sin( then possible values of n for which 0)cos( n is (are) ___

2. JEE (Advanced) 2015(I) – 60(B): In a triangle XYZ, let a, band c be the lengths of

the sides opposite to the angles X, Y and Z respectively. If

YXYX sinsin22cos22cos1 , then possible value(s) of b

a is (are) ________

3. JEE (Advanced) 2013(II) – 48: In a PQR, P is the largest angle and3

1cos P .

Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M

respectively, such that the lengths of PN, QL and RM are consecutive even integers.

The possible length(s) of the side(s) of the triangle is (are)

53


A) 16 B)18 C)24 D) 22

4. JEE (Advanced) 2012(II) – 47: Let PQR be a triangle of area with a = 2, b= 2

7

and c =2

5, where a, b c are the lengths of the triangle opposite to the angles at P, Q

and R respectively. Then PP

PP

2sinsin2

2sinsin2

equals

A) 4

3 B)

4

45 C)

2

4

3

D)

2

4

45

(n + 2)(n + 2)

n

n(n-2)

(n-2)

P Q

R

5

2

7

2

2QR

P

54


5. JEE (Advanced) 2011(I) – 65: The positive integer value of n > 3 satisfying the

equation

nnn

3sin

1

2sin

1

sin

1 is ___________

6. JEE (Advanced) 2010(I) – 36: If the angles A, B and C of a triangle are in an

arithmetic progression and if a, b and c denote the lengths of the sides opposite to A,

B and C respectively, then the value of the expression Aa

cC

c

a2sin2sin is

A) 2

1 B)

2

3 C) 1 D) 3

7. JEE (Advanced) 2010(I) – 37: Let ABC be a triangle such that 6

ACB and let a,

b and c denote the lengths of the sides opposite to A, B and C respectively. The

value(s)of x for which 12 xxa , 12xb and 12 xc is (are)

A) 32 B) 31 C) 32 D) 34

8. JEE (Advanced) 2010(II) – 27: Two parallel chords of a circle of radius 2 are at a

distance 3 + 1 apart. If the chords subtend at the center, angles of k

and

k

2,

30°

2x + 1

x2 + x + 1

x2 - 1

A

C B

55


where k > 0, then the value of [k] is______________

[Note: [k] denotes the largest integer less than or equal to k]

9. JEE (Advanced) 2010(II) – 29: Consider a triangle ABC and let a, b and c denote the

lengths of the sides opposite to vertices A, B and C respectively. Suppose a = 6, b =

10 and the area of the triangle is 15 3 . If ACB is obtuse and if r denotes the radius

of the incircle of the triangle, then 2r is equal to____

3 + 1

2

2

π

k

π

2k

10

6CB

A

56


10. JEE (Advanced) 2009(I) – 30: In a triangle ABC with fixed base BC, the vertex A

moves such that2

sin4coscos 2 ACB .

If a , b and c denote the lengths of the sides of the triangle opposite to the angles A, B

and C respectively, then

A) acb 4 B) acb 2

C) locus of the point A is an ellipse.

D) Locus of the point A is a pair of straight lines.

11. JEE (ADV) 2018(I): In a triangle PQR, let 30PQR and the sides PQ and QR

have lengths 10 3 and 10, respectively. Then which of the following statement(s) is

(are) TRUE?

A) 45QPR

B) The area of the triangle PQR is 25 3 and 120QRP

10

30°

10 3

P

RQ

57


C) The radius of the incircle of the triangle PQR is 10 3 –15

D) The area of the circumcircle of the triangle PQR is 100

ANSWERS

1. n = 1, 3, 5 (given values of n in the question paper)

2. Express in terms of sines, use law of sines to get b

a= 1

3. Use law of cosines. B), D)

4. C). Simplify the given expression and substitute for Pcos . Use Heron’s formula to find area .

5. Let

n. Simplify to get n = 7.

6. Use law of sines. Answer D)

7. Use law of cosines. Answer B)

8. Take cosines of the two given angles. Answer = 3

9. Use area = Cabsin2

1 to find C. Use law of cosines to find c. Use )(

2

1cbar to find radius ‘r’

of the incircle. Answer: 2r = 3

10. B), C)

11.B), C), D) are true

58


COMPLEX NUMBERS - I

A. Express the following complex numbers in the form iba :

1) )86()32(2)23(3 iii Answer: i2011

2) )23)(23)(32( iii Answer: i3926

3) )1(

)3)(2)(1(

i

iii

Answer. i55 4)

i

i

2

1 Answer:

5

31 i

5)i

i

41

32

Answer: i

17

5

17

14 6)

3

2

3

2

1

i Answer: i01

7))31)(2(

1

ii Answer:

50

71 i 8)

i

i

1

1 Answer :0+ i

9)2

2

)32(

)54(

i

i

Answer:

169

92525 i 10) )4)(53( ii Answer: i1717

11) )37)(116( ii Answer: i959 12) i

i

54

3

Answer: i

41

19

41

7

13) 3)74( i Answer: i 7524 14) i

i

41

32

Answer: i

17

5

17

14

15) i

i

2

)32( 2

Answer:5

292 i 16)

i

i

3

)1( 2

Answer: i5

3

5

1

17) 3)1(

1

iAnswer:

44

1 i 18)

2

2

)2(

)21(

i

i

Answer: i

25

24

25

7

59


19)i

i

i

i

2

)2(

2

)2( 22

Answer: i5

22 20)

23

21

4

i

ii Answer: i43

21)i

i

i

i

4

1

23

2 Answer: i

221

54

221

107 22) 10)1( i Answer: 32i

23) i

i

1

32Answer:

22

5 i

24)

161

252

Answer:

17

322 i

25)i

i

31

25

Answer:

4

352532 i 26)

i

i

45

32

Answer:

41

7

41

22 i

27) 44 )1()1( ii Answer: – 8 28))1)(24(

43

ii

i

Answer:

4

31 i

29) Express

2323

53 53

ii

ii

in the form iba . Answer: i

2

27

POLAR FORM OF A COMPLEX NUMBER

sincos iriba

π < θ ≤ π

r = z

O

P(r cosθ, r sinθ)

(r, 0)θ

y

x

60


B. Write the following complex numbers in the polar form:

1) i1 2) i3 3) i3

4) i 3 5) i 3 6)31

16

i

Answers:

1)

4sin

4cos2

i 2)

6sin

6cos2

i

3)

6sin

6cos2

i 4)

6

5sin

6

5cos2

i

5)

6

5sin

6

5cos2

i 6)

3

2sin

3

2cos8

i

MODULUS AND CONJUGATE OF A COMPLEX NUMBER

Modulus of Z = 22z ba , Conjugate of Z = iba

_z

(0, 0)

(a, b)

z= a2+b2

-θ

θ

= a - i bz

z = a + i b

x

y

61


C. Find modulus and amplitude of the following complex numbers:

1) 310sin310cos6 i Answer: 50sin50cos6 i

2) 30sin30cos2 i Answer: )150sin()150cos(2 i

D. 1) For any two complex numbers 1z and 2z , prove that

2

21 zz2

21 zz =2

22

1 22 zz

2) Interpret the locus 532 iz Answer: Circle with center(2, -3), radius 5

3) If 1z and 2z complex numbers, prove that 2121 zzzz

4) Solve the equation: 012 2 x Answer: 2

ix

E. Find the square root of the following complex numbers:

1) i 247 2) i 1024 3) i 4) i 409

5) i 6011 6) i 125 7) i 3847 8) i 2021

9) aia 212 10) ibaab 24 22 11) i43

Answers:

1) i43 2) i51 3)

2

1 i 4) i45

5) i65 6) i23 7) i 341 8) i52

9) ia 10) ibaba )()( 11) i 2

F. 1) If one root of the equation 02 2 cbxx is i23 , find b and c (b and c are real

numbers). Answer: b = –12 , c = 26

62


2) If i 32 a root of the equation 02 qpxx where Rqp , , then find p and q

. Answer: p = –4 , q = 7

63


COMPLEX NUMBERS - II

A. If A.M and G.M. of roots of a quadratic equation are 8 and 5, respectively, obtain the

quadratic equation. Answer: 025162 xx

B. Let iZ 21 and iZ 22 . Find

a)

1

21ReZ

ZZ Answer:

5

2

b)

21

1Im

ZZ Answer: 0

C. Solve for x and y: ii

iyi

i

ixi

3

)32(

3

2)1( Answer: 1 ,3 yx

D. Express the following complex numbers in the form bia

1) 660sin60cos1 i Answer: -27

2))sin(cos1

1

i Answer: cot

2

1

2

1i

3))32)(24(

1

ii Answer:

130

47 i

4) )23)(52)(32( iii Answer: i2665

5) 11)( ii Answer:2

i

6) 77

2

1 iii

Answer: – 1

64


7) i

i

i

i

43

267

86

362

Answer: 8

8) i

i

32

53

Answer: i

13

19

13

9

9) 2020 )1()1( ii Answer: 0

10)

20

22

31

i

i Answer: i

2

3

2

1

E. 1) If iz 32 , evaluate the expression1

12

2

zz

zz. Answer:

229

72147 i

2) the complex number z satisfies the equation izz 82 . Find z.

Answer:. i 815

3) Solve for z: 2

_zz , 0z . Answer: i

2

3

2

1

4) If iz 32 find the value of 41752 234 zzzz . Answer: 6

5) Solve for x and y: 3))(2( iiyxi Answer: 1 ,1 yx

6) Ifi

iz

43

7

, find

14z . Answer: i128

7) Evaluate: terms100 443322 toiiii . Answer: )1( 50 i

8) Find z so that iiz 3243 Answer: 25

18 i

9) Solve for z: iiizizi 71)43)(1()21)(( Answer: i1

65


10) Find the value of the sum

13

1

1 n

nini Answer: i1

11) If 32233)( xxxxF , find F (1 + i). Answer: i38

12) Let iz 31 and iz 22 . Find the locus of the complex number iyxz

such that 21 zzzz . Answer: 0542 yx

13) Let ix

x 21 , find the value of

2187

2187 1

xx .

Hint: Solve for x Answer:

i 2

14) If 023 izziz , show that 1z

15) If iziz 131_

52 , find z. Answer: 1 ,3 yx

16) If biaz is a complex number such that iz 432 and iz 1123 ,

find z. Answer: iz 2

17) Evaluate: 2003 432 iiiii . Answer: – 1

18) Solve for z: 02

zz Answer: i2

3

2

1

F. Express the following complex numbers in the polar form.

1) i31 Answer: )120sin()120cos(2 i

2) i1 Answer: )45sin()45cos(2 i

66


3)

3sin

3cos

1

i

i

Answer:

12

5sin

12

5cos2

i

4)i

ii

3

)22)(31( Answer:

4

3sin

4

3cos22

i

G. 1) If ibaiyx 3 , show that

224 ba

b

y

a

x

2) If idc

ibaiyx

, prove that

22

22222

dc

bayx

3) If 2)1( n

n it , find the value of 25531 tttt Answer:13

4)Express 3108

1694

2 iii

iii

in the form iba Answer: i21

H. Find the modulus and the principal argument of the following complex numbers.

1)

3sin

3cos2

31

ii

i

Answer:

2sin

2cos2

i

2)

5

2cos1

5

2sin

1

i

i

Answer:

20

11sin

20

11cos

5sin 2

1

i

67


JEE QUESTIONS ON COMPLEX NUMBERS

Note: If 1z , then z lies on the unit circle centered at (0, 0). Hence take z as sincos iz .

Also i

ez

1. JEE (Advanced) 2016(II) – 46: Let Rba , and 022 ba . Suppose

0 , ,1

: tRtibta

zCzS , where 1i . If iyxz and Sz , then (x, y) lies on

A) the circle with radius a2

1 and center

0 ,

2

1

afor a > 0 and b 0.

B) the circle with radius a2

1 and center

0 ,

2

1

afor a < 0 and b 0.

C) the x-axis for a 0, b = 0.

D) the y-axis for a = 0 and b 0.

2. JEE (Advanced) 2013 (II)– 51 :Let 321 SSSS where

4:1 zCzS ,

0

31

31Im:2

i

izCzS

0Re:3 zCzS

68


Area of S is

A) 3

10

B) 3

20

C) 3

16

D) 3

32

(ii) 2013(II) – 52 (Continuation of the previous question): ziSz

31min =

A) 2

32 B)

2

32 C)

2

33 D)

2

33

3. JEE (Advanced) 2012(I) – 43: Let z be a complex number such that the imaginary

part of z is non-zero and 12 zza is real. Then a cannot take the value

A) –1 B) 3

1 C)

2

1 D)

4

3

y + 3 x = 0

(1, -3)

z=4

(4, 0)

y

x

69


4. JEE (Advanced) 2011(I) – 67: If z is any complex number satisfying 223 iz ,

then the minimum value of iz 562 is ____________

5. JEE (Advanced) 2011((II) –60 (A):The set

1 ,1znumber,complex a is :

1

2Re

2zz

z

izis

p) ) ,1( )1 ,( q) ) ,0( )0 ,(

r) ) ,2( s) ) ,1[ ]1 ,(

t) ) ,2[ ]0 ,(

O

y

x

(3, 2)

(3, 0)

(3, -2.5)

70


6. JEE (Advanced) 2009(I) – 26: Let sincos iz . Then the value of

15

1

12Im m

mz at 2 is

A) 2sin

1 B)

2sin3

1 C)

2sin2

1 D)

2sin4

1

7. JEE (Advanced) 2009(I) – 28: Let iyxz be a complex number where x and y

are integers. Then the area of the rectangle whose vertices are the roots of the

equation: 33_

)_ ( zzzz = 350

A) 48 B) 32 C) 40 D) 80

8. JEE (Advanced) 2008(II) – 9: A particle P starts from the point iZ 210 . It moves

first horizontally away from the origin by 5 units and then vertically away from the

origin by 3 units to reach a point 1Z . From 1Z the particle moves 2 units in the

direction of the vector j

i and then it moves through an angle 2

in anticlockwise

direction on a circle with center at origin, to reach a point 2Z . The point 2Z is given

by

A) i76 B) i67 C) i67 D) i76

9. JEE (Advanced) 2007(I) – 52: A man walks a distance of 3 units from the origin

towards the north-east (N 45 E) direction. From there, he walks a distance of 4 units

towards the north-west (N 45 W) direction to reach a point P. Then the position of P

71


in the Argand plane is

A) i

e 43

+ 4i B) i

ei 4)43(

C) i

ei 4)34(

D) i

ei 4)43(

10. JEE (Advanced) 2007(II) –46: If 1z and 1z , then all the values of 21 z

z

lie on

A) a line not passing through the origin. B) 2z

C) the x-axis D) the y-axis

11. AIEEE – 2004(4): If iyxz and iqpz 3

1

, then

22 qp

q

y

p

x

is equal to

A) 1 B) –2 C) 2 D) –1

12. JEE (Main) 2016 – E(62): The value of for which

sin 21

sin 32

i

i

is purely imaginary,

is

A) 3

B)

6

C)

4

3sin 1

D)

3

1sin 1

13. JEE (Main) 2016 - 9th

APRIL(62): The point represented by i2 in the Argand

plane moves 1 unit eastward, then 2 units northward and finally from there 22 units

in the south-westward direction. Then its new position in the Argand plane at the

point represented by

A) i22 B) i1 C) i11 D) i22

72


14. JEE (Main) 2017(3): The equation 012

Im

iz

zi, Cz , iz , represents a part of

a circle having radius equal to:

1) 2 2) 1 3) 43 4)

21

15. JE Main – 2018(I): For a non-zero complex number z, let arg (z) denote the

principal argument with )arg(z . Then, which of the following statement(s) is

(are) FALSE?

A) 4

)1arg(

i , where 1i

B) The function ,: Rf , defined by )1arg()( ittf for all Rt , is

continuous at all points of R, where 1i

C) For any two non-zero complex numbers 1z and 2z ,

21

2

1 argargarg ZZZ

Z

is

an integer multiple of 2

D) For any three given distinct complex numbers 1z , 2z and 3z , the locus of the

point z satisfying the condition

123

321argZZZZ

ZZZZ

lies on a straight line.

73


HINTS AND ANSWERS

1. A),C), D). Hint: Express z in the form ibta . Equate real and imaginary part.

2. i) B). ii) C)

3. D). Hint Solve for z: 0)1(2 azz . The discriminant of the equation should not be zero.

4. Answer : 5. Distance is minimum when z is at (3, 0). Red segment shows the minimum value.

5. s); Hint: When simplified the expression becomes sin

1

6. D). Hint: Multiply and divide the imaginary part by 2sin4 . Convert each product to a difference.

7. A). Hint: use zz2

z . Guess the value of y, in order for x to become 4th

root of an integer.

8. D). Just translation and rotation. 90 rotation will interchange coordinates,

9. D). Use parametric equations of the circle.

10. D). Hint: Simplify the given expression. 21 z

z

= ec

icos

20

11. B) 12.D)

13. B) From the new position move 2 units along negative x-axis and 2 units along negative y-axis.

14. Answer: Option 3)

15.A), B) and D) are false statements.

(3, -2.5)

2

(3, 2)

z

74


JEE QUESTIONS ON INEQUALITY

1. If 2 , aa falls inside the angle made by the lines 2

xy , xy 3 , x > 0 and then a

belongs to

A)

3 ,

2

1 B)

2

1 ,3 C)

2

1 ,0 D) ) ,3(

y

x

y = x2

y = 3x

y = x

2

75


2. Determine all values of for which the point 2 , lies inside the triangle formed

by the lines 0132 yx , 032 yx , 0165 yx .

3. JEE (Advanced) 2014(I) – 55: For a point P in the plane, let 1d (P) and 2d (P) be the

distances of the point P from the lines 0 yx and 0 yx respectively. The area of

the region R consisting of all points P lying in the first quadrant of the plane and

5x - 6y - 1 = 0

x + 2y - 3 = 0

2x + 3y - 1 = 0

y

x

α, α2

y = x2

C

B

A

76


satisfying 4)()(2 21 PdPd , is____________

4. JEE (Advanced) 2013(I) – 43: For a > b > c > 0, the distance between

(1, 1) and the point of intersection of the lines 0 cbyax and 0 caybx is less

than 22 . Then

y

x

y = -xy = x

bx + ay + c = 0

ax + by + c = 0

(1, 1)

B

A

77


A) 0 cba B) 0 cba

C) 0 cba D) 0 cba

5. JEE (Advanced) 2011(II) – 58: The straight line 132 yx divides the circular

region into two parts.

If

then the number of point(s) in S lying inside the smaller part is _______

HINTS AND ANSWERS

1. A) 2. Answer: 12

3 , 1

2

1

3. Answer 6 square units. Hint:

422

2

yxyx

, take 2 cases.

622 yx

2x - 3y = 1

x2 + y2 = 6

y

x

4

1 ,

8

1 ,

4

1 ,

4

1 ,

4

3 ,

2

5 ,

4

3 ,2S

78


i) x > y, ii) y > x.

4. A), C). Set AB < 22 .

5. Answer: 2. Count how many of the given points lie in the shaded region.

79


PRINCIPLE OF MATHEMATICAL INDUCTION

Peano’s axioms on the set of natural numbers N.

1. N is not empty.

2. There exists 1 – 1 mapping aa of N into itself.

3. The range of successor mapping is a proper subset of N.

4. Any subset of N that contains an element that is not a successor

and that contains the successor of every element in the set

coincides with N. This is called axiom of induction.

a - successor of a

MODEL ANSWER TO PMI

Problem:

By using principle of mathematical induction(PMI) prove that

222222 132 nn , for all natural numbers n.

Solution:

Let P (n): 222222 132 nn

For n = 1, P (1): 222 11

= 222

= 4 – 2

= 2

80


The statement P (1) is true. Let the statement be valid for any natural

number k.

i.e. P (k): 222222 132 kk is true. …………. (a)

We have to prove the statement:

P (k+1): 2222222 2132 kkk ……………. (b)

Consider LHS of P (k+1)

132 22222 kk =

= 132 22222

kk (separating out the last

term)

= 11 222

kk By induction hypothesis from (a)

222 11 kk grouping like terms

= 222 1 k , adding like terms

22 2 k . law of exponents

= Right hand side of (b)

Hence P (k+1): 2222222 2132 kkk is a true statement

whenever P (k) is true. Since p (1) is true and )1()( kPkP , by principle

of mathematical induction P (n) is true for all n.

81


PRINCIPLE OF MATHEMATICAL INDUCTION – I

1. By the principle of mathematical induction, prove that for all natural number n

3

)2)(1()1(3221

nnnnn


2)12(252321 n =3

)124( nn


46)23()13(

1

118

1

85

1

52

1

n

n

nn

4. Apply the principle of mathematical induction to prove that for all n N

4

)7)(6)(1()6()3(963852741

nnnnnnn

5. Prove by the principle of mathematical induction that 132 n is divisible by 8 for

every natural number n.

6. Prove by using principle of mathematical induction that 123 n is divisible by 7 for


7. Prove by the principle of mathematical induction that 110 12 n is divisible by 11 for


82


8. Prove by the principle of mathematical induction that )2)(1( nnn is divisible by 6

for every natural number n.

9. Prove by the principle of mathematical induction that 1332 327 nnn is divisible

by 25 for any natural number n.

10. Prove by the principle of mathematical induction

2cosec

2sin

2

1coscos3cos2coscos:)(

xx

nx

nnxxxxnP

83


MATHEMATICAL INDUCTION - II

1. Prove by the principle of mathematical induction that

a) 2)12(531 nn .

b) 1)1(

1

43

1

32

1

21

1

n

n

nn

c) 13)13()23(

1

107

1

74

1

41

1

n

n

nn

d) )2)(1( 4

)3(

)2()1(

1

543

1

432

1

321

1

nn

nn

nnn

e) 27

109103333333333333

1

s3' ofnumber

nn

n

Hint: 1

s3' 1)k(

10...10010133333333

k. Use GP for RHS.

2. Prove by the principle of mathematical induction that

a) 1212211 nn

is divisible by 133.

b) nynx is divisible by x – y for all natural numbers n.

c) 524310 nn is divisible by 9 for all natural numbers n.

3. Prove by the principle of mathematical induction to prove that nn 2 for all positive

integers n.

84


4. Prove by induction that2)3(72 nn for all Nn . Using this, prove by induction that

32 2)3( nn for all Nn .

5. If cos21

xx , then prove by using PMI that n

nx

nx cos21

.

Hint:

1

1 1

k

k

xx

k

k

k

k

xxx

x

x

xxx

1

1

1 1

k

k

xx

6. Let 23)12(5 31)( KKKS . Then which of the following is true?

A) S(1) is correct

B) Principle of Mathematical Induction can be used to prove the formula.

D) )1()( KSKS

C) S(K) S(K + 1)

85


PERMUTATIONS AND COMBINATIONS - I

Counting Principle

To find the number of ways a series of successive events can occur, multiply the number of

ways in which each event can occur.

PERMUTATION

A permutation is an arrangement of things in a definite order.

120 permutations of 5 objects taken all at a time

)!(

!

rn

nrPn

86


1. Find the number of ways in which 5 boys and 5 girls be seated in a row so that

a) No two girls may sit together. Answer: 5! 5

6 P 86400

b) All the girls sit together and all the boys sit together. Answer: !52 !5 = 28800

c) All the girls are never together. Answer: 10! !5 6! =3542400

2. How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd

digits always occupy the odd places? Answer: 18

3. How many 7-digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2 and 4?

Answer: 360

4. In how many ways can the letters of the word ARRANGE be arranged so that

a) the two R’s are never together? Answer: 900

b) the two A’s are together but not two R’s Answer: 240

c) neither two A’s nor the two R’s are together? Answer: 660

5. How many numbers greater than 1000, but not greater than 4000 can be formed with

the digits 0, 1, 2, 3, 4 repetitions of digits being allowed? Answer: 375

6. A room is to be decorated with 14 flags; if 2 of them are blue, 3 red, 2 white, 3 green,

2 yellow and 2 purple, in how many ways can they be hung? Answer: 576

!14

7. The letters of the word NUMBER are written in all possible orders and these words

are written out as in a dictionary. What is the rank of the word NUMBER?

Answer: 469

87


8. A group of 2n students consisting of n boys and n girls, are to be arranged in a row

such that the adjacent members are of opposite sex. Find the total number of

arrangements. Answer: 2)! ( 2 n

COMBINATIONS

A subset of a set is also called a combination.

1))!( !

!

rnr

nrCn

2) rnCnrCn

3) rCnrCn

rCn 1 1 (Pascal’s formula)

Ten combinations of 5 objects taken 3 at a time

9. a) In how many ways can a football team of 11 players be selected from 15 players?

Answer: 1365

b) In how many ways a particular player is included? Answer: 1001

c) In how many ways a particular player is excluded? Answer: 364

88


10. An examination paper consists of 12 questions divided into 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 maximum ways

can the candidate select the questions? Answer: 420

11. A committee of 5 men and 3 women is to be formed out of 8 men and 5 women. In

how many ways can the committee be formed? Answer: 560

12. There are 10 points in a plane. Of these ten points only 4 points are in a straight line

and no other 3 points are in the same straight line. Find

a) the number of straight lines. Answer: 40

b) the number of triangles that can be formed by joining these points. Answer: 116

13. Find n.

a) Answer: n = 6

b) 5:8 : 41

3 CC nn

Answer: n = 8

c) 362 Cn

Answer: n = 9

d) 231 2 CC nn

Answer: n = 5

14. a) If 102020 rr CC , find 18

rC Answer: 816

b) If 1:30800: 354

656 rr PP , find r. Answer: r = 41

c) If 5 :3 : 121

12

n

nn

n PP find n. Answer; n =4

3:44 : 232 CC nn

89


15. Out of 7 consonants and 4 vowels, how many words can be made each containing 3

consonants and 2 vowels? Answer: 25200

16. From 12 books in how many ways can a selection of 5 be made when

a) one specified book is always included? Answer: 330

b) one specified book is always excluded? Answer: 462

17. A cricket team of 11 players is to be formed from 16 players containing 4 bowlers and

2 wicket-keepers. In how many different ways can a team be formed so as to contain

at least 3 bowlers and at least one wicket-keeper? Answer: 2472

18. A committee of 6 is chosen from 10 men and 7 women so as to contain at least 3 men

and 2 women. In how many different ways can this be done if two particular women

refuse to serve on the same committee? Answer: 7800

19. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to

be selected and arranged in a row so that the dictionary is always in the middle. Find

the total number of such arrangements. Answer: 1080

90


PERMUTATIONS AND COMBINATIONS - II

PERMUTATIONS

1. How many six-digit multiple of 5 can be formed from the digits 1, 2, 3, 4, 5 and 6

using each of the digits exactly once? Answer: 120

2. How many three digit numbers satisfy the conditions that there is no repetition in the

digits, the number must contain a 5, and is less than 800? Answer: 168

3. How many integers can be created using the digits 5, 6, 7, 8, 9 if no digit is repeated in

any of the integers and if each of the integers is greater than 770 and less than 96,000?

Answer: 252

4. How many distinct rearrangements of the letters in the word CURRENT have both

the vowels first? Answer:120

5. How many ways are there to arrange the letters of the word GARDEN with vowels in

alphabetical order. Answer: 360

6. In how many different ways can the letters of the word ALGEBRA be arranged in a

row

a) if the 2 A’s are together. Answer: 720

b) if the 2 A’s are not together. Answer: 1800

7. Find the number of arrangements of the letters of the word BANANA in which the

two N’s do not appear adjacently. Answer: 40

91


8. How many different nine digit numbers can be formed from the number 223355888

by rearranging its digits so that the odd digits occupy even places? Answer: 60

9. The letters of the word COCHIN are permuted and all the permutations are arranged

in an alphabetical order as in an English dictionary. Find the number of words that

appear before the word COCHIN. Answer: 96

10. All the words that can be using alphabets A, H, L, U, and R are written as in a

dictionary. Find the rank of the word RAHUL. Answer: 74

11. Find the total number of ways in which six ‘+’ and four ‘–’ can be arranged in a

row such that two ‘–’ signs are not together. Answer: 35

12. All arrangements of letters VNNWHTAAIE are listed in dictionary order. If

AAEHINNTVW is the first entry, what entry number is VANNAWHITE?

Answer: 738826

13. Find how many arrangements can be made with the letters of the word MISSISSIPPI.

In how many of them the four I’s do not come together? Answer: !2)!4(

!112

, 33810

14. Find the total number of permutations of the word ASSASSIN in which no two S’s

are together. Answer: 60

15. The integers 1, 2, 3, 4, 5, 6, 7 are arranged to form a seven-digit number. Determine

how many odd numbers can be formed. Answer: 2880

16. What is the total number of ways that a five digit number divisible by 3 can be formed

using the numerals 0, 1, 2, 3, 4 and 5 without repetition? Answer:216

92


17. CMC : The integers 1, 2, 3, 4, 5, 6, 7 are arranged to form a seven-digit number.

Determine how many numbers formed are less than 3, 200, 000. Answer: 1560

18. How many even integers between 4000 and 7000 have four different digits?

Answer: 728

19. In how many ways can four boys and three girls stand in a row such that the girls are

together and the boys are also together? Answer: 288

20. Six papers are set in examination, two of them in mathematics; in how many different

orders can the papers be given, provided only that the two mathematical papers are not

successive? Answer: 480

21. Find the total number of numbers from 1000 to 9999 (both inclusive) that do not have

4 different digits. Answer: 4464

22. How many numbers are there between 100 and 1000 such that at least one of their

digits is 7? Answer: 252

23. Find the number of divisors of 8400 excluding 1 and 8400. Answer: 58

Hint: A divisor of 8400 is utsr 7352 where r = 0, 1, 2,3, 4. s = 0, 1, 2. t = 0, 1 and u = 0, 1.

24. How many odd numbers, greater than 600.000 can be formed from the digits 5, 6, 7,

8, 9, 0 if

a) repetition of the digits are not allowed? Answer: 480

b) repetition of the digits are allowed? Answer:

554

93


COMBINATIONS

25. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can

this be done when the committee consists of

a) exactly 3 girls? Answer: 504

b) at least 3 girls. Answer: 588

26. In a class there are 4 girls and 7 boys. In how many ways can a team of 5 students can

be selected if the team has

a) no girl. Answer: 5

7C 21

b) at least one boy and one girl. Answer: 5

11C 5

7C 441

c) at least 3 girls. Answer: 4

4C 1

7C 3

4C 2

7C = 91

27. Each of the set of 6 parallel lines cuts each of another set of 5 parallel lines. What is

the number of parallelograms formed by the intersection of these lines? Answer: 150

28. In how many ways is it possible to choose 4 distinct integers from 1, 2, 3, 4, 5, 6 and 7

so that their sum is even? Answer: 19

29. For an examination, a candidate has to select 7 subjects from 3 different groups A, B,

C which contains 4, 5, 6 subjects respectively. In how many different ways can a

candidate make his selection if he has to select at least 2 subjects from each group?

Answer: 2700

94


30. A cricket eleven has to be chosen from 13 men of whom only 4 can bowl. In how

many ways can the team be made up so as to include at least 2 bowlers? Answer: 78

31. At a meeting, each person shook hands with every other person. If the total number of

hand shakes was 28, how many persons were present? Answer: 8

32. a) Find the number of diagonals in a regular 17-sided polygon? Answer: 119

b) A polygon has 170 diagonals. How many sides will it have? Answer: 20

33. A student is to answer 10 out of 13 questions in an examination such that he must

choose at least 4 from the first 5 questions. Find the total number of choices available

to him. Answer: 196

34. A woman has 11 friends. She wishes to invite 5 of them for a dinner. In how many

ways can she invite them if two of them are not on speaking terms and will not attend

together? Answer: 378

35. A woman has 11 friends. She wishes to invite 5 of them for a dinner. In how many

ways can she invite them if two of the friends are married and will not attend

separately. Answer: 210

36. From 6 gentlemen and 4 ladies, a committee of 5 is to be formed. In how many ways

can this be done if

a) there is no restriction about its formation? Answer: 252

b) the committee is to include at least one lady? Answer: 246

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37. In how many ways can a committee of 3 ladies and 4 gentlemen be appointed from a

group consisting of 8 ladies and 7 gentlemen? What will be the number of ways if

Mrs. X refuses to serve in a committee in which Mr. Y is a member?

Answer: 1960, 1540

38. A committee of 12 is to be formed from 9 women and 8 men. In how many ways can

this be done if at least five women have to be included in a committee? Answer: 6062

In how many of these committee

a) the women are in majority? Answer: 2702

b) the men are in majority? Answer: 1008

39. In how many ways can 12 men be selected out of 17 if

a) there is no restriction on the choice? Answer: 6188

b) two particular men are always included? Answer: 3003

c) two particular men never are chosen together? Answer: 3185

40. A man has 12 relations, 7 ladies and 5 gentlemen; his wife has 12 relations, 5 ladies

and 7 gentlemen. In how many ways can they invite a dinner party of 6 ladies and 6

gentlemen so that there may be 6 of the man’s relations and 6 of the wife’s?

Answer: 267148

41. With 9 consonants and 7 vowels, how many words can be made, each containing 4

consonants and 3 vowels

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a) when there is no restriction on the arrangements of letters? Answer: 22226400

b) when two consonants are never allowed to come together? Answer: 635040

42. If the number of triangles which can be formed by using the vertices of a regular

polygon of (n + 3) sides is 220, find n. Answer: n =9

43. Find the total number of ways in which one can select three distinct integers between

1 and 30, both inclusive, whose sum is even. Answer:2030

44. If35

13222

12

nn

nn

C

C, find n. Answer: n = 6

45. Prove that: )!1( 3 2 11 33

22

11 nPnPPP n

n

46. A student is allowed to select at most n books from a collection of (2n+1) books. If

the total number of ways in which he can select at least one book is 63, find the value

of n. Answer: 3

47. In how many ways can a lawn-tennis mixed double be made up from 7 married

couples if no husband and wife play in the same set? Answer: 420

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JEE QUESTIONS ON

PERMUTATIONS

1. JEE (Advanced) 2017(II) - 43 :Three randomly chosen nonnegative integers x, y and

z are found to satisfy the equation 10 zyx . Then the probability that z is even,

is

A) 55

36 B)

11

6 C)

2

1 D)

11

5

2. JEE (Advanced) 2017(I):47 : Words of length 10 are formed using the letters A, B,

C, D, E, F, G, H, I, J. Let x be the number of such words where no letter is repeated;

and let y be the number of such words where exactly one letter is repeated twice and

no other letter is repeated. Then, x

y

9_________

3. JEE (Advanced) 2017(II) - 42 : Let S = {1, 2, 3, . . ., 9}. For k = 1, 2, . . . , 5, let Nk

be the number of subsets of S, each containing five elements out of which exactly k

are odd. Then N1 + N2 + N3 + N4 + N5 =

A) 210 B) 252 C) 125 D) 126

4. JEE (Advanced) 2016(I) -38: A debate club consists of 6 girls and 4 boys. A team of

4 members is to be selected from this club including the selection of a captain (from

among these 4 members) for the team. If the team has to include at most one boy, then

98


the number of ways of selecting the team is

A) 380 B) 320 C) 260 D) 95

5. JEE (Advanced) 2015(I) - 41: Let n be the number of ways in which 5 boys and 5

girls can stand in queue in such a way that all the girls stand consecutively in the

queue. Let m be the number of ways in which 5 boys and 5 girls can stand in queue in

such a way that exactly four girls stand consecutively in the queue. Then the value of

n

m is _______

6. JEE (Advanced) 2014(I) -53: Let 2n be an integer. Take n distinct points on a

circle and join each pair of points by a line segment. Color the line segments joining

every pair of adjacent points by blue and the rest by red. If the number of red and blue

line segments are equal, then the value of n is ________

7. JEE (Advanced) 2012(I) - 45:The total number of ways in which 5 balls of different

colors can be distributed among 3 persons so that each person gets at least one ball is

A) 75 B) 150 C)210 D)243

8. JEE (Advanced) 2009(I) – 25: The number of seven digit integers, with sum of the

digits equal to 10 and formed by using the digits 1, 2 and 3 only, is

A) 55 B) 66 C) 77 D) 88

9. JEE (Advanced) 2008(II)-21:

Consider all permutations of the word ENDEANOEL.

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Match the expressions/statements in column I with the expressions/statements in

column II.

I II

A) The number of permutations containing

the word ENDEA is

p) 5!

B) The number of permutations in which the

letter E occurs in the first and last positions is

q) !52

C) The number of permutations in which none of the

letters D, L, N occurs in the last five positions is

r) !57

D) The number of permutations in which the

letters A, E, O occurs only in odd positions is

!521 ) s

10. JEE (Advanced) 2007(II) – 53: The letters of the word COCHIN are permuted and

all the permutations are arranged in an alphabetical order as in an English dictionary.

100


The number of words that appear before the word COCHIN is

A) 360 B)192 C)96 D) 48

11. Let nT denote the number of triangles which can be formed using the vertices of a

regular polygon of n sides. If nn TT 1 = 21, then n equals

A) 5 B) 7 C) 6 D) 4

12. AIEEE 2012 – 74: Assuming the balls to be identical except for difference in colors,

the number of ways in which one or more balls can be selected from 10 white, 9 green

and 7 black balls is:

A) 629 B) 630 C) 879 D)880

13. AIEEE - 2011(81):

Statement – 1:

The number of ways of distributing 10 identical balls in 4 distinct boxes such that no

box is empty is 3

9C .

Statement – 2:

The number of ways of choosing any 3 places from 9 different places is 3

9C .

A) Statement-1 is true, Statement-2 is true; Statement-2 is a

correct explanation for Statement-1.

B) Statement-1is true, Statement-2 is true; Statement-2 is

not a correct explanation for Statement-1 .

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C) Statement-1 is true, Statement-2 is false.

D) Statement-1 is false, Statement-2 is true.

14. AIEEE - 2008 (88): In a shop there are five types of ice-creams available. A child

buys six ice-creams.

Statement – 1:

The number of different ways the child can buy the six ice-creams is 5

10C

Statement – 2:

The number of different ways the child can buy the six ice-creams is equal to the

number of different ways of arranging 6 A’s and 4 B’s in a row.

A) Statement-1 is true, Statement-2 is true; Statement-2 is a

correct explanation for Statement-1.

B) Statement-1is true, Statement-2 is true; Statement-2 is

not a correct explanation for Statement-1 .

C) Statement-1 is true, Statement-2 is false.

D) Statement-1 is false, Statement-2 is true.

15. AIEEE - 2007 (109): The set S = {1, 2, 3, . . . , 12} is partitioned into three sets A,

B, C of equal size. Thus, SCBA , BA = CB = CA = . The number of

ways to partition S is

A) 3)! 4(

! 12 B)

3)! 3(

! 12 C)

3)! 4( ! 3

! 12 D)

4)! 3(! 3

! 12

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16. JEE(Main) 2015 (66): The number of integers greater than 6,000 that can be formed,

using the digits 3, 5, 6, 7 and 8, without repetition, is:

A) 216 B) 192 C)120 D) 72

17. AIEEE - 2008(80): How many different words can be formed by jumbling the letters

in the word MISSISSIPPI in which no two S are adjacent?

A) 4786 C B) 4

84

67 CC

C) 47

468 CC D) 4

876 C

18. JEE(Main) 2016 - 9th

APRIL(66): If the four letters (need not be meaningful) are to

be formed using the letters from the word MEDITERRANEAN such that the first

letter is R and the fourth letter is E, then the total number of such words is

A) 3)! 2(

! 11 B)110 C) 56 D) 59

19. APTITUDE – 2017 (6): An urn contains 5 red, 4 black and 3 white marbles. Then

the number of ways in which 4 marbles can be drawn from it so that at most 3 of them

are red, is:

1) 455 2) 460 3) 490 4) 495

20. RMO – 2012(4): Let X ={1, 2, 3,…,11}. Find the number of pairs {A, B} such that

XA , XB , BA and BA = {4, 5, 7, 8, 9, 10}.

21. RMO - 2007(4) : How many 6-digit numbers are there such that:

(a) the digits of each number are all from the set {1, 2, 3, 4, 5}

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(b) any digit that appears in the number appears at least twice.

(Example: 225252 is an admissible number, while 222133 is not.)

22. RMO –2008(4): Find the number of all 6- digit natural numbers such that the sum of

their digits is 10 and each of the digits 0, 1, 2, 3 occurs at least once in them.

23. RMO –2000 (4): All the 7-digit numbers containing each of the digits 1, 2, 3, 4, 5, 6,

7 exactly once, and not divisible by 5, are arranged in increasing order. Find the 2000th

number in this list.

24. JEE (ADV) 2018(I): The number of 5 digit numbers which are divisible by 4, with

digits from the set {1. 2, 3, 4, 5} and the repetition of digits is allowed, is ________.

25. JEE (ADV) 2018(II): In a high school, a committee has to be formed from a group of

6 boys21 , MM ,

43 , MM , 65 , MM and 5 girls

21 , GG , 43 , GG , 5G .

(i) Let 1 be the total number of ways in which the committee can be formed such

that the committee has 5 members, having exactly 3 boys and 2 girls.

(ii) Let 2 be the total number of ways in which the committee can be formed such

that the committee has at least 2 members, having equal number of boys and

girls.

(iii) Let 3 be the total number of ways in which the committee can be formed

such that the committee has 5 members, at least 2 of them being girls.

(iv) Let 4 be the total number of ways in which the committee can be formed

such that the committee has 4 members, having at least 2 girls and such that

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1M and

1G are NOT in the committee together.

LIST – I LIST – II

P: The value of 1 is 1. 136

Q: The value of 2 is 2. 189

R: The value of 3 is 3. 192

S: The value of 4

is 4. 200

5. 381

6. 461

The correct option is

A) P 4: Q 6; R 2; S 1

B) P 1: Q 4; R 2; S 3

C) P 4: Q 6; R 5; S 2

D) P 4: Q 2; R 3; S 1

HINTS AND ANSWERS

1. B)

Solution: Total number of solution in nonnegative integers is

111 S 11111 S 11

x = 3, y = 5 and z = 2, S- space. 3 + 5 + 2 = 10

131)310(

C = 2

12C

1111S111111S z = 0 Total 11 ways (only red are rearranged)

1111S1111S11 z = 2 Total 9 ways (only red are rearranged)

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SS1111111111 z = 10 Total 1 way (only red are rearranged)

Total = 36 ways z can be even. Probability = 66

36=

11

6

2. Answer 5. Y = Select one letter from given 10 letters and 8 letters from the remaining 9 letters.

3. D). Hint: N3 = 60 because S3 = { 3 odd numbers, 2 even numbers}

4. A). Hint: Captain’s selection can be done in 4 ways.

5. Answer 5.

6. Answer 5.

7. B) Hint: one person can receive 3 balls, other two persons will receive one ball each OR two persons

will get 2 balls each , third person will get 1 ball.

8. C Hint: i) Select five 1’s, one 2, one 3 OR three 2’s , four 1’s.

9. (A, p), (B, s), (C, q), (D, q)

10. C). 11. B)

12. C). White ball selection can be done in 0, 1, 2, 3,…10 (Total 11 ways). Green ball selection is 10 and

Black ball selection is in 8 ways. Total 880. Total = 880 – 1 (no ball is selected)

13. A) 3 places out of 9 places shown in red are to be selected for3 boxes. The balls which are left of the

box will go in to that box. The 4th box will always occupy the last position.

987654321

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14. D) The figure shows the child bought 2 ice creams of first type and one each of the other 4 types.

15. A) Hint: First select 4 numbers for set A. From the remaining 8 numbers select 4 numbers for B

and so on.

16. B)

17. B) Simplify option B.

18.D) We can select 2 E’s, 2 N’s, 2 A’s for 2nd

and third position. 3 ways. We can select 2 letters from

the set {M, E, D, I, T, R, A, N} in 28 ways . They can be arranged in 56 ways.

19. Option 3)

20. Let P = A – B, R = B – A and Q = X – BA

Remaining elements are 1, 2, 3, 6, 11

1 can be put either in P, Q or R i.e., in 3 ways.

Total number of ways of filling 5 elements is 243. If all the 5 elements go to Q, then P = R.

Total number of subsets = 242.

21. Answer: 1405. Consider 4 cases. For example {111, 222}, {11, 22, 33} etc.,

22. Answer: 490 . Hint: Consider 3 cases.

X

BA

10

9875

4

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23. Solution: Numbers starting with 1, 2, 3 are 5! 5 = 600 each.

Total = 1800 numbers.

Numbers starting with 41, 42 are 4! 4each. Total = 1800 + 96 +96.

4,312,567, 4,312,576, 4,312,657, 4,312,756

4,315,267, 4,315,276, 4,315,627, 4,315,672....2000th

number.

24. 625 25. Option C)

June 16 CBSE/JEE 2019 Questions · 2019-06-16 · 4 9. Out of 1600 students in a college, 390...

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Transcript of June 16 CBSE/JEE 2019 Questions · 2019-06-16 · 4 9. Out of 1600 students in a college, 390...