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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
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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
5 JEE Questions 40 - 47
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
13 JEE Questions 97 - 107
2
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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}.
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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
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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
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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
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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
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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
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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.
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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
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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 )(
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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
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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
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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
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sin π -θ( )θ π - θ
y
x
θ
sin π + θ( )
π + θ
y
x
15
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)(
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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
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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
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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
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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
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of the pole. Answer: 12 m
β
αh
2h
4 m
A
B C
D
31
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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
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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
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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
380sin60sin40sin20sin
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
170sin50sin30sin10sin
35. Prove that 178tan66tan42tan6tan
36. Find the value of: 81tan63tan27tan9tan Answer: 2
37. Prove the identity:
8sin8
sin
8cos
4cos
2cos
34
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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
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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
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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
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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
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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(
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83. Prove the identity
2sin
23sin
23
sin26
cos4
84. Prove the identity cos21cos2cos
3cos2coscos12
xxx
xxx
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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
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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
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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
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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
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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
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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
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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
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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
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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 _____
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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2) If i 32 a root of the equation 02 qpxx where Rqp , , then find p and q
. Answer: p = –4 , q = 7
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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.
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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
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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
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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
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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
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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
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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.
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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
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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.
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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. By the principle of mathematical induction, prove that for all natural number n
2)12(252321 n =3
)124( nn
3. By the principle of mathematical induction, prove that for all natural number n
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
every natural number n.
7. Prove by the principle of mathematical induction that 110 12 n is divisible by 11 for
every natural number n.
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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
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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.
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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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1111S11S1111 z = 4 Total 7 ways (only red are rearranged)
11S11S111111 z = 6 Total 5 ways (only red are rearranged)
1S1S11111111 z = 8 Total 3 ways (only red are rearranged)
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)