MAINS - CW -- VOL

JEE ADVANCED VOL -- III

I
LEVEL-VI  
 x1 
9. The domain of f  x   cot  2 2 
SINGLE ANSWER QUESTIONS x   x  
 
(where [.] is G.I.F)
1. If [.] denotes G.I.F , then the domain of
a)R b) R  0
f  x   cos 1  x   x  is
a) (0,1) b) [0,1) c)[0,1] d)[-1,1] 
c) R   n , n  Z and n  0 
d)None
2.
2
 2

If f  x   min . x , x, sgn x  4 x  5  f  x  
1
10. The domain of ln  cos 1 x  where
then the value of f  2  is equal to
[.] is G.I.F is
a) 2 b) 0 c) 1 d) 4
3. Let ‘f’ be an injective mapping with domain a)[0,1] b)  1, cos 2

 x, y, z and range 1, 2,3 such that exactly c) [1,cos3)   cos3,cos 4 

one of the following statements is correct d) [1,cos3)   cos3,cos 2 
and the remaining are false
f  x   1; f  y   1; f  z   2 then f 1 1  11. The range of f  x   sin
1
 x 2  x  1 is 
[IIT 1982]          
a)x b)y c)z d) None a)  0,  b)  0,  c)  ,  d)  , 
 2  3 3 2 6 2
4. The domain of 12. The function
 
f  x   log10 1  log10  x 2  5 x  16  is f  x  | px  q | r | x |, x   ,   where
p>0,q>0,r>0 assumes its minimum value
a) (2,3) b)  0,   c) 1,3 d)  2,3 only at one point if
5. The period of the real - valued function a) p  q b) r  q c) r  p d) p  q  r
satisfying 13. Let f,g,h be real -valued functions defined
2 2

f  x   f  x  4   f  x  2   f  x  6  is on the interval [0,1] by f  x   e x  e  x ;

2 2 2 2
a) 10 b) 8 c) 12 d) 6 g  x   x.e x  e  x and h  x   x 2 .e x  e  x .
6. There are exactly two linear functions which If a,b,c denote respectively the absolute
map from [-1,1] onto [0,2] they are max.values of f,g,h on [0,1] then
a) y  x  1, y  x  1 b) y  x 1, y  x 1 a) a  b and c  b b) a  c and a  b
c) y  2 x  1, y  x  4 d) y  x, y  3 x c) a  b and c  b d)a=b=c
2 2
7. If the graph of a function f(x) is symmetrical e x  e x
about the line x=a then 14. Let f : R  R defined by f  x   2 2
e x  e x
a) f  a  x   f  a  x  then f(x) is
a) one-one but not onto
b) f  a  x   f  x  a  b)neither one-one nor onto
c)many-one but onto d)one-one but not onto
c) f  x   f   x  d) f  x    f  x 
15. f : (, 1]  (0, e5 ] defined by
8. If f  x   f  2a  x  then the graph of f(x) 3
f  x   e x 3 x 2 is
is symmetric about the line a)one-one and into b)one-one and onto
a)y=a b)x=2a c)x=a d)x=-a c)many-one and into d)many-one and onto
 FUNCTIONS
16. If f  x   1 | x  2 |;0  x  4 22. If ' p ' and ' q ' are +ve integers , f is a
g  x   2 | x |; 1  x  3 then  fog  x   function defined for +ve numbers and attains
only +ve values such that
1  x; 1  x  0 1  x;0  x  1
a) x  1 ; 0  x  2 b) x  1 ; 1  x  2 f  xf ,  y    x p . y q then p 2 
a) 2q b) q c)3q d) 4q
c) 1  x; 1  x  2 d) does not exists
23. If the function ‘f’ satisfies the relation

2  2
17. If f  x   sin x  sin   x  f  x  y  f  x  y
3 
  5  2 f  x  f  y  x, y  R and f  0   0
 cos x.cos  x   and g    1
 3 4 then f(x) is an
then the graph of y  g  f  x   is a) even function
a) a circle b) a straight line b)odd function
c) a parabola d) a pair of staraight lines c)Neither even nor odd
 2x2  1  d)can not decide
1
18. The range of f  x   cos  
 x 2  1  is 24. Let f  x   x  2 x  1  2 x  1 . If f  x   k
 
has exactly one real solution, then k equals
   
a)   ,  b)  0,  c) [0, ) d)  0,   a) 3 b) 0 c) 1 d) 2
 2 2 2
25. A real valued function f(x) satisfies
1 the function
19. If f :[1,  )  [2, ) is given by f  x   x 
x
1
f  x  y   f  x. f  y   f a  x f a  y 
then f  x  
where ‘a’ is a given constant and f  0   1,
x  x2  3 x  x2  4
a) b) then the graph of the function is symmetrical
4 2 about [IIT 2005]
x  x2  4 x  x2  4
c) d) a) po int  2a, 0  b) po int  a, 0 
2 2
c) line x  2a d) line x  a
20. If f  x  is an even function and satisfies the
26. The function ‘f’ satisfies the functional
2 1
relation x . f  x   2. f    g  x  where  x  59 
 x equation 3 f  x   2 f    10 x  30
 x 1 
g  x  is an odd function , then the value of
for all real x  1, then the value of f  7  is
f  5  is
a)8 b) 4 c) -8 d)11
37 27. Let ‘f’ be a real valued function defined for
a)0 b) c) 4 d) 5
55 all x  R such that for some fixed a>0,
21. Consider a real valued function f(x)
1 2
satisfying
y
2 f  xy    f  x     f  y  
x
f  x  a   f  x    f  x   for all ' x '
2
x, y  R and f 1  a where a  1 then then the period of f  x  is
n
 a  1  f  i   a a
i 1 a) b) c) 2a d)None
4 3
a) a n 1
 a b) a n 1 c) a n 1
 a d) a n 2
a
 MAINS - CW -- VOL
JEE ADVANCED VOL -- III
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28. Let f  x, y  be a periodic functin satisfying
MULTIPLE ANSWER QUESTIONS
f  x, y   f  2 x  2 y, 2 y  2 x  x, y  R
33. For the function f  x satisfying
defined g  x   f  2 x ,0  then the period of
2. f  sin x   f  cos x   x, x  R
g  x  is
a) 4 b) 6 c) 8 d) 12  2  
a) Domain is [0,1] b) range is  , 
29. Let f(x) be a periodic function with  3 3

 2     
period 3 and f    7 and g  x   c)Domain is [-1,1] d)range is  , 
 3   2 2
x 1
f  t  n  dt where n  3K , K  N then f  x 
0
34. The domain of | | x | 1 | 5
7 is where [.] G.I.F
g1   
3 a)  7,7  b) ( , 7]
2 7 c) ( , 7] d) [7,  )
a) b)7 c)-7 d)
3 3
35. Let f  x   sin x, g  x   ln | x | If the ranges
1 1 1
30. Let f1  n   1         then of fog and gof are R1 and R2 respectively
2 3 n
then [IIT 1994 ]
f1 1  f1  2   f1  3        f1  n  
a) R1  U : 1  U  1
a) n. f1  n   1 b)  n  1 f1  n   n b) R2  V :   V  0
c)  n  1 f1  n   n d) n. f1  n   n c) R1  U : 0  U  1 d)None
n 1 36. The polynomial p(x) is such that for any
31. Let f 1  1 and f  n   2 f  r  then polynomial q(x) we have
r 1

m
p  q  x    q  p  x   then p  x  is
 f  n 
n 1
a) even b)odd
c)of even degree d)of odd degree
a) 3m  1 b) 3m c) 3m1 d) 3m2 37. If a polynomial of degree ‘n’ satisfies
f  x   f 1  x  . f 11  x  n  R then f(x) is
 1  1
32. If
f  x   cos
1
x  x 
2
 1   
2
 | x |   x  1
a) an onto function b) an into function
  c) no such function is possible
d) even function
then domain of f  x  is where [.] is G.I.F is
x
Let f  x   ln  2 x  x   sin 
2
 38.  then
1 5   1 5   2 
a)  2, 2  b)   2, 2 
    a) graph of ‘f’ is symmetrical about the line x=1
b) graph of ‘f’ is symmetrical about the line x=2
 1 5   7  c)max. value of ‘f’ is 1
c)  2, 2  d)  7 , 2  d)min.value of ‘f’ does not exist
   
 FUNCTIONS
39. If f  x   x | x |; 0  x  1 Now answer the following:

 2x ; x  1 then its 43. Domain and range of H  x  are

a)even extension is b)odd extension is a) R and 1 b) R and 0,1
2 2
x ; 1  x  0 x ; 1  x  0
 
2x;   x  1 2x;   x  1 c) R   2n  1  and 1 d)None
 4
c)even extension d) does not exists
n z
x2 ; 0  x  1
44. If F : R   2, 2 then F  x  is
2 x;1  x  
a)one-one b) onto c) into d)None
40. If f  x   3 | x |  x  2 and g  x   sin x 45. Which of the following is correct
a) periods of f(x) ,g(x) and F(x) makes A.P with
then the domain of  fog  x  

 the common difference
 3
a) 2m   ; m  z
 2 b) perod of f(x),g(x) and F(x) are same and is
equal to 2 
 7 11 
b) 2m  , 2m  c)sum of the periods of f(x),g(x) and F(x) is 3
 6 6  m  Z
d) sum of the periods of f(x),g(x) and F(x) is 6
  46. Which of the following is correct
c) 2m   , m  z d) 
 3 a) the domain of G(x) and H(x) are same
b)the range of G(x) and H(x) are same
41. Let f  x   max 1  sin x,1,1  cos x ;
c) the union of the dmain of G(x) and H(x) are all
x   0, 2  and g  x   max 1,| x  1| ; real numbers
d)None
x  R then
a) g  f  0    1 b) g  f 1   1 47. If the solutions of F  x   G ( x )  0 are

c) f  g 1   1 d) f  g  0    sin1 x1 , x2 , x3      xn where x   0,5  then

a) x1 , x2 , x3      xn are in A.P with
42. Let f  x   sin x  cos  
4  a 2 x . Then

the integral values of ‘a’ for which f(x) is a common difference
4
periodic function , are given by b) the no.of solutions of
a)1 b) 2 c) -2 d)0
F  x   G  x   0 is 10 x   0,5 
PARAGRAPH QUESTIONS c)t he sum of all solutions of
F  x   G  x   0 is x   0,5  is 25
(P) Let F  x   f  x   g  x  ;
d) b,c are true
f  x (P) Let f : N  N be a function defined by f(x)
G  x   f  x   g  x  and H  x   g x
  = the biggest +ve integer obtained by
reshuffling the digits of ‘x’. For example
where f  x   1  2sin 2 x and f(296)=962. Now answer the following.
g  x   cos  2 x  f : R   1,1 and 48. f is
a) one-one ,onto b) one-one and into
g : R   1,1 c)many-one and onto d)many-one and into
 MAINS - CW -- VOL
JEE ADVANCED VOL -- III
I
49. The biggest +ve integer which divides 55. The equation f  x   x  f  0   0 have
f  n   nv  N is exactly
a) no solution b) one solution
a)3 b)9 c)18 d)27
c)two solution d)infinite solutions
50. The range of f is
a)N 56. f 1 0 
b)set of +ve integers whose digits are non-
a)0 b)1 c) f  0  d)  f  0 
increasing from left to right
c)set of +ve whoe digits are non-decreaseing 2  1 x 
from left to right (P) If  f  x   . f    64 x; x  0 , then
 1 x 
d)None
(P) A function ‘f’ from a set X to Y is called 57. f  x 
onto if every y  Y x  X such that f(x)=y.. 1 1
2 1
Unless the contrary is specified, a real  1  x 3
3  1  x 3
3
a) 4 x   b) x  
function is onto if it takes all real values,  1 x   1 x 
Otherwise, it is called on into function. Thus, 1 1/ 3
if X and Y are finite sets , then ‘f’ can not be 2
 1  x 3 1/ 3  1  x 
onto if Y contains more elements than ‘x’. c) x 3
 d) 16 x  
1 x   1 x 
Now answer teh following.
51. The polynomial function 58. The domain of f  x  is
n n 1 n 2
f  x   a0 x  a1 x  a2 x  .......  an a) [0,  ) b) R  1,1
where a0  0 is onto , for
c)  ,   d) R  0,1, 1
a) all positive integers n
b) all even +ve integers ‘n’ 9
c)all odd +ve integers’n’ 59. The value of f   is
7
d)no +ve interger
2/3 1/3

x2  2 x  c 7 9
a) 8   b) 4  
52. The function f  x   2 is onto , if 9 7
x  4 x  3c
2/3 1/ 3
a) 0  c  2 b) 0  c  4 9 9
c) 8   d) 4  
1 1 7 7
c)  c  d) 0  c  1
2 2 (P) Based upon each paragraph, three multiple
53. Which of the following is not true choice questions have to be an swered. Each
a)A one-one function from t he set question hads four choices a,b,c and d, out
of which only one is correct.
a, b, c to  ,  ,   is onto also Passage: Consider the functions
b)An onto function from an infinite set to a finite
set can not be one-one  x  1, x 1
f  x   and
c)An onto function is alwasys invertible 2 x  1, 1  x  2
d)the functions tanx and cotx are onto
 x2 , 1  x  2
(P) Let f : R  R be a continuous function g  x  
 x  2, 2  x  3
 x  x
such that f  x   2. f    f    x2 . 60. The domain of the fucntion f  g  x   is
2  4
Now answer the following
a)  0, 2  b)  1, 2
54. f  3 
a) f(0) b) 4+f(0) c) 9+f(0) d) 16+f(0) c)  1, 2  d) None of these
 FUNCTIONS
61. The range of the function f  g  x   is COLUMN-II
p) one-one
a) 1,5 b)  2,3 q) into
c) 1, 2   3,5 d) none of these r) many-one
62. The number of roots of the equation s) onto
67. Match the following:
f  g  x    2 is
LIST-I
a) 1 b) 2
c) 4 d) None of these a) the integral value of x   ,   satisfying
(P) Let f  x   sin x  x cos x, x  R . Now the equation | x 2  1  cos x || x 2  1|  | cos x |
answer the following.
can be
63. The least +ve value of ‘x’ for which f(x)=0
lies in the quadrant 2
b)the no.of solutin of  x   x  2  x is where
a) Q1 B) Q2 C) Q3 D) Q4
[.] is G.I.F {.} is fractional part function
64. The set of all the values of x   0, 2  for
c) If f  x   sin 1 x  cos1 x  tan 1 x then
which f  x   0 , is
 f  x   is
  3 
a)  0,   b)  , 2  c)  ,  d)None
2 2  d)An allowable value of
65. If  is the least +ve value for which
tan    , then the area bounded by f  x   ln  cos  sin x  
y  f  x  , X  axis, x  0 and x  2 , is LIST-II
p)0
a)4 b) 4 1  cos  
q)1
c) 4 1  cos   d) 4  2  2   2  cos  r)2
s)4
MATRIX-MATCHING QUESTIONS t)-1
68. Match the following equations with their no.
66. Match the following: of roots
COLUMN-I
LIST-I LIST-II
  a) x 2 tan x  1; x   0, 2 p) 5
a) f : R   ,   and
4 
b) 2cos x | sin x |; x  0, 2 q) 2
f  x   cot 1
 2x  x 2
 2  then f(x) is
c)If f  x  is a polynomial of r) 3
b) f : R  R and f  x   e ax .sin  bx  where
degree 5 with real coeffs.
a, b  R  then f(x) is
Such that f  | x |  0 has 8
2

c) f : R  [2, ) and f  x   2  3x real roots, then the no.of
then f(x) is
roots of f  x   0
d) f : X  X and f  f  x    xx  X then
d) 7 | x |  1 5  | x |   1 s) 4
f(x) is
 MAINS - CW -- VOL
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69. Match the following: (B) Both S-I and S-II are individually true but
COLUMN-I S-II is not the correct explanation of S-I
a) If f  x   x  1, when 1  x  0 (C) S-I is true but S-II is false
(D) S-I is false but S-II is true
 x 2  1; whenx  0 then  fof  x  for 2
70. Consider two functions f  x   1  e cot x
and
1  x  0 is
1  cos  2 x 
 2 tan x  g  x   2 | sin x | 1 
b) If f  2  1  sin 4 x
 1  tan x 
S-I: The solution of f(x)=g(x) is given by
 cos  2 x   1  sec 2
x  2 tan x  
 then f(x) is x   2n  1 ; n  I
2 2
2 S-II: If f  x  k and
c) If f  x  y  1   f  x  
f  y  x,
g  x   k  where K  R  then the solutions
y  R and f  0   1 then f  x  is
of the equation f  x   g  x  is the solution
x
d) If 4  x  5 and f  x      2 x  2 corresponding to f  x   K
4
71. S-I: f : R   1,1 , f  x   cos x is not a
Where [.] is G..I.F then f 1  x  is
bijection
COLUMN-II S-II: Every even function is not one-one
x 3 72. S-I: The function f  x  | x | is not one-one
p)
2
S-II:The negative real numbers are not the
q) x 2  2 x pre images of any real numbers
r)1+x 73. S-I: If f  x   x5  16 x  2 then f(x) =0
2
s)  x  1 has only one root in [-1,1]
S-II:f(-1) and f(1) are of opp.signs
ASSERTION - REASON QUESTIONS
INTEGER QUESTIONS

The following questions consist of two statements, 74. The period of the function satisfying the
one labelled as ‘Statement-I’ and the other
relation f  x   f  x  3  0 x  R is
‘Statement-II’. You are to examine these two
statements carefully and decide if the cos  sin  nx  
; n  N is 6
Statement-I and the Statement-II are individually  x
75. If the period of tan  
true and if so, whether the Statement-II is the n
correct expalnation for the given Statement-I. then ‘ n’ is
Select your answer to these items using the codes
76. Let f  x   x  2  x  ;0  x  2 .
If the
given below and then select the correct option.
definition of ‘f’ is extended over the set R-
(A) Both S-I and S-II are individually true and
R is the correct expalnation of A [0,2] by f  x  1  f  x  then f is a periodic
function with period.
 FUNCTIONS

77. If f  x  and g  x  are any two real

KEY - LEVEL-VI
valued functions such that | f  x   g  x  |
SINGLE ANSWER QUESTIONS
| f  x  |  | g  x  | and g  x   0;
100
1) B 2) C 3) B 4) A
f  x  .g  x   0 then the value of  f  r   5) B 6) B 7) A 8) C
r 1
9) C 10) B 11) C 12) C
78. The period of f  x   sin  sin  x    e3 x
13) D 14) B 15) A 16) A
is, where {.} is fractional part function. 17) B 18) C 19) C 20) A
79. If f  x   sgn  x 2  2 x  3 , then the value 21) C 22) B 23) A 24) B
25) B 26) B 27) C 28) D
of f  x  is:
29) B 30) C 31) C 32) C
80. Given y  2  x   3 and y  3 x  2  5 ,
MULTIPLE ANSWER QUESTIONS
where . dentoes the greatest integer
function, then the value of  x  3 y  is 37 , 33) B,C 34) C,D 35) A,B 36) B,D
then  is 37)A 38)A,C,D 39)A,B 40)A,B
81. If the range of the function 41)A,B 42)B,C,D

f  x 
1  x  x 1  x  , when
2 4

PARAGRAPH QUESTIONS
x  0 is
x3
[ K ,  ) , then K is 43)C 44)B 45)C 46)C
1 47)D 48)D 49)B 50)B
82. If fractional parts of and x 2 for some 51)C 52)D 53)C 54)D
x
55)C 56)A 57)A 58)B
x  
2, 3 are equal, then the value of
59)C 60)C 61)C 62)B
3 63)C 64)D 65)D
x4  is
x
MATRIX-MATCHING QUESTIONS
83. A non-zero function f  x  is symmetrical
about the line y  x then the value of  66) A-Q,R; B-R,S; C-P,S; D-P,S
(constant) such that
67) A-Q,T; B-S; C-P,Q,R; D-P
2
f 2  x    f 1  x     xf  x  f 1  x  68) A-Q; B-S; C-P; D-S
69) A-Q; B-R; C-S; D-P
3x 2 f  x  x  R  is
84. If x 2  4 y 2  4 and range of f  x, y  is ASSERTION - REASON QUESTIONS

 L, M  , where f  x, y   x 2  y 2  xy , then 70)C 71)A 72)C 73)B

LM is
85. If the periodic function f  x  satisfies the INTEGER QUESTIONS
equation f  x  1  f  x  1  3 f  x 
74)6 75)6 76)2 77)0
x  R and the period of f  x  is 4 then 78)2 79) 1 80) 1 81)6
 82)5 83)3 84)3 85)3
 MAINS - CW -- VOL
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8. Put x=a-x on both sides
HINTS - LEVEL-VI
 f  x  a  f a  x
SINGLE ANSWER QUESTIONS 9. x 2   x 2   0

1. -1  x+[x]  1 x 2   x 2 
when x Z i.e Let x=k x 2  R and x 2  0,1, 2     
1 1
 -1  2k  1 
2
 k   k  0  x=0
2 
 Domain is R   n , n  z and n  0 
when x Z i.e let x=k+f  -1  2k+ f  1  cos 1 x   1 and cos 1 x   0
10.
1  f 1 f
 k  k  0  0  x  1. cos 1 x  [1,2) [cos1 x]  2
2 2
Df  [0,1) 2  cos 1 x   ,cos n  x  1, 1  x  cos 2
2. f  2  = min ( 4, 2, sgn (17) ) = min ( 4,2,1) = 1 2 3 1
11. least valve of x  x  1  at x 
4 2
3. f(x)=1 is false ; f  y   1 is false and f  z   2
is true then Injection is possible f  1  2
i.e f  x   2  or  3 ; f  y   1; f  z   1 or  3 f 1  0 incase of decreasing function
4. 1  log10  x 2  5 x  6   0 and a  b  2
ab  0
x 2  5 x  16  0  b 1
log10  x 2  5 x  6   1 a  1
 f ( x)   x  1
x 2  5 x  16  10
3
x   2,3  lease value of x 2  x  1 is 2
this is true for all x  R ( Disc  0)
Also x2  x  1  1
and leading coeff  0
5. f  x   f  x  4 3
Thus  X 2  X 1  1
2
 f  x  2   f  x  6  x  R
3
Replace x by x  2 and simplify  sin 1 ( )  sin 1 ( x 2  x  1)  sin 1 (1)
2
On solving we get f  x   f  x  8   
  sin 1 ( x 2  x  1) 
6. f  1  2 3 2
12. f  x    px  q  rx; x  0
f 1  0 incase of decreasing function
q
a  b  2  f  x    px  q  rx;0  x 
ab  0 p
 b 1 q
 f  x   px  q  rx; x
a  1 p
 f ( x)   x  1 for r  p ; f 1  x   0; x  0
7. Even functin is symmetrical about the y-axis i.e ,
x=0 q
 0; if 0  x 
 f  x   f  x   f 0  x   f  0  x  p
symmetrical about x  0 q
 0 if x 
Similarly symmetrical about p
x=a  f  a  x   f  a  x from the gaph(i) we have infinitely many points
for min.value of f(x)
 FUNCTIONS
for r  p f 1  x   0 if x  0  fog  x   1  x; 2  x  0; 1  x  0
q  x  1; 0  x  2; 1  x  0
 0 if 0  x 
p  x  1; 0  x  2; 0  x  3
q   x  1; 2  x  0; 0  x  3
 0 if x 
p 5
from the graph(ii) , only one point of minimum 17. f  x  on simplification
4
x=q/p
5
for r  p f 1  x   0 ; if x  0 Now g  f  x    g    1  y  1 is a
4
 0; if x  0 straight line
from the graph(iii) , onley one point of minimum  2x2 1 
1
13.
2
f  x   e x  e x  f 1  x 
2
18. Let y  f  
x  cos  2
 x 1 

 
 2
 2 x e x  e x  0x  0,1
2

 Clearly y   0,   -----(1)

f(x) is a non-decreasing function on  0,1 2 x2  1

Also cos y  1  0
1 x2
Hence max.value of f  x   f 1  e   a cos y  0
e
2 2 
g 1  x    2 x 2  1 e x  2 x.e x y      (2)
2
 0x  0,1  g  x   
is an increasing function [0,1] from (1) and (2) y  0,   R f
 2
2 2
14. e x  e  x  0 and x2  1
19. Let y  f  x   y   x 2  xy  1  0
x2  x2  x6  x
e e  2  x 2   ......  0
 3  y  y2  4
x
f(x)>0 2
 range  co-domain  info function
Also y  y2  4
f 1 
8x 2
f 1  x  2  f 1  x  0
2
and f 1  x   0 x  x2  4

x
e e  x2
 1
f  x 
2
if x  0 if x  0  the range of f 1  x  is [1, ) , ignore the
Not one-one ----> many-one
2
negative sign
15. Clearly f 1  x   e3 x 3 x 2
.3  x 2  1  0  ' f ' x  x2  4
1
is increasing  f  x 
2
f(x) is one-one
1
range of f  (   , 20. x 2 f  x   2. f    g  x   replacing ' x ' by
 x
f  1]   0, e 4   (0, e5 ]  ' f ' is into 1 1 1 1
, then 2 . f    2 f  x   g   (or)
16. f  x   1  x;0  x  2 x x x  x
 x  3; 2  x  4 1 1
2 f    4 x 2 f  x   2 x 2 .g  
g  x   2  x; 1  x  0  x x

 2  x; 0  3 --------(2)
 MAINS - CW -- VOL
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 1 yq/ p
2
 g  x   2 x .g  x   f  y  1/ p
(1)  (2)  f  x     2
   f 1 
 3x  1



  f 1  1/ p
 f 1 
 2  1
 g   x   2 x .g   x    f 1  1
f  x     
 3x 2
 then f  y   y q/ p      3
 
  23. f  x  y  f  x  y
f  x   f  x  2 f  x  f  y       1
( g  x  is an odd and f(x) is an even function) replace ‘x’ by ‘y’ and ‘y’ by ‘x’ in ---(1)
f  x   0  f  5  0 f  y  x  f  y  x
y x  2 f  y f  x     2
21. 2 f  xy    f  x     f  y   replacing ‘y’ by from (1) and (2) , we get
‘1’ , then we get
x
f  y  x   f  x  y  putting y  2 x then
2 f  x   f  x    f 1 
f  x  f x
 f  x  ax
Hence f  x  is an even function.
n
24.
Now  a  1  f  i    a  1
i 1 Y
n
i 2
 a   a  1 a  a  a 3        a n 
i 1
(1, 5)
 a  a n  1 
  a  1    a n1  a ( -1, 3 )
 a 1 
  X
1 q O
22. f  xf  y    x p y q   f  xf ( y )    x. y p p

1/ p
 f ( xf  y  
x     1 Let f  x   x  2  x  1  2 x  1
yq / p
Then
Now let x. f  y   1
 x  2  x  1  2  x  1 , x  1
1 
x .....  2  f  x    x  2  x  1  2  x  1 , x  x  1
f  y

Hence f  x. y q / p   x p . y q  x  2  x  1  2  x  1 , x  1
put y q / p  z 3x, x  1

f  xz    xz 
p f  x    x  4,  1  x  1
5 x, x  1
 f     p      4 
graph of y  f  x  is as shown. Clearly y = k
from  3 and  4  y p  y q / p
can intersect y  f  x  at exactly one point only
q
 p if k = 3.
p 25. replace ‘x’ by ‘a’ and ‘y’ by ‘x-a’
2
 p q  f  a   x  a    f  a  . f  x  a   f  0 . f  x 
 FUNCTIONS
    1 f  x, y   f  212 x, 212 y 
2 2 replace ‘y’ by 0
put x  0; y  0  f  0    f  0     f  a  
f  x,0   f  212 x,0 
 f a  0
replace ‘x’ by 2 y
 f  0    1
f  2 y , 0   f  212 .2 y , 0 
from 1 , f  2a  x    f  x 
 f  212 y ,0 
 x  59 
26. 3 f  x  2 f    10 x  30 f  2 x ,0   f  212 x , o 
 x 1 
put x  7 g  x   g 12  x  for all ' x '
3 f  7   2 f 11  70  30  100 period of ' g ' is 12
put x  11 29. f  x   f  x  3  0     1
3 f 11  2 f  7   140  f  7   4 replace ‘x’ by ‘x+3’
1 2
f  x  3  f  x  6   0      2 
27. f  x  a  f  x    f  x  ,
2 1   2   f  x   f  x  6 
put x  x  a
period of f  x  is 6
1 2
f  x  a  a   f  x  a    f  x  a  g 1  x   f  x  n   f  x   n  3  3
2
1  f  x   n  3   period of ' f ' is 3
f  x  2a    f  x  a 1  f  x  a 
2  f  x  3  f  x   g1  x   f  x 


1

2

 1

2
 f  x    f  x 
2

1

2
 f  x    f  x 
2
 7 7
 g1    f    f 
3 3
 2
 3
  2 
 3  f    7
  3 
2 1
1 1 2 30. In the sum 1 occurs n times, occurs (n-1)
     f  x    f  x  2
2 2 1
times , occurs  n  2  times and so on
1 1 3
  f  x 
2 2  f1 1  f 2  2   f 3  3  ........  f1  n 
 f  x  f  x  2 1 1 1
 n 1   n  1   n  2       1.
f  x  2a   f  x  2 3 n
period of ' f ' is 2a  1 1 1 
 n 1          
 2 3 n
28. f  x, y   f  2 x  2 y, 2 y  2 x  x, 1 2 3
        
n 1 

y  R      1 2 3 4 n 
 1   1   1   1 
f  2 2x  2 y  2 2 y  2x ,2 2 y  2x  2 2x  2 y   n. f1  n   1    1    1        1   
 2   3   4   n 
f  x, y    from 1   n. f1  n    n  f1  n     n  1 f1  n   n
f  x, y   f  8 y ,8 x       2  1

31. f  2   2 f 1  2
 f  8  8 x  , 8 8 y    from  2   r 1
2
f  x, y   f  64 x, 64 y      3 f  3  2 f  r   2  f 1  f  2    2 1  2  6
r 1
 f  64  64 x  , 64  64 y   from  3
 MAINS - CW -- VOL
JEE ADVANCED VOL -- III
I
m
f  4   18   f  n   3m1 f 1  x   f 1  x     symmetric about
n 1

1 the line x  1
32. Take 1  x  x 2  1 and 1  0
|x| max.value of ln  2x  x 2  as well as that of
and x 2  1  1; x 2  2 x
sin   ocurs at x=1
 2 
MULTIPLE ANSWER QUESTIONS Domain  2 x  x 2  0  x 2  2 x  0
x   0, 2  as x  0   or  x  2 
33. 2 f  sin x   f  cos x   x ----(1)
f  x   

replace ‘x’ by '  x ' ,we get 39. Even extension is f(-x)
2 Odd extgensin is -f(-x)

2 f  cos x   f  sin x    x      2  40.  fog  x   3 | sin x |  sin x  2
2
 3 | sin x |  sin x  2  0
from 1 and  2 
If sin x  0  sin x  1

f  sin x   x  sin x  1
6

let sin x  y  x  sin 1 y x  2n 
2
f  y   sin 1 y   / 6 1
sin x  0  1  sin x 
Domain   1,1 2
   7  
  sin 1 x  x   2 n  , 2n 
2 2  6 6 
2   3
  sin 1 x   41. f  x   1  sin x;0  x 
3 6 3 4
34. | | x | 1 | 5  0 3 3
 1  cos x; x
4 2
| x | 1  6  or  | x | 1  5
3
 x  (, 7]  [7, )  1;  x  2
2
35.  fog  x   sin  ln | x |  R1   1,1 g  x   1  x; x  0
 gof  x   ln | sin x | R2  (, 0]  1;0  x  2
 x  1; x  2
36. Let q  x   K where K  R
f  0   1  g  f 1   1
and
p  x   a0  a1x  a2 x 2  a3 x 3      an x n  k f 1  1  sin1
a0  a2  a3      an  0 and a1  1 and f  g  0    1  sin1
p(x)--->odd and of odd degree If f  x  is periodic
42.
37. f  x   f 1  x  . f 11  x 
if 4  a 2
degree of f  x  is 3
is a perfect square i.e 4  a 2 is a rational
n  n 1 n  2    n  3 number.
f (x) is an onto function
38. Clearly  a  0, 2, 2
 FUNCTIONS

PARAGRAPH QUESTIONS 4  3c  0 and

2
16 1  c   4  4  3c 1  c   0
43. H(x)=
0  c 1
f ( x) 1  2sin 2 x 53. concept
  1but cos(2 x )  0
g ( x) cos 2 x  x  x
54. f  x  2 f    f    x2
 2 4
2 x  n 
2 x
 Replace x by
Domain  R  {(2n  1) } 2
4 2
 x  x  x  x
andRange  1 f  2f   f    
44. F(x)=f(x)+g(x)=  2 4 8  2
2
1  2sin 2 x  cos 2 x  2 cos(2 x)  x  x  x  x
f  2f   f    
1  cos(2 x )  1  2  2 cos(2 x)  2  4 8  16   4 
 ontofunction 2
 x   x   x   x 
f  n   2 f  n 1   f  n  2    n 
45. f ( x)  1  2sin 2 x with period  2  2  2  2 
Addin we get
g ( x )  cos(2 x ) with period 
 x  x   x 
F ( x)  2 cos 2 x with period f  x   f    f  n 1   f  n  2 
2 2  2 
        3
46. G(x)=f(x)-g(x)=  1 1 
 x2 1  2         2n 
1  2sin 2 x  cos(2x)  0  2 2 
H ( x)  1, unionoftheirDomainisR As n  
2
47. F  x   G  x   2g  x   0  x  4x
f  x  f   
 cos  2 x   0  x 2 3
repeating we get
  5 7  19 16 x 2
 , , , ,     , f  x  f 0 
4 4 4 4 4 9
No.of solutions=10 and their sum  25
48. As f(296) =f(926)---> many-one f  3  16  f  0 
Further, the number whose digits increase from 55. f  x   f 0  x  0
left to right
ex: 159 have no pre-image---> into 16 x 2
49. Here f(n) and n leave the same remainder , when  x0
9
divided by 9. The remainder , when a
+ve integer is divided by 9 is same as the sum of 9
 x  0,
the digits of the number ( until the sum becomes 16
a single digit number 32 x
1
50. Celar from the definition 56. f  x    f 1 0  0
51. If the degree is odd , then 9
57,58,59.
f     ; f     
 f  x   . f  11  xx   64 x     1
2
x2  2x  c  
52. Let y 
x 2  4 x  3c 1 x 1 y
  y  1 x 2   4 y  2  x   3  y  2    0. put  y and x  we get
1 x 1 y
Here   0 2
  1  y   1 y 
 4  3c  y 2  4  1  c  y  1  c   0 f    f  y   64  
  1  y   1 y 
 MAINS - CW -- VOL
JEE ADVANCED VOL -- III
I
2
  1  x  1 x  Draw the graphs of
 f  x .  f  1  x    64  1  x       2  y  tan x & y  x
    
 
2
from 1 and  2  , we have 64. for x   0,  , f  x   0
 2
2
  1 x 
4 as tan x  x
 f  x f   2  2
  1  x     64 x  65. Re qd Area    f  x  dx  f  x dx
2 0 
  1 x   1 x   2
f  x  f  64 1 x    2 cos x  x sin x 0   2 cos x  x sin x 
  
  1  x 
 4  2  2   2  cos 
3 1 x 
  f  x    64 x 2  
1 x  MATRIX-MATCHING QUESTIONS
1/3
1 x 
 f  x   4.x 2/3 
a    Let y  f  x   cot  2 x  x  2 
1 2
 66.
1 x 
Now solve other two questions clearly y   0,       1
60-62
and
 x  1, x 1 2
f  x   cot y  2 x  x 2  2  1   x  1  1
2 x  1, 1  x  2
cot y  1
 x2 , 1  x  2
g  x    3 
 x  2, 2  x  3 y   ,      into
 4 
 g  x 1, g  x  1 more over f 1  x  changes its sign in the
 f  g  x   
2g  x 1, 1  g  x  2 neighbourhood of x=2
many-one
 x2  1, x2  1, 1  x  2
 b    eax  e ax sin  bx   eax   
 x  2  1, x  2  1, 2 x3
 f  g  x    2 2 many  one, onto
 2x 1, 1  x  2, 1  x  2
2  x  2 1, 1  x  2  2, 2  x  3 c f 1  x   x    f 1  x   0   

2 many  one i.e, f  x  is an increasing function
 x  1, 1  x  1
 f  g  x   2 range   f  0  , f    
 2 x  1, 1  x  2
  2,    co  domain    into
60. Hence, the domain of f  x  is  1, 2 
d---> clearly identity function--->one-one and
61. For 1  x  1 we have onto
x 2   0,1  x 2  1 1, 2 67. a   x 2  1 cos x  0  x 2  1  0 and
For 1  x  2 , we have cos x  0 (or) x 2  1  0 and cos x  0
x 2  1, 2  2 x 2  1   3,5  
  x  1  OR  1  x 
Hence, the range is 1, 2   3,5 2 2
2

62. For f  g  x    2  x 2  1  2 and 2

b   x   x  2  x   x 
 x    x
3
1 2
2
2 x  1  2  x  1 or x    x    x  1
2 Now 0 
 x  1 only. Hence, 2 roots 3
2
f  x  0  2  x   x
63.   x   b   x   x  2  x   x 
 
tan x  x  3 

 FUNCTIONS
  x   1,0,1, 2
2 2 1 8
 x  ,0,0,  x  ,0,1,
3 3 3 3
 
c  1  x  1    tan 1 x 
4 4
  3
 f  x    tan 1 x   f  x  
2 4 4
  f  x    0,1, or 2
d  ln  cos  sin x    0
 cos  sin x   1  cos  sin x  1
d  Draw t he graphs of y  7 |x| and
 ln  cos  sin x    0
y | 5 | x ||
1
68. a  draw the graphs of y  tan x & y  69. a  when 1  x  0 f  x   [0,1) and
x2
from the graph , it is clear that it will have two f  x  x 1
real roots
2
 f  f  x     x  1  1  x 2  2 x

cos  2 x   1 sec 2 x  2 tan x 

b
2
1  tan 2 x  2 tan x

1  tan 2 x
 2 tan x  2 tan x
i.e f  2   1
 1  tan x  1  tan 2 x
b  draw the gaphs y  2cos x and y | sin x |  f  x  1 x
no.of P.O.I=4 c  put x  y  0
2
 f 1  1  1  2 2 ;

put x  0; y  1  f  2   32
d  when 4  x  5  f  x 
x3
 2 x  3  f 1  x  
2
c-->  f  | x |  0 has 8 real roots 4 sin 2  cos   cos   sin 
f   
 f  x   0 has 4 real roots  f  x  is a 4  cos   sin  
plynomial of degree 5 , f(x) cannot have even 1 1
no.of real roots  sin   sin   cos   
2 4
f  x  has all the fire roots real and one roots
1  1 1 
f  x  has all the fire roots real and one root is = sin  2   cos  2     , 
4 2 2 2 2 
negative
 MAINS - CW -- VOL
JEE ADVANCED VOL -- III
I

ASSERTION - REASON QUESTIONS 77. | f  x   g  x  |  | f  x  |  | g  x  |

     1
cot 2 x
70. f  x   g  x   L.H .S  1  e 2 but given that
As 2 | sin x | 1  and | f  x   f  x  | | f  x  |  | g  x  |

1  cos  2 x  2sin 2 x      2

 from (1) and (2)
1  sin 4 x 1  sin 4 x
2 | f  x   g  x  || f  x  |  | g  x  |
 1 this is possible only when
1 2
 sin x
sin 2 x f  x  .g  x   0 but given that
 RHS  2  Equality holds good f  x  .g  x   0
L.H.S=R.H.S =2
 f  x  g  x   0  f  x   0  g  x   0 
Which is possible when
100
cot 2 x  0 and | sin x | 1   f r  0
 r 1
 x   2n  1 ; n  I 78. Conceptual
2
79: b 2  4ac  0 .
 S1 is correct 80. Properties of G.I.F.
Clearly  S 2 need not be correct always 1  x 4  x  x5  x 2  x 6
81. f  x 
 S1 is true and S 2 is false x3
71. Conceptual 1   1  1 
  3  x3    x     x2  2 
72. conceptual x   x  x 
73. f(-1) =17 and f(1)=-13
clearly S1 is also true 1 3 1 1
2 3
.x  2 x.  2 x 2 . 2  6
x x x
INTEGER QUESTIONS
 1 
 use a   2, a  0 
 a 
74. f  x   f  x  3  0 --(1)
 f  x  6
replace ‘x’ by ‘x+3’
 R f  [6,  ) .
f  x  3  f  x  6   0       2 
1   2  f  x   f  x  6  82.  x  2, 3 
 period of f  x  is 6 1 1 1
 2  x 3 or  
 3 x 2
75. Period of cos  sin  nx   is and the period of
n 1  1
    and 2  x   3
 x x x
tan   is n
n or 0  x 2  2  1
  x 2  x 2  2
Now L.C.M of , n is n  n  6
n 1  1
Given     x    x  2
2 2
...  i 
76. f  x  2   f  x  1  1  f  x  1  f  x  x
  x
3
Period =2  x  2x 1  0
 FUNCTIONS

  x  1  x 2  x  1  0 5  13 5  13
  f    .
x 1 2 2
 x2  x  1  0  x2  x  1 5  13 5  13
 x4  x2  2x 1  x  1 2x  1 Hence, L  and M  .
2 2
 x 4  3x  2 .......  ii 
 5  13  5  13  25  13
 x 4  3x  2  LM   
 2   2  4
  
3
 x 4   3x  2  3  x2  2 
x 12
  3.
(from eqs. (i) and (ii)) 4
  3x  2  3  x  1  2 85: We have

 x 2
 x  1 f  x  1  f  x  1  3 f  x  x  R ...(i)
=5. Replacing x by x  1 and x  1 in (i),then
83. Given f  x   0 f  x   f  x  2   3 f  x  1 ...  ii 
 f  x  is symmetrical about the line and f  x  2   f  x   3 f  x  1 ...  iii 
y  x , then
Adding (ii) and (iii), we get
f 1  x   f  x   y  x
2 f  x   f  x  2  f  x  2
2
 f 2
 x    f  x 
1
  xf  x  f 1
 x
 3  f  x  1  f  x  1 
2
3x f  x 
 2 f  x   f  x  2   f  x  2   3. 3 f  x 
 x 2  x 2   x.x.x  3 x 2 . x
[From (i)]
  x3  3x 3 .
   3, x  0 .  f  x  2  f  x  2  f  x  .....  iv 
84.  x2  4 y2  4 Replacing x by x  2 in equation (iv),
2 2 then f  x  4   f  x   f  x  2  ....(v)
 x  y
       1. Adding equations (iv) and (v), we get
 2  1
Let x  2 cos  , y  sin  f  x  4   f  x  2   0 .... (vi)
2 2
 f     2 cos     sin     2 cos   sin   Again replacing x by x  6 in (vi), then

 4 cos 2   sin 2   sin 2 f  x  10   f  x  4   0 ....(vii)

 3cos 2    cos 2   sin 2    sin  Subracting (vi) from (vii), we get

3 3 5 f  x  10   f  x  2   0 ....(viii)
 1  cos 2   1  sin 2  cos 2  sin 2 
2 2 2 Replacing x by x  2 in (viii), then

5 3 2 3
2 f  x  12   f  x   0
      1  cos 2  sin 
2 2 2 or f  x  12   f  x 
2
5 5 3 2 Hence f  x  is periodic function with
      1
2 2 2 period 12.