MATHEMATICS

Q.1 Domain of f(x) = sin–1 [x2– 4]2 is

(where [x] represents greatest integer  x)
8 8
(A) (– 6 , 6 ) Q.7 Given f(x) =  and
1 x 1 x
(B) [ 3 , 6 )
4 4
(C) (– 6 ,– 3 ][ 3 , 6 ) g(x) = + then g(x) is -
f (sin x ) f (cos x )
(D) (– 5 ,– 3 ][ 3 , 5 ) [C]
(A) periodic with period /2
(B) periodic with period 
Q.2 Let f: R  R be a function defined by
(C) periodic with period 2
| x |3  | x | (D) non periodic [A]
f(x) = – , then the graph of f(x) lies
2
1 x
in the – 1 x 
Q.8 If 2f(x–1) – f   = x, then f(x) is -
(A) I and II quadrants  x 
(B) I and III quadrants
1  1 
(C) II and III quadrants (A)  2( x  1) 
3  x  1 
(D) III and IV quadrants [D]
 1 x 
(B) 2 ( x  1) 
Q.3 The value of x satisfying the inequality  x 
[tan–1 x]2 – [tan–1x] – 2  0, where [.] denotes
1
integral parts is– (C) x2 + +3
(A) [–/4, ) (B) [–/4, tan 2] x2
(C) [–tan 1, ) (D) [–tan 1, tan 3] [C] 1  1
(D)  2x   [A]
3  x
Q.4 The range of the function f(x) = cos (2 sin x) is–
(A) [–2, 2] (B) [–1, 1] Q.9 Suppose f is a real valued function satisfying
(C) [cos 2, 1] (D) [0, 1] [C] f(x + f(x)) = 4 f(x) and f(1) = 4. The value of
f(21) is -
Q.5 The domain of the function (A) 16 (B) 64
5x (C) 4 (D) 44 [B]
f(x) = + a sin–1
( x  1) ( x  2) ( x  3)
 y y
 2x  5  Q.10 If f  2 x  , 2 x   = xy then f(x, y) + f(y,
 3  ; a  0 and [.] denotes greatest  8 8
 
x)=
integer function, is– (A) 0 (B) 1
(A) [1, 5] – {2, 3} (B) [1, 2)  (3, 5] (C) 2 (D) None of these [A]
 11 
(C) 1,  (D) None of
 2  Q.11 Let f(x) = x2 + bx + c, where b, c  R. If f (x) is
these [A] a factor of both x4 + 6x2 + 25 and
3x4 + 4x2 + 28x + 5, then the least value of f(x)
Q.6 Let a function f(x) satisfies is -
f(x) + f(2x) + f(2–x) +f(1+x) = x for  x R. (A) 2 (B) 3
Then f(0) equal to – (C) 5/2 (D) 4 [D]
1 1 1 1
(A) – (B) (C) (D) –
4 4 2 2
Q.12 The range of the function f(x) = 3 |sin x| – 2 |cos x|
[A] is -
(A) [2, 13 ] (B) [–2, 3]
(C) [3, 13 ] (D) None of these [B]
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 Q.19 The domain of definition of the function
f(x) = 2 2 x  (64) ( x – 2) / 3 – 2 –1 (72  2 2 x )
Q.13 Let f : R  R be a function defined by
is
e cos x  e  sin x (A) (– , ) (B) (– , –3]
f(x) = sin x , then
e  e sin x 1 
(C) [3, ) (D)  ,1
(A) f is both one-one and onto 9 
(B) f is one-one but not onto [C]
(C) f is onto but not one-one
(D) f is neither one-one nor onto [D]
1
Q.20 If af (x) + bf   = x – 1, x  0 and a  b, then
Q.14 Which one of the following functions has the period x
equal to  ?
(A) f(x) = sin (cos x) f (2) =
(B) f(x) = |sin x | + |cos x| a a  2b
(C) f(x) = cos (sinx) (A) 
(B)
(D) f(x) = tan2x + sin 3x [C] a –b 2
a  – b2
a – 2b
Q.15 Let f : A  B be an invertible function. If (C) (D) None of these [D]
a  – b2
f(x) = 2x3 + 3x2 + x – 1, then f–1(5) =
(A) 2 (B) 1 Q.21 The domain of the function
(C) 6 (D) can not be determined [B] 2x – 1
y=  sin –1 (log 2 x ) is -
Q.16 If 2f(sin x) + 2 f (  cos x ) = – tanx, then 2 x  3x 2  x
3

f(1/2) is equal to 1 
(A)  ,   (B)
3 2 2 
(A) (B)
6 1 
 ,2 
3 2 2 
6 (C) [1, 2] (D) (1, ) [C]
2 3 Q.22 If f (x + 2y, x –2y) = 4xy, then f(x, y) =
(C) (D)
6
(A)
 x  y y (B)
x 2 – y2
3 2 4 4
[C]
6 x  y2
2
x 2 – y2
(C) (D) [D]
Q.17 If [x] denotes the integral part of x, then the 4 2
domain of the function f(x) = sin–1 [2x2 – 3] +
x2 – 2
log2 {log1/2(x2 – 5x + 5) } is Q.23 If f(x) = then Rf is
   
x2 – 3
– 5 5 
(A) , – 1 (B) 1,  (A) (– , 2/3]
 2   2
   
(B) (1, )
   
(C)  –
5
, – 1  1,
5  (D) None of (C) (– , 2/3]  (1, )
2 2 
  
   (D) None of these [C]
these
[D] x 2  2x  3
Q.24 If g(x) = then Rg is -
Q.18 The domain of the function x
(A) [2 – 2 3 , 2 + 2 3 ]
 1– | x | 
f(x) = sec –1   is (B) R – (2 – 2 3 , 2 + 2 3 )
 2 
(C) [2 + 2 3 , 2 – 2 3 ]
(A) (– , –3]
(D) None of these [B]
(B) [3, )
(C) (– , – 3]  [3, ) Q.25 Which of the following graphs are graphs of
(D) None of these [C] functions

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 y (A) 992 (B) 994
(C) 996 (D) 998 [D]
(A)
O
x Q.29 If | f(x) + 6 – x 2 | = | f(x) | + | 4 – x 2 | + 2, then
f(x) is necessarily non negative in
(A) [–2, 2] (B) (– , –2)  (2, )
y
(C) [– 6 , 6 ] (D) None of these [A]

(B) Q.30 The range of the function

x
O f(x) = | x – 1| + | x– 2 |, –1  x  3, is
(A) [1, 3] (B) [1, 5]
y
(C) [3, 5] (D) None [B]

2
Q.31 Range of f(x) = 3tan – x 2 is
(C) x
9
O
(A) [–3 3 , 3 3 ] (B) [0, 3]
(C) [0, 3 3 ] (D) None of these [C]

y
Q.32 Range of f(x) = 16 – x C 2 x –1  20 – 3xC 4 x – 5 is
(A) [728, 1474] (B) {728, 1474}
(D) [B] (C) {0, 728} (D) None of these [B]
x
O
Q.33 The range of the function
f(x) = cos2 x–5 cos x – 6 is
(A) [– 5, 0] (B) [0, 10]
(C) [–10, 0] (D) None of these [C]
log 2 x 3
Q.26 is defined for
cos –1 (2 x – 1) Q.34 The range of the function f(x) = [sin x + cos x]
(A) (0, 1) (where [.] denotes the greatest integer function)
(B) (0, 1/2)  (1/2, 1) is
(C) (1, 2) (A) [–2, 1] (B) {–2, –1, 0, 1}
(D) None of these [B] (C) {–1, 1} (D) {–2, –1, 1} [B]

Q.27 The domain of the function 36

Q.35 If f(x) = + x2, then f() = ……. ; where 
f(x) = 16 – x
C 2 x –1  20 – 3 x
P4 x – 5 , where the x2
symbols have their usual meanings is the set 6
is a root of + x = 2.
(A) {1, 2, 3, 4, 5} x
(B) {2, 3, 4} (A) 8 (B) – 8
(C) {2, 3} (C) 4 (D) – 4 [B]
(D) None of these [C]
Q.36 If f : [–2, 2]  R given by
4x x cos x
Q.28 Let f(x) = , and given that
x
4 2 f(x) =  x  ; where [x] is the greatest
f(x) + f(1 – x) = 1, then  5   0. 5
 
integer function then f(x)
 1   2   3   1996  (A) is an even function
f   f   f   ........  f   (B) is an odd function
 1997   1997   1997   1997 
is - (C) is neither even nor odd function
(D) Maps into [–2, 2] [A]
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 (D) None of these [C]
Q.37 If [.] denotes greatest integer function then
the domain of the real valued function Q.43 Let [x] represents greatest integer x. If
log 1 2 [ n2  ] = [ n 2  1 ] + 2 where , n N,
[ x  ] |x – x – 2| is
2 then can assume
3  (A) (2n + 4) different values
(A)  ,  
2  (B) (2n + 5) different values
3  (C) (2n + 3) different values
(B)  ,2   (2, ) (D) (2n + 6) different values [B]
2 
1  Q.44 Domain of y = 3e x 2 –1 loge (x – 1) is :
(C)  ,2   (2, )
2  (A) (1, ) (B) [1, )
(D) None of these [B] (C) R –{1} (D) None of these [A]
x
4  1 
Q.38 If f(x) = , f(x)+f(1 – x) = a and f   Q.45 Domain of f(x) = log 2 ( x 2 – 6 x  6) is :
4 2x  97 
(A) (– , 3 – 3 ]  [3 + 3 , )
 2   96 
+ f  + …. + f   = b then order pair (B) (– , 3 – 3 )  (3 + 3 , )
 97   97 
(C) (– ,1] [5, )
(a, b) is -
(D) (– , 1)  (5, ) [C]
(A) (2, 20) (B) (1, 48)
(C) (1, 24) (D) (2, 96) [B]
Q.46 Which of the following function is non-
periodic:
Q.39 Let f be a real valued function, defined on
(A) |sin3x| + sin2x (B) cos x + cos2x
(0,1)  (2,4) such that f ' (x) = 0 then -
(C) cos4x + tan2x (D) cos2x + sinx [B]
(A) f is a constant function
(B) f is a constant function if f (1/2) = f(3)
Q.47 Range of f(x) = sin2x + cos2(x + ) + 2sin x sin
(C) f is not a constant function
 cos(x + ) is :
(D) f is a constant function if f (1/2) = 0 [B]
(A) {1}
(B) (0, 1)
Q.40 The function f(x) is defined for all real x. If
(C) {cos2}
f(a + b) = f(a . b) for all real values of a & b,
and f(–1/2) = – 1/2, then f(2009) is equal to – (D) (cos , 1 + cos ) [C]
(A) – 2009 (B) 2009
(C) – 1/2 (D) – 1004 [C] x –1
Q.48 If f(x) = then f(2x) is terms of f(x) is :
x 1
Q.41 Number of solutions of the equation f (x)  1 3f ( x )  1
cos [x] = e2x–1, x  [0, 2] where [x] denotes (A) (B)
f (x)  3 f (x)  3
greatest integer function, is -
f (x)  3 f (x)  3
(A) 0 (B) 1 (C) (D) [B]
(C) 2 (D) 3 [B] f (x)  1 3f ( x )  1

Q.49 Function f : R  R
Q.42 Domain of the function f(x) = x3 + 3x2 + 10x + 2sin x is :
1 (A) One-one but not onto
cos x –
f(x) = 2 is (B) Not-one one but onto
(C) Neither one-one nor onto
6  35x – 6x 2
(D) One-one and onto [D]
(A) [2n, 2n + /3] [2n+ 5/3, 4n]
1 x x
(B) [– , 6] Q.50 Period of function sin + cos is :
6 3 4
1 (A) 16 (B) 24
(C) ( – ,  3 ]  [5/3, 6) (C) 3 (D) Non-periodic [B]
6

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Q.51 Which of the following functions from Z  Z is 1 1
bijective : (A) (B) –
2 2 2
(A) f(x) = x3 (B) f(x) = x + 2
1 1
(C) f(x) = 2x + 1 (D) f(x) = x2 + x [B] (C) – (D) [A]
2 2 2

 4 – x2 

Q.52 Domain of f(x) = sin log  is : [D] Q.59 Which of the function is not even :
 1– x 
 
1 x2 
(A) [–2, 2] (B) (–2, 2) (A) log  2

 (B) sin2x + cos2x
(C) [–2, 1] (D) (– 2, 1) 1– x 
1 x3  (1  2 x ) 2
 (C) log   (D) [C]
  1– x
3 
 2x
Q.53 Range of y = cos   sin  cos( sin x )   ,
 2 
where Q.60 Let f(x) = sin [a ] x (where [] denotes the
x R, is : greatest integer function). If f is periodic with
(A) [–1, 1] (B) [– , ] fundamental period , then a belongs to :
(C) [0, 1] (D) [–1, 0] [A] (A) [2, 3) (B) {4, 5}
(C) [4, 5] (D) [4, 5) [D]
Q.54 The value of n  I for which the period of
sin x Q.61 The domain of the function
function f(x) = is 4, is :
sin x / n f(x) = log( x log 2 x
– 2) is -
(A) – 3 (B) 3
(C) 2 (D) 4 [C]  1
(A) 0,  [4, ) (B)
 4 
Q.55 Let f : A  [3, ), f(x) = x2 – 6x + 12 be a 1 
bijective function then interval A can be :  4 , 1  [4, )
 
(A) [0, ) (B) [3, )
(C) [12, ) (D) [– 6, ) [B]  1
(C)  0,   (4, ) (D) None of these
 4
 x
Q.56 The period of sin [x] + cos where [x] Q.62 Let f(x) =
4 2
0 for x  0
denotes integral part of x, is : 
 2   
x sin   for – 1  x  1; (x
(A) 24 (B) 3 (C) 4 (D) 8 [D]   x 
x | x |
 for x  1 or x  –1

Q.57 The function f(x) = (tan 11 5 11

x ) e x sgn(x ). then -
(A) f(x) is an odd function
 1  (B) f(x) is an even function
 2  , where [.] denotes greatest integer (C) f(x) is neither odd nor even
 3x  2 
(D) f '(x) is an even function
function, is :
(A) even function
Q.63 If f(x) is an odd function then –
(B) odd function
(C) even as well as odd function f (– x )  f ( x )
(i) is an even function
(D) neither even nor odd function [C] 2
 1  (ii) [|f(x)| + 1], is even where [.] denotes greatest
Sol.[C]   = 0  x  R;  f(x) = 0 integer function.
 3x 2  2 
f ( x ) – f (– x )
(iii) is neither even nor odd
2
 x 
Q.58 If tan2   f(x) = 1  cos 2x + | f(x) |, (iv) f(x) + f(–x) is neither even nor odd
 9  Which of these statements are correct
then f(3) is equal to : (A) (i) & (iv) only (B) (i) & (ii) only
(C) (iii) & (iv) only (D) all of these
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 y y
Q.64 Which of the following function(s) is/are even ?

x 2  9 ; if | x | 5
(A) f(x) =  (A) x' x (B) x' o x

 16 ; if | x | 5 o
a x 1 1 x
(B) f(x) = x . n y'
a 1x 1 x y'

ex 1 y y
(C) f(x) = n (x + 1 x2 )
ex 1
1 (C) x' x (D) x' x
(D) f(x) = [x/] + ; where [ . ] denotes o o
2
greater integer function
y' y'

Q.65 The domain set of the function

f(x) =  2 cos 2 x  3 cos x  1 is [B]
  4n  1 4 n  1  
(A)   ,  ; n  z 
 2 2   Q.70 If f (x) = 2 |x – 2| – 3| |x – 3| then for
 ( 2n  1)   2 < x < 3 f (x) is equal to -
(B)  2n, ; n  z  (A) x – 5 (B) 5 – x
 2   (C) 5x – 13 (D) 13 – 5x [C]
  6n  1 6 n  1  
(C)  ,  ; n  z  1
 3 3   Q.71 If f : R  R be given by f (x) = (3 – x 3) 3 , then
(D) None of these
fof (x) is
1
Q.66 Which of the following function have same (A) (B) x3 (C) x (D) (3 – x3)
domain and range.
x3
(A) sinh x (B) cosh x
(C) tanh x (D) sech x [A]

Q.67 Following graph is of the function. [C]

y
(0, 1)
–2, 1 2, 1
x x
Q.72 Period of f ( x )  sin  cos is
  3       3  ( n – 1)! n!
 ,0   ,0   ,0   ,0 
 2   2  O 2   2 
x' x (A) n ! (B) 2 (n !)
(C) 2 (n – 1) ! (D) None of these [B]
–, –1 ( –1)
Q.73 Period of | cos x | + | sin x | + 3 is
(A) sin x (B) cos x (A)  (B) 2
x x 
(C) cos (D) sin [B] (C) (D) None of these [C]
2 2 2

Q.68 Domain of the function f (x) = Q.74 If f(x) = sin 2x + x – [x], where [x] is the
4  x2 .
integral part of x, then f(x) is -
(A) [–2, 2] (B) [–2, 0]
(A) a periodic function with period 
(C) [0, 2] (D) [–2, 2] [A]
(B) a periodic function with period 2
Q.69 The graph of function y = x2 is. (C) a periodic function with period 1
(D) Not a periodic function [D]
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 x x x Q.83 Let f: R  R be a periodic function such that
Q.75 If f(x) = sin x + tan + sin 2 + tan 3 + f (T + x) = 1 + [1 – 3 f (x) + 3 (f (x)) 2 –
2 2 2
(f(x))3]1/3 where T is a fixed positive number,
x x then period of f (x) is
….. + sin n –1
 tan is a periodic
2 2n (A) T (B) 2T
function with period k, then k = (C) 3T (D) None of these [B]
1
(A) 1 (B) 2 (C) 2n (D)
2n
[C]
Q.84 If f (x + f(y)) = f (x) + y
– x, y  R and
V
f (0) = 1, then f (7) =
Q.76 If f(x) = cos x + cos ax is a periodic function, (A) 1 (B) 0 (C) –1 (D) 7 [A]
then a is necessarily
(A) an integer  1  2
(B) a rational number Q.85 If f(x) = [x] +  x     x   –3x + 5 ([.]
 3   3
(C) an irrational number
denotes the greatest integer function) then
(D) an even number [B]
(A) 'f' is a periodic function
(B) 'f' is a non-periodic function
Q.77 Let f : R  R defined by f(x) = x 3 + x2 + 100x
+ 5 sin x, then f is
(A) many-one onto (B) many-one into
(C) 4 < f(x)  5
– x R
V
(C) one-one onto (D) one-one into [C] (D) 'f' is not one-one [D]
2
x  x 1
Q.78 Let f : R  R defined by f ( x )  ,  x 
x 2  ax  5 Q.86 The period of sin [x] + cos +cot [x],
4 2 3
then the set of values of a for which f is onto is
where [x] denotes the integral part of x is -
(A) [0, 1] (B) [–1, 0]
(A) 8 (B) 4 (C) 3 (D) 24 [D]
(C) [–1, 1] (D)R – [– 20 , 20 ]
[D] Q.87 Let f : R  R be a function defined by f(x) =

Q.79 If f(x) = | x – 1 | + | x | + | x + 1| then f(x + 1) is e| x| – e – x

then
(A) an odd function ex  e–x
(B) an even function (A)f is one-one and onto
(C) neither odd nor even function (B) f is one-one but not onto
(D) both odd and even function [C] (C) f is not one-one but onto
(D) f is neither one-one nor onto [D]
Q.80 Period of the function cos {(x + 3) – [x + 3]},
where [x] denotes the integral part of x is Q.88 Period of cosec
(A) 1 (B) 2 (C)  (D)2 [A] x x x
3 (x     ............ ) is
Q.81 If f ( x )  2 sin x  x –[ x ] , where [x] denotes 2 4 8
the integral part of x is a periodic function with (A)  (B) 2 (C) /2 (D) /4 [A]
period
(A)  (B) 2  
Q.89 The minimum value of 5cos x + 3cos   x  +
(C)  (D) None of these [B]  3 
8 is -
Q.82 If [x] denotes the integral part of x, then x – [x]
(A)  (B) 1/2 (C) –1 (D) 15 [A]
+ tan–1 (tan x) is
(A) a periodic function with period 1
Q.90 If n  2 then the number of surjections that can
(B) a periodic function with period 2
be defined from {1, 2, 3,…..n} into {1, 2} is
(C) not a periodic function
(A) 2n–2 (B) 2n+2
(D) a periodic function with period  [A] (C) 2 n
(D) None [A]
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 sin nx
Q.91 Let X and Y be subsets of R, the set of all Q.99 The period of is , where n  z+
cos( x / n )
real numbers the function f : X  Y defined by then n =
f(x) = x2 for x  X is one-one but not onto (A) 1 (B) 2 (C) 3 (D) 4 [B]
if
(A) X = Y = R+ (B) X = R, Y = R+ Q.100 f : R  R f(x) = (x2 + x + 5) (x2 + x – 3)
+
(C) X = R , Y = R (D) X = Y = R [C] (A) f(x) is one-one, into
(B) 'f' is many-one, into
(C) 'f' is one-one, onto
Q.92 f : R  R and g : R  R are defined by f(x) =
(D) 'f' is many-one onto [B]
2x + 3 and g(x) = x2 + 7 then the values of x for
which f [g(x)] = 25 are
Q.101 A polynomial function f(x) satisfies the
(1)  1 (B)  2 (C)  3 (D)  4 [B]
condition f(x) f(1/x) = f(x) + f(1/x). If
Q.93 
If f(x) = 25 – x 4 1/ 4
for 0 < x < 5 then f(10) = 1001, then f(20) =
f(f(1/2))= (A) 2002 (B) 8008
(A) 2–4 (B) 2–3 (C) 2–2 (D) 2–1 [D] (C) 8001 (D) none of these [C]

x|x| 1
Q.102 If f(x) = cos (nx), then f(x) f(y) –
Q.94 If f(x) = – , then f–1(x) equals 2
1 x2
 x 
|x| |x| f    f ( xy)  has the value-
(A) (B) (sgn)  y 
1 – (x) 1– | x |
x 1
(C) – (D) None of these [B] (A) –1 (B)
1– x 2
(C) –2 (D) None of these [D]

ex Q.103 Let f(x) be defined for all x > 0 and be

Q.95 Range of f(x) = , where x  0 is
1  [x ]
x
(A) [0, ) (B) [1, ) continuous. Let f(x) satisfy f   = f(x) – f(y)
 y
(C) (– , ) (D) (–, 0] [B]
for all x, y and f(e) = 1. Then-
x
a –1 (A) f(x) is bounded
Q.96 If the real valued function f(x) = 1
x (a x  1)
n
(B) f    0 as x  0
is even, then n equals x
1 1 2 (C) x f(x)  1 as x  0
(A) – (B) (C) (D) 2 [A] (D) f(x) = n x [D]
3 4 3

Q.104 Let f(x) = (x + 1)2 – 1, (x > – 1). Then the set

 1  1 S = {x : f (x) = f –1(x)} is –
Q.97 If f(x) = cos–1(x – x2) + 1 –   2
 | x |  [ x – 1] (A) Empty
then domain of f(x) is (where [.] is the greatest (B) {0, –1}
integer) (C) {0, 1, –1}
 1 5   1 5  
 3i 3 3i 3

(A)  2 ,  (B)  – 2,  (D) 0,1, , 
 2   2  
 2 2 

 1– 2  [B]
(C)  – 2,  (D) None of these [A]
 2 
Q.105 If f(x + 2a) = f(x – 2a), then f(x) is-
Q.98 The range of log 5
 2 (sin x – cos x )  3 is  (A) a periodic function with period 4a
(B) a periodic function with period 2a
(A) [0, 2] (B) [1, 2]
(C) a periodic function with indeterminate
(C) [0, 3] (D) [1, 3] [A]
period

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 (D) a non periodic function [A] 1
Sol. 2f(x) + 3f   = x2 –1
Sol. f(x + 2a) = f(x – 2a) x
x  x + 2a
1 1 1
f(x) = f(x + 4a) ; T = 4a put x  , 2f   + 3 f(x) = 2 –1
x x x
Solve to get f(x).
1  2 x  [ x ] 
Q.106 Domain of f(x) = sin   , where [.] Q.109 If A > 0, c, d, u, v are non-zero constants, and
 [x]  the graphs of f(x) = |Ax + c| + d and
denotes the greatest integer function, is g(x) = –|Ax + u| + v intersect exactly at 2 points
(A) (– 1) – {0} uc
 4  (1, 4) and (3, 1) then the value of equals
(B)  , 0   {0} A
 3  to -
(C) (–, 0)  I+ (A) 4 (B) –4
(D) (–, ) – [0, 1) [D] (C) 2 (D) –2 [B]
1  2 x  Sol. f(x) = |Ax + c | + d
Sol. f(x) = sin   1 g(x) = –|Ax + u| + v
 [ x ] 
which are sides of parallelogram and the
(I) [x]  0  x  [0, 1)
diagonals bisect each other
2x
(II) 1 11  u  c uc
[x]   A    A  = 3 + 1  =–
    A
2x 2{x}
0 2; 02+ 2 4
[x] [x]
{x} Q.110 The polynomial function f(x) satisfies the
1  0 equation f(x) – f(x – 2) = (2x – 1) 2 for all x.
[x]
If p and q are the coefficients of x2 and
x respectively in f(x), then p + q is equal to-
Q.107 If f(x) is continuous and increasing function (A) 0 (B) 5/6
such that domain of g(x) = f (x)  x be R (C) 4/3 (D) 1 [B]
and
Sol. Let f(x) = ax3 + px2 + qx + r
1
h(x) = , then the domain of Now use f(x) – f(x – 2) = x2 – 4x + 1
1 x On comparing the coefficients, we get
(x) = f (f (f ( x )))  h (h (h ( x ))) is- 2 1
(A) R (B) {0, 1} a= ,p=1;q= 
3 6
(C) R – {0, 1} (D) R – (0, 1) [C]
5
1 Hence, p + q =
Sol. h(x) = ,x1 6
1 x
x 1
h(h(x)) = , x  0, 1  1 
x Q.111 If af(x + 1) + bf   = x, x  – 1, a  b
 x 1 
 h(h(h(x))) = x, x  0, 1
then f(2) is equal is -
Also g(x)  0  x  R
2a  b a
 f(x)  x  f(f(x))  f(x)  x (A) (B) 2
(f(x) is an increasing function) 2(a 2  b 2 ) a  b2
 f(f(f(x)))  f(f(x))  f(x)  x a  2b
(C) (D) None of these [A]
 f(f(f(x))) – x  0  x  R – {0, 1} a 2  b2
 (x) is defined for all x  R – {0, 1}
Q.112 Let x > 0 and let x = [x] + (x) where [x] denotes
Q.108 If 2f(x) + 3f(1/x) = x2 –1 then f(x) is
the greatest integer less than or equal to x and
(A) periodic function (B) an even function
(x) denotes the fraction part of x, the graph of
(C) odd function (D) None of these [B]
y = (x)[x] lies :
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 (A) entirely within the unit circle, center origin fohog(x) = sin2(cos–1 x ) = 1 – x.
(B) entirely within the rectangle bounded by Thus no two composites are equal.
x = 0, y = 0, x = 1, y = 1
x x x
(C) entirely within the strip bounded by Q.119 If f(x) = sin x + tan + sin 2 + tan 3 +
y = 0, y = 1 2 2 2
(D) entirely within the strip bounded by x x
y=x&y=x–1 [C] …..+ sin n 1 + tan is a periodic
2 2n
Q.113 Let f : R  R and g : R  R be two one-one function with period k, then k =
and onto functions such that they are the mirror
1
images of each other about the line y = 2. If h(x) (A) 1 (B) 2 (C) 2n (D)
= f(x) + g(x), then h(0) equal to 2n
(A) 2 (B) 4 (C) 0 (D) 1 [B] x x
Sol.[C] sin x, sin 2 , ….. , sin n 1 are periodic
Q.114 If an is the digit in the unit place of the number 2 2
1  2  3 + …. + n ; then a8 + a9 + a10 + … +
functions with period 2, 23, 25 ,…., 2n
a16 is
x x x
(A) 9 (B) 18 (C) 27 (D) 36 [C] respectively and tan , tan 3 , tan 5 , …..
2 2 2
Q.115 The value of the function y = |2x + 1| + 2|x – 2| x
, tan are periodic functions with period 2,
in the interval –
1
< x < 2, is 2n
2 23, 25 ,…., 2n respectively. L.C.M. of 2,
(A) 4x – 3 (B) 3x – 1 23, 25 ,…., 2n is 2n
(C) 5 (D) 1 [C] Hence f(x) is a periodic function with period
 2 y2 y 2  2n k = 2n.
Q.116 If f  2x  ,2 x 2 
 = xy then f(x, y) is
 8 8  Q.120 If f(g(x)) = | cos x |, g(f(x)) = cos2 x , then -
equal to (if f(x, y) is always positive) (A) f(x) is a periodic function and g(x) is a
(A) 4 x 2
y 2
(B) x 2
y 2 non-periodic function.
1 (B) f(x) is a non-periodic function and g(x) is a
(C) x 2  y2 (D) none of these [B] periodic function.
4
Q.117 Solution set of x – 1 | x | < 0 is (C) Both f(x) and g(x) are periodic functions

(D) Neither f(x) nor g(x) is a periodic function
1 5

(A) 1,  (B) [– 1, 1] Sol.[B] Given, f(g(x)) = | cos x | = cos 2 x ….(i)

 2 
 1 5 g(f(x)) = cos2 x ….(ii)
(C)   1,  (D)
 2
 2  from (i) and (ii), f(x) = x . And g(x) = cos x

 1 5 1 5  Clearly f(x) is a non-periodic function and g(x)

 ,  is a periodic function.
 2 2 

[A] Q.121 Let f(x) = log x and
Q.118 If f(x) = sin2x, g(x) = x and h(x) = cos–1x,
x 4  2 x 3  3x 2  2 x  2
0  x  1, then - g(x) = . The domain
2x 2  2x  1
(A) hogof(x) = gofoh(x)
of the composite function fog(x) is -
(B) gofoh(x) = fohog(x)
(C) fohog(x) = hogof(x) (A)   ,   (B) [0, )
(D) None of these (C) (0, ) (D) [1, )
–1  sin 2 x 
Sol.[D] hogof(x) = cos 



4 3
x  2 x  3x  2 x  2 2
Sol.[A] g(x) =
 2x 2  2x  1
= cos–1(sin x) = –x
2
( x 2  1)( x 2  2 x  2)
gofoh(x) = sin (cos2 –1
x) = 1 x 2 =
2( x 2  x )  1
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 ( x 2  1)[( x  1) 2  1] the inverse function of f(x). Let f(x) and f –1(x)
2
0 intersect each other at x = 0 and x =  ( > 0)
=  1 1
2 x     
 2  2
  f ( x )dx
–1
only and if f ( x )dx = A;
for all x in (–, ) and log x is real for all x > 0. 0 0
It follows that fog(x) = log g(x) is defined for
2
all real x. = B and = C then -
2
(A) A > B > C (B) A > C > B
Q.122 If f(x) = cosec–1(cosec x) and cosec (cosec–1x)
(C) A < B < C (D) A < C < B
are equal functions then maximum range of
Sol.[D]
values of x is.
y = f(x) 
    
(A)   ,1  1, 
 2   2 y = f–1 (x)
    
(B)  ,0  0, 
 2   2
(C) (– , – 1]  [1, )
y=x
(D) [– 1, 0)  [0, 1) [A]
Clearly A < C < B.
Q.123 The maximum value attained by the function  2
2  –1  1  
Q.126 Let I1 = + 2 , I2 =  tan    +
y = 10 – |x – 10|, if – 9  x  9, is 4  e 
(A) 10 (B) 9 (C) +  (D) 1 2e 2
Sol.[B] y = 10 – |x – 10| 2
, I3 = (tan–1 e)2 + 2
, then
e 1 e 1
–9x9
which of the following is true -
– 19  x – 10  – 1 < 0
(A) I1 < I2 < I3 (B) I2 < I1 < I3
 y = 10 – (10 – x) = x
(C) I1 < I3 < I2 (D) I3 < I2 < I1
 maximum value of y = 9
2
Sol.[B] Consider f(x) = (tan–1 x)2 + 2
then
x 1
f '(x) > 0  x  (0, )
Q.124 If f (x) be an invertible function with 1
 f   < f(1) < f(e)  I2 < I1 < I3.
d 2 f 1 ( x ) e
f  (x) f  (x) > 0  x  R, then ,
dx 2
where f–1 (x) is the inverse function of f (x) is Q.127 Let n(A) = 4 and n(B) = 6. Then the number of
(A) positive  x  R one-one functions from A to B is
(A) 120 (B) 360
(B) negative  x  R
(C) 24 (D) None of these
(C) data insufficient
Sol.[B] If n(A) = m and n(B) = m  n. Then the number
d 2 f 1 ( x )
(D) will have sign opposite to of one-one functions from A to B is
dx 2 n!
2 .
d f (x) ( n  m)!
dx 2
d 2 f 1 ( x ) Q.128 Given the function f(x) = 1 /(1 – x), the points
Sol.[B] If f (x) > 0 and f (x) > 0  0 of discontinuity of the composite function
2
dx
y = f 3n (x), where f n(x) = fof …. of (n times)
d 2 f 1 ( x ) are (n  N)
If f (x) < 0 and f (x) < 0  0
dx 2 (A) 0, 1 (B) 2n
Q.125 Consider an invertible function f(x) with (C) 3n (D) 2n + 1
f '(x) > 0 and f ''(x) < 0  x  R. Also f–1(x) is
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Sol.[A] The point x = 1 is a discontinuity of the function Q.135 This graph could be a sketch of the part of the
f(x) = 1/(1 – x). If x  1, then curve -
x 1 y
u(x) = f(f(x)) = . Hence x = 0 is a point
x
of discontinuity of the function u. If x  0, and
/4
x 1, then fofof(x) = x. Hence y = f 3n(x)
= (f 3(x)n) = x is continuous everywhere. x' x
Therefore, 0 and 1 are the only points of O
–1 1 1
discontinuities of y.
2
y'
Q.129 If f(x) = ax + b and g(x) = cx + d, then f[g(x)] = (A) 6y = 13cos–1 x – 5
g[f(x)] if and only if (B) 2y = cos–1 x
(A) f(a) = g(c) (B) f(b) = g(b) (C) y = cos–1 x –/4
(C) f(d) = g(b) (D) f(c) = g(a) [C] (D) 3y = cos–1 x –/4 [C]
Q.130 In which of the following functions, range is
singleton set. Q.136 The domain of f(x) is (0, 1) therefore domain of
(A) f(x) = [x] + [–x] (B) f(x) = {x} + {–x} f(ex) + f(n|x|) is -
(C) f(x) = |sgn (x)| (D) f(x) = 
x  [x]  (A) (–1, e) (B) (1, e)
Where [x], {x} and sgn {x} are greatest integer (C) (– e, –1) (D) (– e, 1) [C]
function, fractional part function and signum
function respectively. [D] Q.137 If f : R  R and for a fixed positive number
c, f(x + c) = 1 + [1 – 5 f(x) + 10 {f(x)} 2 –
n x
Q.131 Range of the function f(x) = is 10 {f(x)}3 + 5{f(x)}4 – {f(x)}5 ]1/5 for all x  R,
x
then f(x) is a periodic function with period
(A) (– , e) (B) (– , e2) (A) c (B) 2c (C) 3c (D) 5c [B]
 2  1
(C)   , (D)   , [C]
 e
  e
 Q.138 Let f and g be two functions both being defined
x | x |
Q.132 Let f : R  R be a function defined by f(x) = from R  R as follows f(x) = and g(x)
2
[x]2 + [x + 1] – 3 {where [] denotes greatest
integer function}, then f(x) is x ; x0
=  2 then
(A) many-one into (B) many-one onto x ; x0
(C) one-one into (D) one-one onto [A] (A) fog is defined but gof is not
(B) gof is defined but fog is not
Q.133 If f(x) is an invertible function, and g(x) =2 f(x) (C) both gof and fog are defined but they are
+ 5, then the value of g–1(x), is unequal
1 (D) both gof and fog are defined and they are
(A) 2f–1(x) – 5 (B) –1
2f ( x )  5 equal function [D]
Q.139 Let f be a real valued function defined by f(x) =
1 1 1  x  5 
(C) f (x)  5 (D) f   [D] e x  e |x|
2  2  then range of f is
e x  e|x|
Q.134 If x, y are real numbers satisfying x > 0, y > 0 (A) R (B) [0, 1]
and x + y  5, then which of the following is not (C) [0, 1) (D) [0, 1/2) [D]
necessarily true ?
(A) 3x + 4y  25 (B) 3x + 4y  18 Q.140 Consider the following equations
(C) 2x + 3y  15 (D) 2x + 3y  18 [B] (i) 2cos (x/3) = 2x + 2–x
(ii) x2 + cos x = 0
(iii) 2|x| = sin x2
(iv) 3|sin x|
| cos x |

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 Which of these have no real roots If f(x) is one-one then set of values of ‘m’ will
(A) (i) & (ii) (B) (ii) & (iii) be
(C) (iv) & (iii) (D) (iv) & (i) [B] (A) (– , 0) (B) (– , 0]
(C) (0, ) (D) [0, )
 x   x   x  11x
Q.141 If          ; x  [0, 500], then Sol. [A]
 2   4   6  12
For f to be one-one, vertex must lie on or to the
number of such x is right of y-axis.
(A) 40 (B) 41 –m0  m0
(C) 42 (D) None of these [C]
 2
x  1; x0
Q.142 Let f : R  R be a function satisfying f(x + y) = for m = 0, f(x) = 
  1;
 x0
f(x) + 2y2 + kxy for all x, y  R. If f(1) = 2 and
which is not one-one.
f(2) = 8, then f(x) is equal to-
(A) 2x2 (B) 6x – 4  m  (– , 0)
(C) x2 + 3x – 2 (D) –x2 + 9x – 6
(sin x  cos x )
Sol.[A] We have, Q.145 If f(x) =  4 , then cos–1(cos
2
f(x + y) = f(x) + 2y2 + kxy for all x, y  R
f(x))is
f ( x  y)  f ( x ) (A) f(x) (B) f(x) – 
 = 2y + kx for all x  R
y (C) 2 –f(x) (D) – f(x)
Sol. [C]
f ( x  y)  f ( x ) ( 2 y  kx )
 lim = ylim
0 1 1
y 0 y Range of f(x) is 4 –  f(x)  4 +
2 2
 f '(x) = kx for all x  R. 3.293  f(x)  4.707
kx 2 from the graph of cos–1 (cos x)
 f(x) = + C for all x  R
2
y= x y=2–x
But, f(1) = 2 and f(2) = 8.
k O 
2
Therefore, 2 = + C and 8 = 2k + C –1
2 cos (cos f(x)) = 2 – f(x)
 k = 4 and C = 0
Q.146 Which of the following is correct for three
Hence, f(x) = 2x2 for all x  R.
function f, g, h defined from R to R
x x (A) (f – g) oh = foh + goh
Q.143 The period of f(x) = sin + cos
n! ( n  1) ! f  foh
(B)  
 oh = goh
x g
 
+ tan is
n! (C) (fg) oh = (foh) (goh)
(A) 2[(n + 1)!] (B) 2 (n !) (D) (fog) oh = (foh) o (goh)
(C) (n + 1) ! (D) (2n + 2) f  foh
! Sol.[B] obviously   oh = while remaining
g goh
x x x
Sol.[A] f(x) = sin + cos + tan choices are incorrect
n! ( n  1)! n!
   Q.147 If f : R  R satisfying f(0) = 1, f(1) = 2 and
2n! 2(n+1)! n! f(x + 2) = 2 f(x) + f(x + 1) then f(6) is
L.C.M of [2n!, 2(n + 1)!, n!] = 2(n + 1)! (A) 8 (B) 32
(C) 16 (D) 64
Sol.[D] f(2) = 2f(0) + f(1) = 4
Q.144 Let a function is defined as f : R  R
f(3) = 4 + 4 = 8

 x 2  2mx  1, x0
f(x) =  f(4) = 8 + 8 = 16

 mx  1, x0 f(5) = 16 + 16 = 32
f(6) = 32 + 32 = 64

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 (B) gof is one-one but g is not one-one
–1
Q.148 Domain of sec (sinx) is - (C) gof is invertible but g is not invertible
(A) R– (–1, 1) (B)  (D) f and g are both one-one [A]

  
(C)  – ,  (D) None of these [D] x2 
 2 2 Q.154 Let f: [–10, 10]  R and f(x) = sin x +  
 a 
cosx; for what value of ‘a’ given function is an
Q.149 Let f : A  [2, 6]; f(x) = 3 sinx + cosx + 4 odd function (Here [x] represents greatest
is bijective function then set A is - integer  x):
 2     5  (A) a  (–100, 100) (B) a < 100
(A)  – ,  (B)  – , 
 3 3  6 6  (C) a > 100 (D) a < – 100 [C]

      Q.155 If 3f (x) –2f (1/x) = x2 + 2, then f(2) equals to:

(C)  – ,  (D)  – ,  [A]
 2 2  3 3 (A) 10 (B) 5/2
(C) 3/2 (D) None of these [B]
Q.150 Let P(x) be a polynomial function with integral
coefficients. if there exist two integers a and b Q.156 Let ‘f’ be a function defined from R+  R+. If
such that P(a) – P(b) = 1, then - (f(xy))2 = x(f(y))2 for all positive numbers x &
y. If f(2) = 6, find f(50) = ?
(A) both a and b must be even
(A) 20 (B) 30 (C) 5 (D) 40 [B]
(B) both a and b must be odd
(C) a & b must be consecutive integers Q.157 If f(x) = 1 + x is bijective then  is:
(D) None of these [C] (A) any real number
(B) any positive real number
Q.151 The value of n for which the equation (C) any non-zero real number
n{x} = [x] + {x}; where [x] & {x} respectively (D) any negative real number [C]
represent integral and fractional parts of x, has
exactly ten solutions, is - Q.158 The inverse of function f(x) = log a (x +
(A) 10 (B) 11 x 2  1 ); (where a > 0 and a  1) is:
(C) 12 (D) None of these [B] 1 x –x 1 x –x
(A) (a –a ) (B) (a + a )
2 2
Q.152 The range of the function (C) does not exist if x  R
f(x) = 8x + 4x + 8–x + 4–x + 5 is: (D) exists only if x  R+ [A]
7  Q.159 If f(x) = sin x + cos ax is periodic, then the value of
(A)  ,   (B)
 4  a is:
(A) any real number (B) rational
7 
4 ,  (C) irrational (D) None of these [B]
 
(C) [9, ) (D) (9, ) [C] Q.160 If [2 sin x ] + [cos x] = – 3 then the range of the
function f(x) = sinx + 3 cosx in [0, 2] is
(where [·] denotes greatest integer function)
Q.153 Let Z denote the set of integers. Let p be a
(A) [–2, – 1] (B) (–2, –1)
prime number and let Z 1  {0, 1}. Let f: Z  Z (C) (–1, –1/2) (D) None of these
and
Sol.[B] [2 sin x] + [cos x]  – 3 only [2 sin x] = – 2
g : Z  Z 1 be defined as follows: [cos x] = – 1
f(n) = pn ; if n  Z and – 2  2sin x < – 1 and – 1 cos x < 0
g(n) = 1; if n is a perfect square.
– 1 sinx < –1/2 and –1 cosx < 0
= 0; otherwise
7 11  3
If we consider the composite function gof, then: <x< and <x<
(A) gof is onto but f is not onto 6 6 2 2

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 7 3 f(x) = 0
common values of x is <x<
6 2
{function is periodic so that consider the interval Q.162 The domain set of the function
0  x  2} f(x) = log7 log5 log3 log2 (2x3 + 5x2 –14x) is
(A) (– ) – {0}
 
For f(x) = sin x + 3 cos x = 2sin   x (B) (0,)
3  (C) (– 41/2)  (2, )
7 3 (D) (– 4– 1/2)  (2, ) [D]
Now <x<
6 2
3  11 Q.163 If 5{x} = x + [x] and [x] – {x} = 1/2 where {x}
< +x<
2 3 6 and [x] are fractional part and integral part of x
then x is
  1
– 1 < sin   x  < (A) 1/2 (B) 3/2
 3  2 (C) 2 (D) None of these [B]
 
– 2 < 2 sin   x < – 1
 3  Q.164 Consider a function g(x) defined as
2008
Range is (–2, – 1) 1)
g(x) ( x ( 2  1) = (x + 1) (x2 + 1) (x4 +1)
2007

Q.161 If f(x) is an even function and satisfies the ….. ( x 2  1) – 1 then value of g(2) equals
(A) 1 (B) 22008 –1
1
relation x2f(x) – 2f   = g(x) where g(x) is (C) 22008 (D) 2
x Sol.[D] R.H.S.
an odd function then f(5) equals =
(A) 0 (B)
50
75
( x  1)( x  1)( x 2  1)........ x 2  2007
1 1
49 x 1
(C) (D) None of these =

 
75
2007
1 ( x 2  1)( x 2  1)( x 4  1)........ x 2 1 1
Sol.[A] x f(x) – 2f   = g(x)
2
...(1)
x x 1

x
1
x
=
2
( x 2  1)( x 2  1)........ x 2
2
 2007
1 1 
x 1

x
1
2
1 1
f   – 2 f(x) = g  
x x g(x) x  ( 2 2008 1)
 1 
x 2 2008
1 1 
x 1
1 1
f   – 2x2 f(x) = x2g  
x
1
x
1

g(2) 2 ( 2
2008
1)
 
 1  22
2008
1 1 
2f   – 4x2 f(x) = 2x2g  
x x
...(2)
g(2)
2 2008
2 2 2 2008
2
Equation (1) + Equation (2) 2
g(2) = 2
1
– 3x f(x) = g(x) + 2x g  
2 2

x x 1 1
 1 2x 2
 g ( x )  2 x 2g   Q.165 Let f(x) = sin x 1 , f(x) be an odd
 x
f(x)= –   x3 3x 4 1
3x 2
  function and its odd value is equal to g(x) then
 
f(1) g(1) is -
1 (A) –1 (B) – 4
g(x) and g   is odd so that f(x) will be odd
x (C) – 5 (D) 1
but given f(x) is even so that f(x) should be zero Sol.[B] f(x) is odd function

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 g(x) = f(–x) = – f(x) (A) – 2003 (B) 2003
x 1 1 (C) –1/2 (D) 1/2
g(x).f(x) = – sin x 2x 2 1 Sol.[C] Let f(0) = k, a=0
x3 3x 4 1 We get f(b) = f(0) = k and again if b = 0 gives
f(a) = k  f(a) = f(b) = k  a, b  f(x) is a
x 1 1 constant function
sin x 2x 2 1 f (2003) = – 1/2
x3 3x 4 1
sin x cos 3x
= – Q.168 Let y = , then -
sin 3x cos x
x2  2 x sin( x )  2 x 2  1 4x 4  1
1 
x sin( x )  2 x  1 2 2
sin (x )  4 x  1 4 3
6 x 6y1  3 ,3
x sin(x )  (A) (B) y < 1/3 or y  3
 
4x 4  1 x 3 sin( x )  6 x 6  1 6
x  9x  14
(C) y  – 3 or y > 1/3 (D) none of these
3 3 5 sin x cos 3x
Sol.[B] y =
f(1) = – 3 5 7 =–4 sin 3x cos x
5 7 11  y (3 sinx – 4 sin3x) cosx = sinx (4 cos3x – 3 cosx)
y (3 – 4 sin2 x) = 4 cos2 x – 3
Q.166 The range of function 4(y – 1) sin2 x + (1 – 3y) = 3
 2 1  2 1 3y  1
f(x) = sin–1  x   + cos–1  x   , where  sin2 x =
 2  2 4( y  1)
[.] is the greatest integer function is - 3y  1
0 < sin2 x  1  0 < 1
    1 4( y  1)
(A)  ,  (B) 0, 
2   2  3y  1 3y  1
 > 0 and –10
  4( y  1) 4( y  1)
(C)  (D)  0, 
 2 (3y – 1) (y – 1) > 0 and (y – 3) (y – 1)  0
y < 1/3 or y < 1 and
 2 1  2 1
Sol.[C] f(x) = sin–1  x   + cos–1 x  2  y < 1 or y 3  y < 1/3 or y  3
 2  
 2 1  2 1  Q.169 The 'x' for which sin x (sin x + cos x) = [x]
= sin–1  x   + cos–1  x  2  1
 2   where [.] denotes greatest integral function is -
(A) [0, 2) (B) [0, 1]  [2, 3)
 2 1  2 1 
= sin–1  x   + cos–1   x    1 (C) [–1, 1)  [1, 2) (D) None of these
 2  2  Sol.[A] sin2 x + sin x cos x = [x]
1 1 1  cos 2x sin 2 x
Since x2 +   = [x]
2 2 2 2
 2 1 sin 2x – cos2x = 2 | x | – 1
So,  x   is defined only for the two
 2 2 {sin 2x cos /4 – cos 2x sin /4}
values. = 2[x] – 1
= 2 {sin (2x – /4} = 2(x) – 1
 2 1
 x  2  = 0  f(x) = sin 0 + cos (–1) = 
–1 –1

  – 2  {2 | x | – 1}  2

 2 1 1 2 1 2
 [x] 
 x  2  = 1  f(x) = sin 1 + cos 0 = 
–1 –1

  2 2
 [x] = 0.1
So, range of f(x) = 
 x  [0, 1)  [1, 2)
Q.167 The function f(x) is defined for all real x. If
Q.170 A function F(x) satisfies the functional equation
f(a + b) = f(ab)  a, b and f (–1/2) = –1/2, then
x2 F(x) + F(1 – x) = 2x – x4 for all real x.
f(2003) equals -
F(x) must be -
Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159 FUNCTION 16
 (A) x2 (B) 1 – x2 domain of the given function is ((–, 1) ~ {0})
(C) 1 + x2 (D) x2 + x + 1  {x : x  –2}= [–2, 1) ~ {0}
Sol.[B] Replacing x by (1 – x) gives x  [x]
(1 – x)2 F (1 – x) + F(x) = 2(1 – x) – (1 – x)4 Q.174 Let f(x) = , x  R, then the range of f
1  x  [x]
Eliminating F(1 – x) from (1) and (2), we get
is :
F(x) = 1– x2
(A) [0, 1] (B) [0, 1/2]
(C) [0, 1/2) (D) (0, 1)
Q.171 The complete set of values of 'a' for which the
Sol. [C]
function f(x) = tan–1 (x2 –18x + a) > 0
f(x) = 0 if x   and for x  R ~ I
 x  R, is –
2(x – [x]) < 1 + x – [x]. Thus f(x) < 1/2.
(A) (81, ) (B) [81, )
(C) (–, 81) (D) (–, 81]
Q.175 Let f : R R be a function defined by
Sol.[A] tan–1 (x2 –18x + a) > 0,  x  R
 x2 –18x + a > 0,  x  R e |x|  e  x
f(x) = . Then :
 (18)2 – 4a < 0 e x  ex
 a > 81 (A) f is both one-one and onto
 a  (81, ) (B) f is one-one but not onto
(C) f is onto but not one-one
Q.172 Let f(x) = [x] + {x} ; where [.] denotes the (D) f is neither one-one nor onto
integral part of x and {x} denotes the fractional Sol.[D] f is not one-one as f(0) = 0 and f(–1) = 0. f is
part of x. Then f–1(x) is- also not onto as for y = 1 there is no x R such
(A) [x] + {x} (B) [x]2 + {x} that f(x) = 1. If there is such a x  R then e|x| –
e–x = ex + e–x. Clearly x  0. For x > 0, this
(C) [x] + {x}2 (D) {x} + {x}
equation gives –e–x = e–x which is not possible.
Sol.[C] Let y = f(x) and [x] = I For x < 0, the above equation gives ex = –e–x
y=I+ xI which is also not possible.
 x  I = (y – I)
 x – I = (y – I)2 Q.176 If f(x) is a polynomial satisfying f(x) . f(1/x) =
 x = (y – I)2 + I f(x) + f (1/x), and f(3) = 28, then f(4) is given
 x = {y}2 + [y] by:
f–1(x) = [x] + {x}2 (A) 63 (B) 65 (C) 67 (D) 68
th
Alternatively Sol.[B] By considering a general n degree polynomial
and writing the expression f(x) . f(1/x) = f(x)
y = [x] + {x}
= f(x) + f(1/x) in terms of it, it can be proved by
 [y] + {y} = [x] + {x} comparing the coefficients of xn, xn–1, ....... and
 {y} = {x} ….. ([y] = [x]) the constant term, that the polynomial satisfying
 {x} = {y}2 the above equation is either of the form xn + 1 or
 {x} + [x] = [y] + {y}2 –xn +1 . Now, from f(3) = 3n + 1 = 28, we get 3n
 x = [y] + {y}2 = 27, or n = 3. But f(3) = –3 n + 1 = 28 is not
possible, as –3n = 27 is not true for any value of
 f–1 (x) = [x] + {x}2
n. Hence f(4) = 43 + 1 = 65.
Q.173 The domain of definition of the function
1 Q.177 Let f(x) = sin–1 (sin x). Then:
y= + x  2 is :
log10 (1  x ) (A) f is periodic with period 
(A) (–3, –2) ~ {–2.5} (B) [0, 1] ~ {0.5} (B) f is periodic with period /2
(C) [–2, 1) ~ {0) (D) None of these (C) f is periodic with period 2
Sol.[C] log10 (1 – x) is defined if 1 – x > 0 and 1/log 10 (D) f is non-periodic
(1 – x) is defined for x (–, 1) ~ Sol.[C] On the interval [–/2,/2], we have y = sin–1
{x : log10 (1 – x) = 0} = (–, 1) = (–, 1) ~ {0}. (sin x) = x by definition of the function
Also x  2 is meaningful if x  – 2. Thus the sin–1 x. To obtain the graph of the function on
the interval /2  x 3/2, put z = x –,

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 then x =  + z, –/2  z 2, (A) {0, 3} (B) (0, 3)
y = sin–1 (sin x) = sin–1 (sin (z + ) = – sin–1 (sin z) (C) {0, –3} (D) [–3, 0] [C]
= –z =  –x and so on.  2 | x | 
Q.184 The domain of sec 1   is
 4 
(A) R (B) R – (–1, 1)
y (C) R – (–3, 3) (D) R – (– 6, 6) [D]
Q.185 The domain of the function -
–   3/2 f(x) = 24 – xC3x – 1 +40 – 6xC8x – 10 is -
– • • • • x
(A) {2, 3} (B) {1, 2, 3}
(C) {1, 2, 3, 4} (D) None of these [A]

Q.178 Suppose f(x) = (x + 1)2 for x  –1. If g(x) is the Q.186 Let f be a real valued function defined by
function whose graph is the reflection of f(x)  1| x | 
w.r.t. y = x, then g(x) equals: f(x) = sin–1   + cos–1
 3 
1
(A) – x –1, x  0 (B) , x > –1  | x | 3 
( x  1) 2  
 5 
(C) x  1 , x  –1 (D) x –1, x  0
Then domain of f(x) is given by
Sol.[D] Clearly g(x) = f–1 (x). Let y = f(x) = (x + 1)2
(A) [–4, 4] (B) [0, 4]
 y –1 = x. Hence f–1 (x) = x –1, x  0 (C) [–3, 3] (D) [–5, 5] [A]
1
 5x  x 2  Q.187 The range of the function y = is :
Q.179 Domain of y  log10  : 2  sin 3x
4 
  1 
(A) (0, 5) (B) [1, 4] (A)  , 1 (B)
3 
(C) (–, 0)  (5, ) (D) (–, 1)  (4, ) [A]
1 
 3 , 1
 
 2x  1 
Q.180 Domain of y   log x  4  log 2  1 
2 
3 x  (C)  , 1 (D) None of
3 
(A) (–4, – 3)  (4, ) (B) (–, –3)  (4, ) these [C]
(C) (–, –4)  (3, ) (D) (–4, – 3)  (3, 4) [D]
Q.188 The range of the function
Q.181 The domain of definition of f(x) = loge(3x2 – 4x + 5) is -
f(x) = sin–1 (|x –1| –2) is:
 11 
(A) [–2, 0]  [2, 4] (B) (–2, 0)  (2, 4) (A)   , log e (B)
 3
(C) [–2, 0]  [1, 3] (D) (–2, 0)  (1, 3) [A]
 11 
Q.182 The domain of definition of log e 3 ,  
 
 log 0.3 ( x  1)  11 11 
f(x) = is: (C)  log e , log e (D) None of these
2
x  2x  8  3 3 
(A) (1, 4) (B) (–2, 4) [B]
(C) (2, 4) (D) [2, ) [D]
Q.189 The value of the function
x 2  3x  2
f(x) = lies in the interval -
x2  x  6
Q.183 The function 1 
(A) (– , )/  , 1 (B) (– , )
f(x) = cot–1 ( x  3) x + cos–1 5 
x 2  3x  1
is defined on the set S, where S is equal to: (C) (– , )/{1} (D) None of these [B]

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 e x  ex e x  ex
Q.190 Find the range of the following function, (C) (D) [C]
2 e x  e x
y  log ( 2 (sin x  cos x )  5)
7 Q.199 Let f : RR be given by f(x) = (x + 1) 2–1, x  –1.
(A) R (B) Z Then the set of values of x for which
(C) [ log7 4, log7 5] (D) Q [D] f(x) = f –1(x) is given by -
(A) {0} (B) {0, –1}
Q.191 The greatest value of the function (C) {–1} (D) None of these [B]
f (x) = cos{xe[x] + 2x2 – x}, x  (–1, ) where Q.200 If f (x) = x3 – 1 and domain of f = {0,1, 2, 3},
[x] denotes the greatest integer less than or then domain of f –1 is -
equal to x, is (A) {0, 1, 2, 3} (B) {1, 0, –7, –26}
(A) 1 (B) – 1 (C) {–1, 0, 7, 26} (D) {0, –1,– 2, –3} [C]
(C) 0 (D) 5 [A]
Q.192 Which of the following function (s) has the ex – e–x
Q.201 The inverse of the function y = is
range [–1, 1] ex  e–x
(A) f(x) = cos (2 sin x) 1 1 x 1 2x
(A) log (B) log
 1 2 1 x 2 2x
(B) g(x) = cos 1  
 1 x  2 1 1 x
(C) log (D) 2 log (1+ x) [C]
(C) h(x) = sin (log2 x) 2 1 x
(D) k(x) = sin (ex) [C] Q.202 The function f(x) is defined in [0, 1] then the
domain of definition of the function f[n (1–
Q.193 Range of f(x) = 4 x + 2 x + 1 is x2)] is given by :
(A) (0, ) (B) (1, ) (A) x  {0}
(C) (2, ) (D) (3, ) [A] (B) x [– 1  e –1]  [1 + 1  e ]
(C) x  (– , )
Q.194 Let f: R  R be a function defined by
(D) None of these [A]
f(x) = (1 – x)1/3 is:
x 1
(A) one-one and onto (B) many one and onto Q.203 If f(x) = then f (2x) is
(C) one-one and into (D) many one and into x 1
[A] f (x)  1 3f ( x )  1
(A) (B)
Q.195 Let f : R  R be a function defined by f (x)  3 f (x)  3
x 2  2x  5 f (x)  3 f (x)  3
f(x) = is : (C) (D) [B]
2
x  x 1 f (x)  1 3f ( x )  1
(A) one-one and into (B) one-one and onto Q.204 If f (x) = cos (log x) then
(C) many-one and onto (D) many-one and into f(x2) f(y2) – [f(x2/y2) + f (x2 y2)] has the value
[D] (A) –2 (B) –1
Q.196 The function f : [2, )  Y defined by (C) 1/2 (D) None [D]
f(x) = x2 – 4x + 5 is both one–one & onto if: Q.205 If f(x) = (x – 1)/x for all real number except
(A) Y = R (B) Y = [1, ) x = 0 and g (u) = u 2 + 1, for all u R then
(C) Y = [4, ) (D) Y = [5, ) [C] f [g (u)] is defined for:
(A) all real number u (B) u = –1
Q.197 Let F : R  R be a function defined by (C) u2 = 1 (D) u = 0 [A]
f(x) = x3 + x2 + 3x + sin x. Then f is :
Q.206 If f : R  R is a function satisfying the property
(A) one– one & onto (B) one –one & into
f(x +1) + f(x + 3) = 2  x R then the period
(C) many one & onto (D) many one & into [C]
(may not be fundamental period) of f (x) is
(A) 3 (B) 4
Q.198 If f (x) = loge (x+ 1  x 2 ) , then f –1(x) =
(C) 7 (D) 6 [B]
e x  e x
(A) log (x– 1  x 2 ) (B) Q.207 Which of the following function is periodic
2 (A) f(x) = x – [x] where [x] denotes the largest
integer less than or equal to the real number x
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 (B) f(x) = sin (1/x) for x  0 f(0) = 0 1
(C) f(x) = x cos x Q.214 If f (x) = 27x3 + and ,  are the roots of
x3
 x cos  / x x  0
(D) f ( x )   [A] 1
0 x0 3x + = 2 then -
x
(A) f () = f () (B) f () = 10
(C) f () = – 12 (D) None of these [A]

Q.215 The set of points for which

Q.208 Fundamental period of f(x) = sec (sin x) is:
f(x) = cos (sinx) > 0 contains -

(A) (B) 2 (A) (– , 0] (B) [–1, 1]
2
(C) (–, ) (D) All are correct [D]
(C)  (D) a periodic [C]

Q.209 The fundamental period of the function:

Q.216 Which of the following function from
f(x) = x + a – [x + b] + sin x + cos 2x
A = {x : –1  x  1} to itself are bijections-
+ sin 3x + cos 4x + …. + sin (2n – 1) x
(A) f(x) = x/2 (B) g(x) = sin (x/2)
+ cos 2 nx for every a, b R is:
(C) h(x) = |x| (D) k(x) = x2 [B]
(where [.] denotes the greatest integer function)
(A) 2 (B) 4 (C) 1 (D) 0 [A]
Q.217 Which of the following functions is absolute
value function -
Q.210 Which of the following is an even function ?
 x: x 0
a x 1 (A)  (B) ( x 2 )
(A) x (B) tan x   x : x 0
a x 1
a x  a x a x 1 (C) Max {x, –x} (D) All of these [D]
(C) (D)
x
2 a 1
[A] Q.218 If [x] stands for the greatest integer
function, then the value of
Q.211 In the following, odd function is - 1 1  1 2 
(A) cos x2 (B) (ex +1) / (ex – 1)  2  1000  +  2  1000  +.........+
   
2
(C) x – |x| (D) None of these [B]
1 999 
 2  1000 
Q.212 The function f(x) = x2 – |x| is -  
(A) An odd function (B) A rational function (A) 498 (B) 499
(C) An even function (D) None of these [C] (C) 500 (D) 501 [C]

Q.213 Which of the following function is an odd Q.219 Let f(x) = sin  a  x (where [ ] denotes the
function greatest integers function). If f is periodic with
(A) f(x) = 1  x  x 2 – 1  x  x 2 fundamental period , then a belongs to -
(A) [2, 3) (B) {4, 5}
 a x 1
(B) f(x) = x  x 

(C) [4, 5] (D) [4, 5) [D]
 a 1  Q.220 Let the function f(x) = 3x2 – 4x + 8 log ( 1 + | x | )
 1 x  be defined on the interval [0, 1]. The even
(C) f(x) = log   extension of f(x) to the interval [–1, 0] is -
1 x2 
(A) 3x2 + 4x + 8 log(1 + | x |)
(D) f(x) = k (constant) [A]
(B) 3x2 – 4x + 8 log(1 + | x |)
(C) 3x2 + 4x – 8 log(1 + | x |)
(D) 3x2 – 4x – 8 log (1 + | x |) [A]

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 sinx
1   
Q.221 The function f(x) =   is - (B)   , 
2  3 6
(A) periodic with period 2 (C)
(B) an odd function 1   3   
(C) a even function   1   ,  1   
(D) None of these [A] 2
 3 3 2 6 6 

1    3   
(D)   1   ,  1   
Q.222 The function f(x) = log10 cos( 2x ) exists  2 3  3  2 6  6 
 
- [C]
(A) for any rational x
(B) only when x is a positive integer Q.228 If A be the set of all triangles and B that of
(C) only when x is fractional positive real numbers, then the mapping
(D) for any integer value of x including zero [D] f : A  B given by
f() = area of , (  A) is -
(A) one one into mapping
Q.223 The domain of the function sec–1[x2 – x + 1], is (B) one one onto mapping
given by– (C) many-one into mapping
where [·] is greatest integer function - (D) many-one onto mapping [D]
(A) [0, 1] (B) (–, 0]  [1, )
1 – 5 1 5   
(C)  ,  (D) None of Q.229 Let f : R  A =  y | 0  y   be a function

 2 2   2
these [B] such that f (x) = tan –1 (x2 + x + k), where k is a
constant. The value of k for which f is an onto
Q.224 The domain of definition of the function function, is -
cot –1 x (A) 1 (B) 0
f(x) = , where [x] denotes the
{x 2 – [ x 2 ]} (C) 1/4 (D) None of these [C]
greatest integer less than or equal to x is -
Q.230 Which of the following functions are not
(A) R
injective map -
(B) R – {± n : n  I+  {0}}
(A) f(x) = |x + 1|, x  [–1, )
(C) R – {0}
1
(D) R – {n : n  } [C] (B) g(x) = x + ; x  (0, )
x
Q.225 The domain of the definition of (C) h(x) = x2 + 4x – 5 ; x  (0, )
f(x) = log{(log x)2 – 5 log x + 6} is equal to- (D) k(x) = e–x ; x  [0, ) [B]
(A) (0, 102) (B) (103, )
Q.231 Let f be an injective map. with domain {x, y, z}
(C) (10 , 10 )
2 3
(D) (0, 102)  (103, )
and range {1, 2, 3}, such that exactly one of the
[D]
following statements is correct and the
Q.226 The domain of the function
remaining are false :
log log .........log x
10 10
y =         
10
is - f(x) = 1 , f(y)  1, f(z)  2
n  times
The value of f –1 (1) is -
(A) [10n, + ) (B) (10n–1, + ) (A) x (B) y
(C) [10n–2, + ) (D) None of these [D] (C) z (D) None of these [B]
  
Q.227 If A =  x :  x   and Q.232 Let f : R  R and g : R  R be two one-one
 6 3
onto functions such that they are mirror image
f(x) = cos x – x (1 + x) then f(A) is equal to-
of each other about the line y = 0, then
  h(x) = f(x) + g(x) is-
(A)  , 
6 3 (A) one-one and onto
(B) one-one but not onto
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 (C) not one-one but onto then -
(D) Neither one-one nor onto [D]  – 1; – 1  x  0 or x  1

(A) fog (x) =  0 ; x  0, 1, – 1
Q.233 If the function y = logx(x + 1) is plotted for all  1;
 0  x 1
real value of x for which it is defined, the graph
looks like  – 1; –1  x  0

Y (B) fog (x) =  0 ; x  0, 1, – 1
 1; 0  x 1

(C) fog (x) =
(A) –1 X
 – 1; – 1  x  0 or x  1

 0; x  0, 1, – 1
 1; 0  x  1 or x  –1

Y (D) fog (x) =
1; – 1  x  0 or x  1

0 ; x  0, 1, – 1
(B)  1;
 0  x  1 or x  –1

1 X
[C]
Y
Q.236 Period of f(x) = e + sin  [x] is (where, [.]
cos {x}

and { } denote the greatest integer function and

fractional part of function respectively).
(C) (A) 1 (B) 2
1 (C)  (D) 2 [A]
X Q.237 If f(x) = cos (ax) + sin (bx) is periodic, then
which of the followings is false -
(A) a and b both are rational
Y (B) non-periodic if a is rational but b is
irrational
(C) non-periodic if a is irrational but b is
rational
(D) [A] (D) none of these [D]
–1 1 Q.238 The function
X f(x) = 2(x – [x]) + sin2 (x – [x]) is -
(Where [.] denotes greatest integer function)
(A) Non periodic
Q.234 Which of the following functions is inverse of (B) periodic with period 1
itself - (C) periodic with period 2
1 x (D) None of these [B]
(A) f (x) = (B) g (x) = 5log x
1 x
(C) h (x) = 2x (x–1) (D) None of these [A]  x2 
Q.239 If f: [–20, 20]  R is defined by f(x) =  
 a 
Q.235 If
sin x + cos x, is an even function, then the set
of values of a is-
 – 1; x0 (A) (–, 100) (B) (400, )
 (C) (– 400, 400) (D) None of these [B]
f (x)   0 ; x  0 and g ( x )  x (1  x 2 ) ,
 1; x0


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Q.240 Let f be a function satisfying f (x + y) = f (x).f (y) (C) {0, 1, –1}
n 
 3i 3 3i 3

for all x, y  R. If f (1) = 3 then  f (r ) is (D) 0,1,

 2
,
2



r 1
equal to - [B]
3 n 3   
(A) (3 – 1) (B) n (n + 1)
2 2 Let f(x) = [x] sin 
Q.246  , where [.] denotes
 [ x  1] 
(C) 3n+1 – 3 (D) None of these
[A]
the greatest integer function. The domain of
f is ....... [IIT 96]
Q.241 If f() =
(A) {x  R| x  [–1, 0)}
(2 cos  – 1)(2 cos 2 – 1)(2 cos 4 – 1) ..... ( 2 cos 2 n –1(B)
 –{x1) R| x  [1, 0)}
n
2 cos 2   1 (C) {x  R| x  [–1, 0)}
(D) None of these [C]
2
for n  N and  2m ± , m  I,
3  
then f(/4) = Q.247 If f(x) = sin2x + sin2  x  
 3
(A) 1 – 2 (B) 2 –
  5
1 + cos x cos  x   and g   = 1, then (gof)
 3 4
(C) 2 +1 (D) None of
these [B] (x) =
[IIT 96]
Q.242 All the values of a for which (A) –2 (B) –1 (C) 2 (D) 1 [D]
2

 [a
2
 ( 4  4a ) x + 4x 3] dx 12 are given Q.248 If g (f(x)) = | sin x | and f(g(x)) = (sin x )2,
1 then [IIT 98]
by- (A) f(x) = sin2x, g(x) = x
(A) a = 3 (B) a  4 (B) f(x) = sin x, g(x) = |x|
(C) 0  a < 3 (D) None of these [A]
(C) f(x) = x2, g(x) = sin x
(D) f and g cannot be determined [A]
Q.243 If f(x) = [x2] – [x]2 where [·] denotes the
greatest integer function and x  [0, 2], the set Q.249 If f(x) = 3x – 5, then f–1(x) is - [IIT 1998]
of values of f(x) is - 1
(A) {–1, 0} (B) {–1, 0, 1} (A) is given by
3x – 5
(C) {0} (D) {0, 1 , 2} [D]
x 5
(B) is given by
Q.244 Let f(x) be defined for all x > 0 and be 3
(C) does not exist because f is not one-one
x
continuous. Let f(x) satisfy f   = f(x) – f(y) (D) does not exist because f is not onto [B]
 y
for all x, y and f(e) = 1. Then - [IIT-1995S] Q.250 If the function f : [1, )  [1, ) is defined by
f (x) = 2x(x–1) , then f –1 (x) is [IIT 99]
(A) f(x) is bounded x ( x 1)
1
x (A)    
(B) f    0 as x  0 2
 y
(C) x f(x)  1 as x  0 (B)
1
2

1  1  4log 2 x 
(D) f(x) = n x [D]
(C)
1
2

1  1  4log 2 x 
Q.245 Let f(x) = (x + 1)2– 1, (x > – 1). Then the set
(D) not defined [B]
S = {x : f (x) = f –1(x)} is – [IIT 95]
(A) Empty
(B) {0, –1}
Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159 FUNCTION 23
Q.251 The domain of definition of the function y(x) R
| 1, x0
given by the equatin 2x + 2y = 2 is – Q.258 Let g (x) = 1 + x – [x] and f (x) = S 0, x0
[IIT Scr. 2000] |T1, x0
(A) 0 < x < 1 (B) 0 < x < 1
Then for all x, f (g(x) is equal to–
(C) – < x < 0 (D) – < x < 1 [D]
[IIT Scr. 2001]
(A) x (B) f (x)
Q.252 Let f() = sin  (sin  + sin 3 ). Then f()-
(C) 1 (D) g (x) [C]
[IIT 2000]
(A)  0 only when   0
Q.259 Suppose f(x) = (x + 1) 2 for x  – 1. If g(x) is the
(B) 0 for all real 
function whose graph is the reflection of the
(C)  0 for all real  graph of f(x) with respect to the line
(D)  0 only when   0 [C] y = x, then g(x) equals– [IIT Scr. 2002]
1
Q.253 Let f (x) = (1 + b2)x2 + 2bx + 1 and let m(b) be (A) – x – 1, x  0 (B) ,x>–1
( x  1) 2
the minimum value of f(x). As b varies, the
range of m(b) is - [IIT 2001] (C) x  1 , x  – 1 (D) x – 1, x  0 [D]
(A) [0, 1] (B) [0, 1/2]
(C) [1/2, 1] (D) (0, 1] [D] Q.260 Let function f : R  R be defined by
f(x) = 2x + sin x for x  R. Then f is–
Q.254 Let E = {1, 2, 3, 4} and F = {1, 2}. Then the [IIT Scr. 2002]
number of onto functions from E to F is- (A) one to one and onto
[IIT 2001] (B) one to one but not onto
(A) 14 (B) 16 (C) 12 (D) 8 [A] (C) onto but not one to one
(D) neither one to one nor onto [A]
x
Q.255 Let f (x) = , x  – 1, then for what value
x 1 x
Q.261 Let f(x) = defined as [0, )  [ 0, ),
of  f { f (x) } = x. [IIT Scr. 2001] 1 x
(A) 2 (B) – 2 (C) 1 (D) –1 [D] f(x) is– [IIT Scr.2003]
(A) one one & onto
(B) one–one but not onto
(C) not one–one but onto
(D) neither one–one nor onto [B]

Q.256 The domain of definition of tan 2 

log ( x  3) Q.262 If f(x) = x x +
2 ,   (0, /2),
f (x) = 2 2 is – [IIT Scr. 2001] x2  x
x  3x  2 x > 0 then value of f(x) is greater than or equal
(A) R – { –2, + 2} to– [ IIT Scr.2003]
(B) (– 2, ) (A) 2 (B) 2 tan 
(C) R– {–1, –2, –3} (C) 5/2 (D) sec  [B]
(D) (–3, ) / {–1, –2} [D]

x2  x  2
Q.257 If f : [1, )  [2, ) is given by Q.263 Find the range of f(x) = is–
1 x2  x 1
f (x) = x + then f–1 (x) equals –
x
[IIT Scr.2003]
[IIT Scr. 2001]
x  11 
(A) x  x  4
2
(B) (A) (1, ) (B) 1, 
1  x2  7 
2
2
(C) x  x  4 (D) 1 + x2  4 [A]
2

Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159 FUNCTION 24
  7 (A) 15 (B) 0
(C) 1,  (D) (C) 5 (D) 10 [C]
 3
 7 Q.269 Let f : R  R be any function . Define
1,  [C]
 5 g : R  R by g (x) = |f (x)| for all x. then g is

Q.264 Domain of f(x) = sin 1 ( 2 x )   / 6 is -
[IIT 2000S]
(A) onto if f is onto
[IIT Scr.2003] (B) one-one if f is one-one
 1 1  1 1 (C) continuous if f is continuous
(A)  ,  (B)   ,  (D) differentiable if f is differentiable [C]
 4 2  2 2
 1 1  1 1 Q.270 Let g (x) = log f (x) where f(x) is a twice
(C)  ,  (D)  ,  [A]
 4 4  2 4 differentiable positive function on (0, ) such
that
Q.265 Let f(x) = sin x + cos x and g(x) = x2 – 1, then f(x + 1) = x f(x). Then, for N = 1, 2, 3, …
g (f(x)) will be invertible for the domain-  1 1
g  N   – g   = [IIT 2008]
[IIT Scr.2004]  2  2
  
(A) x [0, ] (B) x   – , 
 4 4 
 1 1 1 

(A) – 4 1    .....  
     
 9 25 ( 2 N – 1) 2 

(C) x  0,  (D) x   – , 0  [B]
 2  2 

 1 1 1 

(B) 4 1    .... 
2 

 9 25 ( 2 N – 1) 

x , x Q
Q.266 f (x)   ;  
 0, x Q  1 1 1 
(C) – 4 1    .... 
2 

 9 25 ( 2 N  1) 
0 xQ
g (x)  
x x Q 
 1 1 1 

(D) 4 1    ..... 
2  [A]
then (f – g) is [IIT Scr.2005] 
 9 25 (2 N  1) 

(A) one-one , onto

  
(B) neither one-one, nor onto Q.271 Let the function g : (– , )   – ,  be
(C) one-one but not onto  2 2
(D) onto but not one - one [A] 
given by g(u) = 2 tan–1 (eu) – . Then g is -
2
Q.267 If X and Y are two non-empty sets where [IIT 2008]
f : X  Y is function is defined such that (A) even and is strictly increasing in (0, )
f(C) = {f(x): x  C} for C  X (B) odd and is strictly decreasing in (–, )
and f–1(D) = {x : f(x)  D} for D  Y (C) odd and is strictly increasing in (–, )
for any A  Y and B  Y then- [IIT 2005] (D) neither even nor odd, but is strictly
(A) f–1(f(A)) = A increasing in (–, ) [C]
(B) f–1(f(A)) = A only if f(X) = Y
(C) f(f–1(B)) = B only if B  f(x) Q.272 The range of the function f(x) = cos [x], for –/2
(D) f(f–1(B)) = B [C] < x < 2 contains
(A) {–1, 1, 0} (B) {cos 1, 1, cos 2}
Q.268 If f  (x) = –f(x) and g(x) = f (x) and (C) {cos 1, –cos 1, 1} (D) [–1, 1]
2 2 Sol.[B] For the given range of x, we have
  x    x 
F(x) =  f    +  g   and given that [x] = –2 for –/2 < x < –1
  2  2   
 f(x) = cos (–2) = cos 2,
F(5) = 5, then F (10) is– [IIT 2006]
Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159 FUNCTION 25
 [x] = –1 for –1  x < 0 f(x) = cos (–1) = cos 1, Q.278 The range of k for which ||x–1|–5| = k have four
[x] = 0 for 0  x < 1  f(x) = cos 0 = 1, distinct solutions -
and [x] = 1 for 1  x < /2  f(x) = cos 1 (A) [0, 5] (B) (–, 5)
so that the range of f(x) is {cos 1, 1, cos 2} (C) [0, ) (D) (0, 5)
Sol.[D] f(x) = ||x–1| –5 |
Case I x  1 f(x) = |x – 1 – 5|
Questions Add (24–6-09) = |x – 6| =
x – 6 x  6

– x  6 1 x  6

Case II x < 1 f(x) = |x + 4|

Q.273 The image of the interval [–1, 3], under x  4 – 4  x 1
= 
 – ( x  4) x  –4
the mapping specified by the function
f(x) = 4x3 –12x is -
(A) [f(+1), f(–1)] (B) [f(–1), f(3)] (1,5)
(C) [– 8, 16] (D) [– 8, 72] [D]
y=k
Q.274 If b2 – 4ac = 0, a > 0 then the domain of the
function f(x) = log (ax3 + (a + b)x2 + (b + c)x +
c) is –4 1 6
 b 
(A) R –   Q.279 Let f(x) be a second degree polynomial function
 2a 
such that nf(x) > 0  x  R & the equation
 b   f '(x) + 786f(x), has no real roots. If g(x) = e786x
(B) R –    {x | x  1}
  2a   f(x), then -
(A) g(x) is an increasing function
 b   (B) f(x) is a decreasing function
(C) R –    (– ,–1]
 2a   (C) g(x) is an even function
(D) None of these [D] (D) the graph of g(x) cuts x-axis at least once.
Sol.[A] n (f(x)) > 0  f(x) > 1
Q.275 If f(x) is a function such that f(0) = 2, f(1) = 3  f '(x) + 786f(x) > 0,  x  R
and f(x + 2) = 2f(x) –f(x + 1) for xR, then  e786x f '(x) + 786e786x f(x) > 0
f(5) is - d
(A) 7 (B) 13 (C) 1 (D) 5 [B]  (f(x).e786x) > 0  x  R
dx
Q.276 Let f(x) =  g(x) is an increasing function
2n 2 n 1
a 2n x  a 2 n 1x  ...  a 1x  a 0
2n 2 n 1
, Q.280 The range of y = sin 3x + sin x is -
b 2n x  b 2 n 1x  ...  b1x  b 0
 8 8   4 4 
where nN, ai, bi R and b2n  0. If domain of (A)  ,  (B)  , 
f(x) is R, then  3 3 3 3  3 3 3 3
(A) f(x) = 0 has at least one real root  3 3   3 3 
(C)  ,  (D)  , 
(B) f(x) is one-one function  2 2  3 2 2 2
(C) f(x) is many one function
(D) None of these [C] Sol.[A] y = 3 sin x – 4 sin3 x + sin x = 4 (sin x – sin3 x)
dy
Now, = 4 cos x (1 – 3 sin2 x) = 0
dx
1
 sin x = ± we get maxima and minima
4 2  2 2 3
Q.277 If x = and y = , where is a
1 2 1 2 8 8
real parameter, then x2 – xy + y2 lies between value is and – .
3 3 3 3
(A) [2, 6 ] (B) [2, 4]
Q.281 If |f(x) + 6 –x2| = |4 –x2| + 2 + |f(x)| , then f(x) is
(C) [4, 6] (D) None of these [A]
necessarily non- negative in-
(A) [–2, 2] (B) (–, –2)  (2, )
Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159 FUNCTION 26
 (C) [– 6 , 6 ] (D) None of these  9 
(C)  ,0  (D) None
Sol.[A] Equation will be satisfied by f(x)  0  4 
and 4 – x2  0; So x  [–2, 2] x2 1 2
Q.282 The set of integer values of x satisfying Sol.[C] y =
x2 1
  
 1
 ( x  4) ( x  10) 2
 2 tan x  4  < 0 is- y=1–
  x2 1
x !  ( x  1) !
ymax. = 1 Range [–1, 1)
(A) {2, 3} (B) {5, 6, 7, 8, 9} ymin. = – 1
(C)  (D) None of these Now f(y) = y2 + y – 2 take minimum at
Sol.[B] x! – (x –1)!  0  x  I+ – {1} x = –1/2 so that value of f(y) 
1
2  / tan
–1
x
 4  0 as tan x </2   1  
( x  4) ( x  10) f  2 , f (1) 

 <0    
( x  1)! ( x  1)
 9 
 x  {5, 6, 7, 8, 9}   ,0 
 4 
Q. 283 Let A = (x1, x2,…..,x8), B = (y1, y2, y3), the total
x 1
no. of functions f : A  B that are onto and Q.286 The set of values of c so that f(x) =
there are exactly four elements (x) in A such c  x2  1
that does not take any value in the interval [–1, –1/3] is
f(x) = y3, is equal to (A) (–, –1/4] (B) [2, )
(A) 16 × 8C4 (B) 14 × 8C4 (C) (–1/4, 2) (D) (–, –1/4]  [2, )
(C) 16 × C44
(D) None of these x 1
Sol.[B] Four elements of A having image y 3 can be Sol.[A] y = f(x) =
c  x2  1
selected in 8C4 ways. Now, no. of ways in which
remaining 4 elements of A can be connected with 1 
y = – t where t   ,1
two elements of B is 24 – 2 = 14 ways 3 
 Total function are = 8C4 × 14 x 1
–t =
c  x2 1
x x x 1
Q.284 The function f(x) = sin – cos is x2 – c – 1 =
n! n  1! t
(A) not periodic
1 1
(B) periodic with period 2(n!) x2 – x  – c – 1 = 0
(C) periodic with period (n +1) t t
(D) None of these  1
As – t    1,  hence the above must not
x x  3
Sol.[D] f(x) = sin – cos
n! n  1! passes real solution
x 2 So D < 0
period of sin is × n! = 2 × n!
n!  1
2 1 
  –4   c  1 < 0
x 2 t  t 
period of cos is × n + 1! = 2 × (n +
n  1!  1 4
2
 + 4 < – 4c  c < –
1)! t t
period is LCM of (2 × n!, 2 (n + 1)!)
2
period is, 2 (n + 1)! 1 1 
  2
4 t 
x2 1 1 1
Q.285 y = then range of the expression y2 + y – Now t1–1 –21
x 12 3 t
2 is 1 1 1 2
–     2   0
(A) [–1, 1] (B) [0, 1] 4 4 t 
Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159 FUNCTION 27
  1 1 1
So value of c    ,   >x x<
 4 16 16
0 < x < 1/16
Q.287 If f(x + y) = f(x) . f(y) for all real x, y and f(0) 
f (x) Q.290 Range of y = sin 3x + sin x is-
0 then the function g(x) = 2 is  8 8   4 4 
1  {f ( x )} (A)  , ,
 (B)  
(A) even function  3 3 3 3 3 3 3 3 
(B) odd function  3 3 
(C) odd if f(x) > 0 (C)  ,  (D) None of these
2 3 2 3 
(D) neither even nor odd
Sol.[A]  y = sin 3x + sin x
Sol.[A] f(x + y) = f(x) f(y) put x = y = 0 than f(0) = 1
= 4 (sin x – sin3 x)
y = – x then f(0) = f(x) f(–x)  f(–x) = 1/f(x)
dy
f (x)  = cos x (1 – 3 sin2 x) = 0
g(x) = dx
1   f ( x ) 2 1
f (– x ) 1/ f (x)  sin x = ± ,±1
3
g(–x) = 2 
1   f (– x ) 1  1 / f ( x ) 2 1
find maxima/minima at x = ±
f (x) 3
 = g(x)
1   f ( x ) 2
Q.291 Let
A = {n: n is a multiple of 7 and n  200 and n  N}
Q.288 If a, b be two fixed positive integers such that B = {n:n is a multiple of 11 and n  300 & n  N}
f(a + x) = b + [b 3 + 1 – 3b2 f(x) + 3b {f(x)}2 – Then number of ordered pairs (a, b) such that
{f(x)}3]1/3 for all real x, then f(x) is a periodic
a  A, b  B, but neither A nor B belong to A  B -
function with period -
(A) 650 (B) 756 (C) 624 (D) 752
(A) a (B) 2a (C) b (D) 2b
Sol.[B] f(a + x) = b + [1 + b3 – 3b2 f(x) + 3b {f(x)}2  200 
Sol.[A] n(A) =   = 28
– {f(x)}3]1/3  7 
3 1/3
= b + [1 + {b – f(x)} ]  300 
f(a + x) – b = [1 – {f(x) – b}3]1/3 n(B) =   = 27
 11 
(a + x) = [1 – {(x)}3]1/3 x = f(x) – b
A  B = {77, 154}  n (A B) = 2
(2a + x) = [1 – {(x + a)}3]1/3 = (x)
 No. of ordered pairs (a, b)
f(x + 2) – b = f(x) – b  f(x + 2a) = f(x)
= 28–2C1. 27–2C1 = 26 × 25 = 650
f(x) is periodic with period 2a.
Q.292 If graph of y = ax 3 + bx2 + cx + d is symmetrical
Q.289 The domain of the function
about line x = k then-
  1   c
f(x) = log2   log1/ 2 1    1 is - (A) a + + k = 0 (B) k = c
  4 x   2b
(A) (0, 1) (B) (0, 1] c
(C) k = – (D) None
(C) [1, ) (D) (0, 1/16) b
 1  Sol.[A] If a polynomial is symmetrical about x = k then
Sol.[D] For f(x) to be defined log1/2 1   + 1 < 0
it has to be of even degree.
 4 x 
a=0
 1  1
 log1/2 1   < – 1  1 + >2  y = bx2 + cx + d is symmetrical about
 4 x  4 x c
x= 
1 2b
 >1
4 x c
k+ +a=0
Squaring 2b

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 Q.298 Given f(x) as a periodic function with period
Questions Add (1–7-09) 2 and it is defined as
 x 
f(x) = cos + 1; for 0 < x < 1
1 1 1  1 2   2 
Q.293 The sum         
 2   2 2000   2 2000  =2–x ; for 1  x < 2
Here [x] represents greatest integer  x. If
1 3   1 1999  f(0) = 1, which of the following be the graph of
 2  2000   ....   2  2000  y = f(x), for x[–2, 1]
   
y
(where [] denotes the greatest integer function)
is equal to - 1
(A) 1000 (B) 999 (C) 1001 (D) none
Sol. [A] (A) x
–2 –1 O 1
Q.294 Let f(x) = log2(log1/3(log7(sin x + a))) be defined
for every real value of x, then set of all possible
values of a is : y
(A) [2, 6) (B) (2, 6)
(C) [2, 6] (D) None of these 1
Sol. [B] (B) x
–2 –1 O 1
Q.295 The domain of the function
 log 0.3 ( x  1)
f(x) = is y
 x 2  2x  8
1
(A) (– 2, 4) (B) [2, 4)
(C) [1, 5] (D) None of these (C) x
–2 –1 O 1
Sol. [B]
x 1
Q.296 If f(x) = then f(ax) in terms of f(x) is
x 1
y
equal to - 3
f (x)  a 2
(A) (B)
1  af ( x )
(D) 1
x
(a  1) f ( x )  a  1 –2 –1 O 1
(C)
(a  1)f ( x )  a  1
(a  1) f ( x )  a  1
(D) None of these Sol. [D]
(a  1)f ( x )  a  1
Sol. [C]
Q.299 Let f: R  [–1, 1] and g : R  B, where R be
the set of all real numbers g(x) = sin–1
x2
Q.297 Let f : R  R such that f ( x )  . Which  f (x)  
x2 1  4  f 2 ( x )   .If g(x) & f(x) both
of the following not hold good ?  2  3
(A) f(x) is bounded are surjective, then set B is
(B) the line y = 0 is asymptote to f(x)     2 
(C) f(|x|) is not differentiable at x = 0 (A)  – ,  (B) 0, 3 
 3 3  
(D) range of f(x) contains 2 integers
Sol. [D]  
(C) 0,  (D) [0, ]
 3
Sol. [B]

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 (D) None of these
Q.300 Which one of the following statements is not Sol. [D]
correct ? x
(A) A line and a point not on it determine one Q.307 The domain of the function y = 2 sin is
2
and only one plane
(B) If two lines intersect, they lie in different   
(A)  n , ( 2n  1)  ; n  z
planes  2 
(C) Any three non-collinear points determine a
plane
(B)  4n , 2 (2n  1)  ; n  z
(D) If two distinct planes intersect, they   n  
intersect along a line (C)  , (2n  1)  ; n  z 
 2 4 
Sol. [B]
(D) None of these
Q.301 f : {1, 2, 3, 4, 5}  {a, b, c, d}. Total number of
Sol. [B]
onto functions ‘f ’, is equal to
(A) 242 (B) 245 (C) 1024 (D) 240 Q.308 Let f : R  R is defined by f(x) = sin ( n | x | )
Sol. [D]
; x  0 then f(x) 1 – (f ( y)) 2 +
Q.302 The value of 'c' for which the set {(x, y)|x 2 + y2 f ( y ). 1 – (f ( x )) 2 is
+ 2x  1} {(x, y)|x – y + c  0} contains only (A) f (x) + f (y) (B) f (x). f(y)
one point in common is - x
(A) (– , –1]  [3, ) (B) {–1, 3} (C) f (xy) (D) f  
y
(C) {–3} (D) {–1}
Sol. [C]
Sol. [D]
9x
Q.303 Find the set of values of '' for which the Q.309 Let f(x) = then f(x) + f(1 – x) =
9x  3
x 2  6 x – 8 (A) 9x (B) 1/9x (C) 1/9 (D) 1
expression y have a
  6 x – 8x 2 Sol. [D]
common linear factor in numerator and Q.310 The domain set of the function f(x) = log7 log5
denominator log3 log2 (2x3 + 5x2 – 14x) is
(A) {14} (B) {2}
(A) (– , ) – {0} (B) (0, )
(C) {–8, 2, 14} (D) {0, 2, 14}
(C) (– 4, 1/2)  (2, ) (D) (– 4, –1/2)  (2, )
Sol. [C]
Sol. [D]
x x
Q.304 Value of x satisfying = is/are Q.311 The range of the function log2(2 – log 2 (16
|x| |x|
2
sin x + 1)) is
(A) x  R (B) x  R – {0}
(A) (– , 0] (B) [0, )
(C) x  R+ (D) x  R–
Sol. [C] (C) (– , 1] (D) [0, 1]
Sol. [C]
Q.305 If [x] = [], where [.] denotes greatest integer Q.312 Domain of the function
function, then
1
(A) x  (3, 4] (B) x  [3, 4) cos x 
f(x) = 2 is
(C) x = 3 (D) none of these
2
Sol. [D] 6  35x  6 x
 5 
(A) [2n, 2n + /3]  2n  ,4n
  1 2   3 
Q.306 Find the domain of f(x) = sin–1 log 2  x 
  2   1 
(B)  , 6
(A) ( – – 2 ) ( 2, )  6 
(B) [– 1, 1] 1  
  5 
(C) ( – 2, – 1) ( 1, 2) (C)   ,  , 6
 6 3  3 
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 (D) None of these
Sol. [C] Q.319 If f is a strictly decreasing function with range
Q.313 Domain set of the function f(x) = [a, b], its domain is
(A) [b, a] (B) [f–1(a), f–1(b)]
3 log 64 x – 1 –1 –1
is (C) [f (b), f (a)] (D) none of these
3
2 x – 11 Sol.[C] ‘f ’ is one-one so inverse exist.
(A) [4, )
 11  Q.320 The period of the function
(B)  , 
 2   4 x  3   4 x  3 
4 sin4  2
 + 2 cos   is -
 11   11   6   3 2 
(C)  4,    ,
 2   2  3 2 3 3
(A) (B) (C)
(D) None of these 4 4
Sol. [C]
3 3 4 3
Q.314 The domain of the function f(x) = log2log3log4/ (D)
2 3
(tan–1x)–1 is
2
(A) R (B) (4/ )  2  4 x  3  
(C) (0, 1) (D) None of these
Sol.[B] f(x) = 2 sin   + 2 cos
  6 2  
Sol. [C]
 4 x  3 
Q.315 Which of the following sets of ordered pairs
 
 3 2 
define a one to one function ?
(A) f = {(x, y) ; x2 + y2 = 2} 3 1  8x  6 
= + cos  
(B) A = {1, 2, 3} , B = {1, 2, 3, 4, 5} 2 2  3 2 
f = {(x, y) ; 5x + 2y is a prime number,
3 3
xA, yB} T=
4
(C) A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5, 6, 7} and
f = {(x, y) ; y = x2 – 3x + 3, xA, yB} Q.321 Let f : {x, y, z} {1, 2, 3} be a one-one
mapping such that only one of the following
(D) None of these
three statements is true and remaining two are
Sol. [D]
false : f(x)  2, f(y) = 2, f(z)  1, then:
Q.316 Let f(x) = sin23x – cos22x and g(x) = (A) f(x) > f(y) > f(z) (B) f(x) < f(y) < f(z)
(C) f(y) < f(x)< f(z) (D) f(y) < f(z) < f(x)
1
1 tan 1 | x | , then the number of values of Sol.[C] Let f(x)  2 be true and f(y) = 2, f(z)  1 are
2 false
x in interval [– 10, 20] satisfying the equation  f(x)  2, f(y)  2, f(z) = 1
f(x) = sgn(g(x)) is  f(x) = 3, f(y) = 3, f(z) = 1 but then function is
(A) 6 (B) 15 (C) 10 (D) 20 many one, similarly two other cases.
Sol. [B]
Q.322 If a2 + b2 + c2 = 1, then ab + bc + ca lies in the
Q.317 The equation of the image of the pair of rays y = interval :
|x – 1| by the line, x = 0 is  1   1
(A)  – ,1 (B) 0, 
(A) |y| = x + 1 (B) y = |x + 1|  2   2
(C) y = |x| +1 (D) None of these (C) [0, 1] (D) [1, 2]
Sol. [B] Sol.[A] (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)  0
Q.318 Let [x] denotes greatest integer  x, then –1
ab + bc + ca 
equation sin x = [1 + sin x] + [1 – cos x] has - 2
(A) one solution in [0, /2] Also a2 + b2 + c2 – ab – bc – ca
(B) one solution in [/2, ] 1
= [(a – b)2 + (b – c)2 + (c – a)2] 
(C) one solution in [–/2, 0] 2
(D) no solution for any x  R 0
Sol. [D]
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 So ab + bc + ca  a2 + b2 + c2  1 (D) None of these
Q.323 | – 2x2 + 1 + ex + sin x| = |2x2 – 1| + ex + |sin x| if Sol.[B] If y = f(x) is symmetric about x = a,
& only if x belongs to f(x + a)= f(a – x)
 1 
(A) 0,  (B) R
 2 tan 1 x
Q.327 If g(x) = n(1 + x) – , x > 0, then sgn
 1 1 x
1 
(C) [0, ] (D)  ,  (g(x)) is
 2 2
(A) 1 (B) – 1
Sol.[A] In R.H.S. each term is positive & ex > 0 (C) 0 (D) 4
So, 1 – 2x2  0 & sin x  0 Sol.[A] Consider f(x) =(x + 1) n (x + 1) – tan–1x
 1 1  1 1
x  ,  & x [2n (2n + 1)] f (x) = (x + 1) . + n(x + 1) –
 2 2  x 1 1 x2
(nI) x2
= + n(1 + x)
 1  1 x 2
 x 0, 
 2 f (x) > 0  x (0, )  f(x) is increasing
tan 2 x f(0) = 0
Q.324 Domain of function f(x) =
6 cos x  2 sin 2 x f (x)
g(x) = > 0  x  (0, )
is x 1
  sgn (g(x)) = 1
(A) R – (2n + 1) ; nI
2 Q.328 Let f(x) = 2x + 2 ; x  2
 x
(B) R – (2n + 1) ; n I = + 10 ; x < 2
4 2
   If f(x) is onto function, then  belongs to -
(C) R – (2n  1)  (2n  1)  ; nI (A) [1, 4] (B) [–2, 3]
 2 4
(C) [0, 3] (D) [2, 5]
(D) None of these Sol..[C]

Sol.[C] tan 2x is undefined for 2x = (2n + 1) or Q.329 Given the function f(x) = (ax + a–x)/2; (a > 2),
2 then f(x + y) + f(x – y) =
 (A) 2 f(x) . f(y) (B) f(x) . f(y)
x = (2n + 1) also 6 cos x + 2 sin 2x  0 or
4 f (x)
cos x (6 + 4 sin x)  0  cos x  0 (C) (D) None of these
f ( y)
(1  sin x ) t  1 Sol. [A]
Q.325 If f(x) = tLim
 then range
(1  sin x ) t  1
of f(x) is Q.330 Let f : R  [–1, 1] and g : R  B, where R be
(A) [– 1, 1] (B) [0, 1] the set of all real numbers. g(x) = sin –1
(C) {–1, 1} (D){– 1, 0, 1}  f (x)  
 4  f 2 (x)  + . If g(x) & f(x)
 2  3
Sol.[D] when sin (x) > 0 (1 + sin x) > 1 both are surjective, then set B is -
 (1 + sin x)t      2 
for sin x = 0 (1 + sin x)t = 1 (A)   ,  (B) 0,
 3 3  3 
for – 1  sin  x < 0 (1 + sin x)t  0
 
(C) 0,  (D) [0, ]
Q.326 Graph of y = f(x) is symmetrical about the line  3
x = 1 then Sol. [B]
(A) f(x) = f(– x) sin x  sin 3x
(B) f(1 + x) = f(1 – x ) Q.331 Period of f(x) = is
cos x  cos 3x
(C) f(x + 1) = f(x – 1)
(A)  (B) /2 (C) /4 (D) 2
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Sol. [A] 3 t 5/ 2 3
Q.332 Let f : R  R be a continuous & differentiable c=  g(t) = 
5 5/ 2 5
function such that  f ( x 2  1)  x
 5 for x ( 4) 5 / 2 3 67
g(4) =  
4 5/ 2 5 5
  2 
(0,) then the value of  f  16  y
y
 for
  y 2 
   Q.337 If |f(x) + 6 – x2| = |4 – x2| + 2 + |f(x)|, then f(x) is
y(0, ) is equal to- necessarily non-negative in -
(A) 5 (B) 25 (A) [–2, 2] (B) (– , –2) (2,)
(C) 125 (D) 625 (C) [– 6 , 6 ] (D) None of these
Sol. [B]
Sol.[A] Equation will be satisfied by f(x)  0
Q.333 If f(x) = 5 log5x then f –1 ( – ) where  R
is equal to- and 4 – x2 so x  [–2, 2]

f 1 () Q.338 Let f be an even periodic function with period 2

(A) f –1 () – f –1 () (B) 1 and f(x) = x  x [0, 1]. Then f(3.14) equals
f ()
(A) –3.14 (B) –0.14
1 1 (C) 0.14 (D) 0.86
(C) (D)
f (   ) f ()  f () Sol.[D] f(3.14) = f(3.14 – 4) = f(–0.86) = f(0.86) = 0.86
Q.334 If f(x) = 5 log5x then f –1 ( – ) where  R Q.339 If f(x) = x2 + bx + c and f(2 + t) = f(2 – t) for all
is equal to- real numbers t, then which of the following is
f 1 () true
(A) f –1 () – f –1 () (B) (A) f(1) < f(2) < f(4) (B) f(2) < f(1) < f(4)
f 1 ()
(C) f(2) < f(4) < f(1) (D) f(4) < f(2) < f(1)
1 1
(C) (D) Sol.[B] at t = 1
f (   ) f ()  f ()
Sol.[B] f(x) = 5log5x  f–1 (x) = 5x/5

–1
  5 / 5 f 1 ( )
f ( –) = = =
5 5 5 / 5 f 1 () x=2
f(3) = f(1) < f(4) so f(2) < f(1) < f(4)
Q.335 Let [x] represents greatest integer  x. If
Q.340 Let f(x) be a continuous function in R such that
2 2
[ n   ]  [ n  1]  2 where , n N,
then  can assume f(x) does not vanish for all x  R.
(A) (2n + 4) different values 5 5

(B) (2n + 5) different values If  f ( x ) dx =  f ( x ) dx , then in R,

(C) (2n + 3) different values 1 1

(D) (2n + 6) different values f(x) is -

Sol. [B] (A) an even function
Q.336 A function g defined for all real x > 0 satisfies (B) an odd function
g(1) = 1, g (x2) = x3 for all x > 0, then g(4) equals (C) a periodic function with period 5
13 (D) none of these
(A) (B) 3
3
5 5
67
(C)
5
(D) none of these Sol.[D] Using 
1
f ( x ) dx   f ( x ) dx
1
=0 

Sol.[C] g (x2) = x3 1
 g (t) = t3/2, where x2 = t
5/ 2
 f ( x ) dx =0
1
t
 g(t) = +c
5/ 2 sin( [ x ])
Q.341 The range of the function f(x) =
(1) 5 / 2 x2 1
 g(1) = +c (where [·] denotes greatest integer function) is:
5/ 2
(A) 0 (B) R
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 (C) (0, 1) (D) None of these rx
sin [ x ] Sol.[B] f(g(x)) = 1  ( r  1) x , g (f(x)) = rx
Sol.[A] f(x) = [x] I
x2 1 f(g(x)) = g(f(x)) for infinite many real x
r = 0, 1 two values
sin [ x ]
f(x) = =0 Q.347 If f(x) = px + q and f(f(f(x)) = 8x + 21 where p
x2 1
Q.342 The domain of the function and q are real numbers, then p + q equals -
(A) 3 (B) 5 (C) 7 (D) 11
x
f(x) = x 2  2 | x |  sin 1 is Sol.[B] f(f(f(x))) = p (p2x + pq + q) + q
4 = p3x + p2q + pq + q
p =8  p=2
3
 p2q + pq + q = 21
(A) (–, – 2] [2, ) (B) [–4, 4] 5q + 2q = 21  q = 3
(C) [–4, – 2] [2, 4] (D) None of these p + q = 5
Q.348 If f(x) is a function from R  R, we say that f(x)
x
Sol.[C] (i)  [–1, 1] (ii) x2 – 2|x|  0 has property (I) If f(f(x)) = x for all real numbers x
4 and we say that f(x) has property (II) If f(– f(x)) = –
Q.343 Which of the following functions are not x for all real number x. How many linear functions
have both property I and II ?
bounded (A) exactly one (B) exactly two
2x (C) exactly three (D) infinite
(A) f(x) = , [–2, 2] Sol.[B] Two functions f(x) = x and f(x) = –x
1 x2
|x| |x|
1  cos x Q.349 f(x) = e{e sgn x} , g(x) = e[ e sgn x ] , x  R
(B) f(x) = , [– 2, 2] – {0} where {x} and [x] denotes the fractional part
x2
and integral part functions respectively. Also
x 3  8x  6 h(x) = n f(x) + n g(x) then for all real x, h(x)
(C) f(x) = , [0, 5] is -
4x  1
(A) odd function
(D) none of these (B) even function
Sol.[D] f(x) is not bounded if f(x)  or – . (C) neither odd nor even function
(D) both odd as well as even function
Sol.[A] h(x) = e|x| sgnx = ex x>0
Q.344 Which of the following statements are incorrect ? =0 x = 0  odd function
I. If f(x) and g(x) are one to one then f(x) + = – e–x x < 0
Q.350 The ‘x’ for which sinx(sinx + cosx) = [x] where
g(x) is also one to one.
[] denotes greatest integer function is
II. If f(x) and g(x) are one-one then f(x). g(x)
(A) [0, 2) (B) [0, 1] [2, 3)
is also one-one
(C) [–1, 1)  [1, 2) (D) None of these
III. If f(x) is odd then it is necessarily one one Sol.[A] sin2x + sin x cos x = [x]
(A) I and II only (B) II and III only 1  cos 2x sin 2 x
 = [x]
(C) III and I only (D) I, II and III 2 2
sin2x – cos2x = 2| x | – 1
Sol.[D]
2 {sin2x cos /4 – cos2x sin /4} = 2[x] – 1
Q.345 The function f is one to one and sum of all the = 2 {sin (2x – /4)} = 2(x) – 1
intercepts of the graph is 5. The sum of all of 1 2
the intercept of the graph of y = f–1 (x) is - – 2  {2| x | – 1}  2  2  [x] 
1 2 1 2
(A) 5 (B) (C) (D) –5  [x] = 0.1  x [0, 1)  [1, 2)
5 5 2
Sol.[A] Obviously. Q. 351 If f(x + 2a) = f(x – 2a), then f(x) is-
x rx
Q.346 Let f(x) = and g(x) = . Let S be (A) a periodic function with period 4a
1 x 1 x
the set of all real numbers r such that f(g(x)) = (B) a periodic function with period 2a
g(f(x)) for infinitely many real numbers x. The
(C) a periodic function with indeterminate
number of elements in set S is -
(A) 1 (B) 2 (C) 3 (D) 5 period
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 (D) a non periodic function nx
Q.357 Range of f(x) = is
Sol.[A] f(x + 2a) = f(x – 2a) x
x  x + 2a (A) (–, e) (B) (–,
f(x) = f(x + 4a) ; T = 4a e2]
tan n x (C) (–, 2/e] (D) (–, 1/e)
2n
Q.352 Let f(x) = , n N where x  Sol.[C]
 tan
r 0
r
x
Q.358 Which of the following is even function

a x 1 a x 1
  (A) f(x) =  (B) f(x) = x 
0, 2  a x 1 a x 1
 
(A) f(x) is bounded and it takes both of it's a x  a x
(C) f(x) = (D) None of these
bounds and the range of f(x) contains a x  a x
exactly one integral point
a x 1
(B) f(x) is bounded and takes both of it's Sol. [A] f(x) = a x  1
bounds and the range of f(x) contains
a x  1 1 a x
more than one integral point. f(–x) = a  x  1 
1 a x
(C) f(x) is bounded but minimum and = –f(x)
maximum does not exists.
Q.359 The Domain of the function
(D) f(x) is not bounded as the upper bound not
16 –x 20 – 3x
f(x) = C2x – 1 + P4x – 5 where the symbols
exist.
Sol.[A] have their usual meaning, is the set
Q.353 Real values of x for which xnx –x + 1 > 0, are- (A) {2, 3} (B) {2, 3, 4}
(A) (0, ) (B) (0, 1)  (1, ) (C) {1, 2, 3, 4, 5} (D) None of these
(C) (1, ) (D) (0, 1) Sol. [A] 16 – x  2x – 1 & 20 – 3x  4x – 5
Sol.[B] 3x  17 7x  25
Q.354 If f(x) = 4x 3 – x 2 – 2x + 1 and x  17/3 x  25/7
 Min  f ( t ) :0  t  x ; 0  x  1 x  25/7 xz
g(x) =  x = 2, 3 (2x – 1  0, & 4x  0)
3 – x ; 1 x  2
then Q.360 f : {1, 2, 3, 4, 5}  {a, b, c, d}. Total number of

1 3 5 onto function f, is equal to –

g   + g   + g   the value equal to :
4 4 4 (A) 242 (B) 245 (C) 1024
7 9 13 5 (D) 240
(A) (B) (C) (D)
4 4 4 2 Sol. [D]
Sol.[D] Q.361 Graph of f(x) = loga ( x  x 2  1) (a > 0,  1)
Q.355 Number of roots of the equation x2.e2–| x | = 1 is:
(A) 2 (B) 4 (C) 6 (D) zero is symmetric about
Sol.[B] (A) x-axis (B) origin
Q.356 Number of solution of the equation (C) y-axis (D) y = x

  Sol.[B] f(x) is an odd function

3 tan x + x3 = 2 in  0,  is ax  b
 4 Q.362 Let f(x) = then f [f(x)] = x provided
cx  d
(A) 0 (B) 1
that
(C) 2 (D) 3 (A) d = – a (B) d = a
Sol.[B] (C) a = b = 1 (D) a = b = c = d = 1

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  ax  b  Q.368 The range of f(x) = cot–1(–x) – tan–1 x + sec–1 x
a b is :
 cx  d  x
Sol.[A] fof(x) =   3   
 ax  b  (A)   ,  (B)  ,   
c d  2 2  2 
 cx  d 
Solving it  3 
 ,
(ac + dc)x2 + (bc + d2 – bc – a2)x – ab – bd = 0  2 

ac + dc = 0, a2 – d2 = 0, ab + bd = 0    3   
so a = – d (C)  ,  (D)  ,   
 2 2  2 
VID ********** 6/6/10 (363 onward)  3 
 , 
 2 
x 2  3x  2
Q.363 The value of the function f(x) = Sol. [B] f(x) =  – (tan–1x + cot–1x) + sec–1x
x2  x  6

lies in the interval - = + sec–1x
2
1 
(A) (– , ) –  , 1 (B) (– , )   
,     ,
3 
5  Range of f(x) = 
2   2 

(C) (– , ) – {1} (D) None of these
 
Q.364 The function f : [2, )  Y defined by Q.369 Let f : R   0, be defined as
 6

f(x) = x2 – 4x + 5 is both one–one & onto if:
(A) Y = R (B) Y = [1, )  4 
f(x) = sin–1  2
 then f(x) is:
(C) Y = [4, ) (D) Y = [5, )  4 x  12 x  17 
   (A) injective as well as surjective
Q.365 If A =  x :  x   and
 6 3 (B) surjective but not injective
f(x) = cos x – x (1 + x) then f(A) is equal to- (C) injective but not surjective
(D) neither injective nor surjective
    
(A)  ,  (B)   ,   4 
6 3  3 6 Sol. [B] f(x) = sin–1  2


 ( 2 x  3 )  8 
(C)
1 4 1
  3    
 f is surjective
  1   ,  1    ( 2 x  3) 2  8 2
2
 3 3 2 6 6 

(2x – 3)2 = 1 for two values of x so f is not
1    3    injective.
(D)   1   ,  1   
2 3 
 3 2 6 6 

Q.366 If f: [–20, 20]  R is defined by Q.370 Domain of definition of the function
 x2 
f(x) =   sin x + cos x, is an even function, f(x) = 3 cos 1 ( 4 x )   is equal to-
 a 
 1 1 1 
then the set of values of a is- (A)   ,  (B)  ,1
 4 8 8 
(A) (–, 100) (B) (400, )
(C) (– 400, 400) (D) None of these 1 1 
(C)  ,  (D)
Q.367 If A be the set of all triangles and B that of 8 4 
positive real numbers, then the mapping  1
f : A  B given by  1, 8 
 
f() = area of , (  A) is -
(A) one one into mapping 
Sol. [A] cos–1 (4x) 
(B) one one onto mapping 3
(C) many-one into mapping  1
(D) many-one onto mapping 4x    1, 
 2

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  1 1 1
x    , (A) [f(x + y) + f(x – y)]
 4 8  2
1
(B) [f(2x) + f(2y)]
Q.371 Which of the following relations is a function- 2
(A) {(1, 4), (2, 6), (1, 5), (3, 9)} 1
(B) {(3, 3), (2, 1), (1, 2), (2, 3)} (C) [f(x + y). f(x – y)]
2
(C) {(1, 2), (2, 2), (3, 2), (4, 2)}
(D) none of these
(D) {(3, 1), (3, 2), (3, 3), (3, 4)}
1
Sol. [C] (A constant function) Sol. [B] f(x + y)f(x – y) = [(2x+y + 2–x–y)(2x–y + 2–
4
Q.372 If x, y  {1,2,3,4} then which of the following
x+y 1 2x 2y –2y –2x
relations is a function- )] = [2 + 2 + 2 + 2 ]
4
(A) {(x, y) |y = x + 2} (B) {(x, y) | y < x + 5}
(C) {(x, y) | x + y > 4} (D) {(x, y) | x + y = 5} 1  2 2 x  2 –2 x 2 2 y  2 –2 y  1
=   = [f(2x)
Sol. [D] Try yourself (very simple) 2  2 2  2
+f(2y)]
Q.373 The number of functions that can be defined
2x 
from B to A when A = {x1 , x2 , x3 , x4} and Q.376 If f(x) = log   , 0 < a < 2 then
2–x
B = {5 , 6, 7} is –
(A) 4 (B) 64 (C) 81 (D) 1  8a 
105 f =
2  4  a2 
Sol. [B] The number of functions defined from B to
(A) f(a) (B) 2f(a)
A = n(A)n(B) = 43 = 64.
1
(C) f(a) (D) –f(a)
Q.374 Which one of the following is not a function 2
(A) { (x,y) : x , y R, x2 = y }
 8a 
(B) { (x,y) : x , y  R, y2 = x }  2 
(C) { (x,y) : x , y  R, x3 = y } Sol. [A]
1  8a 
f  =
1
log  4  a2 
(D) { (x,y) : x , y  R, y3 = x }
2  4  a2  2  8a 
2– 
Sol. [B] Let R1 = {(x, y) : x , y R, x2 = y }  4  a2 
R1 is a function because for each x  R, x2 is 1 2( a 2  4a  4) 1
uniquely determined. = log 2
= log
2 2( a – 4a  4) 2
Let R2 = {(x, y) : x , y R, y2 = x}
2
we have ( 9 , 3) , (9, –3)  R2 a2
 
 9 has two images 3, –3 a –2
 R2 is not a function 1 a2
= . 2log = f(a)
Let R3 = {(x, y) : x , y R, x3 = y } 2 a–2
R3 is a function because for each x  R, x3 is Q.377 If f(x) + 2 f(1 –x) = x2 + 2, x  R then f(x) is
uniquely determined. Let R4{(x,y): x, y  R, given as -
y3=x}
( x – 2) 2
R4 is a function because for each x R, x1/3 is (A) (B) x2 –2
3
uniquely determined in R
(C) 1 (D) none of these
Sol. [A] f(1 – x) + 2 f(x) = (1 – x)2 + 2
2x  2–x f(x) + 2f (1 – x) = x2 + 2
Q.375 If f(x) = , then f(x + y). f(x – y) is
2
( x – 2) 2
equal to- Solving, we get f(x) = .
3

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Q.378 If f(x + y, x – y) = xy, then the arithmetic mean {(1, 4), ( 2, 19), (3, 67), (4, 259)}
of
f(x, y) and f(y, x) is – Q.381 If f(x) is a function which is odd and even
(A) y (B) x simultaneously, then f(3) – f(2) is equal to –
(C) 0 (D) none of these (A) 1 (B) –1
Sol. [C] Let x + y = a and x – y = b (C) 0 (D) none of these
ab a–b Sol. [C] f(x) = 0; x  R  f(3) – f(2) = 0
x= and y =
2 2
Q.382 If f :R  R, f(x) = ex, then correct statement is–
 f(x + y, x – y) = xy
(A) f(x + y) = f(xy)
a  b a – b a 2 – b2
 f(a, b) = .  (B) f(x + y) = f(x) f(y)
2 2 4
(B) f(x + y) = f(x) + f(y)
x 2 – y2 y2 – x 2 (D) none of these
 f(x, y) = and f(y, x) =
4 4 Sol. [B] Obviously
 Arithmetic mean of f(x, y) and f(y, x) is zero
1 Q.383 Which of the following is/are odd function(s)?
Q.379 If f(x) = +
x  2 2x – 4 (A) x. log {x + 1 x2 }

1 (B) f(x) where f(x + y) = f(x) + f(y); x,y R

for x > 2, then f(11) = (C) x2 sin hx
x – 2 2x – 4
(D) None
7 5 6 5
(A) (B) (C) (D) Sol. [C]
6 6 7 7
1
Sol. [C] f(x) = + Q.384 Function f : C  R, f(z) = |z| is-(where C is set
x  2 2x – 4
of complex number)
1 (A) one-one (B) onto
x – 2 2x – 4 (C) one-one onto (D) many one into
1 1 Sol. [D] |2 + 3i| = |2–3i| many & |z|  0  into
f(11) = +
11  2 18 11 – 2 18
2x 
1 1 3– 2 3 2 Q.385 If f(x)= log   ,
= + = + 2–x
3 2 3– 2 7 7
1  8a 
6 0 < a < 2 then f  =
= 2  4  a2 
7
(A) f (a) (B) 2f (a)
Q.380 If f = {(1 , 1 ), ( 2 , 4), ( 3, 8), (4, 16) } then 1
(C) f (a) (D) –f (a)
f2 + 3 = 2
(A) { (1 , 4) , ( 2, 7), ( 3 , 11), ( 4 , 19) }
 8a 
(B) { (1 , 1) , ( 2, 16), ( 3 , 64), ( 4 , 256) }  2 
(C) { (1 , 4) , ( 2, 19), ( 3 , 67), ( 4 , 259) } Sol. [A]
1  8a 
f  =
1
log 
4  a2 
(D) { (1 , 4) , ( 2, 11), ( 3 , 17), ( 4 , 35) }
2
2  4a  2  8a 
2– 
 4  a2 
Sol. [C] f (x) = [f (x)]2 + 3
1 2( a 2  4a  4) 1
f (1) = [f (1)]2 + 3 = 1 + 3 = 4 = log 2 = log
2 2( a – 4a  4) 2
f (2) = [f (2)]2 + 3 = 42 + 3 = 19
f ( 3)= [f(3)]2 + 3 = 82 + 3 = 67 2
a2
f (4) = [f(4)]2 + 3 = 162 + 3 = 259  
a –2
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 1 a2 1 1 1 1
= . 2log = f(a)   2 >0 –  2 <0
2 a–2 2 x 2 2 x 2
Q.386 If f(x) + 2 f(1 –x) = x2 + 2, x  R then f(x) is 1 1 
  1– < 1 sin–1
given as – 2 2
x 2 6
2
( x – 2)
(A) (B) x2 – 2  1  
3 1 – 2  <
 x 2 2
(C) 1 (D) none of these
Sol. [A] f(1 – x) + 2 f(x) = (1 – x)2 + 2
Q.390 If f(x) = a cos (bx + c) + d , then range of f(x)
f(x) + 2f (1 – x) = x2 + 2
is
( x – 2) 2 (A) [d + a,d + 2a] (B) [a – d,a + d]
Solving, we get f(x) = .
3 (C) [d + a,a – d] (D) [d – a, d + a]
Q.387 If f(x + y, x – y) = xy, then the arithmetic mean Sol. [D] f(x) = a cos (bx + c) + d ….. (i)
of f(x, y) and f(y, x) is For minimum cos (bx + c) = – 1
(A) y (B) x from (i), f(x) = – a + d = (d – a) ,
(C) 0 (D) none of these for maximum cos(bx + c) = 1
Sol. [C] Let x + y = a and x – y = b from (i), f(x) = a + d = (d + a)
ab a–b  Range of f(x) = [d – a, d + a].
x= and y =
2 2
 f(x + y, x – y) = xy  2 
Q.391 The range of f(x) = sec  cos x  , –  < x <
ab a –b a –b 2 2  4 
 f(a, b) = . 
2 2 4  is -
x –y 2 2
y –x 2 2 (A) [1, 2 ]
 f(x, y) = and f(y, x) =
4 4 (B) [1, )
Arithmetic mean of f(x, y) and f(y, x) is zero (C) [– 2 , –1][1, 2 ]
(D) (– ,–1][1,)
Q.388 Domain of the function  2 
1 Sol. [A] f(x) = sec  cos x 
 4 
f(x) = 10 contains the points
C x –1 – 3.10 C x We know that, 0 cos2 x  1 at cos x = 0,
(A) 9, 10, 11 (B) 9, 10, 12 f(x) = 1 and at cos x = 1, f(x) = 2
(C) all natural numbers(D) none of these 1x 2  x  [1, 2 ]

Sol. [D] Given function is defined if 10Cx–1 > 310Cx

Q.392 The range of cos2 x + sin4 x is –
1 3
 >  4x > 33 3 
11 – x x (A)  ,1 (B) [ 0 , 1]
4 
x  9 but x  10  x = 9, 10.
 3
(C) 0,  (D)
 4
 x 1  2
Q.389 Range of sin–1  2  is:

x 2  – 3
 – 1, 4 
(A)[0, /2] (B) (0, /6)  
(C) [/6, /2) (D) none of these
Sol [A] f(x) = cos2x + sin4x
2
x 1 1 = cos2x + sin2x (1 – cos2x)
Sol. [C] Here, =1–
x2  2 x2  2 1 3
=1– sin2x  f(x)  1
Now, 2  x2 + 2 <  for all x  R 4 4

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Q.393 Let f : R – {n}  R be a function defined by
Q.396 Let f(x) be a function whose domain is [–5, 7].
x–m Then the domain of f(|2x + 5|) is –
f(x) = , where m  n . This function is-
x–n (A) [– 6, 1] (B) (– 6, 1]
(A) one-one onto (B) one-one into (C) [– 6, 1) (D) None of these
(C) many-one onto (D) many one into Sol. [A]  – 5  |2x + 5|  7
x–m |2x + 5|  – 5 and |2x + 5|  7
Sol. [B] We have f(x) = , x R – {n}
x–n Always true and – 7  2x + 5  7
Let f(x1) = f(x2) for x1 , x2 R – {n} –6x1
x1 – m x2 – m Hence domain is [– 6, 1].
 =
x1 – n x2 – n Hence (A) is the correct answer.
 x1x2 –nx1 – mx2 + mn = x1x2 –nx2 – mx1 + mn 2x  2 – x
Q.397 If f(x)= ,then f(x + y). f(x–y) is equal
 (m–n)x1 = (m – n)x2  x1 = x2 2
 f is one-one to-
x–m 1
f(x) = k  =k (A) [f(x + y) + f(x – y)]
x–n 2
kn – m 1
x – m = kx – kn  x = (B) [f(2x) + f(2y)]
k –1 2
1  R and for k = 1, x is not a real no. 1
(C) [f(x + y). f(x – y)]
 f is not onto 2
(D) none of these
Q.394 If a function f : [2 , )  B defined by Sol. [B]
f(x) = x2 – 4x + 5 is a bijection, then B is equal 1
to f(x + y)f(x – y) = [(2x+y + 2–x–y)(2x–y + 2–x+y)]
4
(A) R (B) [1, )
1 2x 2y –2y –2x
(C) [4, ) (D) [5, ) = [2 + 2 + 2 + 2 ]
4
Sol. [B] We have f(x) = x – 4x + 5, x  [ 2 , )
2

 f(x) = (x – 2)2 + 1  1  R (f)  [1 , ) 1  2 2 x  2 –2 x 2 2 y  2 –2 y 

=   
Let k  [1 , ) and f(x) = k 2  2 2 
x2 – 4x + 5 = k x2 – 4x + 5 – k = 0 1
= [f(2x)+f(2y)]
 (x–2)2 = k –1 x – 2 = ± k –1 2
x=2± k –1  x = 2 + k – 1  [2, ) cos x
 [1, )  R(f)  R(f) = [1, ) Q.398 f(x)=  2 x  1 , where x is not an integral
   2
 
   multiple of  and [.] denotes the greatest integer
Q.395 If 1 + 2x is a function having  – ,  as
 2 2 function is -
domain and (– , ) as codomain, then it is - (A) an odd function
(A) onto but not one-one (B) even function
(B) one-one but not onto (C) neither odd nor even
(C) one-one and onto (D) none of these
(D) neither one-one nor onto  2x  1   2x  1 
Sol. [A] Clearly  –  =–     
      2    2 
Sol. [B] Let f(x) = 1 + 2x, x   – , 
 2 2  f(x) is an odd function.
f(x1) = f(x2) = 1 + 2x1 = 1 + 2x2  x1 = x2
 f is one-one
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 1
Sol.[B] log must be + ive and |sin x|  0 or sin
Q.399 Let the function f : R  R be defined by | sin x |
f(x) = 2x + sin x, x  R. Then f is -
x  0 or x  n.
(A) one-to-one and onto
(B) one-to-one but not onto  D = R – {n}.
(C) onto but not one-to-one
(D) neither one-to-one nor onto Q.403 If [x] and {x} represent integral and fractinal
Sol. [A] f(x) = 2x + sin x  f '(x) = 2 + cos x > 0
parts of x then the expression
 f(x) is strictly increasing in R more over
lim f(x) = ±  Range is R so it is one-one 2000
{x  r}
x 
[x] +  2000
is equal to -
onto r 1

Q.400 Let f(x) = cos p x, where p = [a] = The (A) 2001x/2 (B) x + 2001
greatest Integer less than or equal to a. If the (C) x (D) [x] + 2001/2
period of f(x) is  then- Sol.[C] Let x – [x] = P
(A) a  [4, 5] (B) a = 4, 5 {x + r} = x + r – [x + r] = x + r – [x] – r = x – [x] = P.
(C) a  [4, 5) (D) None 2000
{x  r} 2000
x  [x]
 [x] +  = [x] + 
2 r 1
2000 r 1
2000
Sol.[C] The period of f(x) = =  (from the
p
( x  [ x ]) 2000
question) [x] +
2000
 1 = [x] + x – [x] = x.
r 1
 p =2
p=4 Q.404 Find the domain of the definition of the function
[a] = 4  4  a < 5 f(x) = log4 (log5 (log3 (18x – x2 – 77)))
(A) x  (12, 20) (B) x  (8, 10)
Q.401 The domain of definition of the function f(x) = (C) x  (20, 25) (D) None of these
Sol.[B] Since log x is defined for x > 0. Therefore,
f(x) = log4 (log5 (log3 (18x – x2 – 77))
sin 1 x  x2 1  x  [ x ]  log x
log5 (log3 (18 x – x2 – 77)) > 0
  1  is -
e sin x  cos x
 log sin  
 log3 (18x – x2 – 77) > 50  (18x – x2 – 77) > 31
  
   x2 
 x2 – 18x + 80 < 0  (x – 8) (x – 10) < 0
(A) (–1, 1) (B) (0, 1)
 8 < x < 10  x  (8, 10)
(C) (1, 0) (D) None of these
Hence, the domain of definition of the given
1
Sol.[D]  is not defined for any real ‘x’, function is (8, 10).
 x2

 domain of the function is null set.

Q.405 A polynomial function f(x) satisfies the
condition f(x) f(1/x) = f(x) + f(1/x).
Q.402 The domain of the function
If f(10) = 1001, then f(20) =
f(x) = log(1 / | sin x |) is - (A) 2002 (B) 8008
(A) R – {2n, n  z} (B) R – {n, n  z} (C) 8001 (D) None of these
(C) R – {–, } (D) None of these Sol.[C] f(x) f(1/x) = f(x) + f(1/x)  f(x) = xn + 1
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 Now, f(10) = 1001  10n + 1 = 1001
Sol.[D] Checking the options
 n = 3,  f(x) = x3 + 1
(a) LHS = x sgn x = x × (+1) = x if x > 0
 f (20) = 203 + 1 = 8001.
x×0 =0 if x = 0
x × (–1) = –x if x < 0
Q.406 If f(x) = (ax2 + b)3 , then the function g such that
= |x| = RHS
f(g(x)) = g(f(x)) is given by -
 correct
 b  x1 / 3 
(A) g(x) =   (B) g(x) = (b) LHS = |x| sgn x = (+x) × (+1) = x if x > 0
a 
  0×0=0 if x = 0
(–x) × (–1) = x if x < 0  correct
1
(c) LHS = (x(sgn x)) × sgn x = |x| sgn x (from (a))
(ax 2  b) 3
=x (from b)
(C) g(x) = (ax2 + b)1/3 (D) g(x) =
= RHS  correct
1/ 2
 x1 / 3  b 
(d) LHS = (|x| sgn x) (sgn x) = x (sgn x)2 2

 
 a  (from (b))
 
1/ 2
= |x| × (sgn x) (from (a))
 x1 / 3  b 
Sol.[D] Obviously, g(x) =   satisfies the = x (from (b))
 a 
 
 RHS  incorrect
relation fog(x) = gof(x).
x
Q.409 If f(x) = , then (fofof) (x) =
(1  x 2 )
Q.407 If g(x) is a polynomial function satisfying
3x x
g(x)g(y) = g(x) + g(y) + g(xy) – 2 for all x, (A) (B)
2
(1  x ) (1  3x 2 )
y  R and g(2) = 5, then g(5) is -
(A) 26 (B) 25 (C) 4 (D) 2 3x
(C) (D) None of these
Sol.[A] Putting x = 2 and y = 1 in the given relation, (1 – x 2 )

we obtain f (x)
Sol.[B] (fof) (x) = f(f(x)) =
g (2) g (1) = g (2) + g (1) + g (2) – 2 1  (f ( x )) 2
 5g (1) = 5 + g (1) + 5 – 2  g (1) = 2 x x
2
1 x 1 x2
Putting y = 1/x in the given relation, we get
= 2 =
g (x) g (1/x) = g (x) + g (1/x) + g (1) – 2  x  1 x 2  x2
1  
 
 g (x) g (1/x) = g (x) + g(1/x) [g (1) = 2]  1 x
2
 1 x2
 g (x) = xn + 1 x
=
 g (2) = 2 + 1  5 = 2 + 1  n = 2.
n n
1  2x 2

 g (x) = x2 + 1  g (5) = 25 + 1 = 26. (fof )( x )

(fofof) (x) = f(fof) (x)) =
1  (fof ( x )) 2
Q.408 Which of the following statement is incorrect -
(A) x sgn x = |x| (B) |x| sgn x = x
(C) x (sgn x) (sgn x) = x (D) |x| (sgn x)3 = |x|
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 x Q.412 The number of solutions of the equation
1  2x 2 x sin x = |log| x|| is -
= 2 =
 x  1  2x 2  x 2 (A) infinite (B) 8
1  
 2  (C) 6 (D) 0
 1  2x 
Sol.[C] The graph of y = sin x and y = |log | x||
x
=  (B) is correct. intersect the 6 points (four positive and two
1  3x 2
negative roots) as clear from the following
 y y
Q.410 If f  2 x  , 2 x   = xy, then f(m, n) + f (n, figures:
 8 8
y
m) = 0
(A) only when m = n (B) only when m  n
(C) only when m = –n (D) for all m and n 1

y y x
–3 –2 –1 1 2 e3
Sol.[D] Let 2x + =  and 2x – = , then x =
8 8
–1
 
and y = 4 ( – ) Q.413 If x and y satisfy the equation max (|x + y|, |x –
4
y|) = 1 and |y| = x – x [x], then the number of

 y y ordered pairs (x, y) is -

Given, f  2 x  , 2 x   = xy  f(, ) =
 8 8 (A) 0 (B) 4 (C) 8 (D) infinite

2 – 2 Sol.[D] Fist consider max (|x + y|, |x – y|) = 1

 f (m, n) + f (n, m) = m 2 – n2 + n2 – m2 = 0 for If |x + y|  |x – y| then |x + y| = 1

all m, n . i.e. if 4xy  0 then x + y = ± 1,

 In first and third quadrants x + y = ± 1
Q.411 Consider a real valued function f(x) satisfying If |x + y|  |x – y| then |x – y| = 1
2f (xy) = (f(x))y + f((y))x  x, y  R and f(1) = a i.e. if 4xy  0, then x – y = ±1
 In second and fourth quadrants x – y = ± 1
n
where a  1, then (a – 1)  f (i) equals - Also |y| = x – |x|  y = ± (x – [x]) as x – [x]  0
i 1 The graph of the curves is shown:
n n+1 n–1 n+1
(A) a (B) a (C) a + a (D) a –a
x – y = –1
Sol.[D] Putting y = 1 in the given relation. 1 y = x – |x|
x+y=1
2f(x) = f(x) + [f(1)]x  f(x) = (f(1))x = ax
–1
n x–y=1
Now,  f (i) = a + a2 + a3 + …. + an = –1 y = –(x – [x])
i 1 x + y = –1

We note that –1  x < 0 two curves concide. So,

a (a n  1)
a 1 there infinitely many solutions

n Q.414 A function f : R  R is defined by f(x + y) –

 (a – 1)  f (i) = an+1 – a.
kxy = f(x) + 2y2  x, y  R and f(1) = 2 ; f(2) =
i 1
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 8, where k is some constant, then f(x + y) . f y = 1 + sin x y = 1 – cos x
2
 1 
  = (x + y  0) -
xy
1
(A) 1 (B) 4 (C) k2 (D) k2 + 4
Sol.[B] Given f(x + y) – kxy = f(x) + 2y 2 . Replace y
by –x, then /2 3/4  3/2 7/4 2

f(0) + kx2 = f(x) + 2x2  f(x) = f(0) + kx2 – 2x2 Y

… (1)
y=1–x
Now f(1) = f(0) + k – 2 = 2  f(0) = – k + 4 y=x–1
and f(2) = f(0) + 4k – 8 = 8  f(0) = –4k + 16
Which give k = 4 and f(0) = 0 X
1 2
Thus, from (1) f(x) = 2x2
 1 
 f(x + y) f   = 4 = k. 1  x , x  0
 x  y  
= 1,0  x  2
 x  1, x  2

Q.415 Let f(x) = max {1 + sin x, 1, 1 – cos x}, x  [0,  f(0) = 1  g(f(0)) = 1 and f(1) = 1 + sin 1
2] and g(x) = max {1, |x – 1|} x  R, then -
(A) g(f(0)) = 1 (B) g(f(1)) = 1  3 
0  1  
 4 
(C) f (g(1)) = 1 (D) f(g(0)) = sin 1
Sol.[A, B]  g(f(1)) = 1 (1 < 1 + sin 1 < 2)
f(x) = max {1 + sin x, 1, 1 – cos x} = Again g(1) = 1  f(g(1)) = 1 + sin 1 and g(0) =
1  f(g(0)) = 1 + sin 1.

Q.416 Let ƒ be a function defined on [–2, 2] and is

 3
1  sin x , 0x
4  – 1, –2x0

 3 3 given by ƒ(x) =  and
x  1, 0x2
1  cos x , x
 4 2 g(x) = ƒ(|x|) + |ƒ(x)| . Then g(x) is equal to –
 1, 3
 x  2

 2   x, 2  x  0

g(x) = max {1, |x – 1|} (A)  0, 0  x 1
x  1, 1 x  2


  x, 2 x  0

(B)  0, 0  x 1
2( x  1), 1 x  2

 – x, –2x0
(C) 
x  1, 0x2

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 (D) None of these  2 1  2 1
 x  2  = 0, –1   x   = 1, 0
2
 1 2 x  0   
Sol.[B] We have, ƒ(x) = 
x  1, 0x2
 ƒ(x) = sin–1 (1) + cos–1 (0)
Since x  [–2, 2], therefore |x|  [0, 2] or ƒ(x) = sin–1 (0) + cos–1(–1)
Therefore,  Range of ƒ(x) = {}
ƒ(| x |) = |x| – 1,  x  [–2, 2] Hence (B) is correct answer.
 x  1 x  [–2,0]   
=  Q.418 If A = x :  x   and ƒ(x) = cos x – x (1 + x),
 x  1, x  [0, 2]  6 3
… (i) then ƒ (A) is equal to -
Also, 1  2 3  2 
(A)    ,   
 1, 2  x  0  2 3 9 2 6 36 

|ƒ(x)| = 1  x 0  x 1
x  1, 1  2 3  2 
 1 x  2 (B)    ,   
 2 3 9 2 6 36 
… (ii)
 1  2 3   2 
From (1) and (2), we get (C)    ,  
2 3 9 2 6 36 
g(x) = ƒ(| x |) + |ƒ(x) | =
(D) None of the above

  x  1  1, 2 x  0 Sol.[A] We have ƒ(x) = cos x – x (1 + x)


x  1  1  x , 0  x 1
 x  1  x  1, 1 x  2  
 Since, in the interval  ,  , cos x decreases
6 3
= and x (1 + x) increases.

 
  x, 2 x  0  ƒ(x) decreases in  ,  .
 6 3
 0, 0  x 1 .
2( x  1), 1 x  2
    
 ƒ    ƒ(x)  ƒ   , x   , 
3
  6
  6 3
 2 1  2 1 Hence ƒ(A) =
Q.417 Range of sin–1  x   + cos–1  x  
 2  2

where [ . ] denotes the greatest integer function, is - 1    3   

  1  ,  1   
   2 3  3 2 6 6 
(A)  ,  (B) {}
2  1  2 3  2 
=    ,   .
  2 3 9 2 6 36 
(C)   (D) None of these
2 Hence (A) is the correct answer.

 2 1  2 1   2 1  3 2 1 
Sol.[B]  x   =  x   1 = 1 +  x   . Q.419 If ƒ(x) = maximum x , x ,   x  [0,
 2  2   2  64 

Thus for domain ), then -

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 
x 2 , 0  x 1 1  x , x  1
(A) ƒ(x) =  3 
 x 1 (B) ƒ(x) =  1, 1  x  1
x ,
1  x , x 1

1 1
 64 , 0x 1  x , x  1
4 

 2 1 1  x  1
 x 1 (C) ƒ (x) =  2,
(B) ƒ(x) =  x , 1  x ,
 3 4  x 1
x , x 1

 (D) None of the above

1 1
 64 , 0x Sol.[C] For x  –1, 1 – x  2 and 1 – x  1 + x
8

 3 1
(C) ƒ(x) =  x  x 1  max {(1 – x), 2, (1 + x)} = 1 – x
 3 8
x , x 1 For –1 < x < 1, 0 < 1 –x < 2 and 0 < 1 + x < 2.


 max {(1 + x), 2, (1 + x)} = 2.
1 1 For x  1, 1 + x  2, 1 + x > 1 – x
 64 , 0  x  8
(D) ƒ(x) =   max {(1 – x), 2, (1 + x)} = 1 + x
1
 x3, x 1  x , x  1
 8 
Hence, ƒ(x) =  2, 1  x  1 .
Sol.[C] Clearly 1  x , x 1

1 1
 64 , 0x
8

 2 1 Q.421 If f : (0, )  R, defined by
ƒ(x) =  x ,  x 1
 3 8 n
x , x 1 f(x) =  [1  sin kx ] , where [x] denotes the

 k 1

integral part of x, then the range of f(x) is -

y = x3 y = x2
(A) {n – 1, n + 1} (B) {n – 1, n , n + 1}
(C) {n, n + 1} (D) None of these
n

1
Sol.[C] ƒ(x) =  (1  [sin kx ]) = n + [sin x] + [sin
y = 64 k 1

O 2x] + [sin nx] … (i)

1 1
1
8 4 
Case I : When kx  for k = 1, 2, 3,….., n
2
Hence (C) is the correct answer.

Since 0 < kx < and kx   0 < sin kx <
2
Q.420 Let ƒ(x) = max. {(1 – x), (1 + x), 2},  x  R . 1, for k = 1, 2, ….., n
Then -  [sin kx] = 0, for k = 1, 2, 3, …., n  From
1  x , x  1 (i), ƒ(x) = n

(A) ƒ (x) =  2, 1  x  1
1  x , x 1


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 Case II : When exactly one of x, 2x, 3x,…., nx We get ƒ(b) = ƒ(0) = k and again b = 0 gives
ƒ(a) = k  ƒ(a) = ƒ(b) = k  a, b  ƒ(x) is a

is . Here not more than one of x, 2x, 3x, …., constant function.
2
1
 ƒ(2007) = – .
 2
nx can be .
2 Q.424 The number of solutions of the equations
In this case one of sin x, sin 2x, ….., sin nx is 1 [y] = sin x and x2 + y2 = 4 is -
and others lie between 0 and 1. (A) 1 (B) 2
 From (i), f(x) = n + 1. Hence range of f = (n, n + 1). (C) 3 (D) No solution
Sol.[B]
–1
Q.422 If [x] denotes the integral part of x and k = sin y

2 x2 +y2 = 4
1 t 2
> 0, then the integral value of  for [y] = sin x
2t
1
which the equation (x – [k]) (x + ) – 1 = 0 has 2
–2 x
integral root is - – –/2 –1 0 1 /2 

(A) –1 (B) 1
–2
(C) 2 (D) None of these

1 t 2 1 t2
Sol.[A] For sin–1 to be defined 1 The two curves at two points, therefore two
2t 2t
solutions.
 1 + t2  2 | t |  (1 – | t |)2  0  (1 – | t |)2
= 0  t = ±1
Q.425 The number of solutions of the equation 5{x} =
2 
1 t 4x + [x] is (Here [ ] denotes greatest integer
 k = sin–1 > 0  k = sin–1 1 =
2t 2
function) -
(A) 0 (B) 1

 [k] =   = 1. The given equation then (C) 2 (D) None of these
2
become Sol.[B] 5(x – [x]) = 4x + [x]  x/6 = [x]. Now plot
(x – 1) (x – ) = 1. For integral values of  and x, we
the graphs of y = x/6 and y = [x].
have either x – 1= 1 and x +  = 1  x = 2 and
=–1 They intersect at one point.
or x – 1 = –1 and x +  = –1  x = 0 and  = –1.  Only one solution which is x = 0.

Q.423 The function ƒ(x) is defined for all real x. If Q.426 ƒ(x) = (tan x5) e x
3
sgn x 7 is -
ƒ(a + b) = ƒ(ab)  a and b and (A) an even function
 1 1 (B) an odd function
ƒ    =  then ƒ(2007) equals -
 2 2 (C) neither even nor odd function
1 1 (D) None of these
(A) –2003 (B) 2003 (C) – (D)
2 2
Sol.[C] Let ƒ(0) = k. Let a = 0

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 1
 
3
f ( x )  (tan ( x 5 )) ex sgn (x 7 ) (C) 1 – 1  4 log 2 x (D) not defined
Sol.[B]      2
O (O)  O  O
eO Sol.[B] y = 2x(x – 1)  x(x – 1) – log2 y = 0

= O × e O × O = O × eE 1  1  4 log 2 y
 x =
= O × E = O. 2

1  1  4 log 2 y
Q.427 Let ƒ : R  R be a function such that But x > 1,  x =
2
ƒ(x) = x3 + x2 + 3x + sin x. Then -
1
(A) f is one-one and into  f–1(x) = [1 + 1  4 log 2 x ]
2
(B) f is one-one and onto
(C) f is many-one and into
2x (sin x  tan x )
(D) f is many-one and onto
Q.429 The graph of f(x) =  x  21  is
Sol.[B] ƒ(x) = 3x2 + 2x + 3 + cos x = 3(x2 + (2/3)x + 1) + cos x 2  – 41
  
= 3
symmetric about -
(A) x-axis (B) y-axis
 2   2 (C) origin (D) None of these
  x  1   1  1   cos x  3  x  2    8  cos x
 3 9  3  3 Sol.[C] Numerator = 2x (sin x + tan x) = O × (O + O)
           
8 5
0   ( 1)  =O×O=E
3 3

 ƒ is an strictly increasing function x 

Denominator = 2   21 – 41 = g(x) (say)
  
 ƒ is one – one function
Let x/ = n + f n  I & f  [0, 1)
lim f(x) = lim x3 1  1  3  sin x  = then g(x) = 2 [n + f + 21] – 41 = 2(n + 21) – 41
x  x   x 
 x2 x3 
= 2n + 1
+
 x 
lim f(x) = – g(–x) = 2   21 – 41
Similarly, x  –   

Moreover f(x) is a continuous function. = 2[–n – f + 21] – 41 = 2[(21 – n) – f] – 41

 Rf = R = codomain = 2 (20 – n) – 41 = –2n – 1

Hence f is an onto function. g(–x) = –g(x)  g is an odd function

 g(x) = E/O = O

Q.428 If the function f : [1, + )  [1, +) is defined Hence (C) is correct.

by f(x) = 2x(x–1) then f–1 (x) is -

x ( x 1) 1
1 Q.430 If ƒ(x) = 2x + | x |, g(x) = (2x – | x |) and
(A)   (B) 3
2
h(x) = ƒ(g(x)), then domain of
1
2

1  1  4 log 2 x  sin 1 ( h ( h ( h ( h....h ( x )....))))
is -
n times

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  1 i.e., 3x2 + 6x + 4  c sin x – b cos x,  x  R
(A) [–1, 1] (B)   1, 2  
 
i.e., 3x2 + 6x + 4  b2  c2 ,xR
1  i.e., b2  c2  3(x2 + 2x + 1) + 1,  x  R
 2 ,1
 
b2  c2  3(x + 1)2 + 1,  x  R
 1 1 
(C)   1,  (D)  ,1
 2 2 
b2  c2  1,  x  R
2 x  x , x0  b2 + c2  1,  x  R
Sol.[A] Since, ƒ(x) =  =
2 x  x , x0
Hence (C) is the correct answer.

3x , x0
 Q.432 Which pair of functions is identical?
 x, x0

1 2 x – x , x0 (A) sin–1(sin x) and sin (sin–1 x)

and g(x) =  =
3 2 x  x , x0
(B) loge ex, e log e x

(C) loge x2 , 2 loge x

x
 , x0 (D) None of these
3
 x, x  0 Sol.[D] Here, (A) sin–1(sin x) is defined for x 

 x
3 , x  0   
  2 , 2  , while sin(sin x) is defined only
–1
f(g(x)) =   3 
 

 x, x0
for
f(g(x)) = x,  x  R
x  [–1, 1]

h(x) = x
(B) loge ex, is defined for all x, while e log e x is
 sin–1 (h(h(h…..(h(x)….))) = sin–1 x defined for x > 0.
(C) loge x2 is defined for all x  R – {0}, while
 Domain of sin–1 (h(h(h(h…h(x)….)))) is [–1, 1]
2 log x is defined for x > 0
Hence (A) is the correct answer.
 None is identical
Hence (D) is the correct answer.
Q.431 ƒ(x) = x3 + 3x2 + 4x + b sin x + c cos x,  x  R
n
is a one-one function then the value of b2 + c2 is - Q.433 If  f ( x  ka )  0 , where a > 0, then the
k 0
(A)  1 (B)  2
(C)  1 (D) none of these period of f(x) is -

Sol.[C] Here, ƒ(x) = x3 + 3x2 + 4x + b sin x + c cos x (A) a (B) (n + 1) a

ƒ(x) = 3x2 + 6x + 4 + b cos x – c sin x a

(C) (D) f(x) is non-periodic
n 1
Now for ƒ(x) to be one-one only possibility is Sol.[B] Given f(x) + f(x + a) + ….. + f(x + an) = 0… (1)
ƒ(x)  0,  x  R Replace x by x + a,
i.e., 3x + 6x + 4 + b cos x – c sin x  0,  x  R
2
f(x + a) + f(x + 2a) + ….. + f {x + a(n + 1)} = 0 … (2)

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 Subtracting (2) from (1) we get Sol.[D] Since f(x) and g(x) are mirror images of each
f(x) – f{x + a(n + 1)} = 0 other about the line y = a, f(x) and g(x) are at
equal distances from the line y = a. Let for some
 f(x) is periodic with period a (n + 1).
particular x0
Q.434 If ƒ(x) = ex – [x] + |cos x| + |cos 2x| + ….. + |cos nx| ; then
f(x0) = a + k, then g(x0) = a – k, then
period of f(x) is -
(A) 1 (B) 1/n h(x0) = f (x0) + g(x0) = 2a

n 2 1  h(x) = 2a  x  R. So, h(x) must be a

(C) (D)
n 2 1 constant function, which is many-one into.

1 Q.437 If f(x) is a periodic function having period 7 and

1.2.3.....n
g(x) is periodic having period 11 then the period
Sol.[A] Function : x – [x] |cos x|cos 2x|
…………. |cos nx|
1 2 1 2 1 x
FP : 1 × =1 × = f (x) f 
2  2 2 2 of D(x) =
 3  is -
x
1 2 g(x ) g 
….. × = 5
2 n
(A) 77 (B) 231 (C) 385 (D) 1155
1
x x
n Sol.[D] We have, D(x) = f(x)g   – g(x) f  
5 3
FP of f(x) is LCM of 1, 1, 1/2, 1/3, ……, 1/n = 1.
x
x x Now period of f(x) g   is 7 × 55 = 385
Q.435 The function f(x) = sin – cos 5
n! ( n  1)!
is- x
Period of g(x) f   is 11 × 21 = 231
(A) not periodic 3
(B) periodic with period 2(n!)  Period of D(x) = LCM of (385, 231) = 1155.
(C) periodic, with period (n + 1)
(D) none of these
ax
x 2 Q.438 If f(x) = x
(a > 0) then
Sol.[D] FP of sin is = 2 n! a  a
n! 1/ n!
x 2 2 n 1
FP of cos = = 2 (n +  r 
( n  1)! 1 /( n  1)!  2f   is -
 2n 
1)! r 1

FP of f(x) is LCM of 2 n! & 2(n + 1)! = 2 (n + 1)! (A) n2 (B) 2n2 – 1

 (D) is correct. (C) 2n – 1 (D) None of these
Q.436 Let f : R  R and g : R  R be two one-one
ax
and onto functions such that they are the mirror Sol.[C] Given, f(x) = …(i)
ax  a
images of each other about the line y = a. If h
a1 x a
(x) = f(x) + g(x) , then h (x) is - Now f(1 – x) = 1 x
= …(ii)
a  a a  ax
(A) one-one onto (B) one-one into
From (i) and (ii) ; we have f(x) + f(1 – x) = 1… (iii)
(C) many-one onto (D) many-one into
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  r   2n  r  ƒ(f(2)) = ƒ(5) = 52 + 1 = 26.
f   + f  =1
 2n   2n 
Q.440 If f(x) is continuous and increasing function
2 n 1 2 n 1
 r   2n  r 
  ƒ
 2 n
 +  ƒ
 2n 
 = 2n – such that the domain of g(x) = f ( x)  x be
r 1 r 1 

1 1
R and h(x) = then the domain of (x) =
2 n 1 2 n 1 1 x
 r   t 
  ƒ
2 n
 +  ƒ  = 2n – 1
2n  f (f (f ( x )))  h ( h ( h ( x ))) is -
r 1   t 1 

(putting 2n – r = t) (A) R (B) {0, 1}

2 n 1 (C) R – {0, 1} (D) R+ – {1}

 r 
Hence, 2 
ƒ  = 2n – 1.
2n 
r 1  1 x 1
Sol.[C] h(x) = , x  1  h(h(x)) = ,x
1 x x
0, 1
 h(h(h(x))) = x, x  0, 1
Also, g(x)  0  x  R  f(x)  x
Q.439 If ƒ is a polynomial function satisfying 2 + ƒ(x)
 f(f(x))  f(x)  x [f(x) is increasing]
· ƒ(y) = ƒ(x) + ƒ(y) + ƒ(xy),  x, y  R and if
 f(f(f(x)))  f(x)  x  f(f(f(x))) – x  0
ƒ(2) = 5, then find ƒ(f(2)) -
 x  R – {0, 1}
(A) 26 (B) 28
 (x) is defined for all x  R – {0, 1}.
(C) 6 (D) None of these
2
Sol.[A] Given 2 + ƒ(x) ƒ(y) = ƒ(x) + ƒ(y) + ƒ(xy). Q.441 The inequality < 3 is true, when x belongs
x
or 1 – ƒ(x) – ƒ(y) + ƒ(x) ƒ(y) = ƒ(xy) – 1 to-
or (1 – ƒ(x)) (1 – ƒ(y)) = ƒ(xy) – 1 2 
(A)  ,   (B)
The above result holds if and only if, 3 
ƒ(x) = 1 + xn
 2
   
If ƒ(x) = anxn + an–1 xn–1 + ….. + a0  3
Then consider (1 + ƒ(x)) (1 – ƒ(y)) = ƒ(xy) – 1 2 
(C)  ,    (–, 0)(D) none of these [C]
Compare constant term on either side, we have  3 
1 – a0 = a0 – 1  a0 = 1
Comparing coefficient of xnyn, we get x4
Q.442 < 2 is satisfied when x satisfies-
x 3
a 2n = an  an = 1 or otherwise polynomial
(A) (–, 3) (10, ) (B) (3, 10)
would not be of n degree.
(C) (–, 3) [10, ) (D) none of these [A]
Comparing coefficient of x, x1, …., xn–1 on
either sides, we have a1 = a2 = ….. = an–1 = 0 x7
Q.443 Solution of > 2 is-
 an = 1 and ƒ(x) = xn + 1 x 3
Given ƒ(2) = 5 i.e., 2n + 1 = 5 (A) (–3, ) (B) (–, –13)
n = 2 (C) (–13, –3) (D) none of these [C]

Thus, ƒ(x) = x2 + 1

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 2x  3 (C) y = x – |x| (D) y = x2 + 1 [C]
Q.444 Solution of  3 is-
3x  5
 12 
(A) 1,  (B)
 7 

 5 12 
 , 
3 7 

 5 12 
(C)   ,  (D)  , [B]
 3  7 

Q.445 Solution of (x – 1)2 (x + 4) < 0 is-

(A) (–, 1) (B) (–, –4)
(C) (–1, 4) (D) (1, 4) [B]

Q.446 Solution of (2x + 1) (x – 3) (x + 7) < 0 is-

 1 
(A) (– , –7)    , 3 
 2 
1 
(B) (– , – 7)   , 3 
2 
 1 
(C) (–, 7)    , 3 
 2 
(D) (–, –7)  (3, ) [A]

Q.447 If x2 + 6x – 27 > 0 and x2 – 3x – 4 < 0, then-

(A) x > 3 (B) x < 4
7
(C) 3 < x < 4 (D) x = [C]
2

Q.448 If x2 – 1  0 and x2 – x – 2  0, then x line in the

interval/set
(A) (–1, 2) (B) (–1, 1)
(C) (1, 2) (D) {– 1} [D]

Q.449 Which of the following relation is a function ?

(A) {(1,4), (2,6), (1,5), (3,9)}
(B) {(3,3), (2,1), (1,2), (2,3)}
(C) {(1,2), (2,2,), (3,2), (4,2)}
(D) {(3,1), (3,2), (3,3), (3,4)} [C]

Q.450 If x, y  R, then which of the following rules is

not a function-
(A) y = 9 –x2 (B) y = 2x2

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Q.

Corporate Office: CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-2434159 FUNCTION 53