IIT-JEE Mathematics Black Book by MC Sir | Complete Theory, Advanced Problems & Detailed Solutions | JEE Main & Advanced Preparation | Comprehensive Maths Study Guide

Maths IIT-JEE ‘Best Approach’ (MC SIR) Functions
Chapter-1_Function
EXERCISE - 1
Single Choice Problems :

1. Range of the function f(x) = log2 (2 – log 2
(16sin2 x + 1)) is-

(A) [0,1] (B) (–,1) (C) [ –1,1 ] (D) (–,)


2. The value of a and b for which e x – b – a = 2, has four distinct solutions, are :

(A) a  (–3, ), b = 0 (B) a  (2, ), b = 0
(C) a  (–3, ), b  R (D) a  (2, ), b = a

3. The range of the function :
1
f (x) = tan–1 x + sin–1 x
2
(A) (– /2, /2) (B) [– /2, /2] – {0}
(C) [– /2, /2] (D) (–3/4, 3/4)


4. Find the number of real ordered pair (s) (x,y) for which :
2 2
16x y  16xy = 1
(A) 0 (B) 1 (C) 2 (D) 3
x
1
5. The complete range of values lf ‘a’ such that   = x2 – a is satisfied for maximum number of values
2
 

of x is :
(A) (–, –1) (B) (–,) (C) (–1,1) (D) (–1, )


6. For real number x, let [x] denotes the greatest the integer less than or equal to x. Let f : R  R be
defined by f (x) = 2x + [x] + sin x cos x. Then f is :
(A) One -one but not onto (B) Onto but not one-one
(C) Both one-one and onto (D) Neither one-one nor onto


 7 – 5( x 2  3) 
7. The maximum value of sec–1   is :
 2 ( x 2
 2) 

5 5 7 2
(A) (B) (C) (D)
6 12 12 3

, Maths IIT-JEE ‘Best Approach’ (MC SIR) Functions
x3 a 2
8. Number of ordered pair (a,) from the set A= {1, 2, 3, 4, 5} so that the function f (x) = + x + bx + 10
3 2
is an injective mapping  x  R :
(A) 13 (B) 14 (C) 15 (D) 16


9. Let Abe the greatest value of the function f (x) = logx [x], (where [.] denotes greatest integer function)
and B be the least value of the function g (x) = sin x  cos x then :

(A) A > B (B) A < B (C) A = B (B) 2A + B = 4


10. Let A = [ a, ] denotes domain, then f : [ a, ]  B, f (x) = 2x3 – 3x2 + 6 will have an inverse for the
smallest real vlaue of a, if :
(A) a = 1, B = [ 5, ] (B) a = 2, B = [ 10, ]
(C) a = 0, B = [ 6, ] (D) a = – 1, B = [ 1, )

11. Solution of the inequation {x} ({x} – 1) ({x} + 2)  0
(where {·} denotes fractional part function) is :
(A) x  (– 2, 1) (B) x  I ( I denote set of integers)
(C) x  (0, 1) (D) x  [–2 , 0]

12. Let f (x), g (x) be two real valued functions then the function h (x) = 2 max { f (x) – g (x), 0} is equal to:
(A) f (x) – g (x) – |g (x) – f (x)| (B) f (x) + g (x) – |g (x) – f (x)|
(C) f (x) – g (x) + |g (x) – f (x)| (D) f (x) + g (x) + |g (x) – f (x)|


13. Let R = {(1,3), (4,2), (2, 4), (2,3), (3,1)} be a relation on the set A = {1,2,3,4}. The relation R is :
(A) a function (B) reflexive (C) not symmetric (D) transitive


 1  = K
14. The true set of values lf ‘K’ for which sin –1 1 sin 2 x  6 may have a solution is :
 

1 1 
1 1 (B) [1, 3] (C) ,  (D) [2, 4]
(A)  ,   6 2 
 4 2 

15. A real valued function f (x) satisfies the functional equation f (x – y) = f (x) f (y) – f (a – x) f (a + y) where
‘a’ is a given constant and f (0) = 1, f (2a – x) is equal to :
(A) –f (x) (B) f (x) (C) f (a) + f (a – x) (D) f (–x)


16. Let g : R  R be given by g (x) = 3 + 4x if gn (x) = gogogo........... go (x) n times. Then inverse of gn
(x) is equal to :
(A) (x + 1 – 4n) · 4–n (B) (x – 1 + 4n) 4–n (C) (x + 1 + 4n) 4–n (D) None of these

, Maths IIT-JEE ‘Best Approach’ (MC SIR) Functions
x 2  2x  a
17. Let f : D  R be defined as : f (x) = where D and R denote the domain of f and the set of
x 2  4x  3a
all real numbers respectively. If f is surjective mapping, then the complete range of a is :
(A) 0  a  1 (B) 0 < a  1 (C) 0  a < 1 (D) 0 < a < 1


18. If f : (–,2]  (–,4] where f (x) = x (4 – x), then f–1 (x) is given by :

(A) 2 – 4– x (B) 2 + 4– x (C) – 2 + 4– x (D) –2 – 4– x


19. If [5 sin x] + [cos x] + 6 = 0, then range of f (x) = 3 cos x + sin x corresponding to solution set of the
given equation is : (where [·] denotes greatest integer function)

 3 3  2   3 3  4 
(A) [– 2, –1) (B)  – ,–1 (C) [–2, – 3 ] (D)  – ,–1
 5   5 


20. If f : R  R , f (x) = ax + cos x is an invertible function, then complete set of values of a is :
(A) (–2, –1]  [1,2] (B) [–1, 1]
(C) (– , –1)  [1, ) (D) (– , –2]  [2, )

 x  x  x
21. The range of function f (x) = [1 + sin x] + 2 sin + 3 sin +.......+ n sin x  [0,],
   






2  3   n 
n  N([·] denotes greatest integer function) is:

n 2  n – 2 n (n 1)  n (n 1) 
(A)  ,  (B)  
 2 2   2 

n (n 1) n 2  n  2 n2  n  4)  n (n 1) n2  n  2 
(C)  , ,  (D)  , 
 2 2 2   2 2 

 x 2  ax 1
22. If f : R  R, f (x) = 2 , then the complete set of values of ‘a’ such that f (x) is onto is :
x  x 1

(A) (–,) (B) (–,0) (C) (0,) (D) Not possible


23. If f (x) and g (x are two functions such that f (x) = [x] + [– x] and g (x) = {x}  x  R and h (x) = f (g(x));
then which of the following is incorrect ?
([·] denotes greatest integer function and {·} denotes fractional part function)
(A) f (x) and h (x) are identical functions (B) f (x) = g (x) has no solution
(C) f (x) + h (x) > 0 has no solution (D) f (x) –h (x) is a periodic function

, Maths IIT-JEE ‘Best Approach’ (MC SIR) Functions
24. Number of elements in the range set of f (x) =  x  – 15  x (0,90); (where[·] denotes greatest
15  x  

integer function) :
(A) 5 (B) 6 (C) 7 (D) Infinite


25. The graph of function f (x) is shown below :



O




1
Then the graph of g (x) = f (| x |) is:




O
(A) (B)




(C) (D)




26. Which of the following function is homogenous ?
y x
(A) f (x) = x sin y + y sin x (B) g (x) = xex  yey

xy x – y cos x
(C) h (x) = x  y2 (D)  (x) = ysin x  y