MATHS

(TOPIC – Functions, limits & continuity of functions & Differentiability)

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1. The domain of the function f  x   (where {.} denotes the fractional part ) is
sin x  sin   x 
  
a)  0,  b)  2n  1 : n is an integer 
 2 
 n 
c)  0,  d) R   : n is an intiger 
 2 
KEY: D
2. Let X  1, 2,3, 4 . The number of functions f : X  X satisfying f  f  i    i for all 1  i  4 is
a) 1 b) 6 c) 9 d) 10
KEY: D
3. Let f  x   x   x (where {.} denotes the fractional part of x) and X,Y be its domain and range
respectively. Then
 1 1   1 1 
a) X   ,  and Y   ,   b) X   ,   and Y   ,  
 2 2   2 2 
 1  1
c) X   ,     0,   and Y   0,   d)  ,   and  0,  
 2  2
KEY: C
4. Let f :    and g :    be two one – one and onto functions such that they are the mirror images
of each other about the line y  a . If h  x   f  x   g  x  , then h  x  is
a) one – one and onto b) only one – one and not onto
c) only onto but not one-one d) Neither one-one nor onto
KEY: D
1
5. Let f  x   lt 2n
. Then the set of values of x for which f  x   0 is
n 
3 1 
 tan 2 x   5
 
a) 2 x  3 b) 2 x  3 c) 2 x  3 d) 2 x  3
KEY: A
1
6. lt (where  x denotes the fractional part of x )
x 1
x   x
1
a) Does not exist b) equals1 c) equals 0 d) equals
2
KEY: A
7 If the following functions are defined from  1,1 to  1,1 , identify these which are bijective.

a) sin  sin 1 x  b)
2 1

sin  sin x  
c) sgn  x  ln  e x   d) x 3  sgn  x  
KEY: A
8. Which of the following functions is / are periodic?
1, x is rational
a) f  x   
0, x is irrational

1
 b) f :    is defined by f  x   x  2n for x   2n  1, 2n  1 , where n  

2x
c) f  x    1    , where [.] denotes the greatest integer function.

x 
d) f  x   ax   ax  a   tan   ,where [.] denotes the greatest integer function and a is a rational
 2 
number.
KEY: A, B, C, D
9. Statement I: If g  x   f  x   1 and f  x   f 1  x   2 , x  R then g  x  is symmetrical about the
1 
point  , 0 
2 
Statement II:- If g  a  x    g  a  x  , x  R then g  x  is symmetrical about the point (a,0)
KEY: A
10. Statement I: Function f  x   x 2  tan 1 x is a non periodic function.
Statement II: The sum of two non periodic functions is always non periodic.
KEY: C

PASSAGE-I :
 x  1, 0  x 1  x2 , 1  x  2
Consider the function f  x    and g  x  
2 x  1, 1  x  2  x  2, 2 x3
11. The domain of the function f  g  x   is
a)  0, 2  b)  1, 2 c)  1, 2  d) [ 0, 5]
KEY: C
12. The range of the function f  g  x   is
a) 1,5 b)  2,3 c) 1, 2   3,5 d) [3 ,5]
KEY: C
13. The number of roots of the equation f  g  x    2 is
a) 1 b) 2 c) 3 d) 4
KEY: B

PASSAGE – II :
Let f :    be a function satisfying
f  2  x   f  2  x  and f  20  x   f  x  , x   . For this function f answer the following
questions.
14. If f  0   5 . Then the minimum possible number of values of x satisfying f  x   5 , for x   0,170 is
a) 21 b) 12 c) 11 d) 22
KEY: C
15. The graph of y  f  x  is
a) symmetrical about x  16 b) symmetrical about x  5
c) symmetrical about x  8 d) symmetrical about x  20
KEY: A
16. If f  2   f  6  , then the
a) Fundamental period of f  x  is 1 b) Fundamental period of f  x  is 4

2
 c) Period of f  x  cannot be 1 d) Fundamental period of f  x  is 8
KEY: C
17 A function f from the set of integers  to is defined as follows:
n  3 if n is odd

f n   n
 2 if n is even

 
Suppose k is odd and f f  f  k    27 . Then the sum of the digits of k is
a) 3 b) 6 c) 9 d) 12
KEY: B
18 Suppose f is a function defined on the set N of natural numbers as follows:
f  n   log8 n, if log8 n is rational
2010
= 0, otherwise. Then  f n 
n 1

55
a) log 2 2010 b) 6 c) 10 d)
3
KEY: D
19. Let  be a non-empty set of real numbers and f :    be such that
f  f  x    x for all x in  .Then f is
a) a bijection b) one- one but not onto
c) onto but not one-one d) neither one-one nor onto
KEY: A
x 2  4 x  30
20. STATEMENT-1::The function f :   , f  x   . is injective.
x 2  8 x  18
STATEMENT-2:A strictly monotonic increasing function is always injective.:
KEY: D
1
21. Consider those functions f that satisfy f  x  4   f  x  4   f  x  for all real x and f  2   .
2
1
STATEMENT-1: f  50  
2
STATEMENT-2: f is a periodic function.
KEY: A
22. Suppose f : A  B and g : B  C are such that gof is onto (Here gof is the composition
of f and g)
STATEMENT-1: g is one-one  f is onto
STATEMENT-2:: g is onto
KEY: B
Paragraph for Questions Nos. 17 to 19
For a finite set A, let A denote the number of elements in A. Let F denote the set of all functions
f : 1, 2,3,   , n  1, 2,3,     k  n  3, k  2  satisfying
f  i   f  i  1 for every i, 1  i  n  1 and C  n, k  denote the number of functions in F satisfying
f  n   f 1 . Then
23. F =
a) k  k  1 b) k n  k  1 c) k n 1  k  1 d) k  k  1
n n 1

3
 KEY: D
24. For n  4, C  n, k  
a) k  k  1  C  n  1, k  b) k n 1  k  1  C  n  1, k 
n 1

c) k n  k  1  C  n, k  d) k  k  1  C  n, k 
n

KEY: A
25. For n  3, C  n, k  =
a)  k  1   1  k  1 b)  k  1   1  k  1
n 1 n n n

c)  k  1   1  k  1 d)  k  1   1  k  1
n n 1 n 1 n 1

KEY: B
26. Mach the following

Column I Column II
If f  x   ax  b and f 1  x   bx  a where a, b
(A) (P) 2
are real, the value of a  b is less than
A function f : C  C is defined by
f  Z   iZ ,where i  1 and Z is the complex
(B) conjugate of z. The number of complex number z (Q) 4
that satisfy both Z  5 and f  Z   Z is not more
than
The function f  x  satisfies f  2  x   f  2  x  for

(C) all real x. If the equation f  x  =0 has exactly four (R) 6

different roots, then the sum of the roots is more
than
Let A  1, 2,3 and f : A  A .The number of
(D) maps g : A  A such that f  x   g  x  for exactly (S) 8
two elements in A is not less than
KEY: A-Q,R,S;
B-P,Q,R,S;
C-P,Q,R;
D-P,Q,R
27. Match the following
Column I Column II
Let f :    be defined by
 x, if x is rational One - one
(A) f  x   (P)
1  x, if x is irrational
Then f is
f :    is defined by
n 1   1
n Onto
(B) f n   for n   . Then f (Q)
2 4
is
(C) f :   , f  x   x 3  7 x  6 is (R) Bijection

4
 f :   ,    0  1,1 defined by
(D) (S) Neither one-one nor onto
sin x
f  x  is
sin x
KEY: A-P,Q,R; B-Q; C-Q; D-S

28. Let a function “f” be defined on the integers by setting

f  m  n, n  if m  n
f  m, n   
 m if m  n
then

Column I Column II
(A) 
f 3 2007

,4  (p) 14

(B) f 2 2006
, 25  (q) 5
(C) f 5 2005

,7  (r) 3

(D) f 16 3
 173  183  193 , 70   (s) 0

1   1   x  
29. If 2f  x   xf    2f  2 sin   x     4 cos 2    x cos , x  R  {0} then which of the
x   4   2  x
following statement(s) is/are true? (L2)
1 1 1
a) f  2   f    1 b) f  2   f 1  0 c) f  2   f 1  f   d) f 1 f   f  2   1
2 2 2
Ans: a,b,c
1 1
2f  2   2f    2f 1  4  f  2   f    2  f 1      (1)
Replace x by 2, 2 2

Replace x by 1, f 1  1      (2)

1 1 1 1 5
Replace x by , 2f    f    2       (3)
2 2 2 2 2
1
Solve (1) & (3)  f    0;f (2)  1
2

Paragraph for Question Nos. 33 to 35

For a finite set A, Let A denote the number of elements in the set A. Also let F denote the set of all
functions f : 1, 2,..., n  1, 2,...., k  n  3, k  2  satisfying f  i   f  i  1 for every i, 1  i  n  1 .

30. F  (L3)

a) k n  k  1 b) k  k  1 c) k n 1  k  1 d) k  k  1
n n 1

31 If c  n, k  denote the number of functions in F satisfying (L3)

f  n   f 1 , then, for n  4, c  n, k  =

5
 a) k  k  1  c  n  1, k  b) k  k  1  c  n  1, k  1
n 1 n

c) k n 1  k  1  c  n  1, k  d) k n  k  1  c  n  1, k 

32. For n  k , c (n, k ) , where c  n, k  has the same meaning as in Q.34, equals (L3)
a) k n   1  k  1 b)  k  1   1  k  1
n n n 1

c)  k  1   1  k  1 k n   1  k  1
n n n 1
d)
Key: D-A-C
Hint 30. The image of the element 1 can be chosen in k ways and for each of the remaining (n - 1)
elements, the image can be defined in (k-1) ways, since f  i   f  i  1
 Total number of mappings in F = k (k - 1)n-1
31. Out of the total number of mappings in F, the number of mappings which satisfy f (n) = f (1) is
same as the number of maps which satisfy f  n  1  f (1) and this number is c(n – 1, k)
 c  n, k   F  c  n  1, k 
c  n, k   k  k  1  c  n  1, k 
n 1
32.
  k  1   k  1  c  n  1, k 
n n 1


 c  n, k    k  1   1 c  n  1, k    k  1
n n 1

  1
n 3
 c 3, k    k  1 
3

But c(3,k) = number of maps f in F for which f  3  f 1

=k(k-1) (k-2)
 c  n, k    k  1   1  k  1  k  k  2    k  1 
n n 1 2
 
  1  k  1
n

 c  n, k    k  1   1  k  1
n n

33. Let A = {1, 3, 5, 7} and B = {2, 4, 6, 8} and f:A  B, then the number of functions ‘f’ possible for (L2)
Column-I Column-II
(A) i  f  i   10, i  1, 3, 5, 7 (p) 16
(B) f  i   i  2, i  1, 3, 5, 7 (q) 24
(C) f  i   6, i  1, 3, 5, 7 (r) 0
(D) f  i   i  1, i  1, 3, 5, 7 (s) 81

Key: A  q; B  r ; C  p ; D  s
Hint. A(q), B(r), C(p), D(s)
(A) For 1, 3, 5, 7  A we have 4, 3, 2, 1, choices respectively.
(B) Image of f(7) should be greater than 9.
(C) For 1, 3, 5, 7  A we have 2, 2, 2, 2 choices respectively.
(D) for any i  A, we have 3 choices.

34. The number of the functions f from the set X = {1, 2, 3} to the Y = {1, 2, 3, 4, 5, 6, 7} such that f(i) 
f(j) for i < j and i, j  X is (L2)
(A) 6C3 (B) 7C3 (C) 8C3 (D) 9C3
Key: D
7
Hint C3 + 2 × 7C2 + 7C1 = 9C3 .

35. If x, y, z are distinct positive numbers, then xlny – lnz + ylnz – lnx + zlnx – lny  (L2)
(A) (0,) (B) (1, ) (C) (3, ) (D) (1, 3)
6
Key: C
x ln y ln z  y ln z ln x  z ln x ln y
Hint. p =  x ln y ln z .y ln z ln x .z ln x ln y
3
 xlny – lnz + ylnz – lnx + zlnx – lny  3 [as xlny = ylnx and x  y  z]
Hence p(3,)

36. f : R  [–1, ) and f(x) = ln( [|sin2x| + |cos2x| ] ) (where [] is greatest integer function). (L2)
(A) R–  range of f is null set
(B) f(x) is periodic but fundamental period not defined
 
(C) f(x) is invertible in  0,  (D) f(x) is into function.
 4
Key: A,B,D
Hint. Period of f(x) = |sin2x| + |cos2x| is /4
but f(x) = ln ([|sin2x| + |cos2x|])
Max. value of |sin2x| + |cos2x| = 2
f(x) = ln ([ 2 ]) = ln (1) = 0
 it is periodic function but fundamental period not defined.
f(x) is many one and into function

37. Let X = {1, 2, 3, ... 100} and Y be a subset of X such that the sum of no two elements in Y is divisible
by 7. If the maximum possible number of element in Y is 40 +  then  is
Ans: 5
Hint Let Yi be the subset of X such that yi = 7m + i, mI
Y0 = {7, 14, ..... 98}, n(Y0) = 14
Y1 = {1, 8, 15 ...99}, n(Y1) = 15
Y2 = {2, 9, 16 ... , 100}, n(Y2) = 15
Y3 = {3, 10, 16 .... 94}, nY3) = 14
Y4 = {4, 11, 18 ... 95}, n(Y4) = 14
Y5 = {5, 12, .... 96}, nY5) = 14
Y6 = {6, 13, .... 97}, n(Y6) = 14
The largest Y will consist of (!) an element of Y0 (ii) Y1 (iii) Y2 (iv) Y3 or Y4
 The maximum possible number of elements in Y = 1 + 15 + 15 + 14 = 45.

  
38. Let f :  0, 3    0,
   loge2  defined by f  x   log e x 2  1  tan 1 x then f  x  is (L1)
 3 
A) one – one and onto B) one – one but not onto
C) onto but not one – one D) neither one – one nor onto
Key: A
x 1
Hint: f ' x   f  x  is increa sin g in 0, 3 
x2  1
39. If the range of f ( x)  2  3 x , 3  x  1 is 0, 3 n  where n  N then n= (L1)
2
 x3 , 1  x  2
(A) 1 (B) 2 (C) 4 (D) 6
Key: C
Hint: The given function has local maximum at x= -1 ,minimum at x=0 and f(0) =0,f(-1) =1,
2
F(-3)= 2  3 3 f (2)  2 3 34
 RANGE OF F(X) =[0, 3
4]

7
40. Let f  x   2  cos x, x  R (L2)
STATEMENT-1: For each t  R  a point c in t , t    such that f 1  c   0 .
STATEMENT-2: f  t   f  t  2  t  R.
Key: B
Hint: S2 is true.
f 1  x    sin x
 every interval of length  t , t    contains at least one point , 0,  , 2
 f 1  x = 0 in t , t    : t  R
 S1 is true.
41. Match the following. (L2)
b
f : R  0  R . Let f  x   ax   ab  0  ,
x

Column I Column II

(A) a  0, b  0 (P) f  x  is a bijective

(B) (Q) f  x  is onto function

a  0, b  0
(C) a  0, b  0 (R) f  x  is one-one function
a  0, b  0 f  x  is neither onto nor one-
(D) (S)
one function
KEY : A – S, B – Q, C – Q, D – S
Hint Conceptual
42. A function f : R  R satisfies the equation f(x) f(y) – f(xy) = x + y  x, y  R and f(1) > 0, then (L2)
(A) f (x) f 1 (x)  x 2  4 (B) f (x) f 1 (x)  x 2  6
(C) f (x) f 1 (x)  x 2  1 (D) none of these
Key: C
Hint: Taking x = y = 1, we get
f 1 f 1  f 1  2
 f 2 1  f 1  2  0   f 1  2   f 1  1  0
 f 1  2 (as f(1) > 0)
Taking y = 1, we get
f(x). f(1) – f(x) = x + 1
 f(x) = x + 1  f 1 (x)  x  1
 f (x).f 1 (x)  x 2  1
 (C) is the correct answer.
43. STATEMENT–1: Reflection of function y  log e x in the straight line x + y = 0 is y = e x . (L1)
STATEMENT–2: The image of any point  ,   in the line x  y  0 is  ,    .
Key: A
Hint: Since the point  ,    lies on y  log e x
   log e       e
  = e   y  e  x .
44. Match the following (L2)
8
 Column1 Column2
(A) 2 3 (P) 0
 x 1 x
If f  x   4 x  8 3  72  4 2 is
defined  x  a, then f(a) =
(B)  1 (Q) 4
If 2f (x 2 )  3f  2   x 2  1,  x  0 then
x 
f(1) =
(C)
f (x)   x   
  , where [] denotes (R) 3
398 x  r

r 1 398
the greatest integer function and {.}
denotes the fractional part of x, then
f(3) =
(D) The number of points of discontinuity of (S) 2
  
f  x    tan x  cot x  ; x   ,  where [.]
12 2 
denotes the greatest integer function.
Key: A
Hint: A)  P, (B)  P, (C)  R, (D)  R
2 3
 x 1 x
(A) 4x  8 3  72  4 2  0
 2 2
2x 2x  2
 72  22x  3  0
 1 1
 22x 1     72
 4 8
22x  64 and x  3
 f(3) = 0
398
x  r  398
(C)   x   x
r 1 398 398
 f(x) = [x] + {x} = x
 f(3) = 3
(D)  f(x) is discontinuous when cot x  Integer
 
As x
12 2
 0  cot x  2  3
 Number of points of discontinuous = 3.
45. Let f and g be functions satisfying the conditions that (L3)
f  0   g  0   1, g  x   f '  x  , g '  x    f  x  then
a) f(x) is periodic function b) f ''  x    f  x 
c) Range of f(x) is [-1, 1] d) Range of f(x) is   2, 2 
Key: A, B, D
Hint:
g 1  x   f 11  x  .   f  x  , g 0  f 1 0  1  0
f 11  x  . f 1  x    f  x  . f 1  x    f 11  x  f 1  x  dx    f  x  . f 1  x  dx

9
  f  x    f  x 
1 2 2

   c  c 1
2 2
f 1  x  dx f  x
f 1  x   2   f  x      1dx  sin 1 f  0   1  c1   / 4
2
 x  c1 ,
2   f  x 
2
2

 
f  x   2 sin   x   sin x  cos x
4 
46. (L2)
Column I Column II
A The interval containing the complete set of P  1
1  x2 1  0, 2 
solution of the equation x  are
x x
B The interval containing the complete set of Q  1,1
values of ‘a’, for which  a  1 x  ay  1  0
is a normal to the curve xy =1, are
C Complete set of values of ‘a’ for which R 0,1
equation a sin 2 x  cos x  2a  0 has atleast
one solution belongs to the interval
D The interval containing the range of the S  1, 0
function
1
f  x 
1  2 cos x  3cos x  4 cos 6 x  .....
2 4

T  1
 1, 2 
Key: A – Q; B – Q, S, T;
C – P, Q, R,T; D – Q, R
1  x2 1  x2 1 1  x2
Hint:  a  x  x   .x  0  x 2  1  0, x  0  x   1,1  0
x x x x
  a  1
 b  slope  0   0  a  1, 0 
a
cos x 1
 c  a  a cos 2 x  cos x  2a  0  a  
1  cos x cos x 
2
1
cos x
1  1
cos x   0,1  cos x    2,    a   0, 
cos x  2
 d  if x  n
S  1  2 cos 2 x  3cos 4 x  ............. 1
cos 2 xS  cos 2 x  cos 4 x  ..........  2  , eq. 1  eq.  2 
sin xS  1  cos x  cos x  ..........
2 2 4


1 1
sin 2 xS  S  f  x   sin 4 x
1  cos x
2
sin 4 x
for x  n f  x  is not difined , Range of f  x    0,1

10
47 Column I (Function) Column II (Type) (L2)
(A) f : R  R (p) one-one
f(x) = x3 + 3x - 7
(B) f:RR (q) Onto
f(x) = x3 - 7x
(C) f : R  [2, 6] f(x) = 3 sinx - cosx + 4 (r) non periodic
(D) f : R  R (s) odd

f(x) = ln x  1  x 2 
Key: (A-p, q, r), (B-q, r,s), (C-q), (D-p, q, r, s)
Hint: A) f (x) = 3x2 + 3 > 0
 f one – one
B) f (x) = 0  x = 0,  7
f is not one – one
C) f (x) = 2(sin(x   / 6))  4
Range = [2, 6]
D) f (x) = Sinh-1 (x)
5m  3  5m  3 
48. If   =  (mN, m  3) and [ ] denote the G.I.F., then  can take (L2)
40  40 
(A) two values (B) one value
(C) infinite values (D) four values
Key: A
5m  3 1 1 7
Hint:  (5 + 52 + 53 + ... + 5m-1 + 2)   = ,
40 10 5 10
Paragraph for Question Nos. 58 to 60
Let f : A  B be a function, the f is said to be one-one, if for any x, yA x y  f(x)  f(y) and f is
said to be onto, if for any yB, there exist at least one xA, such that f(x) = y.

 x 3 , if x  Q
49. Let f : R  R define by f(x) =  3 , then f is (L1)
 x , if x  Q
c

(A) one-one and onto (B) one-one and into

(C) many-one and onto (D) many-one and into
Key: A
Hint: Let x1, x2 R, such that x1  x2
Now x13 = x23 or x13 = -x23 is possible only when x1 = x2 or x1 + x2 = 0, which is not possible, so f is
one-one and f is onto also.
50. f : R  R (L1)
 x  7, x Q
f(x) =  , then f is (L1)
 7  x, x  Q
c

(A) one-one & onto (B) one-one and into

(C) many-one & onto (D) many-one and into
Key: D
Sol : f ( – 7) = f ( 7)  0
 f is not one – one
There is no x  R such that f (x) = 7  1

51. f :R  R (L1)
11
  x  5, x  Q
f(x) =  , then f is
 5  x, x  Q
C

(A) one-one & onto (B) one-one and into

(C) many-one & onto (D) many-one and into
Key: A
The sum of all positive integral values of 'a ', a  1,500 for which the equation  x   x  a  0 has
3
52.
solution is ([.] denote G.I.F) (L2)
(A) 462 (B) 512 (C) 784 (D) 812
Key: D
Hint: a is integer then x must be integer, i.e., [x] = x
3
a x x
1  a  500  1  x  7, x  I

 
2
7 3  7.8   7.8 
 ai   x  x       812
x 1  2   2 

53 Let R   x, y ; x, y  R, x 2 2
 

4 2
 y  25 and R '  x, y  : x, y  R, y  x  then
9 
(A) domain of R  R '   3,3 (B) Range of R  R '   0, 4
(C) Range of R  R '   0,5 (D) R  R ' defines a function

Key: A,C
2 2
Hint: x  y  25
2
9y  9x
2 9
x  y
4
2 9y
x 
4
9y 2
 y  25  0
4
2
4y  9y  1  0
 y  4  4y  25  0
y4 y>0
Domain R  R ' is 3,3
Range of R  R '   0,5

Number of real values of x, satisfying the equation  x   5  x   6  sin x  0 . denoting the greatest
2
54.
integer function is (L2)
Key: 1
5  25  4sin x  24
Hint: x 
2.1
5  1  4sin x

2
1  sin x  1
4  4sin x  4
12
 3  1  4sin  5
0  1  4sin x  5
  x  is an integer  sin x  0
 x  3
 x 
ax  b a
55. Statement-1: The linear fractional function  ad  bc  0  cannot attain the value .
cx  d c
b  dy
Statement-2: The domain of the function g  y   is R ~ {a/c}. (L1)
cy  a
Key: a
ax  b b  dy
Hint: If f  x   y  then x 
cx  d cy  a
b  dy
Thus f 1  y   g  y   and dom g = R ~ {a/c}
cy  a
Hence f cannot attain the value a/c by definition.

f :  0,1  R is a differentiable function such that f  0   0 and f x  k f x

1
56. for all x

  0,1 ,  k  0  , then which of the following is/are always true ? (L2)

(A) f  x   0, x  R (B) f  x   0, x   0,1
(C) f  x   0, x   0,1 (D) f 1  k
Key: B
Hint:  f  x  2  k 2  f  x  2  0
  f '  x   kf  x    f '  x   kf  x    0

 f xe 
 kx '
f xe  0
kx '

 kx
 Exactly one of the functions g  x   f  x  e or
1

 x   f  x  e is non decreasing.
kx
g
2
But f  0   0  both function g and g have a value zero at x = 0
1 2
x   0,1 ,g  0   0 and g increasing  g  x   0  f  x   0
1 1 1
g  0   0 and g decreasing  g  x   0  f  x   0
2 2 2
 f  x   0x   0,1
  
 sin  x  1  sin   x  1 
57. If [.] denotes the integral part of x and f  x    x    , then (L1)
 1 x 
 
(A) f(x) is continuous in R (B) f(x) is continuous but not differentiable in R
(C ) f’(x) exists x  R (D) f  x  is discontinuous at all integral points in R
Key: D
Hint: At x = n, f  n  
n   
sin 
n 1  n 1
f n

 

13
 n 1 
f n  sin
n n
 f(x) is discontinuous at all n  1
 x   x   x  31
58. If 0  x  1000 and          x, where [x] is the greatest integer less than or equal to x,
 2   3   5  30
the number of possible values of x is (L1)
(A) 34 (B) 33
(C) 32 (D) none of these
Key : B
Sol :  LHS is an integer
 RHS is must be an integer for which x is multiple of 30.
 x  30, 60,90,120,.....,990
 Number of possible values of x is 33.
59. (L1)
 Column I   Column II
(A) Solution of (p)  ,1
x 2  1  sin x  x 2  1  sin x in
 2, 2
(B) Domain of f(x) (q)  
1  , 
 4 2
log1/ 2  x  7x  13
2

(C) Domain of single valued function (r)  3, 4 

y  f  x  given by 10 x  10 y  10 is
(D)   (s) x   2,    1, 0  1,    2 
Let x   0,  , then solution of f(x)
 2
1
 is
 log sin x tan x
Key : As; Br; Cp; DDELETED
Sol : A. x 2  1  sin x  x 2  1  sin x

exist only if  x 2  1 and (sin x) are of the same sign

 a  b  a  b  only if (ab  0)
 (x2 –1) sin x  0

  x  1 x  1 sin x  0.
thus x   2,    1, 0
x   2,    1, 0  1,   2
1
B. f  x  
log 1  x  7x  13
2

exists if log 1  x 2  7x  13  0

2
14
 x 2  7x  13  1 and x 2  7x  13  0
2
 7 3
 x  7x  13  0   x     0
2

 2 4
which is true for all x  R
again x 2  7x  13  1  x 2  7x  12  0
  x  3 x  4   0
3 x  4
thus Df is (3,4)
C. since 10 x  10 y  10
 10 y  10  10 x
 y  log10 10  10 x 
Now y is defined if 10  10 x  0
 101  10x
1  x i.e. x  1
 Df   ,1
 
D. Here x   0, 
 2
 0  sin x  1
again log a x  b  x  a b , if 0  a  1 and x  a b if a >1
1
Thus f  x   exists
 log sin x tan x
if  log sin x tan x  0
 log sin x tan x  0
 tan x   sin x   1
0

 tan x  1
     
 x   ,   x   0,  
 4 2   2 
 
 reqd. sol. is x   , 
4 2
60. (L-2) Let X = {1, 2, 3, ... 100} and Y be a subset of X such that the sum of no two elements in Y is
divisible by 7. If the maximum possible number of element in Y is 40 +  then  is
Key: 5
Hint: Let Yi be the subset of X such that yi = 7m + i, mI
Y0 = {7, 14, ..... 98}, n(Y0) = 14
Y1 = {1, 8, 15 ...99}, n(Y1) = 15
Y2 = {2, 9, 16 ... , 100}, n(Y2) = 15
Y3 = {3, 10, 16 .... 94}, nY3) = 14
Y4 = {4, 11, 18 ... 95}, n(Y4) = 14
Y5 = {5, 12, .... 96}, nY5) = 14
Y6 = {6, 13, .... 97}, n(Y6) = 14
The largest Y will consist of (!) an element of Y0 (ii) Y1 (iii) Y2 (iv) Y3 or Y4
15
  The maximum possible number of elements in Y = 1 + 15 + 15 + 14 = 45.

61. The number of functions f from the set A={0,1,2} in to the set
B={0,1,2,3,4,5,6,7} such that f (i )  f ( j ) for i<j and i, j  A is (L-II)

a) 8 C3 b) 8 C3  2(8 C2 ) c) 10 C3 d) 10
C4

KEY : C
HINT :
Hint: 0  1  2
 f (0)  f (1)  f (2)
f(0)<f(1)<f(2) 8 C3
f(0)<f(1)=f(2) 8 C2
f(0)=f(1)<f(2) 8 C2
f(0)=f(1)=f(2) =8C1

62.(L-1) Given two real sets A  a1 , a 2 ,.....a 2n  and B  b1 , b 2 ,.....b n  . If f : A  B

is a function such that every element of B has pre-image and f  a1   f  a 2  ,....  f  a 2n  , then

number of such mappings is

2n 2n 2n 1
a) Cn b) Cn 1 c) Cn 1 d) none of these

Key : c
Sol : Given every element of set B has been mapped to some element of set A.
Without loss of generality we can assume
b1  b 2  b3 ....  b n

Let a1 , a 2 ,.....a 2n  be divided into n subsets, so that each bi is the image of all the elements of one

subset alone.
Therefore the correct answer is the number of ways in which 2n elements can be divided into n
2n 1
non-empty subsets i.e., Cn 1 .

 
63. (L-2) f : R  R is f  x   ln x  x 2  1 , then find the number of solutions to the equations

f 1  x   e
x
.

Key : 2

Sol : 
Let f 1  x  be g  ln g  g 2  1  x 
 g  g 2  1  e x ….. (i)

and  g  g 2  1  e x ….. (ii)

x
Now, e x  e  x  2e

16
 x
Case I : x  0; e  e  x and e x  e  x

 e x  e x  2e x  e x  3e x
x
Case II : x < 0; e  e x and e x  e x
 e  x  e x  2e x  e  x  3e x
 Two solutions

 
64. (L-2)If a function satisfies  x  y  f  x  y    x  y  f  x  y   2 x 2 y  y3 x, y  R and f(1) = 2, then

a) f(x) must be polynomial function b) f(3) = 12

c) f(0) = 0 d) f(x) may not be differentiable
Key : a, b, c
Sol :  x  y  f  x  y    x  y  f  x  y   2y  x  y  x  y  
Let x – y = u; x + y = v
uf  v   vf  u   2uv  v  u 

f  v f u 
  vu
v u
 f  v   f u 
  v    u   constant
 v   u 
f x
Let x 
x


 f  x   x  x 2 
f(1) = 2
 1  2    1

f  x   x2  x

 x   y  x 2  y2
65. (L-2)If f '   .f    x, y  R  and f(1) = 1, then f 2  x  is
y x xy

a) x  ln x b) x 2  2 ln x c) x 2  ln x d) x  2 ln x
Key : b

17
 x x x y 1
Sol : f '   f      f  x  .f '  x   x 
y y y x x
2
 2f  x  f '  x   2x 
x
Integrating

  f  x    x 2  2 log x  c
2

but f(1) = 1  c= 0  f  x    x 2  2 log x

2

x 2  bx  1 1
66. (L-2) f : R  R, f  x   , if the function f(x) and have the same bounded set as their
x  2x  b
2
f x

range, then the value of b is

a) 2 3  2 b) 2 3  2 c) 2 2  2 d) 2 2  2
Key: a

x 2  bx  1
Sol : y  x 2 y  2xy  by  x 2  bx  1
x  2x  b
2

 x 2  y  1  x  2y  b   by  1  0

 2y  b 2  4  y  1 by  1  0

  4  4b  y 2  4y  b 2  4  0 
1
For f(x) and to have the same range thus equation must have reciprocal roots
f x

 b 2  4  4  4b

 b 2  4b  8  0
 b  2  2 3
=1
67. (L-2)
COLUMN – I COLUMN – II

A) f : N 1 f x  p) one-one function

2
 x  12  x    x  12   x  1  , then f(x)
 
will be (where [.] is greatest integer function)
B) log x q) many one function
Let f  x   , and solutions of f(x) = k be
x
denoted by g(k) {where k < 0},

18
 g :  , 0    0,1 , then y = g(x) will be

C) Let f : R  R be a continuous function such r) onto function

that f  x   f  y  x  y for all x, y  R ,
.
then f(x) will be
D)    s) into function
f : R ~ 0    ,  such that
 2 2
x 1/x
1 1
f x   dt   1  t2  x  0 , now f(x)
0 1  t 2
0

will be
t) neither one-one nor onto

Key :
Sol : A) If x  N then f(x) = x
 It is one-one into function
B) log x = kx  for K < 0  only one point
 g  x  :  , 0    0,1 is a bijection

C) f 1 (x)  0  f (x) is in incr. fn  f (x) is one – one

x 1/x
D) f  x    tan 1  t     tan 1  t  
0 0

1
 tan 1 x  tan 1  
x

 if x  0
2

  if x  0
2
It is many one & into function
 1 
y  sin 1  2x   tan 1  2
  2x  
68. (L-2)Number of common points for the curves and
 
y  cos 1  2x  5   1 is (where . denotes greatest integer function)

a) 0 b) 1 c) 3 d) none of these
Key : a
 1  1 
Sol : As domain of first function is   , 0     and domain of 2nd function [-3, -2] there is no
 2  2
common point.
19
 PASSAGE
x f  t  .y 2
A line    t z   is perpendicular to the line of intersection of the planes
2 t
1
 
t.f  t  x  f  2  z  f   t   0 and ty  f   t  z  f t 2  0 , where t  R  0
t 
69. f(t) is
a) even function b) odd function
c) neither even nor odd function d) both even and odd function
  
70. If t  tan , where   R   2n  1 , n  ; n  I , then
 2 

 
a) f  tan     tan  2  .f cot 2  
b) f  tan    sin  2  .f cot 2  
c) f  tan    sin  2  .f  sec  
2
d) f  tan     tan .f  cot  

71. f 1 .f  2  .f  3 .......f  n  is equal to

a) n! b) 2n c) 1 d) 0
Key : b-a-d
Sol : The normals to the planes and the given line are coplanar. Hence applying the condition, the
t 1
2 
f t t2
1
functional equation obtained is tf  t  0 f 2  0
t 
0 t f  t 

1
 2tf  2   t 2f   t   f  t   0 ….. (i)
t 
1
Also, 2tf  2   t 2f  t   f   t   0 ….. (ii) replacing t by –t
t 

69.  
Adding (i) and (ii)   t 2  1 f  t   f   t   0  f  t   f   t 

f(t) is odd function

2t  1 
70. (i) – (ii)  f  t   f 
1 t2  t2 
71. For t = 1, in equation (i), f(t) = 0
72. (L-1)Statement-1 : cos 1  cos x   x is true for all x   0, 

Statement-2 : If f :A  B is an invertible function for all f 1 : B  A , then

 
f f 1  x   x, x  Band f 1  f  x    x, x  A

20
Key : a
Sol : By definition of inverse function.
PASSAGE
(L-3)A cubic function f  x    x 3  ax 2  bx  c . If f  x  is an odd function and f  x   0 at x = -1. Now

the domain of function is reduced so as to make f(x) invertible such that f 1  x  remains in

2nd and 4 th quadrant. Then

73. f 1  x   f 1  x   0 has

a) no solution b) exactly one solution

c) infinite solutions d) exactly three solutions
74. Range of f 1 x is

a) (, 0] b) (, 1] c) [0, ) d) [1, )

75.  
max f 1 x , f 1  x  is identical with

a) f 1  x  b) f 1  x  c) f  x  d) f 1  x 

Key : c – b - b
Sol : f 1  x  remains in 2nd and 4th quadrant

So f(x) defined as

f 1  x  defined as

21
76. (L-2)

Column – I Column – II
A) The number n = xyzxyz will always be 1. 7
divisible by
B) The numbers of the form  6n  1 in the 2. 11

2
64  x
domain of f  x   x ln  x  1  is /
sin x
are
C) Maximum of max 3. 13
12sin x  5cos x;  13  2x will be (where

{.} is fractional part function)

D) Period of the function 4. 5
 2x 
f  x   tan  x    cos 
x  11

 7  sin  2x 
 
 7 
will be (where {.} is fractional part function)
5. 77

Key : A – 1, 2, 3, 5; B – 1, 4; C – 3; D – 1, 5
Sol : A – 1, 2, 3, 5
Consider number n = xyzyxz
= x  105  y  104  z  103  x  102  y  101  z  1

22
  
 1001 102 x  10y  z  7  11 13   integer 

Hence n is divisible by 7, 11, 13

B – 1, 4
Domain of f(x) is (1,8]  , 2

In this set numbers of the form 6n  1 are 7, 5

C–3
max 12sin x  5cos x  , 13  2x  12sin x  5cos x 

 Required maximum value is 122  52  13

D – 1, 5
Period of f(x) will be LCM of (1, 7, 1)  7
2 x  2 x
77. (L-1) f  x   x
2  2 x

Column – I Column – II
A) If 1  x  2, then f  x  satisfies p) f(x) > 1

1
B) If 0  x  , then f  x  satisfies q) f  x   1
2
1 r) f(x) > 3
C) If   x  0, then f  x  satisfies
2
D) If 2  x  1, then f(x) satisfies s) f(x) < -3

t) 1  f  x   1

Key : A – 1; B – 1, 3; C- 2, 4; D – 2
Sol : f x 1 x  0

f  x   1  x  0

1
f x  3  0  x 
2
1
f  x   3   x0
2
78. (L-2)If A  0 , c, d, u, v are non zero constants, and the graphs of f  x   Ax  c  d and

uc
g  x    Ax  u  v intersect exactly at two points (1, 4) and (3, 1), then the value of equals
A
a) 4 b) -4 c) 2 d) -2
Key : b
Sol : The figure is a parallelogram and the diagonals bisect each other

23
   u   c 
      3 1
 A A
uc
  4
A


79. (L-1)If the mapping f  x   ax  b, a  0 maps  1,1 onto [0, 2], then cot cot 1 7  cot 1 8  cot 1 18 is 
a) f  1 b) f(0) c) f(1) d) f(2)

Key : d
Sol : f(x) = ax + b f '  x   a  0

 f(x) is an increasing function

 f  1  0 and f 1  2  a  b  0;a  b  2  f  x   x  1


Now cot cot 1 7  cot 1 8  cot 1 18 
 1 1 1  3  1 
 cot  tan 1  tan 1  tan 1   cot  tan 1    tan 1   
 7 8 18    11   18  

  1 
 
 cot  tan 1     cot cot 1 3  3  f  2 
  3 
200 200
80. (L-2)Consider two polynomials f  x  and g(x) as g(x) =   r x r and f  x    r x r . Given (i)
r 0 r 0

200
r  1r  100 (ii) f(x + 1) = g(x). Let A =   r . Find the remainder when A is divided by 15.
r 100

Key : 1
200 200
  r x r   r 1  x 
r
Sol :
r 0 r 0

 0  1x   2 x 2  ....   200 x 200

 0  1 1  x   ....  200 1  x 
200

24
 Equating coefficient of x100 , we get 100  100C100  101C100  .....  200C100  201C101

Similarly we can find 101 ,.......,  200 .

200
  r  201C101  201C102  .....  201C201
r 100

A = 2200
When A is divided by 15 remainder is 1.
81. (L-2)Which of the following function (is) are injective
1
a) f  x   x  1 , x   1,   b) g  x   x  , x   0,  
x
c) h  x   x 2  4x  5, x   0,   d) h  x   x 2  4x  5, x   0,  

Key ; a, c, d
Sol : For choice (A), we have
f  x   f  y  : x, y  [1, )

 x 1  y 1

 x 1  y 1  x  y
So, f is an injection.
For choice (B),
5 1 5
g  2  and g   
2 2 2
1 1
2  but g  2   g  
2 2
Thus, g(x) is not injective.
It’s easily seen that choices (C) and (D) are also correct.
x2
82. (L-1)Let y  g  x   , then
x 1
a) g 1 1  3 b) x = g(y)

c) y increases with x for x > 1 d) none of these

Key ; B
x2
Sol : y  gx 
x 1
 yx  y  x  2

 x  y  1  y  2

y2
x  x  g  y
y 1

25
 Again, g(1) doesn’t exist [ domain  R  1 ]

Also, g is a rational function of x.

83. (L-1)
COLUMN – I COLUMN – II

A) Domain of the function p) 1 

 2 ,1
1
y  x2
log10 1  x 

B) The range of f  x   log 2  x 4  6x  q)  2, 0    0,1

C)
If f  x  
 2 tan x  sec x  1  cos 2x  .
2
Then
r) (-1, 0]

2
the range f(t) is
D) ex  e
x s) [0,2]
The range of the function f  x  
ex  e
x

Key : A – q; B – p; C – s; D - r
Sol : A–q
For y to be defined
i) log10 1  x  must be defined  1  x  0  x  1

ii) log10 1  x   0  1  x  10  1  x  1  x  0

iii) x  2  0  x  2
 from (i), (ii) and (iii), we get 2  x  1 and x  0
 2  x  0 or 0  x  1


-2 0 1

Hence domain =  2, 0    0,1

B–p

f  x   log e  
x  4  6  x log 2 e

1  1 1 
f ' x      log 2 e
x 4  6x  2 x 4 2 6x 


1
.
6x  x4log 2 e

x 4  6x 2 x 4 6x  
26
= a positive number  6x  x4 
6x  x4  0 x  5

Sign scheme for f '  x  is for  6  x  x  4 is 


-2 0 1

Put x = 6
Greatest value of f  x   f  5   log 2 2  1

1 1
f  4   log 2 2  log 2 2 
2 2
1 1
f  6   log 2 2  log 2 2 
2 2
1
 Least value of f  x  
2
1 
Hence, Range =  ,1
2 
C–s
 1  2
RHS =  2 tan x   cos x
 cos 2 x 
 2sin x cos x  1  1  sin 2x
If sin 2x = t, then we have
f  t   1  t,  where, t  sin 2x &  1  t  1

 domain is [-1, 1]
Adding 1 throughout 0  1  t  2 or 0  f  t   2

 Range of f(t) is [0, 2]

D–r

ex  e ex  e
x x
f x   y  say   y 
ex  e ex  e
x x

if x  0

ex  e
x
0 e x  e x
then y    0  y  0 and if x < 0, then y 
ex  e
x
2e x e x  e x

e2x  1 y 1
y  e 2x 
e 1
2x
1 y

27
 y 1 y 1
0 1 0   1
1 y y 1

 y   1,1 and y   , 0   1,   and y = 0

 Range of f(x) is (-1, 0]

84. (L-1)Let f : [ -1, 1] upto [3, 5] be a linear polynomial. Which of the following can be true ?
 1  7  15  1  1   1 
a) f    b) f    c) f (0)  4 d) f    f    8
 2  2  4 4 2  2 
Key : a, b, d
sol : Let f (x) = ax + b
Case-I : f is increasing
f (-1) = 3 and f (1) = 5  f (x) = x + 4
Case-II : f is decreasing
F(-1) = 5 and f (1) = 3  f (x) = 4 – x
85. (L-1)Consider the following statements
S1 : If f(x) and g(x) both are non-zero odd functions, then the composite function fog, if exists, is
even.
S2 : If f(x) and g(x) both are strictly decreasing functions, then the composite function fog, if
exists, is increasing.
S3 : If f(x) and g(x) both are one-one functions, then the composite function fog, if exists, is one-
one
State, in order, whether S1 ,S2 ,S3 are true or false

a) TFT b) FFT c) FTT d) TTT

Key : d
Sol : S1 : f   x   f  x  , g   x   g  x 

Now fog   x   f  g  x    f  g  x    fog  x 

fog if exists it is even

S1 is true

S2 : S2 is also true x1  x 2  g  x1   g  x 2   f  g  x1    f  g  x 2  

S3 : is also true

passage
(L-2)Let f be a real valued non constant function such that
f 10  x   f 10  x  and f  20  x   f  20  x  for every real number x, then

28
86. Consider the statements (i) f is an odd function, (ii) f is an even function, (iii) f is a periodic
function, then which of the following options is correct
a) (i) and (ii) are true b) (ii) and (iii) are true c) (i) and (iii) are true d) only (i) is true
Key : c
Sol : f 10  x   f 10  x  ;f  20  x   f  20  x 

 f  20  x   f   x   f  40  x   f   x 

f  40  x   f  20  x   f  20  x   f  x 

f  40  x   f  20  x   f  20  x   f  x 

 f is periodic function and f(-x) = -f(x) hence f is odd.

40
87.  f  x  dx =
0

40 20 10
a)  f  x  dx b) 2  f  x  dx c) 2  f  x  dx d) 0
0 0 0

Key : d
Sol : f  40  x   f  20   20  x    f  x 
40
  f  x  dx  0
0

88. Consider the statements (i) lim f  x  must exist, (ii) lim f  x  must exist, (iii) lim f  x  must
x 10 x 20 x 0

exist, then which of the following options is correct

a) (i) and (ii) are correct b) (ii) and (iii) are
correct
c) (i), (ii) and (iii) are false d) only (i) is correct
Key : d
Sol : Lt f  x   Lt f  x   Lt f 10  h 
x 10 x 10 h h 0

Lt f  x   Lt f  x   Lt f 10  h 
x 10 x 10 h h 0

Since f(10 – h) = f(10 + ), Lt f  x  must exist

x 10

But Lt f  x  may not exist

x 20

Since f is odd function Lt f  x  may not exist.

x 0

29
89. (L-2)Let f  x  , f '  x  and f ''  x  are all positive x   0, 7  . If f 1  x  exists, then

 29 
f 1  5   4f 1  6   5f 1   is
 5 
a) always positive b) always negative c) non-negative d) non-positive
Key : b
f x  0 

Sol : f '  x   0  x   0, 7 
f ''  x   0 

x 0 is a point which divides line segment (5, 0) and (6, 0) in 4 : 1 internally

29
x0 
5

 f 1  6  .4  f 1  5  
now P  x 0 ,
 5  
 and Q x 0 , f  x 0 
1

 
y-coordinate of P < y-coordinate of Q

4f 1  6   f 1  5   29 
 f 1  
5  5 
 29 
 4f 1  6   f 1  5   5f 1    0
 5 
 1 1
90. (L-3) S1 :  x     x  if x 
 2 2
1
S2 :  nx   n  x  if x 
n
Where [ ], { } stands for greatest integer and fraction part functions respectively and n is a
natural number
a) S1 and S2 both are correct b) S1 is correct and S2 is INCORRECT

c) S1 is INCORRECT and S2 is correct d) S1 and S2 both are INCORRECT

Key ; a
Sol : Let x   x   F When 0  F  1

30
 1
If  F  1 then
2
x  F   x  2F   x  F    x  1
Since 1  2 F  2
1
If 0  F  then
2
x  F   x  2F   x  F    x
Since 0  2 F  1
1
 F   x 
2
 S1 is true
S2 is also true
91. (L-2)The condition on the parameter a, b, c and so that f (x, y) = a sin (x + y) + b cos (x + y) + a sin (x
– y) + d cos (x – y) can be written as f (x, y) = g (x) . h (y)
a) a2 + b2 = c2 + d2 b) a2 + c2 = b2 + d2 c) a2 + d2 = b2 + c2 d) a2 + b2 = c2 + d2= 1
Key: b
Sol : f  x, y    a  c  sin x cos y   b  d  cos x cos y   a  c  cos x sin y   d  b  sin x sin y
 sin x  a  c  cos y   d  b  sin y   cos x  b  d  cos y   a  c  sin  y 
  a  c  cos y   d  b  sin y   b  d  cos y   a  c  sin y
 ac bd
a  c  d  b  a 2  c2  b2  d 2
92. (L-2)Which of the following statement(s) are correct for the following function
 1 
f (x)  n 
 [cos x]  [sin x] 
[.] denotes greatest integer function.
 
a) f (x) is not injective mapping
b) f (x) is continuous in its domain
c) f (x) is derivable at all points in its domain
d) f (x) is non-derivable at two points in its domain exhibits sharp corner
Key : b,d
Y

x
O

Sol :
93. The solution setof x  1  x  5  6 is:

a)  5, 1 b) 1,   c)  ,  5  d) R

Key:A
Hint:

31
-5 1
94. (L-2)if it is known that a1  ........  a n  0 , then the value of a1  a 2 ........a n  0 are
a) a1  a 2 ,  .....a n  1 b) a1  a 2 ,  .....a n  0
c) a1  a 2 ,  .....a n  1 d) None of these

Key: D
Hint: | a1 +| a2 | + ……..+ an |  | a1| + | a2 | + ……+ | an |
2
95. (L-1)The sum of all the real roots of the equation x  2  x  2  2  0 is

a) 2 b) 3 c) 4 d) 1
Key: C
Hint: x2 

t2  t  2  0
t 1 t  2
  2  1  x3
 1

96. If x  3x  2 2  3 , then

7 3 3 7 7 3
a) x   , b) x   ,  c) x  ,  d) None of these
4 2 2 4 4 2
Key: A
Hint: 3x  2  0
2
x
3
x  7 / 4
2
x
3
x  3/ 2
97. (L-2)Roots of the equation x 2  x  6   x  2  are

a) -2, 1, 4 b) 0, 2, 4 c) 0, 1, 4 d) – 2, 2, 4
Key: D

32
 x2  x  6  0
Hint: x2
 x  3 x  2   0
x  2 x  3 xf  2,3
x  x  6  x  2 x  x  6  x  2
2 2

x2  2x  8  0 x2  4  0
 x  4  x  2   x  2  x  2   0
x  x  4 x  2 x  2

98. (L-1)Solution set of the inequation x  1  x  0 is

 1  5   1  5   1  5 
a)  1,  b)  1,1 c) 1   d) 1  
 2   2   2 

Key: A
Hint:


t 5
2

10x 2
99. (L-1)The equation x  2  x2
3x
has solutions which are

a) Two positive & two negative b) One positive & three negative
c) Three positive & one negative d) Two positive and one zero
Key: d
Sol : 10 x2 = 3x  x = 0, 3/10, by Mod x = 2 also
100. (L-1)Solution of the equation x  2  4  x  6  x is
4 4 4
a) x  4  b) x  4  c) x  4  d) x = 4
5 5 5
Key: D
Hint:By substitution x = 4
101. (L-2)If the equation x  1  x  3  2k  1 has exactly two solutions then k contains:

 1  1 3 3 
a)  2,   b)   ,   c)   ,  d)  ,  
 2  2 2 2 
Key: B,D
Hint:

33

2
102. (L-2)Conceder the equation x  3 x  2  0, x  R . Then
2

a) it has four real roots b) sum of its roots is zero

c) product of its root is 2 d) it has atleast two equal roots
Key: A, B

x 3 x  2  0
2
Hint:

x  1,  2
x  1, 2

103. (L-2)For x  0 , the expression  x x is

a) Equal a non –negative number b) dose not exits

c) equal a positive number d) equals – x
Key: A, C, D
Hint: Put x = – 1
104. Let y  f  x  , y  g  x  , y  b  x  be three invertible functions defined from R to R

COLUMN – I COLUMN – II

A) y  f g  x  p) may not be onto

B) 
y  f g  h  x   q) always one-one

C) y  f x  gx  h x r) always onto

D) f xgx s) may not be one-one

y ,  h  x   0
h x

Key : A – qr; B – qr; C- ps; D- ps.

Sol : using the concept

 a   2a   a  b  1 
105. (L-3) Let H  a, b         ...    where a  R, b  N, b  1 , then the value of
b  b   b 

 1 H  50 49  
  is equal to …….. (where [.] denotes the greatest integer function).
 7 6 

Key : 2
34
  50  100   49  49  
Sol : H  50, 49         .....   
 49   49   49 
50  49
  49   25  49
2
 1 25  49 
 2
7 6 
 5 
106. (L-3)The inverse of the function f  x   sin x  sin x x   ,3  is
2 
x 1 
a) sin 1   x   0, 2 b) sin 1 x, x   ,1
2 2 

c) sin 1 x x   1,1 d) does not exists

Key : a
 y
Sol : y  2sin x  x  sin 1  
2
 x
 f 1  x   sin 1   for x   0, 2
2
2
107. (L-2)The values of x satisfying the inequality  tan 1 x    tan 1 x   2  0 where [.] denotes GIF is

      
a) [ tan1, ) b)   , tan 1 2  c   tan1, tan 2 d)  , 
 4   4 4
Key : a

Sol :  tan 1
 
x   2  tan 1 x   1  0  1   tan 1 x   2  x  [ tan1, )
108. (L-1)Statement-1 : Let f : R  R and g : R  R be two one-one and onto functions such that they are
the mirror images of each other about the line y = a. If h(x) = f(x) + g(x) then h(x) is many – one
and into.
Statement-2 : A constant function on R to R is many one into
Key : a
Sol : Let f(x) = ax + b then the image in y = a is f(x) = -ax + k
h(x) = k + b is a constant function . It is many one into.
109. (L-2)Let f '  1,1   1,1 be defined by Column match the functions with their nature in Column-II

Column – I Column – II

A) x 3 sgn x p) One-one

B) e x sgn x q) Many – one


C) sin sin 1 x  r) Onto

2 1 s) Into
D) sin  sin x 

35
Key : A – q,s; B – p, s; C – p, r; d – p, s
x
Sol : a) x 3sgn x  x 3  x 2 x It is even  many one into
x
x
e x
b) e x sgn x  for x  0  xe x   xe x for x  0
x
It is one as it is monotonic . It is into
  
c) x   1,1  sin 1 x      sin  sin 1 x    1,1
 2 2
It is a bijection both one – one and onto
2
d) sin 1  sin x  clearly it is one – one into

110. (L-1)If y = f(x) is a linear function satisfying the relation f  x.y   f  x  .f  y  , x, y  R then the
x
 
curve y 2   sin t  a 2 t 3  bt dt  ,   R  cuts y  f 1  x 
0

a) at one point b) at least at two points

c) exactly at two points d) exactly at three
points
Key : b
x
Sol : f  x   x  f 1  x   x it cuts y 2   sin t  a 2t 3  bt dt   at least two points.
0

 1  2 1  2 x 
111. (L-1)Let f : R  0,1  R satisfying f  x   f   then
 1  x  x 1  x 

 9e 
4
(A)  f  x  dx  In  4 
3

(B) the graph of y  f  x  crosses x-axis at x  1

(C) f  2   f  3  5

(D) f  2   f  3  6

Key : A,B,C
 1  2 2
Hint : f  x   f   
 1 x  x 1 x
1  1 x 1
Replacing x by and by  1   in equation (1) one by one and on solving, we get f  x  
1 x  x x 1

112. (L-1)Match the following :

2 x  2 x
f  x 
2 x  2 x

36
 Column – I Column – II

(A) If 1  x  2 , then f  x  satisfies (p) f  x  1

(B)
If 0  x 
1
, then f  x  satisfies
(q) f  x   1
2

(C) 1
If   x  0 , then f  x  satisfies
(r) f  x  3
2

(D) If 2  x  1 , then f  x  satisfies (s) f  x   3

(t) 1  f  x   1

Key : A  p; B  p,r; c  q,s; D  q

Hint : f  x   1  x  0
f  x   1  x  0
1
f  x  3  0  x 
2
1
f  x   3   x0
2
113. If f (x) = x + tan x and g (x) is inverse of f (x), then g (x) is equal to
1 1 1
a) b) c) d) none of these
1  (g(x)  x) 2
2  (g(x)  x) 2
2  (g(x)  x) 2
Key : d
Sol : fog(x) = x  f1 (g(x)) g1(x) = 1
1 1
 g1 (x)  1 
f (g(x)) 1  sec 2 (g(x))

114.(L-1)Given two real sets A  a1 , a 2 ,.....a 2n  and B  b1 , b 2 ,.....b n  . If f : A  B is a function such

that every element of B has pre-image and f  a1   f  a 2  ,....  f  a 2n  , then number of such

mappings is
2n 2n 2n 1
a) Cn b) Cn 1 c) Cn 1 d) none of these

Key : c
Sol : Given every element of set B has been mapped to some element of set A.
Without loss of generality we can assume
b1  b 2  b3 ....  b n

Let a1 , a 2 ,.....a 2n  be divided into n subsets, so that each bi is the image of all the elements of one

subset alone.

37
 Therefore the correct answer is the number of ways in which 2n elements can be divided into n
2n 1
non-empty subsets i.e., Cn 1 .

 
115. (L-2) f : R  R is f  x   ln x  x 2  1 , then find the number of solutions to the equations

f 1  x   e
x
.

Key : 2

Sol : 
Let f 1  x  be g  ln g  g 2  1  x 
 g  g 2  1  e x ….. (i)

and  g  g 2  1  e x ….. (ii)

x
Now, e x  e  x  2e

x
Case I : x  0; e  e  x and e x  e  x

 e x  e x  2e x  e x  3e x
x
Case II : x < 0; e  e x and e x  e x
 e  x  e x  2e x  e  x  3e x
 Two solutions

 
116. (L-2)If a function satisfies  x  y  f  x  y    x  y  f  x  y   2 x 2 y  y3 x, y  R and f(1) = 2, then

a) f(x) must be polynomial function b) f(3) = 12

c) f(0) = 0 d) f(x) may not be differentiable
Key : a, b, c
Sol :  x  y  f  x  y    x  y  f  x  y   2y  x  y  x  y  
Let x – y = u; x + y = v
uf  v   vf  u   2uv  v  u 

f  v f u 
  vu
v u

38
  f  v   f u 
  v    u   constant
 v   u 
f x
Let x 
x


 f  x   x  x 2 
f(1) = 2
 1  2    1
f  x   x2  x

 x   y  x 2  y2
117. (L-2)If f '   .f    x, y  R  and f(1) = 1, then f 2  x  is
y x xy

a) x  ln x b) x 2  2 ln x c) x 2  ln x d) x  2 ln x
Key : b
x x x y 1
Sol : f '   f      f  x  .f '  x   x 
y  y y x x
2
 2f  x  f '  x   2x 
x
Integrating

 f  x    x 2  2 log x  c
2

but f(1) = 1  c= 0  f  x    x 2  2 log x

2

x 2  bx  1 1
118. (L-2) f : R  R, f  x   , if the function f(x) and have the same bounded set as their
x  2x  b
2
f x

range, then the value of b is

a) 2 3  2 b) 2 3  2 c) 2 2  2 d) 2 2  2
Key: a

x 2  bx  1
Sol : y  x 2 y  2xy  by  x 2  bx  1
x  2x  b
2

 x 2  y  1  x  2y  b   by  1  0

 2y  b 2  4  y  1 by  1  0

  4  4b  y 2  4y  b 2  4  0 
1
For f(x) and to have the same range thus equation must have reciprocal roots
f x

 b 2  4  4  4b

39
  b 2  4b  8  0
 b  2  2 3
=1
119. (L-2)
COLUMN – I COLUMN – II

A) f : N 1 f x  p) one-one function

2
 x  12  x    x  12   x  1  , then f(x)
 
will be (where [.] is greatest integer function)
B) log x q) many one function
Let f  x   , and solutions of f(x) = k be
x
denoted by g(k) {where k < 0},
g :  , 0    0,1 , then y = g(x) will be

C) Let f : R  R be a continuous function such r) onto function

that f  x   f  y  x  y for all x, y  R ,

then f(x) will be

D)    s) into function
f : R ~ 0    ,  such that
 2 2
x 1/x
1 1
f x   dt   1  t 2  x  0 , now f(x)
0 1 t
2
0

will be
t) neither one-one nor onto

Key :
Sol : A) If x  N then f(x) = x
 It is one-one into function
B) log x = kx  for K < 0  only one point
 g  x  :  , 0    0,1 is a bijection

C) f 1 (x)  0  f (x) is in incr. fn  f (x) is one – one

x 1/x
D) f  x    tan 1  t     tan 1  t  
0 0

1
 tan 1 x  tan 1  
x

 if x  0
2
40
 
  if x  0
2
It is many one & into function
 1 
y  sin 1  2x   tan 1  2
  2x  
120. (L-2)Number of common points for the curves and
 
y  cos 1  2x  5   1 is (where . denotes greatest integer function)

a) 0 b) 1 c) 3 d) none of these
Key : a
 1  1 
Sol : As domain of first function is   , 0     and domain of 2nd function [-3, -2] there is no
 2  2
common point.
121. Which of the following is true? ({.} denotes the fractional part of the function)
 1 x
a) lt  x sin   1 b) lt  2 
x 
 x x  2 x x2
x  1
c) lt  2   d) lt  x sin   
x 1 x  x  2 x 
 x
KEY: A, B, C
122. Given a real valued function f such that
tan 2  x
f  x  , for x  0

x2   x
2

= 1 for x  0
=  x cot  x for x  0 , where [x] is integral part of x and {.} is Fractional part of x. Then
a) lt  f  x  1 b) lt  f  x   cot1
x 0 x 0

 lt  1  lt  
2
c) cot 1 
f  x d) tan 1 
f  x 
x 0 x 0 4
KEY: A,C,D
x b  b
123. Statement I : If a and b are positive and [x] denotes greatest integer  x, then lt  
x 0 a  x  a

Statement II: lt
 x  0 , where {x} denotes Fractional part of x.
x  x
KEY: A
 1x 
e 1 
124. Statement I : lt  x   1 (where [.] represents greatest integer function) does not exist.
x 0  
 ex 1
 
 x1

 e 1 
Statement II: lt does not exist
x 0  1 
 ex 1
 
KEY: B
Paragraph for Questions
PASSAGE – III :

41
 a sin x  bx  cx 2  x 3
Given that lim is finite and equals K.
x 0 2 x 2 log 1  x   2 x 3  x 4

125. The value of a is

a) 0 b) 6 c) 12 d) 4
KEY: B
126. The value of c is
a) 1 b) 2 c) 4 d) 0
KEY: D
127. The value of K is
1 1 3 1
a) B) c) d)
4 8 40 12
KEY: C
n2  n
 1 
128. lim 1  2  
n 
 n  cos n 
1
a) 1 b) e c) d) e 2
e
KEY: B

 
n
129. lim  cot 1 r 2  r  1 
n 
r 1

 
a) b) c)  d) 0
4 2
KEY: A
1
130.
x 

lim e x  x  x 
1
a) 0 b) 1 c) e d)
e
KEY: C
 sin 2 x b 
131 If lim  3  a  2   0 , then
x 0
 x x 
4 4
a) a  , b  2 b) a  c) a  , b  2 d) a
4 4
,b  2 , b  2
3 3 3 3
KEY: C
132. Let f :    be a function such that f  0   1 and for any
x, y  , f  xy  1  f  x  f  y   f  y   x  2 holds. Then lim f  x  
x 0

a) 0 b) 2 c) – 1 d) 1
KEY: D
 xa
x

133. The value of ‘a’ for which lim 

x  x  a
  e is
 
1
a) 1 b) – 1 c) d) 2
2
KEY: C
f  3x 
134. Suppose f :    is a positive increasing function with lim 1
x  f  x
f  2x
STATEMENT-1: lim 1
x  f  x
42
 STATEMENT-2: x  0  0  f  x   f  2 x   f  3 x  :
KEY: A
Paragraph for Questions
Suppose f, g, h are real- valued functions defined on some deleted neighbourhood N of ‘a’ such that
g  x   f  x   h  x  for all x  N . If lim g  x  and lim h  x  both exist and each equals  , then
xa xa

lim f  x  also exists and equals  . (This is called the squeezing principle or sandwich theorem )
xa

135. If f  x   sin  x  1  sin   x  , then lim f  x 

x 

1
a) does not exist b) equals 0 c) equals 1 d) equals
2
KEY: B
If f  x    tan x  sin x   , where[.] is the greatest integer function, then lim f  x  
2
136.
  x 0

1
a) 0 b) 1 c) 2 d)
2
KEY: A

 
sin 2 x
lim 1cos ec x  2cos ec x       ncos ec x
2 2 2
137.
x 0

a) equals 0 b) equals 1 c) equals n d) does not exist

KEY: C
138. Mach the following
Column I Column II
lim 1  x 2 
cot x
(A) x 0
(P)
0
x

(B) 1  e  1  cos x (Q)

lim 1
x  0 sin x
  x x
(C) lim x  ln 1    ln  (R) 2
x 
  2 2
Let  be a rational number and
(D) an  sin  n !  , n  1, 2,3    (S)
1
Then lim an 2
n 

KEY: A-Q, B-Q, C-R, D-P

 tan 2 x  2sin 2 x  3sin x  4  sin 2 x  6sin x  2  , x   / 2


139. Define f :  0,    R by f  x      is
k ,x  /2

continuous at x  ,then k =
2
1 1 1 1
A) B) C) D)
12 6 24 32
KEY: A
 sin 3 x   sin 1 x 2   tan 1 x 2
 , x  0 is continuous at x=0, then k equals
140. f  x   e5 x  1

k ,x 0

43
 A) 3/5 B) -3/5 C) 6/5 D) -6/5
KEY: A
1 1
141. If f  x   and g  x   , then the number of discontinuities of the composite function
 x  1 x  2  x2
f  g  x   is
A) 2 B) 3 C) 4 D) 5
KEY: D

1

Let f  x    x  e  2
 2 e x
, x  e . Which of the following statements is true?
142.
0 ,x e
A) The function is continuous at x = e but not differentiable at x = e.
B) The function is not continuous at x = e
C) The function is differentiable at x = e
D) The function is neither continuous nor differentiable at x = e.
KEY: A
143. Consider the function f defined by
0 , x  0 or x is irrational

f  x  1 m m
 n , where x is a non  zero rational number n , n  0 and n i s in lowest term
Which of the following statements is true?
A) Any irrational number is a point of discontinuity of f
B) Any irrational number is a point of continuity of f
C) The points of discontinuities of f are rational numbers
D) The points of discontinuities are non-zero rational numbers.
KEY: B,D
144. Suppose that f : R  R is continuous and satisfying the equation f(x) .
f (f(x)) = 1, for all real x.
Let f (1000) = 999, then which of the following is true ?
1 1
A) f  500   B) f 199  
500 199
1 1
C) f  x    x  R  0 D) f 1999  
x 1999
KEY: A,B
145. The function
x2 
f  x  ,0  x 1 
a 
a ,1  x  2 
2b  4b
2 
 , 2  x  
x 2

is continuous for 0  x   . Then which of the following statements is correct?
A) The number of all possible ordered pairs (a, b) is 3
B) The number of all possible order pairs (a, b) is 4
C) The product of all possible values of b is – 1
D) The product of all possible values of b is 1.
KEY: A,C
Paragraph for Questions Nos. 33 to 35
Comprehension: 1
A real function f has the intermediate value property on an interval I containing [a, b] if f(a) < v < f(b)
or f(b)< v < f(a) ; that is, if v is between f(a) and f(b), there is between a and b some c such that f(c) =
v.
44
146. Which of the following statements is false?
A) Any continuous function defined on a closed and bounded interval [a, b]
possesses intermediate value property on that interval.
B) If a function is discontinuous on [a, b] then it doesn’t possess intermediate
property on that interval.
C) If f has a derivative at every point of the closed interval [a, b], then f takes
on every value between f(a) and f(b) .
D) If f has a derivative at every point of the closed interval [a, b], then f' takes
on every value between f'(a) and f'(b) .
KEY: B
147. Consider the statements P and Q
P: If f : (a, b)  R is continuous, then given x1 , x 2 , x 3 , x 4 in (a, b), there exist
1
x 0  (a, b) such that f(x 0 )  (f(x1 )  f(x 2 )  f(x 3 )  f(x 4 )) .
4
Q: If f and g have the intermediate value property on [a, b], then so has f+g on
that interval. Which of the following is correct?
A) P is false but Q is true B) P is true but Q is false
C) Both P and Q are false D) Both P and Q are true
KEY: B
148. Consider the statements P and Q
P: For a non zero polynomial p, the equation p(x)  e x has at least one
solution.
Q: There exists a continuous function f : R  R which attains each of its values
exactly two times.
A) P is false but Q is true B) P is true but Q is false
C) Both P and Q are false D) Both P and Q are true
KEY:B
   a  n  nx  tan x  sin nx

149. If f  x    x2 at x  0 , where n is a non-zero real number, and f is
0 at x  0

continuous at x= 0, then a is equal to
n 1
A) 0 B) C) n D) n 
n 1 n
KEY: D
150. Let f be a function defined on   / 2,  / 2  as follows
 2 x e x  x  x ln 2  1
 ,x  0
f  x   x tan x .The value of k so that f is continuous
k ,x 0

at x =0 is
1 1 1
A)  ln 2   ln 2  1 B)  ln 2    ln 2   1
2 2

2 2 2
1 1 1
C)  ln 2    ln 2   D)  ln 2   ln 2 
2 2

2 2 2
KEY: D
151. The function f  x    x    x , where [.]denotes the greatest integer function and {.} denotes the
fractional part function respectively, is discontinuous at
A) all x B) all integer points
C) no x D) x which is not an integer.
KEY: C
45
152. Let a1 sin x  a2 sin 2 x  ...........  a8 sin 8 x  sin x for x  R
Define P  a 1  2a 2  3a 3  ....  8a 8 .Then P satisfies
A) P  1 B) P  1 C) P  1 D) P  1
KEY: A

153. Let f be a function defined on (-1, 1) by f  x  

 2

cos 1 1   x sin 1 1   x 
, x  0 .{.} is the fractional
 x   x
3

part function. Which of the following statements is correct?


A) lim f  x  exists and equals
x 0  2
B) lim f  x  exists and equals  / 4
x 0 
C)f is continuous
D) lim f  x  exists and equals lim f  x 
x 0 x 0 
KEY: A,B
1
154. Consider the function f  x   . Which of the following statements is correct about the composite
x 1
 
function g  x   f f  f  x   .
A) the number of discontinuities of g is 3
B) g is continuous at x = 2
C) g is not continuous at x = 3/2
D) g is continuous at x = 0
KEY: A,C,D
x 2  232 x  416  1 
32

f  x  ,x 1
 x  1
2
155. Let  , the value of k so that the function is continuous at x=1 is

k , x  1
A) 263  231
B) 265  233
C)  216  1 28  1 24  1 22  1 232  231 
D)  232  1 216  1 28  1 24  1 22  1 233  231 
KEY: A,C
156. Match the items of Column – I with those of Column II
 1   2  8 
A) f(x) = x        ........    ,x  0 p)1
x  x x
= 9k, x0
The value of k such that f is continuous at x=0 is
([.]denotes the greatest integer function)
2
e1 / x
 2 1 
B) f(x) =  1  xe 1/x sin 4  , x  0 q)2
 x 
= k, x=0
The value of k such that f is continuous at x=0 is

C) f: [0, )  R ; r)3
x
 1
f(x) =  2 sin x  x sin  , x  0
 x
=k ,x=0
46
 The value of k such that f is continuous at x=0 is
 1  sin x n sin x 
 (  2x)2 . n(1   2  4 x  4 x 2 ) ;x  2
d) f:  0,    R ; f(x)   s)4
k 
;x 
 2

The value of 8 k such that f is continuous at x= is
2
KEY: A—S; B—P; C—P; D—P

 sin  cos 2  tan  sin x  


,x  0
f  x    x2
157. k .The value of k such that f is continuous at x=0,is
 ,x 0

KEY: 1
  1 
 2  x 
x  r ;x  0
158. Let f(x)   r 0 ([.]denotes the greatest integer function)
k
 ; otherwise
2
The value of k such that f become continuous at x=0 is
KEY:1
159. Let f : (, )  [0, ) be a continuous function
such that f(x  y)  f(x)  f(y)  f(x)f(y), x, y  R. Also f '  0   1 .
 f(4) 
Then   equals ( [] represents greatest integer function)
 f(2) 
KEY: 8
2
160. Let f(x) = x 2 e  x , x  1
1
= , x  1 then the value of f'(1) equals
e
KEY: 0
161. Let f(x) = [x2] sin x, x  R , the number of points in the interval  0, 3  at which the function is
discontinuous is_____
KEY: 6
162. Let f(x) = tan 1 x, x  1
 x 1
sgn x 
= , x 1 ,(where sgn denotes signum function)
4 2
Then the value of 4 f'(1 ) equals
KEY: 2

163. (L-1)If Lt
  a  n  nx  tan x  sin nx  0 where n is a non zero real number then a is equal to
x 0 x2
n 1 1
a) 0 b) c) n d) n 
n n
Key: D
 tan x  sin nx
Hint Lt   a  n  n   0
x 0
 x  x
   a  n  n  1 n  0
47
 1
a n
n
 1 1
164. STATEMENT-1: lim x 2  1  cos  
x 
 x 2
 1
STATEMENT–2: lim cos   does not exist
x 0
x
Key: B
 1 
2x 2  sin2  
 1  1   2x 
Hint: lim x 2  1  cos   lim x 2  2sin2    lim
x 
 x  x   2x  x  1
 4x 2
2
4x
1
as x  , 0
2x
 1 1 1
So, lim x 2  1  cos    12 
x 
 x 2 2
 1  1
Also, lim cos   does not exist because cos   does not approach to a definite value as x  0
x 0
x x

 
1
  sec x  x 
x
165. Let x  0 then Lt tan x
x 0

1
(A) 1/ e (B)1 (C) (D) 2
e2
Key: D

   Lt   cos x 
x 1/ x
Hint: Lt  tan x
x 0 x 0

log e 
tan x    
 e0  1 , Lt   cos x 
1/ x
 1 as 0  cos x  1
Lt   
x 0 1    x 0
e x
 x n  sin  x n 
 Lt , if x  0, x  1
166. Let f  x    n x n  sin  x n  . Then, at x = 1,

 1 , if x 1
A) f is continuous
B) f has removable discontinuity (i.e., Lt f  x  exists, but this limit is different from f(1) )
x 1

C) f has finite (jump) discontinuity (i.e., f (1+) and f (1-) both exist finitely, but they are
different)
1 
D) f has infinite or oscillatory discontinuity (for eg like sin at x  0 and tan x at x  )
x 2
Key: C
Hint: 0  x  1  x n  0 as n    f  x   0 and x  1  x n   as n    f  x   1
 f has a jump (finite) discontinuity at x  1
| ax 2  bx  c |
167. ,  are roots of equation ax2 + bx + c = 0 where 1 <  <  if lim  1 then which of the
x  m ax 2  bx  c

following are true

(A) a < 0 and  < m <  (B) a > 0 and m < 1
(C) a > 0 and  < m <  (B) a > 0 and m > 1
Key: A, B (DELETED)
48
 | a(x   )(x  ) |
Hint: lim 1
x m a(x   )(x  )
WHEN A > 0, M < 1  (M  ) (M – ) > 0
 A(M  ) (X – ) > 0
a(x   )(x  )
 lim 1
x m a(x   )(x  )
WHEN A < 0,  < M <   (M  ) (M  ) < 0
 A(M  ) (M  ) > 0
a(x   )(x  )
 lim 1
x m a(x   )(x  )

n
 1  n  1  
168. Lt 1     1    
 n   n  
n 

1
1) 1 2) 3) 1  e 1 4) 0
e 1
Ans: 4
 1  n  1  
Hint: Lt  1     1     e  1  1
 n   n  
n 

x p  x p 1  1
169. Lt
x  x q  x q  2  2
  p  0, q  0 
1) 0 if p  q 2) 1 if p  q 3) infinite if p  q 4) 1 if p  q
Key: 1,2,3
Hint: Conceptual
 1 
170. Let f (x) 
tan x
x
, then log e  lim  f (x)   x
 x 0
2 f (x)

 is equal, (where [] denotes greatest integer

function and {} fractional part)
(A) 1 (B) 2 (C) 3 (D) 4
Key: C
 tan x 
Hint: lim  f  x    lim  1
x 0 x 0  x  
1 1
   
lim  f (x)   x 2 f (x) = lim 1  x 2 f (x) 1 form
x 0 x 0
 
3
x 2
x  x5  . . .
tan x 3 15
Again, f(x) = 
x x
x2 2 4
= 1  x . . .
3 15
x2 2 4
 
f (x)   x . . .
3 15
(i) becomes,
x2
lim
 lim x 2  1  x 0 2
x 2 4
 x  ...
log e  e
x 0 f (x)  e 3 15 3
 
 
 (C) is the correct answer.

49
171. Match the following:- (DELETED)
Column I Column II
(A)  n 4  
n 
5 
6
 (p)
    
x  x
  x1  4
lim x1  , then t can be

n  n  n  5
  x   x 
t 9t
 x1   
x1 
(B) If m be the slope of tangent to the curve 2
(q)
y x
x y at e, e , then 3  m can be
(C) 1 1 4
If f : R  0  R, f  x  f  y  f  xy  3   , then (r)
 x y 
 1
2 f   can be
 2 
 n!k 
 
(D) 3
lim  2  1   1, then k can be (s)
n   
(where {.} is a fraction part function)
(t) 7

Key: A q,t B q,sC q,s D p,q,r,s,t

 n 4  n  1 1
  5  4 5
    
x x  y dy  y dy
  x1  t  110  t  4
Hint: A) lim x1  10 0  
x  n t   
n
9 t

t
1
9t 30 5
  x   x 
   y dy  y dy
 x1  x1  0 0
2
 t  9t  14  0, t  2t  7  0, t  2, 7
y x
B) x  y  y ln x  x ln y
2
 x ln y  y y ln x 1
Differentiating with respect to x we get y '  
 y ln x  x  x 2 ln y 1
2
y ln x 1 y
y '  lim  lim
xe x 2  ln y 1 x e xy '
ye ye
 So m  1 and 3  m is equal to 2 or 4
C) Put x = y = 1
2
  f 1  f 1  6  f 1  3, 2
3 1  x  1  x 
Put y  1, we get f  x    
 or f  x    
2  x   x 
m m
D) If m is even  2 1   2 1   2I
 
n!k   
n!k   
n!k 
 lim  2  1
x  
 
  lim 2 I  2 1
 x 

  lim 
 x 
  
2 1   1

 n ! k must be positive integer  k is integer.
50
  (1) n 
172. STATEMENT- 1: lim 1   , ([ ] denotes the G.I.F.) does not exist.
n 
 n 1 
 (1) n 
STATEMENT 2: lim 1   =1
n 
 n 1 
Key: B
Hint: Conceptual
   sin  sin 
173. If  ,     .0  such that  sin   sin     0 and  sin   sin    1 and
 2  sin  sin 

 =
lim 1   2sin  2n then
n 
 2sin  2n
 
(A)    (B)   2 (C)    (D)   1
6 3
Key: A,B
Hint:  sin   sin  2  1
 sin   sin   1
1 1
 sin   sin  
2 2
0
    30

x gx  hx  7
p
174. If f  x   lim p
; x  1 and f 1  7,f  x  ,g  x  and h  x  are all continuous function at
p7x  3x  1
x  1 . Then which of the following statement(s) is/are correct
(A) g 1  h 1  70 (B) g 1  h 1  28
(C) g 1  h 1  60 (D) g 1  h 1  28

Key: A,B

Hint: When x  1 When x > 1

p  hx 7 
x g  x   p  p 
h 1  7  x 
f 1  f 1 
Lt x
3 1 p p  3x  1 
x 7  P 
 x 
h 1  7 g 1
7 7
4 7
h 1  21  g 1  49
 g 1  h 1  28
g 1  h 1  70

 1  x    
  4   1 then ordered pair(s)   ,   can be
Lim
x  
175. If x tan 
  x  
(A) (2000,2011) (B) (0,1) (C) (5,3) (D) (1,0)
Key: C

51
 1  x    
tan  
Lim  x    4
Hint: 1
x ki 1
x
Apply L’ hospital rule and simplifying we get
    x
2
Lim
1
x 
2x  2x      
2
 2
 
2


 1
2
 2
   ,   can be (5,3)

Paragraph for Questions Nos. 58 to 60

If f, g and h are functions having a common domain D and h  x   f  x   g  x  , x  D and if
lim h  x   lim g  x   l then lim f  x  = l . This is known as Sandwich Theorem.
x a x a x a

 1 
176. lim x 4 sin   is
x 0
3 x 
1
(a) 0 (b) 1 (c) (d) does not exist
3
Key: A
 1   1 
Hint: Since 1  sin   1, so  x 4  x 4 sin    x . But lim
4
x4  0 .

 3x  3x
x 0

 1 
 lim x 4 sin   =0
x 0
3 x 
e1/ x  e 1/ x
177. Let f  x   x 2 , x  0 and f(0) = 1 then
e1/ x  e 1/ x
(a) lim f  x  does not exist (b) lim f  x  does not exist
x 0 x 0

(c) lim f  x  exists (d) f is continuous at x = 0.

x 0
Key: c
e1/ x  e 1/ x 2 1 e
2 / x
Hint: 0  x2 1/ x
 x 2 / x
 x 2 for x > 0
e e
1/ x
1 e
e2 / x  1
So lim f  x   0 . Also lim f  x   lim x 2 2 / x 0
x 0 x 0 x 0 e 1
lim f  x   0
x 0

1
178. Let f  x   x 5  3  , x  0 and f(0) = 0 ([.] denotes the greatest integer function)
x 
(a) lim f  x  does not exist (b) f is not continuous at x = 0
x 0

(c) lim f  x  = 1 (d) lim f  x  = 0

x 0 x 0
Key: d
Hint: Since x  1   x   x for x  R
1 1 1  1  1
1   3   3  x 5  3  1  x 5  3   x 2
x  x x  x 
3
x
52
 1
so lim x 5  3   0
x 0
x 
1 1
179 The function f (x)  where u 
u u2
2
x 1
3
a) has a removable discontinuous at x=1 b) has irremovable discontinuous at x  0,
2
c) discontinuous at u  2, 1 d) discontinuous at u  1, 2
Key: A,B,D
1
Hint The function u  is discontinuous at x = 1
x 1
1
f x  is discontinuous at u  1, 2
 u  1 u  2 
3
i.e. at x  0,
2
also we have Lt f  x   Lt f  x   
x 0 u 1

Lt f  x   Lt f  x   
x
3 u 2
2

Lt f  x   Lt f  x   0
x 1 x 

 x 2  232.x  416  1
32

 ,x 1
180. Let f  x     x  1 .
2


k , x 1
Then value of k so that the function is continuous at x = 1 is
(A) 263  231 (B) 265  233

   
(C) 216  1 28  1 24  1 22  1 232  231  
    
(D) 232  1 216  1 28  1 24  1 22  1 233  232  
Key: A,C
Hint Conceptual
 x 
  x  3  
181. Let f  x   cos  where  . denotes greatest integer function  x  then
 
Statement I : f  x  is continuous at atleast one integer in the domain of f  x 
Statement II : f  x  is discontinuous at all integers in its domain.
Key: C
Hint: Domain =  , 3  [4, )
Let k  Domain of f(x), k  I
 k 
Lt  f  x   cos  
xk  k 3
 k 
Lt  f  x   cos  
xk k 4
 k 
f  k   cos  
 k 3
53
 At k = 4, 0 f(x) is continuous

182. If
1  sin 3 x 
f  x  x
3cos 2 x 2

a x
2
b 1  sin x  
x
  2 x 
2
2
 b
If f  x  is continuous x  then =
2 a
Ans: 8
1 b
Hint: LHL  .RHL 
2 8
1 b
 a
2 8
px
 ; x2
183. Consider the function f  x    x  2 where P(x) is a polynomial such that p '''  x  is identically
 7 ; x2

equal to 0 and p  3  9. If f(x) is continuous at x = 2, then p(x) is
2 2 2 2
(A) 2x  x  6 (B) 2x  x  6 (C) x  3 (D) x  x  7
Key: B

Hint: Since P"'  x  = 0

2
Let p  x   ax  bx  c
p  2  0
4a  2b  c  0 ……..(1)
9a  3b  c  9 ………(2)
p '  2  7
 4a  b  7
Solve 1,2 and 3 to get a,b,c
 x 3 , x 1
 2
184 The function f  x    x   3x   13  is
 4    2    4  , x  1
     
(A) continuous at x = 1 (B) differentiable at x =1
(C) continuous at x=3 (D) differentiable at x = 3
Key: A,B,C

Hint: f 1  h   2  f 1  h 
f 3    f 3   0
Continues at x  1,3
f ' 1    f ' 1    1
Not differentiable at x = 3

54
 1
(n !) n
185. lim equals
n  n
a) e b) e 1 c) e-2 d) e2
KEY : B
HINT
52
1
(n !) n
let P
n
1
 (n !)  n
= n 
 n 
1 n r
log P=  log  
n r 1 n
186. Which of the following is true? ({.} denotes the fractional part of the function)
 1 x
a) lt  x sin   1 b) lt  2 
x 
 x x2 x  x  2

x  1
c) lt  2   d) lt  x sin   
x 1 x  x  2 x 
 x
KEY: A, B, C
187. Given a real valued function f such that
tan 2  x
f  x  , for x  0

x2   x
2

= 1 for x  0
=  x cot  x for x  0 , where [x] is integral part of x and {.} is Fractional part of x. Then
a) lt  f  x  1 b) lt  f  x   cot1
x 0 x 0

 lt  1  lt  
2
c) cot 1 
f  x d) tan 1 
f  x 
x 0 x 0 4
KEY: A,C,D
x b  b
188. Statement I : If a and b are positive and [x] denotes greatest integer  x, then lt  
x 0 a  x  a

Statement II: lt
 x  0 , where {x} denotes Fractional part of x.
x  x
KEY: A
 1x 
 e 1 
189. Statement I : lt  x  1 (where [.] represents greatest integer function) does not exist.
x 0  
 ex 1
 
 x1

e 1 
Statement II: lt  1 does not exist
x 0  
 ex 1
 
KEY: B

55
 PASSAGE – III :
a sin x  bx  cx 2  x 3
Given that lim is finite and equals K.
x 0 2 x 2 log 1  x   2 x 3  x 4

190. The value of a is

a) 0 b) 6 c) 12 d) 4
KEY: B
191. The value of c is
a) 1 b) 2 c) 4 d) 0
KEY: D
192. The value of K is
1 1 3 1
a) B) c) d)
4 8 40 12
KEY: C
n2  n
 1 
193. lim 1  2  
n 
 n  cos n 
1
a) 1 b) e c) d) e 2
e
KEY: B

 
n
194. lim  cot 1 r 2  r  1 
n 
r 1

 
a) b) c)  d) 0
4 2
KEY: A
1
195.
x 

lim e x  x  x 
1
a) 0 b) 1 c) e d)
e
KEY: C
 sin 2 x b 
196. If lim  3  a  2   0 , then
x 0
 x x 
4 4
a) a  , b  2 b) a  c) a  , b  2 d) a
4 4
,b  2 , b  2
3 3 3 3
KEY: C
197. Let f :    be a function such that f  0   1 and for any
x, y  , f  xy  1  f  x  f  y   f  y   x  2 holds. Then lim f  x  
x 0

a) 0 b) 2 c) – 1 d) 1
KEY: D
 xa
x

198. The value of ‘a’ for which lim 

x  x  a
  e is
 
1
a) 1 b) – 1 c) d) 2
2
KEY: C
f  3x 
199. Suppose f :    is a positive increasing function with lim 1
x  f  x

56
 f  2x
STATEMENT-1: lim 1
x  f  x
STATEMENT-2: x  0  0  f  x   f  2 x   f  3x  :
KEY: A
Paragraph for Questions Nos. 14 to 16
Suppose f, g, h are real- valued functions defined on some deleted neighbourhood N of ‘a’ such that
g  x   f  x   h  x  for all x  N . If lim g  x  and lim h  x  both exist and each equals  , then
xa xa

lim f  x  also exists and equals  . (This is called the squeezing principle or sandwich theorem )
xa

200. If f  x   sin  x  1  sin   x  , then lim f  x 

x 

1
a) does not exist b) equals 0 c) equals 1 d) equals
2
KEY: B
If f  x    tan x  sin x   , where[.] is the greatest integer function, then lim f  x  
2
201.
  x 0

1
a) 0 b) 1 c) 2 d)
2
KEY: A

 
sin 2 x
lim 1cos ec x  2cos ec x       ncos ec x
2 2 2
202.
x 0

a) equals 0 b) equals 1 c) equals n d) does not exist

KEY: C
203. Mach the following
Column I Column II
lim 1  x 2 
cot x
(A) x 0
(P)
0
x

(B) 1  e  1  cos x (Q)

lim 1
x  0 sin x
  x x
(C) lim x  ln 1    ln  (R) 2
x 
  2 2
Let  be a rational number and
(D) an  sin  n !  , n  1, 2,3    (S)
1
Then lim an 2
n 

KEY: A-Q, B-Q, C-R, D-P

 tan 2 x  2sin 2 x  3sin x  4  sin 2 x  6sin x  2  , x   / 2


204. Define f :  0,    R by f  x      is
k ,x  /2

continuous at x  ,then k =
2
1 1 1 1
A) B) C) D)
12 6 24 32
KEY: A

57
  sin 3 x   sin 1 x 2   tan 1 x 2
 , x  0 is continuous at x=0, then k equals
205. f  x   e5 x  1

k ,x 0
A) 3/5 B) -3/5 C) 6/5 D) -6/5
KEY: A
1 1
206. If f  x   and g  x   2 , then the number of discontinuities of the composite function
 x  1 x  2  x
f  g  x   is
A) 2 B) 3 C) 4 D) 5
KEY: D

1

Let f  x    x  e  2
 2 e x
, x  e . Which of the following statements is true?
207.
0 ,x e
A) The function is continuous at x = e but not differentiable at x = e.
B) The function is not continuous at x = e
C) The function is differentiable at x = e
D) The function is neither continuous nor differentiable at x = e.
KEY: A
208. Consider the function f defined by
0 , x  0 or x is irrational

f  x  1 m m
 n , where x is a non  zero rational number n , n  0 and n i s in lowest term
Which of the following statements is true?
A) Any irrational number is a point of discontinuity of f
B) Any irrational number is a point of continuity of f
C) The points of discontinuities of f are rational numbers
D) The points of discontinuities are non-zero rational numbers.
KEY: B,D
209. Suppose that f : R  R is continuous and satisfying the equation f(x) .
f (f(x)) = 1, for all real x.
Let f (1000) = 999, then which of the following is true ?
1 1
A) f  500   B) f 199  
500 199
1 1
C) f  x    x  R  0 D) f 1999  
x 1999
KEY: A,B
210. The function
x2 
f  x  ,0  x 1 
a 
a ,1  x  2 
2b 2  4b 
 , 2  x   
x2 
is continuous for 0  x   . Then which of the following statements is correct?
A) The number of all possible ordered pairs (a, b) is 3
B) The number of all possible order pairs (a, b) is 4
C) The product of all possible values of b is – 1
D) The product of all possible values of b is 1.
KEY: A,C
Paragraph for Questions Nos. 33 to 35
58
Comprehension: 1
A real function f has the intermediate value property on an interval I containing [a, b] if f(a) < v < f(b)
or f(b)< v < f(a) ; that is, if v is between f(a) and f(b), there is between a and b some c such that f(c) =
v.
211. Which of the following statements is false?
A) Any continuous function defined on a closed and bounded interval [a, b]
possesses intermediate value property on that interval.
B) If a function is discontinuous on [a, b] then it doesn’t possess intermediate
property on that interval.
C) If f has a derivative at every point of the closed interval [a, b], then f takes
on every value between f(a) and f(b) .
D) If f has a derivative at every point of the closed interval [a, b], then f' takes
on every value between f'(a) and f'(b) .
KEY: B
212. Consider the statements P and Q
P: If f : (a, b)  R is continuous, then given x1 , x 2 , x 3 , x 4 in (a, b), there exist
1
x 0  (a, b) such that f(x 0 )  (f(x1 )  f(x 2 )  f(x 3 )  f(x 4 )) .
4
Q: If f and g have the intermediate value property on [a, b], then so has f+g on
that interval. Which of the following is correct?
A) P is false but Q is true B) P is true but Q is false
C) Both P and Q are false D) Both P and Q are true
KEY: B
213. Consider the statements P and Q
P: For a non zero polynomial p, the equation p(x)  e x has at least one
solution.
Q: There exists a continuous function f : R  R which attains each of its values
exactly two times.
A) P is false but Q is true B) P is true but Q is false
C) Both P and Q are false D) Both P and Q are true
KEY:B
   a  n  nx  tan x  sin nx

214. If f  x    x2 at x  0 , where n is a non-zero real number, and f is
0 at x  0

continuous at x= 0, then a is equal to
n 1
A) 0 B) C) n D) n 
n 1 n
KEY: D
215. Let f be a function defined on   / 2,  / 2  as follows
 2 x e x  x  x ln 2  1
 ,x  0
f  x   x tan x .The value of k so that f is continuous
k ,x 0

at x =0 is
1 1 1
A)  ln 2   ln 2  1 B)  ln 2    ln 2   1
2 2

2 2 2
1 1 1
C)  ln 2    ln 2   D)  ln 2   ln 2 
2 2

2 2 2
KEY: D

59
216. The function f  x    x    x , where [.]denotes the greatest integer function and {.} denotes the
fractional part function respectively, is discontinuous at
A) all x B) all integer points
C) no x D) x which is not an integer.
KEY: C
217. Let a1 sin x  a2 sin 2 x  ...........  a8 sin 8 x  sin x for x  R
Define P  a 1  2a 2  3a 3  ....  8a 8 .Then P satisfies
A) P  1 B) P  1 C) P  1 D) P  1
KEY: A

218. Let f be a function defined on (-1, 1) by f  x  

 2

cos 1 1   x sin 1 1   x 
, x  0 .{.} is the fractional
 x   x
3

part function. Which of the following statements is correct?


A) lim f  x  exists and equals
x 0  2
B) lim f  x  exists and equals  / 4
x 0 
C)f is continuous
D) lim f  x  exists and equals lim f  x 
x 0 x 0 
KEY: A,B
1
219. Consider the function f  x   . Which of the following statements is correct about the composite
x 1
 
function g  x   f f  f  x   .
A) the number of discontinuities of g is 3
B) g is continuous at x = 2
C) g is not continuous at x = 3/2
D) g is continuous at x = 0
KEY: A,C,D
x 2  232 x  416  1 
32

f  x  , x  1 
 x  1
2
220. Let  , the value of k so that the function is continuous at x=1 is

k , x  1
A) 263  231
B) 265  233
C)  216  1 28  1 24  1 22  1 232  231 
D)  232  1 216  1 28  1 24  1 22  1 233  231 
KEY: A,C
221 Match the items of Column – I with those of Column II
 1   2  8 
A) f(x) = x        ........    ,x  0 p)1
x  x x
= 9k, x0
The value of k such that f is continuous at x=0 is
([.]denotes the greatest integer function)
2
e1 / x
 2 1 
B) f(x) =  1  xe 1/x sin 4  , x  0 q)2
 x 
= k, x=0
The value of k such that f is continuous at x=0 is
60
 C) f: [0, )  R ; r)3
x
 1
f(x) =  2 sin x  x sin  , x  0
 x
=k ,x=0
The value of k such that f is continuous at x=0 is
 1  sin x n sin x 
 (  2x)2 . n(1   2  4 x  4 x 2 ) ;x  2
d) f:  0,    R ; f(x)   s)4
k 
;x 
 2

The value of 8 k such that f is continuous at x= is
2
KEY: A—S; B—P; C—P; D—P

 sin  cos 2  tan  sin x  


,x  0
f  x    x2
222. .The value of k such that f is continuous at x=0,is
k ,x 0


KEY: 1
  1 
 2  x 
x  r ;x  0
223. Let f(x)   r 0 ([.]denotes the greatest integer function)
k
 ; otherwise
2
The value of k such that f become continuous at x=0 is
KEY:1
224. Let f : (, )  [0, ) be a continuous function
such that f(x  y)  f(x)  f(y)  f(x)f(y), x, y  R. Also f '  0   1 .
 f(4) 
Then   equals ( [] represents greatest integer function)
 f(2) 
KEY: 8
2
225. Let f(x) = x 2 e  x , x  1
1
= , x  1 then the value of f'(1) equals
e
KEY: 0
226. Let f(x) = [x2] sin x, x  R , the number of points in the interval  0, 3  at which the function is
discontinuous is_____
KEY: 6
227. Let f(x) = tan 1 x, x  1
 x 1
= sgn x  , x 1 ,(where sgn denotes signum function)
4 2
Then the value of 4 f'(1 ) equals
KEY: 2

xn
228.(L-2)If x1  3 and x n 1  , n  N , then lim 2n x n equals to
n 
1 1  x 2n

61
 3 2 2 3
a) b) c) d)
2 3 3 2
Key : c
Sol : Let x n  tan n

xn tan n
Now, tan n 1  x n 1  
1  1  x 2n 1  1  tan 2 n

tan n sin n 
 tan n 1    tan n
1  sec n 1  cos n 2

n 
 n 1   n  n 1
2 2
   2 
Now, 1   n  n 1
 x n  tan  n 
3 3.2  3.2 

 lim 2 x n  lim
n    2
tan 2 / 3.2n
n  n 
1/ 2 n 3

 e x  e 2x  e3x 
ln  

 
1/x  3
Let f  x   x , g  n, x   sin    n    n    , then
 h x 
2
229.
  x
(where [.] represent greatest integer function


a) lim lim f g  n, h  x    1
n  x 0
  
b) lim lim f g  n, h  x   does not exist
n  x 0

c) lim lim g  n, h  x   does not exist d) lim lim g  n, h  x   = 1

n  x 0 n  x 0

KEY:
Sol :
230. (L-3)
COLUMN – I COLUMN – II

A)  n 4  n 5  p) 6
  x   x 
lim  xn1   nx 1   , then t can be
4
n  
t  9 t  5
  x   x 
 x 1   x 1 
B) If m be the slope of tangent to the curve q) 2
x y  y x at (e, e), then (3 – m) can be
C) If r) 3
1 1
f : R  0  R, f  x  f  y   f  xy   3    ,
x y

62
  1
then 2 f    can be
 2
D)
lim
n   2 1 
n! k
  1 , then K can be (where
s) 4

{.} is a fractional part function)

t) 7

Key : A – q, t; B – q, s; C – q,s; D – p,q,r,s,t

1 1

 y dy  y dy
4 5

Sol : A) given that  0 0


 t  110  t   4
1 1
9 t
30 5
 y dy  y
t
dy
0 0

 t  2 or t  7

y 2  ln x  1
B) x  y  y ln x  x ln y Differentiating y ' 
y x
at (e, e) m  1  3  m  2 or 4
x 2  ln y  1

C) Put x = y = 1  f 1   f 1  6  f 1  3, 2

2

3  1 x   1 x 
Put y  1  f  x     or f  x     
2 x   x 

   
m m
D) If m is even 2 1 2 1  2I

lim
n 
 2 1 
n! k
 Lt 2I 
n 
  2 1 
n! k
 
 lim  2  1n! k  1
n 

 n! k must be positive integer of k is an integer
 1
1   1  1 2x  1 
1   lim 
1

1 3
  1  2x  x 2
  2
 x  13x  x  x  0 1 3x  x x
lim
x  01 3x
 e   xlim e e  e3 .
231.(L-1)Statement – 1 : lim  x e
x 0 
  1  3x    0
 
Statement – 2 : If lim f  x   l1 and lim g  x   l2 , then lim  f  x  g  x    l1l2 (where l1 and l2 are
x a x a x a

finite)
Key : a
1 1
Sol : Lt 1  3x  x 2 Lt e x  e3
x 0 x 0

 1 
232.(L-2)If a1 is the greatest value of f  x  ; where f  x    (where . denotes greatest integer
 2  sin x  
 

 1
n2

function) and a n 1   a n , then lim  a n  is

 n  1 n 

63
 a) 1 b) e 2 c) 1n 2 d) none of these
Key : c
Sol : a1  1

1 1 1
 a2  1  a3  1 
2 2 3
……………..
……………..
1 1 1
a   1     .... = ln 2
2 3 4
 min  sin x, x  
233.(L-2) lim   where  is root of equation sin x  1  x (here . represent greatest
x 
 x 1 

integer and 
. represent fractional part function)

a) 0 b) 1 c) does not exist d) none of these

key : c
Sol : In the neighborhood of x = a
min  sin x, x 

 x  1; x  a 
 
sin x; x  a 
y
x-1
1
x
1

 x  1
LHL lim    1
x   x  1 

 sin x 
RHL lim  
x   x  1 
0

234.(L-1)The value of lim


cot 1 x  a log a x   a  1 is equal to
x  sec 1 a x
log x a


a) 1 b) 0 c) d) does not exist
2
Key : a

64
  log x 
cot 1  aa 
 x  as  log a x   0 and  ax 
Sol : I  lim  a     
1  a
x 
x   x   log a x 
sec  
 log a x 
 
I  1
2 2
1 1
x tan   ln   f  x  dx
235.(L-3)If f  x   lim
n 
e n n and  3 sin11 x cos x  g  x   C , then
 3
a) g    b) g(x) is continuous for all x
4 2
   15
c) g   
4 8
d) g(x) is non differentiable at infinitely many points
Key : c, d
1 1
Sol : lim tan   ln    0 [By L’ Hospital’s rule]
n  n n
 f x 1

The given integral =  11/3

dx sec 4 x 1  tan 2 x sec 2 x  
sin x cos1/3 x  tan11/3 x
 dx   dx
tan11/3 x
3 3
  tan x 8/3   tan x 2/3  C
8 2
 15
g   
4 8

236.(L-1) lim x x is equal to

x 0

a) 1 b) 0 c) e 1 d) doesn’t exist
Key : b
Sol : y  xx
ln y  x ln x

y  e x ln x
ln x
1  ln x  
 ex xlim 1
is form 
 0 x
 
1/x 
e

 1/x 2   e x

65
 1
lim y  lim 0
x 0 x 0 ex

a 0 x m  a1x m 1  ....  a k x m  k
237.(L-1) Let   x   , where a 0  0, b0  0 then lim   x 
b0 x n  b1x n 1  ....  b1x n 1 x 0

COLUMN – I COLUMN – II

A) m>n p) 

B) m=n q) -

C) m < n and n – m is even, a 0 / b0  0 r) a 0 / b0

D) m < n and n – m is even, a 0 / b0  0 s) 0

Key : A – s; B – r; C – p; D - q
a 0 x m n  a1x m 1n  .....  a k x m  k n
Sol : If m > n then   x   so lim   x   0
b0  b1x  ....  be x l x 0

a 0  a1x  .....  a k x k a0
If m = n then   x   so lim   x  
b0  b1x  ...  bl x l x 0 b0

a0 a
If m < n and n – m is even then   x    or   according as  0 or 0  0 respectively.
b0 b0

ex sin x x
238.(L-1)If   x  
 
, then value of lim is
cos x ln 1  x 2 x 0 x

a) 0 b) 2 c) -1 d) 1
Key : c
sin x
lim e x lim
x x 0 x 0 x 1 1
   1
 x
Sol : lim
x 0 x l n 1 x 2
1 0
lim cos x lim
x 0 x 0 x2
 e1/x   1 
239. (L-1) lim  1/x  is equal to
x  0  e   1 
 
a) 1 b) -1 c) zero d) doesn’t exist
Key : d
Sol : For x > 0 and x  0
 1 
 e1/x   1   1  1/x  
 , as x  0 , e    1/x  0  1
 1/x 1
RHL  lim  1/x   lim  e
x  0  e   1  x  0  1  e 

   1  1/x  
 e 

66
 For x  0, x  0

 1 
 e1/x   1   1  1/x  
LHL = lim  1/x   lim  e 
x 0   
 1  x  0  1  1 
e  
 e  
1/x

 1  e1/ y  
Let x = -y, so as to x  0 , y  0  lim  , y  0
y  0  1  e1/ y  
 
 1 
 1/ y   1 
 lim  e   1
y0  1 
 1/ y   1 
 e 
 e1/x   1   e1/x   1 
 lim  1/x   lim    limit doesn’t exist
x  0  e   1  x  0  e1/x   1 
   
x
 x 2  5x  3 
240. (L-1)If f  x    2  then lim f  x  is
 x x2  x 

a) e4 b) e3 c) e2 d) 24
Key ; a
x
 x 2  5x  3   4x  1 
x
Sol : lim f  x   lim  2   lim 1  2 
x  x  x  x  2  x   x x2
 

 x2 x2 
 4x  1  4x 1 
 lim 1  2   …… (1)
x   x x2
 

1
x  4x  1 4
Where,    x  4 as x  
x2  x  2 1
1 2

x x2

 1  lim f  x   e 4
x 

tan x 2   x  1 sin x
n
241. (L-1)Consider f  x   lim , n  N , then
x 2   x  1
n  n

a) f(x) is continuous at x = 0 b) lim f  x   0

x 0

c) lim f  x   1 d) lim f  x   0
x 0 x 0

Key : c, d
Sol : f  0    1, f  0    0

67
 x mf  x   h  x   3
242. (L-2)If g  x   lim when x  1 and g 1  e3 such that f(x), g(x) and h(x) are
m 2x  4x  1
m

continuous functions at x = 1, then

a) f 1  2e3 b) h 1  5e3  3 c) f 1  h 1  7e3  5 d) f 1  h 1  7e3  5

Key : a,b

 x m f  x   h  x   3 
Sol : lim g  x   lim  lim 
x 1 x 1  m 
 2x m  4x  1 

 x m f  x   h  x   3 
lim lim  
 2x  4x  1 
m   x 1 m

 h x  3 
 f  x   
lim lim  x m   f 1

m   x 1
 2  4x  1  2
 x m


 f 1  2e3

h 1  3
Similarlity lim g  x  
x 1 5

 h 1  5e3  3
1/x
 n  n i  r
243. lim   1 n Cr   n r Ck 2k x k  x 2  x
x 0 
0
  

 en , then  is equal to
 r 0  k 0  
Key :
Sol :
lim  sin 1 x 
244.(L-3) x  0   (where . denotes greatest integer function) is
 x 

(A) left hand limit is – 2 (B) left hand limit is -

1
(C)right hand limit is 1 (D) limit exists and
both are equal to 1
Key : A,C
 sin 1 h   sin 1 h 
Hint : LHL  lim    lim   2
h 0  h  h0   h 
 
lim  sin h 
1

RHL  h  0   1
 h 

Passage : Let f (x) = max (a, b, c) where

68
 lim lim  n | sin x |   n | cos x |
a
n    1 n  n
lim  n | sin x |   n | cos x |
lim
b
n    1 n  n
lim    2 (n  1) 
c 1  cos  cos  .......  cos . Then
n   4n  2n 2n 2n 
245. The value of a is
1
a) 2 | sin x | b) | cos x | c) | sin x | d)
2
Key : c
lim lim  n | sin x |   n | cos x |
sol : a
n     1 n  n
lim | sin x |  2n | cos x |
lim
a | sin x |
n     1 1   2n
1
246. The value of b + c - is
2
a) | cos x | b) 2 | cos x | = 1 c) | sin x | + 1 d) | sin x | + | cos x |
Key : a
1  1
sol : b  c  | cos x |  c 
2  2
247. Range of f (x) is
1   1  1 
a) [0, 1] b)  ,1 c)  ,1 d)  , 2 
2   2  2 
Key : b
1 
sol :  ,1
2 

Lim   x  2 x  ae a
248.(L-2) If the value of the limit, 2e     equals , where is a rational in its
x     x  1   b b
lowest form, then value of a4 + b5 is
a) 259 b) 113 c) 97 d) 48
Key : b
 1 x1 1   3 1 
x 
x loge 1
1  x   .....  1  ......
 x2 

x 1 
 x 1 2(x 1)2 3(x 1)3 
 2(x 1) 2(x 1)2 
Sol :    e  e   e 
 x 1 
Lt   x  2 x  Lt  3 
 x e     x e  x  (1   ......) 
x     x  1   x    2(1  x) 
3
e  a4 + b5 = 81 + 32 + 113
2
Lim  1  1  1   1 
249.(L-2) If a1 = 1 and an = n (1 + an – 1 ) n  2, then 1  1   1   ......  1   \
n    a1  a 2   a 3   an 
1/2
a) e b) loge 2 c) e d) log2 e
Key : a
 1  1   1  a 1 an 1 an 1 1
Sol : 1   1   .........  1    n 1       ....  1
 a1  a 2   a n  (n  1)! n! n! n! n! (n  1)!
69
 Lt  1  1   1  Lt 1 1 1
1    1   ....... 1   1    .....   e
n    a1   a 2   an  n   2! 3! n!


1
250. The value of  n(n  1)(n  1) is less than or equal to
n 1

a) 3 – e b) e – 2 c) e  d) 4 – e
2
Key : a, b, d
  
Sol : f  x  is defined when  cos x   sin x   x   0
 2 
F(x) is not injective f(x) is continues
F(x) is derivable at all points from graph
251.(L-2) Statement – 1 : f  x    x   x2 is discontinuous at all integers (where . denotes greatest
2

integer function and 

. denotes fractional part function).

Statement – 2 : f(x) = [x] is discontinuous at all integers

Key : d

f  x   x  x2
2
Sol : is continuous at x = 1
n 1
x
252. (L-2)Let f  x   lim  , then
r 0  rx  1  r  1 x  1
n 

a) f(x) is continuous but not differentiable at x = 0

b) f(x) is both continuous and differentiable at x = 0
c) f(x) is neither continuous nor differentiable at x = 0
d) f(x) is a periodic function
Key : c

Sol : t r 1 
x

 r  1 x  1   rx  1  1  1
 rx  1  r  1 x  1  rx  1  r  1 x  1  rx  1  r  1 x  1
n 1  1
1  ,x  0
 Sn   t r 1   nx  1
r 0  0 ,x  0

1, x  0
f  x   lim s n  
n  0, x  0
 lim f  x   1 and f  0   0
x 0

Hence f(x) is neither continuous nor differentiable at x = 0

Clearly f(x) is not a periodic function.
1  sin x   
253.(L-1)Let f  x   , x  . If f(x) is continuous at x  , then f   should be equal to
sin 2x 2 2 2

70
 1 1
a) 1 b) 0 c) d) 
2 2
Key: b
 
1  sin   h 
1  sin x 2   lim 1  cosh
Sol : RHL  lim   lim
x    /2  sin 2x h 0   h  0  sin 2h
sin 2   h 
2 

2sin 2  h / 2   h  2
 lim  
2sin 2  h / 2  h 0  h / 2 2  2  2 h
  lim    . lim  0
h 0 sin 2h sin 2h 2 h 0 8
lim .2h
h  0 2h

 
1  sin   h 
1  sin x 2   lim 1  cos h
LHL  lim   lim
x    /2  sin 2x h 0   h  0 sin 2h
sin 2   h 
2 

2sin 2  h / 2   h  2
lim  
sin 2  h / 2  h 0  h / 2 2  2  2 h
 lim 2   . lim  0
h 0 sin 2h sin 2h 2 h 0 8
lim .2h
h  0 2h

LHL  RHL  0


f    0 (since f has to be continuous)
2

254. (L-3)Statement – I : If f  x   x for all x  R then f  x  is continuous at 0.

Statement – II : If f(x) is continuous then f  x  is also continuous.

Key : b
Sol : The Statement-1 is true. Let us show it.
Since f  x   x for all x.

We have f  0   0

But f  0   0

 f  0  has to be zero.

Now f  x   x

 lim f  x   lim x  0  lim f  x   0

x 0 x 0 x 0

But lim f  x  must be  0  lim f  x  must be 0

x 0 x 0

71
 Thus, lim f  x   f  0   f  x  is continuous at 0.
x 0

Statement-2 is also true.

1  cos 4x
 x0
 x2

255. (L-1)Let f  x    a x  0 . Then the function f(x) is

 x
x0
 16  x  4

a) Not continuous at x = 0 b) Continuous at x = 0, if a = 6
c) Continuous at x = 0, if a = 12 d) Continuous at x = 0, if a = 8
Key : d
Sol : f(x) will be continuous at x = 0 if
lim f  0  h   lim f  0  h   f  0  …. (1)
h 0 h 0

0h
Now, lim f  0  h   lim
h 0 h 0
16  0  h  4

 lim
h
 lim
h  16  
h 4
h 0
16  h  4 h 0 16  h  16

=  lim
h 0
 16  h 4 8 
1  cos 4  0  h 
lim f  0  h   lim
h 0 h 0
 0  h 2
1  cos 4h 2sin 2 2h
lim  lim
h 0 h2 h 0 h2
2
 sin 2h 
 lim 2.   4  8
h  0  2h 

f(0) = a
by 1 ,8  8  a; a  8

256. (L-1)Statement – I : If f  x  
e ax

 1 sin ax
, x  0 and f(0) = 9 and f(x) is continuous at x = 0 then
4x 2
a  6 .
Statement – II : For continuity R.H.L. = L.H.L = f(a).
Key : a

1 eax  1 sin ax 1 a2
Sol : Lt f  x   Lt .  .a.a 
x 0 x 0 4 x x 4 4

72
 By expansion or by L Hospital’s rule
a2
  9 or a 2  36 a  6
4
 x n  sin x n
 lim n , x  1, x  0
257.(L-1) Consider the function f  x    n  x  sin x n
 ,x 1
 0
Which of the following is correct statement
a) f(x) has non-removable discontinuity at x = 1 with jump of discontinuity being 2.
b) f(x) has non-removable discontinuity at x = 1 with jump of discontinuity being 1.
c) f(x) has removable discontinuity at x = 1
d) f(x) is continuous at x = 1
Key : b
Sol : At x = 1

L.H.L  lim lim

1  h   sin 1  h 
n n
1
   
h  0 n  1  h n  sin 1  h n

1  cos x
 ;x  0
 x2

258. (L-1) f  x   a ; x  0 . The value of ‘a’ for which f(x) is continuous at x = 0 is

 x
;x  0
 16  x  4

1
a) b) 2
2
c) 8 d) no value of ‘a’ is possible
Key ; d
1  cos x
 ;x  0
 x2

Sol : f  x   a ;x  0

 x
;x  0
 16  x  4

If f(x) is continuous at x = 0

 
f  0   f  0 '  f 0

1  cos h 1
 
f 0  lim
h 0 h2

2

h
f  0 '  lim 0
h 0
16  h  4

73
  
 f 0  f  0 '  , so no value of a is possible.

x
 , x0
259. (L-2)Let f  x    x  
and g  x   x 1  x 2 then
0, x  0

a) fog is continuous function b) gof is continuous at x = 0
c) gof is continuous function d) fog is discontinuous at x = -1
Key : b, c
Sol : gof  x   0.x  R  _____________ x axis 
 1 x  1
0 x  1, 0,1

fog  x    1 1  x  0 it is discontinuous at x = -1, 0, 1
1 0  x 1

 1 x 1

260. (L-2) Statement-1 : Minimum number of points of discontinuity of the function

f  x    g  x    2x  1 x   3, 1 . Where [.] denotes the greatest integer function and

g  x   ax 3  x 2  1 is zero

Statement-2 : f(x) can be continuous at a point of discontinuity, say x  c1 of  2x  1 if g  c1   0

Key : D
5 3
Sol : [2x – 1] is discontinuous at three points x  , and -2
2 2
5 3
f(x) may be continuous if g  x   ax 3  x 2  1  0 at x  , and -2
2 2
g(x) can be zero at only one point for a fixed value of
 minimum number of points of discontinuity = 2.
 x
1  x , x  1

261 Let f  x    . Then which of the following statements is correct?
 x , x 1
1  x
A) f is continuous but not differentiable on R  1,1
B) f is both continuous and differentiable on R  1,1
C) f is not differentiable on R  1, 0,1
D) f is not continuous on R  1, 0,1
KEY: B
a x  1, 0  x  3
262. Let g be a real valued function defined by g(x)  
 bx  2, 3  x  5

74
 If g be differentiable on (0, 5) then (a+b) equals
A)1 B)2 C)3 D)4
KEY: B
 x2 
263. Let f(x) = min  x ,  , x   5, 5  .
 2
The number of points where f(x) is not derivable is
A)1 B)2 C)3 D)4
KEY: B

264. The number of points where f(x) = cos x  cos 1  sgn x   nx is not differentiable on (0,2  ) is (sgn
denote signum function)
A) 1 B) 2 C) 3 D) 4
KEY: C
Paragraph for Question

L’ Hopital’s rule has many versions. One of them is this.

Suppose f, g: (a, b)  R are differentiable on (a, b). Suppose further that
(i) g '(x)  0 for x  (a, b) (ii) lim g(x)   (or  )
x a
f'(x) f(x)
(iii) lim L Then lim L
x a g '(x) x  a g(x)
(This rule can be extended to cover the case when a or b tends to infinity or L tends to infinity)
265. Let f be a differentiable function on (o, )
     
If lim  sin   .f(x)  f'(x)   sec   , then lim f(x) equals
x 
  10   5 x 

1 3 5
A) B) 4 C) 3  5 D)
4 4
KEY:B

266. Let f be a differentiable function on (0, ) .

   
If lim  tan   .f(x)  2 xf'(x)   cot , then lim f(x) equals
x 
 8  12 x 

A) 8  6  4  3 B) 8  6  4  3
C) 3  4  6  8 D) 8  6  4  3
KEY: C
267. Let f be three times differentiable on (0, ) and such that f(x)  0, f'(x)  0, f"(x)  0 for x > 0
f'(x)f'''(x)  xf''(x)
If lim  tan , then lim equals
x  (f"(x))2 12 x  f'(x)
3 1 3 1
A) 2  3 B) 2  3 C) D)
2 2
KEY: D
268. Let f be a polynomial of degree 4 over reals satisfying
1
f'(0)  f'(1)  f'(1)  0 and f(0)  4, f"   1
2
Match the items in Column – I with those in Column II

Column – I Column - II
A) f(x)=0 has p) root at x = 2

75
 B) 4 – f(x)=0 has q) root at x = 1
C) f'(x) +x - 1=0 has r) 2 equal real roots
D) x f'(x) – 4f(x) = 0 has s) no real roots
KEY: A—S; B—R; C—Q,R; D--P

269. Column I lists some functions and Column II lists its properties. Match the items of Column A with
those of Column B.
Column – I Column – II
x
n n
x
A) f(x) = lim x x
,xR P)Continuous at all points in its domain
n  n  n

1
B) f(x) = lim n 4n  x 2 n  2 n , x  R  {0} Q) Discontinuous at finitely many
n  x
points in its domain
n ( e  x )
n n
C) f(x) = lim , x  0, x  R R) Not differentiable at finitely many
n  n
points in its domain.
D) f(x) = lim 2n
cos 2 n x  sin 2 n x, x  R S) Not differentiable at infinitely many
n 
points in its domain.
KEY: A—Q,R; B—P,R; C—P,R; D—P,S
270. Which of the following statements are true?
f(c  h)  f(c  h)
A) If f is differentiable at x = c, then lim exists and equals f '  c  .
h 0 2h
f(c  h)  f(c  h)
B) Given a function f and a point c in the domain of f, if the lim exists, then the
h 0 h
function is differentiable at x = c
 2 1
 x sin 2 , x  0
C) Let g(x)=  x , then g' exists
 0, x  0
 2 1
 x sin 2 , x  0
D) Let g(x)=  x , then g' exists and is continuous.
 0, x  0
KEY: A,C
271. Let f be a function given by
 1 1
 xn2  2 x  1 ;x  0
f(x)   , Then
 1 ;x  0
 2
A) f is continuous on R
n2
B) f is differentiable on R and f'(0) equals
12
C) f is not differentiable at x = 0
n2
D) f is differentiable on R and f'(0) equals
6
KEY: A,B
272. Match the items in Column – I with those in Column - II

76
  x12
e , x  0
A) f(x) =  p) first derivative exists
 0, x  0
 x1
e ,x  0
B) f(x) =  q) first derivative is continuous
 0, x  0

e x  e x   , x  (e, )
1 1

C) f(x) =  r) second derivative exists

 0, x  (e, )

 2 1
x sin , x  0
D) f(x) =  x s) second derivative is continuous
 0, x  0
KEY:A—P,Q,R,S; B—P,Q,R,S; C—P,Q,R,S D--R
2
d 2 y  dy  dy
273 If the independent variable x is changed to y, then the expression x 2      0 is
dx  dx  dx
2
d 2 x  dx  dx
transformed to x 2      , then  equals
dy  dy  dy
KEY: 1

274. Let f and g be continuously differentiable functions such that f(0) = 0, f' (0)=2 and g(x) = f (- x+f(f(x))).
The value of g' (0) equals
KEY: 6

x
275.(L-3)Let f  x    e  dt  x  0  , where [x] denotes greatest integer less than or equal to x, is
t t

a) continuous and differentiable x  (0,3]

b) continuous but not differentiable x  (0,3]
c) f(1) = e
d) f(2) = 2(e =-1)
Key : b, d
x x
We have f  x    e  dt   etdt, so
t t
Sol :
0 0


x t
  e dt if x  [0,1)
0  x
1 x e  1 if x  [0,1)
f  x     e t dt   e t 1 if x  [1, 2)


 
 f  x   (e  1)  e x 1  1 if x  [1, 2)

0
 
1

 2 2  e  1  e x 2  1 if x  [2,3)

1 t 2
t 1
x
t 2
  e dt   e dt   e dt if x  [2,3)
 0 1 2

77
 Clearly f(x) is continuous x  0 but not differentiable x  N  B
Also f(2) = 2  e  1  0  2  e  1  D

276(L-1)Consider the function y  f (x)  1  1  x 2 . Then the true statements among the following is/are
A) f is continuous in its domain B) f is differentiable in (-1, 1)

sin
1 1 3 2
C) Rf '(0)  and Lf '(0)   D) If     then f '(sin ) 
2 2 2 2 cos 
Key: A,C,D
Hint f is continuous in its domain [1,1]
x
f 1 (x)  , x  0, x  1
2 1 1 x2 1 x2

277. (L-1)The function f  x   x 2  3x  2  cos x is not differentiable at how many values of x.

Key : 2
Sol :  f  x   x 2  3x  2  cos x

  x  1  x  2   cos x

 x 2  3x  2  cos x, x  0
 2
 x  3x  2  cos x, 0  x<1
f x   2
 x  3x  2  cos x,1  x  2
 x 2  3x  2  cos x, x  2

2x  3  sin x, x  0
2x  3  sin x, 0  x  1

f  x   
2x  3  sin x,1  x  2
2x  3  sin x, x  2
it is clear f(x) is not differentiable at x = 1.
 f  1   1  sin1

and f  1   1  sin1.

Let f  x    et sin  x  t  dt and g  x   f  x   f ''  x  for all real x. Which of the following statements
x
278.
0

is / are correct ?
a) g  x   0 for all x  R b) g (1) = e

c) g '  x   g  x  for all x  R d) range of g is  0,  

Key; A, B, C
Hint
1 x

f  x 
2
e  sin x  cos x  and g  x   e x
279 Statement - 1: If f is twice differentiable function and f (a)  0, f (b)  1, f (c)  1, f (d)  0 where
a  b  c  d then the minimum number of roots of the equation
f ' (x)  f (x)f ''(x)  0 in  a, d  is 4.
2

78
 Statement - 2: If f is continuous in [, ] and f ( )f ()  0 then r  (, ) such that f (r)  0 and
if further function f is differentiable in (, ) and f ()  f () then   (, )
such that f '()  0 .
KEY : A
HINT
Conceptual Question

 x
1  x , x  1

280. Let f  x    . Then which of the following statements is correct?
 x , x 1
1  x
A) f is continuous but not differentiable on R  1,1
B) f is both continuous and differentiable on R  1,1
C) f is not differentiable on R  1, 0,1
D) f is not continuous on R  1, 0,1
KEY: B
a x  1, 0  x  3
281. Let g be a real valued function defined by g(x)  
 bx  2, 3  x  5

If g be differentiable on (0, 5) then (a+b) equals

A)1 B)2 C)3 D)4
KEY: B
 x2 
282. Let f(x) = min  x ,  , x   5, 5  .
 2
The number of points where f(x) is not derivable is
A)1 B)2 C)3 D)4
KEY: B

283. The number of points where f(x) = cos x  cos 1  sgn x   nx is not differentiable on (0,2  ) is (sgn
denote signum function)
A) 1 B) 2 C) 3 D) 4
KEY: C
Paragraph for Question Nos. 36 to 38

L’ Hopital’s rule has many versions. One of them is this.

Suppose f, g: (a, b)  R are differentiable on (a, b). Suppose further that
(i) g '(x)  0 for x  (a, b) (ii) lim g(x)   (or  )
x a
f'(x) f(x)
(iii) lim L Then lim L
x a g '(x) x  a g(x)
(This rule can be extended to cover the case when a or b tends to infinity or L tends to infinity)
284. Let f be a differentiable function on (o, )
     
If lim  sin   .f(x)  f'(x)   sec   , then lim f(x) equals
x 
  10   5 x 

1 3 5
A) B) 4 C) 3  5 D)
4 4
KEY:B
79
285. Let f be a differentiable function on (0, ) .
   
If lim  tan   .f(x)  2 xf'(x)   cot , then lim f(x) equals
x 
 8  12 x 

A) 8  6  4  3 B) 8  6  4  3
C) 3  4  6  8 D) 8  6  4  3
KEY: C
286. Let f be three times differentiable on (0, ) and such that f(x)  0, f'(x)  0, f"(x)  0 for x > 0
f'(x)f'''(x)  xf''(x)
If lim 2
 tan , then lim equals
x  (f"(x)) 12 x  f'(x)

3 1 3 1
A) 2  3 B) 2  3 C) D)
2 2
KEY: D
287. Let f be a polynomial of degree 4 over reals satisfying
1
f'(0)  f'(1)  f'(1)  0 and f(0)  4, f"   1
2
Match the items in Column – I with those in Column II

Column – I Column - II
A) f(x)=0 has p) root at x = 2
B) 4 – f(x)=0 has q) root at x = 1
C) f'(x) +x - 1=0 has r) 2 equal real roots
D) x f'(x) – 4f(x) = 0 has s) no real roots
KEY: A—S; B—R; C—Q,R; D--P

288. Column I lists some functions and Column II lists its properties. Match the items of Column A with
those of Column B.
Column – I Column – II
x
n n
x
A) f(x) = lim x x
,xR P)Continuous at all points in its domain
n  n  n

1
B) f(x) = lim n 4n  x 2 n  2 n , x  R  {0} Q) Discontinuous at finitely many
n  x
points in its domain
n ( e  x )
n n
C) f(x) = lim , x  0, x  R R) Not differentiable at finitely many
n  n
points in its domain.
D) f(x) = lim 2n
cos 2 n x  sin 2 n x, x  R S) Not differentiable at infinitely many
n 
points in its domain.
KEY: A—Q,R; B—P,R; C—P,R; D—P,S
289. Which of the following statements are true?
f(c  h)  f(c  h)
A) If f is differentiable at x = c, then lim exists and equals f '  c  .
h 0 2h
f(c  h)  f(c  h)
B) Given a function f and a point c in the domain of f, if the lim exists, then the
h 0 h
function is differentiable at x = c

80
  2 1
 x sin 2 , x  0
C) Let g(x)=  x , then g' exists
 0, x  0
 2 1
 x sin 2 , x  0
D) Let g(x)=  x , then g' exists and is continuous.
 0, x  0
KEY: A,C
290. Let f be a function given by
 1 1
 xn2  2 x  1 ;x  0
f(x)   , Then
 1
;x  0
 2
A) f is continuous on R
n2
B) f is differentiable on R and f'(0) equals
12
C) f is not differentiable at x = 0
n2
D) f is differentiable on R and f'(0) equals
6
KEY: A,B
291. Match the items in Column – I with those in Column - II
 x12
e , x  0
A) f(x) =  p) first derivative exists
 0, x  0
 x1
e ,x  0
B) f(x) =  q) first derivative is continuous
 0, x  0

e x  e  x   , x  (e, )
1 1

C) f(x) =  r) second derivative exists

 0, x  (e, )

 2 1
x sin , x  0
D) f(x) =  x s) second derivative is continuous
 0, x  0
KEY:A—P,Q,R,S; B—P,Q,R,S; C—P,Q,R,S D--R
2
d 2 y  dy  dy
32. If the independent variable x is changed to y, then the expression x    0 is
dx 2  dx  dx
2
d 2 x  dx  dx
transformed to x 2
    , then  equals
dy  dy  dy
KEY: 1

292. Let f and g be continuously differentiable functions such that f(0) = 0, f' (0)=2 and g(x) = f (- x+f(f(x))).
The value of g' (0) equals
KEY: 6

293. f :  0,1  R is a differentiable function such that f(0) = 0 and f '  x   k f  x  for all x

 0,1 ,  k  0  , then which of the following is/are always true ?

81
 a) f  x   0, x  R b) f  x   0, x   0,1 c) f  x   0, x   0,1 d) f 1  k

Key : b

f ' x   k2 f  x   0
2 2
Sol :

  f '  x   kf  x    f '  x   kf  x    0

 
 f  x  e  kx ' f  x  e kx '  0 
 Exactly one of the functions g1  x   f  x  e  kx or g 2  x   f  x  e kx is non decreasing

But f(0) = 0  both functions g1 and g 2 have a value zero at x = 0

x   0,1 , g1  0   0 and g1 increasing  g1  x   0  f  x   0

g 2  0   0 and g 2 decreasing  g 2  x   0  f  x   0

 f  x   0 x   0,1

294. (L-3)
COLUMN – I COLUMN – II

A) f(x) is defined for x  0 and has a continuous p) Cubic polynomial function

derivative such that f(0) = 1; v and
1  f  x   f ''  x   1  x, then f(x) is a/an
B) If f  x  .f  y   f  x  y  , x, y  R and q) Increasing function

f  0   0, f 1  e, f '  0   1, then f(x) is a/an

C) If f  x  is polynomial function satisfying r) Periodic function

 
f  0   0 and f x 2  1  f 2  x   1 , then f(x) is

a/an
D) If f is a polynomial function such that s) linear polynomial function
f : R   R  and f  f  x    6x  f  x  , then

f(x) is a/an
t) Quadratic polynomial function

Key : A-q; B – q; C-q,s; D – q,s

Sol :
e1BC MB  CN e1
A) MB  CN  e1  PB  PC     is increasing at x =0
e3 BC e3

 for some   0 f '  x   0, x   0,   . since f '  0   0

 f     f  0   f     1 ….. (i)
82
 1 
f ''      0  f '  x  is increasing at x =   f '     0
1  f 

Hence f(x) is increasing at x  

Repeat the same argument on interval right to x  
It comes out f '  x   0, x  0

Hence f(x) is increasing x  0 .

B) f  x   e x which is increasing function.

C) If f(x) is polynomial function of degree n, then

h(x) = f(x) – x is also a polynomial degree n

 
Now, f x 2  1  f 2  x   1

x  0  f 1  1  0  h 1  0

x  1  f  2  2  0  h  2  0

……………………….
……………………….
x  r  f  r   r  0  h(r)  0 for r  N

Keep on repeating it until r  n

 h(x) = 0 has more than n roots
Hence h(x) = 0, x  R (it’s an identity)
 f  x   x, x  R i.e., linear and increasing

D) From given functional equation it is obvious that f(x) has to be a linear polynomial.
Let f(x) = ax + b
 f  ax  b   6x   ax  b   a  ax  b   b  6x  ax  b

 
 x a 2  6  a  0 and ab  2b  0

Solving these equation, a = 2 and b = 0 or a = -3 and b = 0

 f  x   2x or f  x   3x

But f  x   3x is not R   R 

Hence f(x) = 2x i.e., linear and increasing.

295.(L-3) f : 1, 4   7,14 is a subjective, twice differentiable function such that f ''  x   0 , x  1, 4  ,

then the equation  f '  x   

2 49
has
9
a) at least one root in (1, 4) b) exactly one root in (1, 4)
c) at least two roots in (1, 4) d) at most two roots in (1, 4)

83
Key : a, d
Sol : Since f ''  x   0 graph of f(x) is concave upward.

7 7
Now f '  x    having a solution would mean line having slope  touching the curve.
3 3
The graph will strictly lie in the rectangle formed by x = 1, x = 4, y = 7, y = 14
If f is monotonic the line parallel to either of the diagonals of rectangle will touch the graph
exactly once and if f is non-monotonic, then lines parallel to diagonals will touch the graph exactly
once.
296. (L-3) f : R  R be a twice differentiable function satisfying f ''  x   5f '  x   6f  x   0 x  0 if f(0) =

1 f '  0   0 . If f(x) satisfies f  x   ah  bx   bh  ax  , x  0 , then find (a + b) h (0).

Key : 5
Sol : Given inequality can be written as :
f ''  x   2f '  x   3  f '  x   2f  x  
Let f '  x   2f  x   g  x 
 g '  x   3g  x   0 Multiply e 3x

 
 g  x  e 3x '  0  g  x  e 3x is non-decreasing
Now g  0   f '  0   2f  0   2
g  x  e 3x  2, x  0
f '  x   2f  x   2e3x , x  0 Multiply e 2x

 
 f  x  e 2x '  2e x , x  0

  f  x  e  2e  '  0
2x x

 f  x  e 2x  2e x  3
 f  x   3e 2x  2e3x , x  0
Comparing ah(bx) – bh(ax) with 3e2x  2e3x we get h  x   e x , a  3, b  2
  a  b h 0  5

297. If f  x  
x g  x  is a periodic function with period
1
, where g(x) is a differentiable function,
x g  x  4

then (where {.} represents fractional part of x)

1 5 k
a) g '  x   0 has exactly three roots in  ,  b) g(x) = 0 at x  , where k  I
4 4 4
c) g(x) must be non-zero function d) g(x) must be periodic function
Key :
Sol : Question Wrong
III. (L-3)Let x = 2t  t  1 , y  2t 2  t t . Elimination of ‘t’ between these two equations leads us to define y

as a function of x, i.e., y = f(x)

84
298. f(x) =
1 1
a)  x  12 , for x  1 b)  x  12 , for  1  x  2
9 3

c) 3  x  1 , for x  2
2
d) none of these

Key : a, b, c
Sol : Consider t < 0. Then x = 2t – (-t + 1) = 3t – 1
y  2t 2  t 2  t 2

i.e., x  3t  1, y  t 2 (where t = 0, x = -1)

1
Eliminating ‘t’ between x  3t  1 and y  t 2 , we get y   x  12
9
1
i.e., f  x    x  12 for x  1   A 
9
Similarly (B) and (C)
299. Thus,
a) f(x) is continuous at x = -1, 2 b) f(x) is differentiable at x = -1
c) f(x) is not differentiable at x = 2 d) none of these
Key ; a, b, c
Sol: Consider the case (t = 1) x = 2
To prove that f(x) is continuous at x = 2
1
 3  x  1 ,  1  x  2
2

Now f  x   
3  x  12 , x  2

(by solution of Q.No. 68)
1
Now lim f  x   lim  x  12  3
x2 x2 3

lim f  x   lim 3  x  1  3
2
x  2' x 2

Hence f(x) is continuous at x = 2

Similarly other cases.
300.(L-1)The number of points at which the function f (x) = (x – | x | )2 (1 – x + | x | )2 is not
differentiability in the interval (– 3, 4) is
a) zero b) one c) two d) three

16x 4  16x 3  4x 2 , x  0
sol : We have f (x) = 
 0, x0

Clearly f (x) is continuous as well as derivable x  R

301. (L-3)Consider the following statements :
85
  
Consider f  x   sin 1 min  x , y  , defined on x   1,1 , where (x, y) lies on the curve

y  1  x 2 . Then

S1 : Number of points where f(x) is continuous but not differentiable is 2.

S2 : The equation 2f(x) – 1 = 0 has four solutions.

S3 : f(x) is one-one

State, in order, whether S1 ,S2 ,S3 are true or false

a) FTF b) TTF c) TTT d) FFF

Key : a
Sol : f  x   sin 1{min  x , y 

 1 1
sin x , 0  x  2

sin 1 1  x 2 , 1  x  1
 2
Graph is

y  /4
y  1/ 2
-1 x  1/ 2 x  1/ 2

302. (L-2)f(x) is a differentiable function such that f  x  y   f  x   f  y   2xy  1x, y  R and

f '  0   cos  , then value of f ' 1 is

a) cos  b) cos   1 c) cos   2 d) cos   3

Key : c
Sol : f  x  y   f  x   f  y   2xy  1

x=x
y=h
f  x  h   f  x   f  h   2xh  1

f x  h  f x f  h   2xh  1
 lim  lim
h 0 h h 0 h
f '  h   2x
 f '  x   lim
h 0 1
f '  0   cos   f '  x   2x  cos 
86
 f ' 1  2  cos 

  
303. (L-3)Let f : 0,    0,1 be a differentiable function such that f  0   0, f    1 , then
 2 2
   
a) f '     1   f     for all    0, 
2
b) f '    
2
for all    0, 
 2   2
1   8  
c) f    f '     for at least one    0,  d) f '     for at least one    0, 
  2  2
 2
Key : c, d
Sol : A) consider g(x) = sin 1 f  x   x


Since g  0   0, g    0
2
 
 there is at least one value of    0,  such that
 2
f '
g '  -1 = 0
1  f  
2

f '  
i.e., g '     for atleast one value of  but may not be for all    0, 
1  f    2
2

 false
2x
B) Consider g  x   f  x  


Since g  0   0, g    0
2
  2
 there is at least one value of    0,  such that g '     f '      0
 2 

2  
i.e., f '     for atleast one value of  but may not be for all    0, 
  2
 false

C) consider g  x    f  x   
2 2x


Since g  0   0, g    0
2
 
 there is at least one value of    0,  such that
 2
2
g '     2f    f '     0


87
 1
f    f '    

 true
4x 2
D) consider g  x   f  x  
2

since g  0   0, g    0
2
 
 there is at least one value of    0,  such that
 2
8
g '  f '  0
2
8
f '    
2
 true
304. (L-2)Statement-1 : If f (x) be a continuous function in [a, b] then for all c  (a, b), f (c)  (f (a), f (b))
Statement-2 : Intermediate theorem states that for a continuous function in closed interval all
intermediate values are obtained by the function.
A) Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.
B) Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation for
Statement-1.
C) Statement-1 is true, Statement-2 is false.
D) Statement-1 is false, Statement-2 is true.
Key : d
Sol : Conceptional A is false R is true

305. (L-3)
Match the following :

 Column I   Column II
(A) If f (x) = (p) 0
lim  lim 1 
 (sin(x  h  t)  sin(x  h)  sin(x t)  sin x 
h  0 t  0 t 
lim f (x)
then is
x0 x
(B)  n(ex 2 )   3  6 nx  (q) 1
Let f (x) = cot 1  2 
 tan 1   then
 n(e / x )   3  6nx 
d2 y
at x = e is
dx 2
(C) n lim (r) -1
Let Pn   cos(x.2 k ) and g(x) = Pn then
k 1 n
lim
g(x) is
x0

88
 (D) lim n x n  [x] (s) Does not exist
where n  N and [.] denotes greatest
x [x]
integer function

Key : (A) – p, s; (B) – p, s; (C) – p,q; (D) – p,r

1  sin  x  n  t   sin  cos ec  sin  x  t   sin x 
Sol : A) f  x   lt  lt  lt 
x 0 x x 0 x 0
 t t 
 lt
1
 cos  x  n   cos x   sin x
x 0
x
f  x sin x
lt   lt   1
x 0 x x 0 x
 1  2 log x  1  3  2 log x 
B) f  x   cot    tan  
 1  2 log x   3  6 log x 
 2 
1  1  2 log x  1
 1  3 log x 
 tan    tan  
 1  2 log x   1  2 log x 
 
 Tan 1  Tan  2 log x   Tan 1  Tan 1  2 log x 
1 1 1

dy d2y
 f  x   / 2  0 2 0
dx dx
 x
C) Pn  k  cos  
2

 2k 
 x  x   x 
 cos   cos  2     cos  x 
2 2  2 
 x  x   x 
lt px  lt cos   cos      cos  
x  x 
2  22   2x 
 x 
sin  2 x 
 lt  2 x  sin x

x 
x  x  x
2 sin  
 2x 
sin x
 lt 1
x 0 x
log x n   x 
D) l t if x is an intergar then  x   x
x   x
1
x  1
x log x  x x
lt  Lt  1
x  x x  1
cos x  sin x cos x  sin x
306. (L-3)The number of points at which f  x    x   0, 2 is not differentiable
2 2
is
a) 2 b) 3 c) 1 d) none
Key : a

89
Sol : It is not differentiable when cosx + sinx = 0

 tan x  1  x  n 
4
3 7
x ,
4 4
No . of points is 2 .


307. (L-1)Total number of points of non-differentiability of f  x   min 1,1  x 3 , x 2  3x  3 are 
a) 2 b) 3 c) 4 d) none of these
Key : a
Sol : f  x   1  x3 , x  0
= 1 0  x 1
= x 2  3 x  3 1  x  2 clearly the non diff pts are 1,2
=1 x>2

2 dn
308. (L-1)Statement-1 : Let f  x   e x  e  x  2sin x  x 3 then the least value of n for which f x x 0
3 dx n
is non-zero is 7.
Statement-2 : If f(x) is an algebraic polynomial in x such that f  x   x 3  x 4  x 5  .... then the least

dn
value of ‘n’ for which f x x 0 is non-zero is 3.
dx n
Key : a
Sol : By finding derivatives successively we get at x = 0 , y1  y2  y3  0 from y4 on words they are non
zeros hence least value of n is 4 not 7.
 Assertion is false. Reason is clearly true.
309. (L-3) If a function satisfies  x  y  f  x  y    x  y  f  x  y   2  x 2 y  y3  x, y  R and f 1  2 , then

(A) f(x) must be polynomial function (B) f(3) = 12

(C) f(0) = 0 (D) f(x) may not be differentiable
Key : a, b, c
Hint :  x  y  f  x  y    x  y  f  x  y   2y  x  y  x  y  
Let x  t  u; x  y  v
uf  v   vf  u   2uv  v  u 
f  v f u 
  vu
v u
 f  v   f u 
  v    u   constant
 v   u 
f x
Let x 
x
 f  x   x  x2 
f 1  2

90
  1  2    1 f  x   x2  x


310. (L-1)If x = a cos t, y = sin t, then at t 
4
dy d2 y 2 2 dx d2x 2 2
a) 1 b)  c)  1 d) 
dx dx 2 a dy dy 2 a
Key : c
 a a 
sol : x2 + y 2 = a 2  , 
 2 2
dx
2x  2y  0
dy
dx  y
  1
dy x

91