Question bank on function limit continuity & derivability
There are 105 questions in this question bank.
Select the correct alternative : (Only one is correct)
Q.13 If both f(x) & g(x) are differentiable functions at x = x0, then the function defined as,
h(x)=Maximum{f(x),g(x)}:
(A) is always differentiable at x = x0
(B) is never differentiable at x = x0
(C*) is differentiable at x = x0 provided f(x0)  g(x0)
(D) cannot be differentiable at x = x0 if f(x0) = g(x0) .
[Hint: Consider the graph of h(x) = max(x, x2) at x = 0 and x = 1]
Q.25 If
0
x
Lim

(x3 sin 3x + ax2 + b) exists and is equal to zero then :
(A*) a =  3 & b = 9/2 (B) a = 3 & b = 9/2
(C) a =  3 & b = 9/2 (D) a = 3 & b = 9/2
[Sol. b
x
a
x
x
3
sin
Lim 2
3
0
x



= 0
x
Lim
 3
3
x
bx
ax
x
3
sin 

= 0
x
Lim
 2
2
x
bx
a
x
3
x
3
sin
3 

for existence of limit 3 + a = 0  a = – 3
 l = 0
x
Lim
 3
3
x
bx
x
3
x
3
sin 

= b
t
t
t
sin
.
27 3


= 0 (3x = t)
= b
6
27

 = 0  b =
2
9
]
[ OR use L' Hospital's rule ]
Q.36 A function f(x) is defined as f(x) =
x x m N
if x
m
x
sin ,
1 0
0 0
 



 . Theleast value of m forwhich f(x)is
continuous at x =0 is
(A) 1 (B) 2 (C*) 3 (D) none
[Sol. f ' (0+) =
h
x
1
sin
h
Lim
m
0
h
must exist
 m > 1
for m > 1 h ' (x) = 


0
x
if
0
0
x
x
1
cos
x
x
1
sin
x
m 2
m
1
m


 

now
h
1
cos
h
h
1
sin
h
m
Lim
)
x
(
h
Lim 2
m
1
m
0
h
0
h






limit existif m>2
 m  N  m = 3]

Q.410 For x > 0, let h(x) =




 
 integers
prime
relatively
are
0
q
&
p
where
irrational
is
x
if
0
x
if q
p
q
1
then which onedoes not hold good?
(A*) h(x)isdiscontinuous for all xin (0,)
(B)h(x)iscontinuous for eachirrationalin (0, )
(C)h(x)isdiscontinuous for each rationalin (0, )
(D) h(x) is not derivable for all x in (0, ).
[Sol. Let x =
2
3
which is rational 
3
1
3
2
h 






Lim h t
t

F
H
G I
K
J 
0
2
3
0 discontinuous at x Q
Let x = 2 
 Q
h 2
d i = 0 consider 2 = 1.41401235839
h 2
d i = h
1414023583
1010
.
F
H
G I
K
J=
1
1010  0
Hencehiscontinuous forallirrational  A
A ]
Q.512 The value of n
e
1
x
e
1
x
x x
3
2
Limit
x
n
x
n















(where N
n ) is
(A) ln 





3
2
(B*) 0 (C) n ln 





3
2
(D)not defined
[Hint: l = n
e
x
e
x
x x
3
2
Limit
x
n
x
n



but 0
e
x
Limit x
n
x



 l = 0 ]
Q.613 For a certain value of c,


x
Lim [(x5 + 7x4 + 2)C - x] is finite & non zero. The value of c and the value
ofthelimitis
(A*) 1/5, 7/5 (B) 0, 1 (C) 1, 7/5 (D) none
[Sol. 


x
Lim
















 x
x
2
x
7
1
x
c
5
c
5
= 


x
Lim


















1
x
2
x
7
1
x
x
c
5
1
c
5
This will beof theform  ×0onlyif
5c – 1 = 0  c = 1/ 5 substituting c = 1/5 , l becomes
Hence l = 


x
Lim  
 
1
X
1
x
5
/
1

 where 5
x
2
x
7
X 

= 


x
Lim








 1
.....
5
X
1
x = 


x
Lim
5
1
.
x
2
x
7
x 5 





 =
5
7
Hence c =
5
1
and l =
5
7
]

Q.714 Considerthepiecewisedefinedfunction
f (x) =





4
x
if
4
x
4
x
0
if
0
0
x
if
x






choose theanswerwhichbest describes thecontinuityofthis function
(A)The functionis unbounded and thereforecannot be continuous.
(B)The functionis right continuous atx =0
(C) Thefunction has a removablediscontinuityat 0 and4, but is continuous on the rest ofthe real line.
(D*)Thefunctionis continuous ontheentirereal line
[Hint: Refer tothe graph ]
Q.817 If ,  arethe roots of thequadraticequation ax2 +bx+ c= 0 then 

x
Lim  
1 2
2
  

cos
( )
ax bx c
x 
equals
(A) 0 (B)
1
2
()2 (C*)
a2
2
()2 (D) 
a2
2
()2
[Sol. Lim
ax bx c
ax bx c
ax bx c
x
x
 
 
F
H
G
I
K
J
 

0
2
2
2
2
2 2
2
2
2
2
4
sin
( )
c h
c h
c h

; Lim
a x
b
a
x
c
a
x
b
a
x
c
a
x x
x
 
F
H
G I
K
J  
F
H
G I
K
J
 
  
2
4
2 2 2
( ) ( )
Lim
a x x
x
x
 


 

2 2 2
2
2
( ) ( )
( )
; Lim
a
C
x
 

 
2
2
2
( ) ]
Q.918 Which one ofthe followingbest representsthe graph ofthefunctionf(x)=  
nx
tan
2
Lim 1
n


 
(A*) (B) (C) (D)
Q.1019 1
x
Lim

 
4 1 3
1
3 1
2 1
2
3
1 4
3 1
x x
x x
x
x
x x


 






 













.
=
(A)
1
3
(B*) 3 (C)
1
2
(D) none

Q.1124 ABCis anisoscelestriangleinscribedinacircleofradiusr.IfAB=AC &histhealtitudefromAtoBC
and Pbe the perimeter ofABC then 0
h
Lim
 3
P

equals (where  is the area of the triangle)
(A)
r
32
1
(B)
r
64
1
(C*)
r
128
1
(D) none
[Sol. AB = 2
2
h
rh
2
h 


 = rh
2
P = rh
2
2 + 2
h
rh
2
2 
P = 




 
 2
h
rh
2
rh
2
2
 =
2
h
.
h
rh
2
2 2



= 2
h
rh
2
h 
3
0
h p
Limit



= 3
2
2
0
h
h
rh
2
rh
2
8
h
rh
2
h
Limit





 



=
 3
2
/
3
2
/
3
0
h
h
r
2
r
2
h
8
h
r
2
h
Limit




=
r
128
1
)
r
2
(
)
r
2
(
8
.
8
r
2
2
/
1
  (C) ]
Alternative: Note that as h  0 , b =
2
a
or 2b = a
Hence 3
p

= 3
)
b
2
a
(
.
R
4
)
b
(
ab

(using R =

4
abc
)
= 3
3
b
64
R
4
b
2
= R
128
1
 (C) ]
Q.1232 Let the function f, g and h be defined as follows :
f (x) =



0
x
for
0
0
x
and
1
x
1
for
x
1
sin
x











g (x) =



0
x
for
0
0
x
and
1
x
1
for
x
1
sin
x2











h (x) = | x |3 for – 1  x  1
Which ofthesefunctions aredifferentiableatx =0?
(A) f and g only (B) f and h only (C*) g and h only (D) none
Q.1333 If [x] denotes the greatest integer  x, then Limit
n  
1
4
n
     
 
1 2
3 3 3
x x n x
  
...... equals
(A) x/2 (B) x/3 (C) x/6 (D*) x/4

Q.1436 Let f (x) = )
x
(
)
x
(
h
g
, where g and h are cotinuous functions on the open interval (a, b). Which of the
followingstatements is true for a < x < b?
(A) f is continuous at all x for which x is not zero.
(B) f is continuous at all x for which g (x) = 0
(C) f is continuous at all x for which g (x) is not equal to zero.
(D*) f is continuous at all x for which h (x) is not equal to zero.
[Hint: Bytheorem,ifgandharecontinuous functions ontheopen interval (a,b),then g/h is alsocontinuous at
all x in the open interval (a, b) where h(x) is not equal to zero. ]
Q.1539 The period of thefunction f (x) =
|
x
cos
x
sin
|
|
x
cos
|
|
x
sin
|


is
(A) /2 (B) /4 (C*)  (D) 2
Q.1645 If f(x) = 2
x
x
x
2
cos
e
x 

, x  0 is continuous at x = 0, then
(A) f (0) =
2
5
(B) [f(0)] = – 2 (C) {f(0)} = –0.5 (D*) [f(0)] . {f(0)} = –1.5
where [x] and{x} denotes greatest integerand fractional part function
[Hint: 2
x
0
x x
)
x
2
cos
1
(
1
e
x
Lim





= –
2
1
– 2 = –
2
5
; hence for continuity f (0) = –
2
5
 [f (x)] = – 3 ; {f (0)} =







2
5
=
2
1
; hence [f (0)] {f (0)} = –
2
3
= – 1.5 ]
Q.1746 The valueof thelimit 









2
n
2
n
1
1 is
(A) 1 (B)
4
1
(C)
3
1
(D*)
2
1
[Hint: 









 
2
n
2
2
n
1
n
= 










 





 
2
n
2
n
n
1
n
n
1
n
= n
·
)
1
n
......(
4
·
3
·
2
)
1
n
........(
3
·
2
·
1


=
2
1
n
·
n
1 
=
2
n
1
1
Lim
n



=
2
1
]
Q.1847 The function g (x) =



0
x
,
x
cos
0
x
,
b
x



can be made differentiable at x = 0.
(A) if b is equal to zero (B) if b is not equal to zero
(C) if b takes anyreal value (D*) for no value of b

[Hint: g' (0+) =
h
1
cosh
Lim
0
h


= 0
g ' (0–) =
h
1
b
h
Lim
0
h 




for existence of line b = 1 thus g ' (0–) = 1
Hence gcan not be madedifferentiable for anyvalueof b.]
Q.1949 Let f be differentiable at x = 0 and f ' (0) = 1. Then
h
)
h
2
(
)
h
(
Lim
0
h



f
f
=
(A*) 3 (B) 2 (C) 1 (D) – 1
[Hint: reduces to 3 f ' (0) ]
Q.2050 If f (x) = sin–1(sinx) ; R
x then f is
(A)continuousanddifferentiableforallx
(B*) continuous for all x but not differentiable for all x = (2k + 1)
2

, I
k
(C) neither continuous nor differentiable for x = (2k – 1)
2

; I
k
(D)neithercontinuousnordifferentiablefor ]
1
,
1
[
R
x 


[Hint:




2
3
k
2
x
2
k
2
for
x
)
1
k
2
(
2
k
2
x
2
k
2
for
k
2
x





















]
Q.2153









 )
x
3
sin
x
sin
3
(
4
1
cos
x
sin
Limit
1
2
x
where [ ] denotes greatest integer function , is
(A*)

2
(B) 1 (C)

4
(D) doesnot exist
[Hint:
]
x
[sin
cos
x
sin
Limit 3
1
2
x



ax x  /2 , [sin3x]  0 and sinx  1
 l =

2
 (A) ]
Q.2260 If
0
x
Lim
 x
)
x
3
(
n
)
x
3
(
n 

 l
l
= k , the value of k is
(A*)
3
2
(B) –
3
1
(C) –
3
2
(D) 0

[Sol.
0
x
Limit
 x
)
x
3
(
n
)
x
3
(
n 

 l
l
=
0
x
Limit

x
3
x
1
n
3
n
3
x
1
n
3
n 















 l
l
l
l
=
0
x
Limit

x
/
1
x
/
1
3
x
1
n
3
x
1
n 













 l
l = 







3
x
x
1
3
x
.
x
1
=
3
2
3
1
3
1

  (A) ]
Q.2364 The function f (x) =
1
x
1
x
Lim n
2
n
2
n 



isidentical withthefunction
(A) g (x) = sgn(x – 1) (B) h (x) = sgn (tan–1x)
(C*) u (x) = sgn( | x | – 1) (D) v (x) = sgn (cot–1x)
[Hint: f (x) =





1
|
x
|
if
1
1
or
1
x
if
0
1
x
1
if
1







and g (x) = sgn(|x| – 1) is same ]
Q.2469 The functions defined by f(x) = max {x2, (x  1)2, 2x (1  x)}, 0  x  1
(A) isdifferentiableforallx
(B) is differentiableforall x excetp at one point
(C*) is differentiable for all x except at two points
(D) is not differentiable at morethan two points.
Q.2573 f (x) =
nx
x
l
and g (x) =
x
nx
l
. Then identifythe CORRECT statement
(A*)
)
x
(
g
1
andf (x) areidentical functions (B)
)
x
(
f
1
and g(x)are identical functions
(C) f (x) . g (x) = 1 0
x 
 (D) 1
)
x
(
g
.
)
x
(
f
1
 0
x 

[Hint: (A)
nx
x
)
x
(
f
;
x
/
nx
1
)
x
(
g
1
l
l

 x > 0 , x  1 for both
(B)
x
nx
)
x
(
g
;
nx
/
x
1
)
x
(
f
1 l
l

 
)
x
(
f
1
is not defined at x =1 but g (1) = 0
(C) f (x) . g (x) =
x
nx
.
nx
x l
l
= 1 if x > 0 , x  1  N. I.
(D) 1
x
nx
.
nx
x
1
)
x
(
g
.
)
x
(
f
1


l
l
only for x > 0 and x  1 ]
Q.2674 If f(3) = 6 & f(3) = 2, then Limit
x  3
x f f x
x
( ) ( )
3 3
3


is given by :
(A) 6 (B) 4 (C*) 0 (D) none of these

[Sol. f (3) = 6 ; f  (3)= 2
Limit
x  3
x f f x
x
( ) ( )
3 3
3


, put x = 3 + h
=
h
)
h
3
(
f
3
)
3
(
f
·
)
h
3
(
Limit
0
h




= )
3
(
f
h
]
)
3
(
f
)
h
3
(
f
[
3
Limit
0
h





= – 3f  (3) + f (3) = – 6 + 6 = 0 Ans ]
Q.2776 Which oneof the followingfunctionsis continuous everywhere inits domain but has atleast onepoint
whereitisnotdifferentiable?
(A*) f (x) = x1/3 (B) f (x) =
x
|
x
|
(C) f (x) = e–x (D) f (x) = tan x
[Hint: x1/3 is not differentiable at x = 0]
Q.2886 The limitingvalue of thefunction f(x) =
x
2
sin
1
)
x
sin
x
(cos
2
2 3



when x 
4

is
(A) 2 (B)
1
2
(C) 3 2 (D*)
3
2
[Hint: put x =
4

+ h or use L'Hospital's rule ]
Q.2989 Let f (x) =

















2
x
if
2
x
3
x
4
x
2
x
if
2
2
6
2
2
2
x
1
x
x
3
x
then
(A) f (2) = 8  f is continuous at x = 2 (B) f (2) = 16 f is continuous at x = 2
(C*) f (2–)  f (2+)  f is discontinuous (D) f has a removable discontinuityat x = 2
[Hint: f (2+) = 8 ; f (2–) = 16 ]
Q.3090 On the interval I = [2, 2], the function f(x) =
 
( ) ( )
( )
| |
x e x
x
x x
 






 
1 0
0 0
1 1
then which oneof the followingdoesnot holdgood?
(A*) is continuous for all values of x I
(B) is continuous for x  I  (0)
(C) assumes all intermediate values from f(2) & f(2)
(D) has amaximum value equal to 3/e.
[Sol. f (x) = 







 
0
x
if
0
0
x
if
1
x
0
x
if
e
)
1
x
( x
/
2
the graph off (x) is
hence f can assume all values for f (– 2) to f (2) ]

Q.3193 Whichofthefollowingfunctionissurjectivebutnotinjective
(A) f : R  R f (x) = x4 + 2x3 – x2 + 1 (B) f : R  R f (x) = x3 + x + 1
(C) f : R  R+ f (x) = 2
x
1 (D*) f : R  R f (x) = x3 + 2x2 – x + 1
[Sol. (A) f (x) = x4 + 2x3 – x2 + 1 Apolynomial of degree even will always be into
say f (x) = a0x2n + a1 x2n–1 + a2 x2n – 2 + .... + a2n



x
Limit f (x) =



x
Limit [x2n











 n
2
n
2
2
2
1
0
x
a
....
x
a
x
a
a = 







0
a
if
0
a
if
0
0
Hence it will never approach  / – 
(B) f (x) = x3 + x + 1  f (x) = 3x2 + 1 – injective as well as surjective
(C) f (x) = 2
x
1 – neitherinjectivenor surjective
f (x) = x3 + 2x2 – x + 1  f (x) =3x2 + 4x – 1  D > 0
Hencef(x) is surjectivebutnot injective.]
Q.3294 Consider the function f (x) =






3
x
2
if
x
6
2
x
if
1
2
x
1
if
]
x
[
x






where [x] denotes step up functionthen at x =2 function
(A)has missingpointremovablediscontinuity
(B*)hasisolatedpointremovablediscontinuity
(C)has nonremovablediscontinuityfinitetype
(D)iscontinuous
[Hint: fromthefigurethefunctionhasanobvious removableisolatedpointdiscontinuous.
Q.3395 Suppose that f is continuous on [a, b] and that f (x) is an integer for each x in [a, b]. Then in [a, b]
(A) f is injective
(B) Range of f mayhavemanyelements
(C) {x} is zero for all x  [a, b] where { } denotes fractional part function
(D*) f(x) is constant
[Hint: Explanation : Let f (x1) = n and f (x2) = m, x1, x2  (a, b) with n > m (say). According to the
intermediate value theorem, between x1 and x2 there must be some value x for which f (x) = m +
2
1
which is impossiblesince m +
2
1
is not aninteger.]
Q.34101 Thegraphof function f contains the point P (1, 2) andQ(s, r). Theequation ofthe secant line through
P and Q is y = 










1
s
3
s
2
s2
x – 1 – s. The value of f ' (1), is
(A) 2 (B) 3 (C*) 4 (D)non existent

[Sol. I Bydefinition f '(1) is the limit of the slope of the secant line when s  1.
Thus f '(1) =
1
s
3
s
2
s
Lim
2
1
s 



=
1
s
)
3
s
)(
1
s
(
Lim
1
s 



= )
3
s
(
Lim
1
s


= 4  (D)
II Bysubstitutingx = s intothe equation of thesecant line, and cancellingbys – 1 again, we get
y = s2 + 2s – 1. This is f (s), and its derivative is f '(s) = 2s + 2, so f ' (1) = 4.]
Q.35108 The range of the function f(x) =
12
x
11
x
2
)
10
x
7
x
(
5
x
n
e
2
2
)
2
x
(
x 2





l
is
(A*) )
,
( 
 (B) )
,
0
[  (C) 






,
2
3
(D) 





4
,
2
3
[Sol. f (x) =
)
4
x
(
)
3
x
2
(
)
5
x
(
)
2
x
(
.
5
nx
e )
2
x
(
x 2





l
Note that at x = 3/2 & x = 4function is not defined andin open interval (3/2,4)function is continuous.











)
ve
(
)
ve
(
)
ve
(
)
ve
(
)
ve
(
Lim
2
3
x









 )
ve
(
)
ve
(
)
ve
(
)
ve
(
)
ve
(
Lim
4
x
In the openinterval (3/2,4) the function iscontinuous & takes up all real values from (– ,)
Hence range of the function is (– , ) or R]
Q.36113 Consider f(x) =
 
 
2
2
3 3
3 3
sin sin sin sin
sin sin sin sin
x x x x
x x x x
  
  








, x 

2
for x  (0, )
f(/2) = 3 where [ ] denotes the greatest integer function then,
(A*) f is continuous &differentiable at x = /2
(B) f is continuous but not differentiable at x = /2
(C) f is neither continuous nor differentiable at x = /2
(D) none of these
[Sol. In the immediate neighborhood of x = /2 , sinx > sin3x  |sinx – sin3x|= sinx – sin3x
Hence for x  /2 , f (x) = 











x
sin
x
sin
)
x
sin
x
(sin
2
x
sin
x
sin
)
x
sin
x
(sin
2
3
3
3
3
= 3
x
sin
x
sin
x
sin
3
x
sin
3
3
3



Hence f is continuous and diff. at x = /2 ]
Q.37114 The number of points at which the function, f(x) = x – 0.5 + x – 1 + tan x does not have a
derivativeintheinterval (0,2) is :
(A) 1 (B) 2 (C*) 3 (D) 4

Q.38117 Let [x] denote the integral part of x  R. g(x) = x  [x]. Let f(x) be any continuous function with
f(0) =f(1) then the function h(x) =f(g(x)) :
(A) hasfinitelymanydiscontinuities (B) is discontinuous at some x = c
(C*) is continuous on R (D) is a constant function .
[Sol. g(x) = x – [x] = {x}
fis continuouswithf(0)=f(1)
h(x) = f (g(x)) = f ({x})
Let the graphof f is as shownin the figure
satisfying
f(0) = f(1)
now h(0) = f ({0}) = f(0) = f(1)
h(0.2) = f ({0.2}) = f(0.2)
h(1.5) = f ({1.5}) = f(0.5) etc.
Hence the graphof h(x) will be periodicgraph as shown
 h is continuous in R  C
]
Q.39118 Given the function f(x) = 2x x3
1
 + 5 x 1 4
 x + 7x2 x  1 + 3x + 2 then :
(A) thefunction is continuous but not differentiable at x = 1
(B*) thefunction is discontinuous at x = 1
(C) the function is both cont. & differentiable at x = 1
(D) the range of f(x) is R+.
Q.40119 If f (x + y) = f (x) + f (y) + | x | y + xy2,  x, y  R and f ' (0) = 0, then
(A) f need not be differentiable at everynon zero x
(B*) f is differentiablefor all x R
(C) f is twice differentiable at x = 0
(D) none
[Hint: f '(x) =
h
xh
h
|
x
|
)
h
(
Lim
h
)
x
(
)
h
x
(
Lim
2
0
h
0
h







f
f
f
 f (0) = 0  f ' (x) = 









xh
|
x
|
h
)
0
(
)
h
(
Lim
0
h
f
f
f ' (x) = f ' (0) + | x | = | x | ]
Q.41131 For
}
x
10
{
}
10
x
sin{
Lim
8
x 


(where { } denotes fractional part function)
(A)LHLexist but RHLdoes not exist (B*) RHLexist but LHLdoes not exist.
(C) neitherLHLnor RHLdoes not exist (D) both RHLand LHLexist and equals to 1

[Hint:
}
x
{
}
x
sin{
Lim
8
x 


= 0 as {I + x } = {x} ; as 0
}
x
{
Lim
I
x



and 1
}
x
{
Lim
I
x




}
x
{
}
x
sin{
Lim
8
x 


  as sin{x}  sin(1) and {–x}  0 ]
Q.42134 

n
Lim
3
3
3
3
2
2
2
2
n
......
3
2
1
1
.
n
.....
)
2
n
(
3
)
1
n
(
2
n
1










is equal to :
(A*)
1
3
(B)
2
3
(C)
1
2
(D)
1
6
[Sol. Lim
n n n n n n
n
n
       

1 2 1 3 2 1
2 2 2 2
3
( ) ( ) ............ ( ( ))
Nr. = n n n n
( ....... ) ( . . . .......... ( ) )
1 2 12 2 3 34 1
2 2 2 2 2 2 2
        
= n n r r
r
n
2 2
2
1
 


 ( ).
= n n r r
r
n
2 3 2
1
 


 ( ) = n n n n
2 3 2
  


= ( )
n n n
  

1 2 3  
 




 Lim
n n n
n
n
( )
1 2 3
3
Lim
n n n n
n n n n
n
  
 

( ) ( )( )
( ) ( )
1 1 2 1
6 1 1
1
4
3
1
1
3
   A
Alternatively:l =




 




 

1
n
......
3
2
1
1
·
n
......
)
1
n
(
2
n
1
3
2
3
3
2
2
2








– 1 =







3
2
2
2
n
)
1
n
(
n
......
)
1
n
(
2
)
1
n
(
1
– 1
l =



3
2
n
n
)
1
n
(
– 1 ]
Q.43135 The domain of definition of the function f (x) = |
6
x
x
|
log 2
x
1
x









+ 16–xC2x–1 + 20–3xP2x–5 is
(A*) {2} (B) }
3
,
2
{
,
4
3







 (C) {2, 3} (D) 






 ,
4
1
Where [x] denotesgreatest integerfunction.
Q.44136 If f (x) =
10
x
7
x
25
bx
x
2
2




for x  5 and f is continuous at x = 5, then f (5) has the value equal to
(A*) 0 (B) 5 (C) 10 (D) 25

[Hint: f (x) =
)
5
x
)(
2
x
(
25
bx
x2




. for existence of limit 25 – 5b + 25 = 0  b = 10
now f (x) =
)
5
x
)(
2
x
(
)
5
x
)(
5
x
(




and )
x
(
f
Lim
5
x
= 0  (A) ]
Q.45139 Let f beadifferentiablefunctionontheopen interval(a,b). Whichof thefollowing statementsmustbe
true?
I. f is continuous on the closedinterval [a, b]
II. f is bounded on the open interval (a, b)
III. If a<a1<b1<b, and f (a1)<0< f (b1), then there is a number c such that a1<c< b1 and f (c)=0
(A) I and II only (B) I and III only (C) II and III only (D*) onlyIII
[Sol. I and IIare false. The function f (x) = 1/x, 0 < x <1, is a counter example.
Statement IIIis true.Applythe intermediate valuetheorem to f on the closed interval [a1, b1]
]
Q.46145 The value of
 
 
a
log
a
sec
x
log
x
cot
x
x
1
a
a
1
x
Limit





(a > 1) is equal to
(A*) 1 (B) 0 (C) /2 (D) does not exist
[Hint:


















x
log
a
sec
x
x
log
cot
Limit
a
x
1
a
a
1
x ; 0
x
x
log
as a
a
x









and 









x
log
a
a
x
(using L’opital rule)
 l =
2
/
2
/


= 1 ]
Q.47148 Let f : (1, 2 )  R satisfies the inequality
2
x
|
8
x
4
|
x
)
x
(
f
2
33
)
4
x
2
cos( 2






, )
2
,
1
(
x 
 . Then )
x
(
f
Lim
2
x 

is equal to
(A) 16
(B*) –16
(C)cannotbedeterminedfromthegiven information
(D) doesnot exists
[Hint: )
x
(
f
Lim
2
x 

16
2
33
)
4
x
2
cos(




; )
x
(
f
Lim
2
x 

16
)
4
(
4
2
x
|
2
x
|
4
.
x2






Bysandwich theorem )
x
(
f
Lim
2
x 

= –16 ]

Q.48163 Let a = min [x2 + 2x + 3, x  R] and b = x
x
0
x e
e
x
cos
x
sin
Lim 
 
. Then the value of 


n
0
r
r
n
r
b
a is
(A) n
1
n
2
·
3
1
2 

(B) n
1
n
2
·
3
1
2 

(C) n
n
2
·
3
1
2 
(D*) n
1
n
2
·
3
1
4 

[Sol. a = (x + 1)2 + 2  a = 2
b =
x
2
x
2
).
1
e
(
2
x
2
sin
Lim x
2
0
x 

=
2
1
now 


n
0
r
r
n
r
b
a = 







r
n
r
2
1
2 = 

n
0
4
r
2
n
2
2
1
= n
2
1
[1 + 22 + 24 + ... + 22n ]
= n
2
1









1
2
1
)
2
(
2
1
n
2
= n
2
1





 

3
1
4 1
n
= n
1
n
2
.
3
1
4 

]
Q.49165 Period of f(x) = nx + n  [nx + n], (n  N where [ ] denotes the greatest integer function is :
(A) 1 (B*) 1/n (C) n (D) none of these
Q.50171 Let f beareal valued functiondefined by f(x)=sin1
1
3







x
+cos1 x 






3
5
.Then domain of f(x)
isgivenby:
(A*) [ 4, 4] (B) [0, 4] (C) [ 3, 3] (D) [ 5, 5]
Q.51172 For the functionf (x) =
)
x
(
sin
n
1
1
Lim 2
n 



, which of thefollowing holds?
(A) The range of f is a singleton set (B)f iscontinuous on R
(C*) f is discontinuous for all x I (D) f is discontinuous for some x R
[Hint: f (x) = 


I
x
if
0
I
x
if
1


 fis discontinuous for all x I ]
Q.52174 Domain of the function f(x) =
x
cot
n
1
1

l
is
(A) (cot1 ,  ) (B) R – {cot1} (C) (– ,0) (0,cot1) (D*) (– , cot1)
[Sol. ln (cot–1x) > 0
cot–1x > 1
x < cot 1
 domain (– , cot1) Ans. ]

Q.53175 The function 








Q
x
,
5
x
2
x
Q
x
,
1
x
2
)
x
(
f
2
is
(A)continuousnowhere
(B)differentiablenowhere
(C)continuousbut not differentiableexactlyatonepoint
(D*)differentiableandcontinuous onlyatonepointand discontinuouselsewhere
[Hint: Diff. & cont. onlyat x = 2 which is the point of intersection of y= 2x + 1 and y= x2–2x +5 ]
[Sol. y = (x – 1)2 + 4
= x2 – 2x + 5 = 2x + 1
= x2 – 4x + 4 = 0
= (x – 2)2 = 0  x = 2
 line is tangent at x = 2 ]
Q.54180 For the function f (x) =
)
2
x
(
1
2
x
1


, x  2 which of the following holds?
(A) f (2) = 1/2 and f is continuous at x =2 (B) f (2)  0, 1/2 and f is continuous at x = 2
(C*) f can not be continuous at x = 2 (D) f (2) = 0 and f is continuous at x = 2.
Q.55184 )
x
tan(sin
1
)
x
cos(sin
x
Lim 1
1
2
1
x 

 

is
(A)
2
1
(B*) –
2
1
(C) 2 (D) – 2
[Hint: put sin–1x = 







 tan
1
cos
sin
Lim
4
= 




cos
Lim
4 = –
2
1
]
Q.56186 Which oneof the followingis not bounded on theintervals as indicated
(A) f(x) = 1
x
1
2  on (0, 1) (B*) g(x) = x cos
1
x
on (–)
(C) h(x) = xe–x on (0, ) (D) l (x) = arc tan2x on (–, )
[Sol. (A) 
0
x
Limit f (x) =
2
1
2
Limit 1
h
1
0
h



; 
1
x
Limit f (x) = 0
2
Limit h
1
0
h



 






2
1
,
0
)
x
(
f  bounded
(C) 0
h
Limit

x e–x = 0
h
Limit

h e –h = 0 ; 

x
Limit x e–x = 

x
Limit 0
e
x
x

 Also x
e
x
y  y = x
2
x
x
e
xe
e 
e x (1 – x)  






e
1
,
0
)
x
(
h ]

Q.57190 The domainof thefunction f(x)=
 
arc x
x x
cot
2 2

, where [x] denotesthe greatest integernotgreater than
x,is:
(A) R (B) R  {0}
(C*) R   
  

n n I
: { }
0 (D) R  {n : n  I}
Hint: x2 – [x2] = {x2} > 0; but 0  {y} < 1  {x2}  0  x  ± n ]
Q.58192 If f(x) = cos x, x =n  , n = 0, 1, 2, 3, .....
= 3, otherwise and
(x) =
x when x x
when x
when x
2
1 3 0
3 0
5 3
  







,
then Limit
x  0 f((x)) =
(A) 1 (B*) 3 (C) 5 (D) none
Q.59195 Let
0
x
Lim

sec–1
x
x
sin





 = l and
0
x
Lim

sec–1
x
x
tan





 = m, then
(A*) l exists but m does not (B) m exists but l does not
(C) l and m both exist (D) neitherl norm exists
Q.60200 Range of the function f (x) = 2
2
x
1
1
)
e
x
(
n
1









l
is , where [*] denotes the greatest integer
function and e =




 /
1
0
)
1
(
Limit
(A) 




 
e
1
e
,
0 {2} (B) (0, 1) (C) (0, 1]  {2} (D*) (0, 1)  {2}
[Sol. 





 )
e
x
(
n
1
2
l
– 




0
x
1
0
x
0








0
x
x
1
1
0
x
2
)
x
(
f
2
Hence range of f (x) is (0, 1)  {2}]
Q.61201 ]
x
[tan
sin
Lim 1
0
x

 
= l then { l } is equal to
(A) 0 (B)
2
1

 (C) 1
2


(D*)
2
2


where [ ] and { } denotes greatest integer andfractional part function.
[Hint:
2
]
x
[tan
sin
Lim 1
0
x




 
; hence





 

2
=








 2
2
2 = 2 –
2

Ans.]

Q.62206 Number of points where the function f (x) = (x2 – 1) | x2 – x – 2 | + sin( | x | ) is not differentiable, is
(A) 0 (B) 1 (C*) 2 (D) 3
[Hint: not derivable at x = 0 and 2 ]
Q.63219
 























 x
1
1
x
1
x
1
x
2
sec
x
1
x
cot
Limit is equal to
(A*) 1 (B) 0 (C) /2 (D)nonexistent
[Hint: x
1
x
Limit
x




= 0  cot–1(0) = /2












x
x 1
x
1
x
2
Limit  sec–1 () = /2  l = 1 Ans ]
Q.64223 If f (x) = 





0
0
2
x
x
if
b
ax
x
x
if
x
derivable R
x
 then the values of a and b are respectively
(A*) 2x0 , – 2
0
x (B) – x0 , 2 2
0
x (C) – 2x0 , – 2
0
x (D) 2 2
0
x , – x0
Q.65231 Let f(x) =
1 2
1
1
2
1
2
2 1
4 2 1 2
1
2





  









cos
sin
,
,
,


x
x
x
p x
x
x
x
. If f(x) is discontinuous at x =
1
2
, then
(A*) p  R  {4} (B) p  R 
1
4






(C) p  R0 (D) p  R
[Hint: f
1
2








 = 4 = f
1
2








 ]
Q.66235 Let f(x) beadifferentiable functionwhichsatisfies theequation
f(xy) = f(x) + f(y) for all x > 0, y> 0 then
f (x) is equal to
(A*)
f
x
'( )
1
(B)
1
x
(C) f (1) (D)f (1).(lnx)
[Sol. f (x) = Limit
h
f x h f x
h

 
0
( ) ( )
= Limit
h
f x
h
x
f x
h













 
0
1
. ( )
=
 
Limit
h
f x f
h
x
f x
h

 





 
0
1 ( )
= Limit
h
f
h
x
x
h
x








0
1
=
 
1
0
1 1
x
Limit
t
f t f
t

  ( )
(note that f(1) = 0) =
f
x
'( )
1
Ans. ]

Q.67240 Given f(x) = b ([x]2 + [x]) + 1 for x  1
= Sin ((x+a) ) for x < 1
where [x] denotes the integral part of x, then for what values of a, b the function is continuous at
x = 1?
(A*) a = 2n + (3/2) ; b  R ; n  I (B) a = 4n + 2 ; b  R ; n  I
(C) a = 4n + (3/2) ; b  R+ ; n  I (D) a = 4n + 1 ; b  R+ ; n  I
[Hint: f (–1) = b (1 – 1) + 1 = 1
)
h
1
(
Lim
0
h



= 1
 
a
)
h
1
(
Lim
)
h
1
(
Lim
0
h
0
h










forcontinuous
sin a = – 1 = sin 




 


2
3
n
2  a =
2
3
n
2


  a =
2
3
n
2 
hence a =
2
3
n
2  , n  I and b  R]
Q.68247 Let f(x) =
)
e
x
(
n
)
e
x
(
n
x
2
4
x
2


l
l
. If Limit
x  
f(x) = l and Limit
x   
f(x) = m then :
(A*) l = m (B) l = 2m (C) 2 l = m (D) l + m = 0
[Sol. 

x
Lim  
 


n x e
n x e
x
x
2
4 2


=


x
Lim


















x
2
4
x
2
x
2
x
e
x
1
e
n
e
x
1
e
n
l
l
=


x
Lim




















x
2
4
x
2
e
x
1
n
x
2
e
x
1
n
x
l
l
= x
/
1
x
2
4
x
/
1
x
2
e
x
1
n
2
e
x
1
n
1




















l
l
note that 

x
as 0
e
x
x
2
 and 

x
as 0
e
x
x
2
2

=


x
Lim






















1
e
x
1
x
1
2
1
e
x
1
x
1
1
x
2
4
x
2
=


x
Lim
x
2
3
x
e
x
2
e
x
1


|||ly
2
1
Limit
x




]
Q.69248 

n
Lim cos 




 
 n
n2
when n is an integer :
(A) is equal to 1 (B) is equal to  1 (C*) is equal to zero (D) does not exist
[Hint:  n 1
1
1 2







n
/
= n  1
1
2
1
2
1
2
1
1
2
1
2
  





 






n n
!
......
=  n
n
  





 






1
2
1
2
1
2
1
1
2
1
!
......
as n  ;

2
. 2 1
1
2
1
1
2
1
n
n
  





 






!
....... = (2n + 1)

2

Alternatively (1): Take A = (2n +1)
2

and B =
2














 ....
n
1
!
2
1
1
2
1
now cos (A + B) = cosA cosB – sinA sinB
= 0 – (1) 

n
Limit sin 
















...
n
1
!
2
1
1
2
1
2
= 0 ]
Alternatively (2) Best l = 




 






n
n
n
cos
Limit 2
n
 










 





n
n
n
cos
Limit 2
n
 













 n
n
n
)
n
(
cos
Limit
2
n
=

















n
1
1
n
n
n
cos
Limit
n
=

















n
1
1
1
cos
Limit
n
= cos
2

 0 ]
Q.70250 0
x
Limit
 x
sin
3
)
x
(sin
)
x
.(tan
7
x
)
x
2
cos
1
(
)
x
tan
x
(sin
5
6
1
7
1
5
4
2







 is equal to
(A) 0 (B)
7
1
(C*)
3
1
(D) 1
[Hint: DivideNr. andDr. byx5 andevaluate limit]
Q.71264 Range of the function , f (x) = cot1
 
log / ( )
4 5
2
5 8 4
x x
  is :
(A) (0 , ) (B*)


4
,





 (C) 0
4
,






 (D) 0
2
,







Q.72266 Let Limit
x  0
[ ]
x
x
2
2 = l & Limit
x  0
[ ]
x
x
2
2 = m , where [ ] denotes greatest integer , then:
(A) l exists but m does not (B*) m exists but l does not
(C) l & m both exist (D) neither l nor m exists .
[Hint:
[ ]
x
x
2
2 =
0 0 1
1 1 0
2
if x
x
if x
 
  




 l does not exist .
[ ]
x
x
2
2
=
0 0 1
0 1 0
if x
if x
 
  


  m exists and is equal to zero. ]
Q.73268 The value of Limit
x  0
 
 
 
tan { } sin { }
{ } { }
x x
x x


1
1
where {x} denotes the fractional part function:
(A) is 1 (B) is tan 1 (C) is sin 1 (D*) is non existent
[Sol. Limit
x  0
 
 
 
tan { } sin { }
{ } { }
x x
x x


1
1
= 
0
x
Limit
f (x) = )
1
h
(
h
sinh
.
)
1
h
tan(
Limit
0
h




= 1
tan
1
)
1
tan(



1
1
sin
)
1
h
1
(
)
h
1
(
)
h
1
sin(
)
1
)
h
1
tan((
Limit
0
x









= sin 1
Hence limit x  0 f (x) does not exist ]

Q.74270 If f(x) =
n
x
e x
x2
2



 


tan
is continuous at x = 0 , then f (0) must be equal to :
(A) 0 (B) 1 (C) e2 (D*) 2
[Hint: f (0) = Limit
x  0
n
x
e x
x2
2



 


= Limit
x  0 l n e x
x x
2
1
2



 


= Limit
x  0
1
x
e x
x2
2 1
 


 

 = Limit
x  0
ex
x
2
1
2










= 2 ]
Q.75272 

x
Lim
x
sin
e
)
x
2
sin
x
2
(
x
2
sin
x
2
2



is :
(A) equal to zero (B) equal to 1 (C) equal to 1 (D*)non existent
[Sol. Limit
x   x
sin
e
x
x
2
sin
2
x
x
2
sin
2
x
2









as x 
l = Limit
x   x
sin
e
.
2
2
= oscillatory between
e
1
to 1
e
1
  non existent ]
Q.76275 The value of  
x
ec bx
ax
0
2
lim cos
cos is
(A) e
b
a









8 2
2
(B) e
a
b









8 2
2
(C*) e
a
b









2
2
2
(D) e
b
a









2
2
2
[Sol. l =
)
1
ax
(cos
bx
2
ec
cos
0
x
Limit
e


now –
bx
sin
ax
cos
1
Limit 2
0
x


= –
ax
cos
1
1
.
bx
sin
ax
sin
Limit 2
2
0
x 

= – 2
2
b
2
a
l = 2
b
2
2
a
e
 ]
Select the correct alternative : (More than one are correct)
Q.77503 c
x
Lim

f(x) does not exist when :
(A) f(x) = [[x]]  [2x 1], c = 3 (B*) f(x) = [x]  x, c = 1
(C*) f(x) = {x}2  {x}2, c = 0 (D) f(x) =
tan (sgn )
sgn
x
x
, c = 0 .
where[x] denotes step up function& {x} fractional part function.

Q.78505 Let f(x) =
tan { }
[ ]
{ } cot { }
2
2 2
1
0
0
0
x
x x
x x
for x
for x
for x











where [x] is the step up function and {x} is the fractional
part function of x , then :
(A*) Limit
x  
0
f(x) = 1 (B) Limit
x  
0
f (x) = 1
(C*) cot -1 Limit f x
x  






0
2
( ) = 1 (D) f is continuous at x = 1 .
Q.79506 If f(x) =  
x n x
n x
x
x
. (cos )

 1 2
0
0 0








then :
(A*) f is continuous at x = 0 (B)fiscontinuous atx=0 but notdifferentiableatx=0
(C*) f is differentiable at x = 0 (D) f is not continuous at x = 0.
Q.80508 Whichofthefollowingfunction(s)is/areTranscidental?
(A*) f (x) = 5 sin x (B*) f (x) =
2 3
2 1
2
sin x
x x
 
(C) f (x) = x x
2
2 1
  (D*) f (x) = (x2 + 3).2x
[Hint: functions whicharenotalgebraicareknownastranscidental function]
Q.81513 Which ofthefollowingfunction(s) is/areperiodic?
(A*) f(x) = x  [x] (B) g(x) = sin (1/x) , x  0 & g(0) = 0
(C) h(x) = x cosx (D*) w(x) = sin1 (sinx)
Q.82515 Which offollowingpairs offunctionsareidentical :
(A) f(x) = e n x
 sec1
& g(x) = sec1 x
(B*) f(x) = tan(tan1 x) & g(x) = cot(cot1 x)
(C*) f(x) = sgn(x)& g(x) = sgn(sgn(x))
(D*) f(x) = cot2 x.cos2 x & g(x)= cot2 x  cos2 x
Q.83516 Whichofthefollowingfunctionsarehomogeneous ?
(A) x siny+ ysinx (B*) x ey/x + y ex/y (C*) x2  xy (D) arc sin xy
Q.84521 If  is small & positivenumber then which of the following is/are correct ?
(A)
sin

= 1 (B)  < sin < tan (C*) sin <  < tan (D*)
tan

>
sin

Q.85522 Let f(x) =
x x
x
x
.
cos
2
1


& g(x) = 2x sin
n
x
2
2





 then :
(A) Limit
x  0 f(x) = ln 2 (B) Limit
x   g(x) = ln 4
(C*) Limit
x  0 f(x) = ln 4 (D*) Limit
x   g(x) = ln 2

Q.86524 Let f(x) =
x
x x

 
1
2 7 5
2
. Then :
(A*) Limit
x  1
f(x) = 
1
3
(B*) Limit
x  0
f(x) = 
1
5
(C*) Limit
x   f(x) = 0 (D*) Limit
x  5 2
/
does not exist
Q.87527 Whichofthefollowinglimitsvanish?
(A*) Limit
x   x
1
4 sin
1
x
(B*) Limit
x  /2 (1  sinx) . tanx
(C) Limit
x  
2 3
5
2
2
x
x x

 
. sgn(x) (D*) Limit
x  
3
[ ]
x
x
2
2
9
9


where [ ] denotes greatest integer function
Q.88528 If x is a real number in [0, 1] then the value of Limit
m  
Limit
n   [1 + cos2m (n!  x)] is given by
(A) 1 or 2 according as x is rational or irrational
(B*) 2 or 1 according as x is rational or irrational
(C) 1 for all x
(D*) 2 for all x .
[Sol. 


 n
m
Limit
Limit 1 + cos2m(n ! n)
Let ]
1
,
0
[
x is a rational
then n!. x if n  becomes integral
 cos2 (n ! x)  = 1
 

m
Limit ( cos2 n ! x  )m = 1
 l = 1 + 1 = 2 ifxisrational
if x is irrational , then n ! x is not an integer
 cos2( n! x)  is less than 1
cos2 ( ( n! x)  )m  0
 l = 1 + 0 = 1  the result ]
Q.89529 If f(x) is a polynomialfunction satisfyingthe condition f(x) . f(1/x) = f(x) + f(1/x)and f(2) = 9 then :
(A) 2 f(4) = 3 f(6) (B*) 14 f(1) = f(3) (C*) 9 f(3) = 2 f(5) (D) f(10) = f(11)
[Hint: f (x) = 1 + xn or 1  xn  n = 3  f(x) = x3 + 1 ]
Q.90531 Which ofthe followingfunction(s) not defined at x = 0 has/have removablediscontinuityat x =0 ?
(A) f(x) =
1
1 2
 cotx
(B*) f(x) = cos 





x
|
x
sin
| (C*) f(x) = x sin

x
(D*) f(x) =
1
n x
Q.91535 The function f(x) =
     
x x
x
x x
 
  




3 1
1
2
4
3
2
13
4
,
,
is :
(A*) continuous at x = 1 (B*) diff. at x = 1
(C*) continuous at x = 3 (D) differentiableat x = 3

[Sol. f (x) =















1
x
if
4
13
2
x
3
4
x
3
x
1
if
x
3
3
x
if
3
x
2
f (1+) =
h
)
1
(
f
)
h
1
(
f
Limit
0
h



= 1
h
2
)
h
1
(
3
Limit
0
h






f (1–) =
h
2
4
13
)
h
1
(
2
3
4
)
h
1
(
Limit
2
0
h 






=
h
4
5
)
h
1
(
6
)
h
1
(
Limit
2
0
h 





=
h
4
h
6
h
2
h
Limit
2
0
h 



= – 1
 f is continuous at x =1 ]
Q.92539 If f(x) = cos

x





 cos  

2
1
x 





 ; where [x] is the greatest integerr function of x, then f(x) is
continuous at :
(A) x = 0 (B*) x = 1 (C*) x = 2 (D) none of these
[Hint: Not defined at x = 0 ; f(x) = cos 3 ; f(2) = 0 and both the limits exist
 B and C ]
Q.93540 Identifythepair(s) of functions which are identical .
(A*) y = tan (cos 1 x) ; y =
1 2
 x
x
(B*) y = tan (cot 1 x) ; y =
1
x
(C*) y = sin (arc tan x) ; y =
x
x
1 2

(D*) y = cos (arc tan x) ; y = sin (arc cot x)
[ Ans. : (A) T domain [ 1, 0)  (0, 1] (B) T (C) T (D) T ,
1
1 2
 x
]
(A) y= tan(cos–1x)
note that 0 < sin–1x < p
but tan(/2) is not defined
hence cos–1x 
2

 x  0
hence domain is [–1,1] – {0}
and range is R.
The graphis as shown.

(B) y= tan(cot–1x)
note that 0 < cot–1x < 
 tan(/2) is not defined
hence cos–1x 
2

 x  0
hence domain = R – {0}
since cot–1x  0 ,   y  0
hence range is R – {0}
The graphis as shown.
(C) y=sin(tan–1x)
–
2

< tan–1x <
2

 y  (–1, 1)
since tan–1x is defined for all x  R
and sin is also defined for all x  R
 domain is R
y = 2
x
1
x

The graphis as shown.
(D) y= cos(tan–1x)
tan–1  




 


2
,
2
 y  (0, 1]
DomainisR
y = 2
x
1
1

Note that sin–1(cot–1x) also has the same graph
The graphis as shown.]
Q.94544 The function, f(x) = [x][x] where [x] denotes greatest integer function
(A*)is continuousforallpositiveintegers
(B*)is discontinuous forallnonpositiveintegers
(C*)hasfinitenumberofelementsinits range
(D*) is such that its graph does not lie above the xaxis .
[Sol. [ | x | ] – | [x] | =














2
x
1
0
1
x
0
0
0
x
1
1
1
x
0
 range is {0, –1}
Thegraphis ]
Q.95545 Let f (x + y) = f(x) + f(y) for all x , y  R. Then :
(A) f(x) must be continuous  x  R (B*) f(x) may be continuous  x  R
(C) f(x) must be discontinuous  x  R (D*) f(x) maybe discontinuous  x  R

[Hint: Limit
h0
f (x + h) = Limit
h0
f(x) + f(h) = f (x) + Limit
h0
f (h)
Hence if h  0 f (h) = 0  'f' is continuous otherwise discontinuous ]
Q.96548 The function f(x) = 1 1 2
  x
(A*) has its domain –1 < x < 1.
(B*) has finite one sided derivates at the point x = 0.
(C)iscontinuous and differentiable atx = 0.
(D*)is continuous but not differentiable atx = 0.
[Hint: f (0+) =
1
2
; f (0–) = –
1
2
]
Q.97553 Let f(x) be defined in [–2, 2] by
f(x) = max (4 – x2, 1 + x2), –2 < x < 0
= min (4 – x2, 1 + x2), 0 < x < 2
Thef(x)
(A)iscontinuousatallpoints
(B*)has apoint ofdiscontinuity
(C) isnot differentiable onlyat onepoint.
(D*) is not differentiableat morethan onepoint
[Hint: f(x) max. (4–x2, 1+x2), –2 < x < 0
min. (4 - x2, 1+x2), 0 < x < 2]
Q.98555 Thefunctionf(x)=sgnx.sinx is
(A*)discontinuousno where. (B*) an evenfunction
(C*) aperiodic (D)differentiableforallx
[Hint: f (x) =



sin
sin
x x
x x

 
0
0
]
Q.99556 The function f(x) = x
n
1
xl
(A*)is aconstant function (B) has a domain (0, 1) U (e, )
(C*) is such that limit
x1
f(x) exist (D)is aperiodic
[Hint: y = x
logx e
= e (constant)  Aand C
as limit
x1
f(x) = e]
Q.100557 Which pair(s)of function(s) is/are equal?
(A*) f(x) = cos(2tan–1x) ; g(x) =
1
1
2
2


x
x
(B*) f(x) =
2
1 2
x
x

; g(x) = sin(2cot–1x)
(C*) f(x) = e n x
 (sgn cot )
1
; g(x) =
 
 
e
n x
 1
(D) f(x) = a
X , a > 0; g(x) = a x
1
, a > 0
where {x}and [x] denotes thefractional part &integral part functions.

Fill in the blanks:
Q.101801 Afunction f is defined as follows, f(x) =
sinx if x c
ax b if x c

 



where c is a known quantity. If f is
derivable at x = c, then the values of 'a' & 'b' are _____ &______ respectively .
[ Ans. : cos c & sin c  c cos c ]
Q.102802Aweight hangs byaspring& is caused to vibrate by a sinusoidal force. Its displacement s(t)at time
t is given by an equation of the form, s(t) =
A
c k
2 2

(sin kt  sin ct) where A, c & k are positive
constants with c  k, then the limiting value of the displacement as c  k is ______.
[Ans. :
At kt
k
cos
2
]
Q.103805
Limit
x  4
(cos ) (sin ) cos
  
x x
x
 

2
4
where 0 <  <

2
is ______ .
[Ans. : cos4  ln cos  sin4  ln sin ]
[Hint: put cos 2  = cos4   sin4  ]
Q.104809
Limit
x  0  
cos
/
2
3 2
x
x
has the value equal to ______ . [ Ans. : e6 ]
Q.105812 If f(x) = sin x, x  n, n = 0, ±1, ±2, ±3,....
= 2, otherwise
and g(x) = x² + 1, x  0, 2
= 4, x = 0
= 5, x = 2
then Limit
x0
g [f(x)] is ______ [Ans. : 1 ] [ IIT’86,2 ]