The document discusses the concepts of continuity and differentiability of functions, highlighting how a function is continuous if it has no breaks or gaps and providing criteria to determine the continuity at a point. It also details various examples of continuous and discontinuous functions, as well as the properties of continuous functions, emphasizing that while differentiability implies continuity, continuity does not necessarily imply differentiability. The final sections address differentiability at points and intervals, listing examples of differentiable and non-differentiable functions.

1. CONTINUITYANDDIFFERENTIABILITY
In Chapter Examples 14
Solved Examples 17
Total No. of questions................................................ 31
Total No. of questions in Continuity and Differentiability are-

2. 1. INTRODUCTION
The word 'Continuous' means without any break
or gap. If the graph of a function has no break
or gap or jump, then it is said to be continuous.
A function which is not continuous is called a
discontinuous function.
In other words,
If there is slight (finite) change in the value of a
function by slightly changing the value of x then
function is continuous, otherwise discontinuous,
while studying graphs of functions, we see that
graphs of functions sin x, x, cos x, ex etc. are
continuous but greatest integer function [x] has
break at every integral point, so it is not continu-
ous. Similarly tan x, cot x, secx, 1/x etc. are
also discontinuous function.
Ex.
(Continuous function)
Y
X
O
f(x)= 1/x
(Discontinuous at x = 0)
(Discontinuous)
For examining continuity of a function at a point,
we find its limit and value at that point, If these
two exist and are equal, then function is continu-
ous at that point.
2. CONTINUITY OF A FUNCTION AT A POINT
A function f(x) is said to be continuous at a point
x = a if
(i) f (a) exists
(ii) x a
Lim
 f(x) exists and finite
so x a
Lim
  f(x) = x a
Lim
  f(x)
(iii) x a
Lim
 f(x) = f(a) .
or function f(x) is continuous at x = a.
If x a
Lim
  f(x) = x a
Lim
  f(x) = f(a).
i.e. If right hand limit at 'a' = left hand limit
at 'a'= value of the function at 'a'.
If x a
Lim
 f(x) does not exist or x a
Lim
 f(x)  f(a),
then f(x) is said to be discontinuous at x= a.
Continuity of a function at a point
Ex.1 Examine the continuity of the function
f (x) =
x
x
when x
when x
2
9
3
3
6 3




R
S
|
T
|
,
,
at x = 3.
Sol. f (3) = 6 ( given)
x
Lim
 3 f(x) = x
Lim
 3
x x
x
 

3 3
3
b g
b g
b g = 6
 x
Lim
 3 f(x) = f(3)
 f (x) is continuous at x = 3.

3. Ex.2 If f(x) =
log ( ) log
,
1 2 1
0
0
  


R
S
|
T
|
ax bx
x
x
k x
b g
If function is continuous at x = 0 then the
value of k is –
(A) a + b (B) 2a +b
(C) a – b (D) 0
Sol. x
Lim
 0
log
1 2
1


F
H
G I
K
J
ax
bx
x
= x
Lim
 0
1
1 2


F
H
G I
K
J
bx
ax
.
1 2 1 2
1
2
   

bx a ax b
bx
b g
b gb g
b g
b g
 x
Lim
 0
2
1 1 2
a b
bx ax

 
b g
b g
=
2
1 1
a b

b g
b
g
bg= 2a + b
Ans.[B]
Ex.3 If f(x) =
1 4
0
0
2



R
S
|
T
|
cos
,
,
x
x
x
a x
is continuous then
the value of a is equal to –
(A) 0 (B) 1
(C) 4 (D) 8
Sol. Since the given function is continuous at x= 0
x
Lim
 0
1 4
2
 cos x
x
= a
x
Lim
 0
2 2
2
2
sin x
x
x
4
4
= a
x
Lim
 0 2
sin 2
2
2
x
x
F
H
G I
K
Jx 4 = a
 2 x 1 x 4 = a
 8 = a Ans.[4]
3. CONTINUITY FROM LEFT AND RIGHT
Function f(x) is said to be
(i) Left continuous at x= a if
x a
Lim
  0 f(x) = f(a)
(ii) right continuous at x = a if
x a
Lim
  0 f(x) = f(a)
Thus a function f(x) is continuous at a point
x = a if it is left continuous as well as right
continuous at x = a.
Continuity From Left and Right
Ex.4 Examine the continuity of the function
f(x) =
x when x
x when x
2
1 2
2 2
 

R
S
|
T
|
,
,
at the point x = 2.
Sol. f(2) = 22 + 1 = 5
f(2– 0) = h
Lim
 0
2 1
2
 
h
b g = 5
 f (2+ 0) = h
Lim
 0 2( 2+ h) = 4
f(2– 0)  f(2+ 0)  f(2)
 f(x) is not continuous at x = 2.
Ex.5 Check the continuity of the function
f(x) =
x x
x
x x
 

 
R
S
|
T
|
2 3
5 3
8 3
at x = 3.
Sol. f (3) = 5
Left hand limit x
Lim
 
3 (3+h) + 2
h
Lim
 0 5+ h = 5
Right hand limit x
Lim
 
3 8–(3–h)
h
Lim
 0 5+ h = 5 = LHL
 f(3) = RHL = LHL
 function is continuous.
Ex.6 If f(x) =
| |
| |
x
x
a x
a b x
x
x
b x


 
 


 
R
S
|
|
T
|
|
1
1
1
1
1
1
1
is continuous at
x = 1 then the value of a & b are respectively-
(A) 1,1 (B) 1,–1
(C) 2,3 (D) None of these
Sol. f(1) = a+ b
f(1+h) =
| |
1 1
1 1
1
 
 
   
h
h
a a
b g

4.  given function is continuous
 f(1) = f (1+h)
= a+b = – 1+a  b= –1
Now f(1-h) =
| |
1 1
1 1
 
 
h
h
b g
+ b =
h
h
b
 = 1+ b
 a+ b = 1+ b  a= 1 Ans.[B]
Ex.7 Function f(x) = [x] is a greatest integer func-
tion which is right continuous at x = 1 but
not left continuous.
Sol.  f (1) = [1] = 1
[1+0] = 1 and [1-0] =0
 x
Lim
 
1 f(x) = f(1) = 1,
and x
Lim
 
1 f(x) = 0  f(1)
so function f(x) = [x] is right continuous but
not left continuous.
4. CONTINUITY OF A FUNCTION IN AN INTERVAL
(a) A function f(x) is said to be continuous in an
open interval (a,b) if it is continuous at every
point in (a, b).
For example function y = sin x, y = cos x ,
y = ex are continuous in (– , ).
(b) A function f(x) is said to be continuous in the
closed interval [a, b] if it is-
(i) Continuous at every point of the open
interval (a, b).
(ii) Right continuous at x = a.
(iii) Left continuous at x = b.
continuity of a function in
an interval
Question
based on
Ex.8 Check the continuity of the function
f(x) =
5 4 0 1
4 3 1 2
2
x x
x x x
  
  
R
S
T
,
,
in an interval [0,2]
Sol. The given function is continuous in the interval
[0, 2] because it is right continuous at x = 0
and left continuous at x = 2 and is continuous
at every point of the interval (0, 2).
Ex.9 For what value of a and b the function
f(x) =
x a x x
x x b x
a x b x x
  
  
  
R
S
|
T
|
2 0 4
2 4 2
2 2
sin , /
cot , / /
cos sin , /

 
 
is
continuous in an interval [0,  ].
Sol.  f(x) is continuous in an interval [0,  ]
So it is also continuous at x =  / 4 ,x=  / 2 .

x
Lim
 
 / 4
f(x) =
x
Lim
 
 / 4
f(x)
  / 4 + a =  / 2 + b ...(1)
and
x
Lim
 
 / 2
f(x) =
x
Lim
 
 / 2
. f(x)
 0+b = –a– b ...(2)
Solving (1) and (2)  a =  / 6 , b = –  / 12 .
5. CONTINUOUS FUNCTIONS
A function is said to be continuous function if it
is continuous at every point in its domain. Fol-
lowing are examples of some continuous func-
tion.
(i) f (x) = x (Identity function)
(ii) f(x) = C (Constant function)
(iii) f(x) = x2
(iv) f(x) = a 0xn + a1xn-1+ ....+ an
(Polynomial).
(v) f(x) = |x|, x+ |x|, x-|x|, x|x|
(vi) f(x) = sin x, f(x) = cos x
(vii) f(x) = ex, f(x) = ax, a> 0
(viii) f(x) = log x, f(x) = logax a> 0
(ix) f(x) = sinh x, cosh x, tanh x
(x) f(x) = xm sin (1/x), m> 0
f(x) = xm cos (1/x), m> 0
6. DISCONTINUOUS FUNCTIONS
A function is said to be a discontinuous function
if it is discontinuous at at least one point in its
domain. Following are examples of some discon-
tinuous function-
(i) f(x) = 1/x at x = 0
(ii) f(x) = e1/x at x = 0
(iii) f(x) = sin 1/x, f(x) = cos 1/x at x = 0
(iv) f(x) = [x] at every integer
(v) f(x) = x– [x] at every integer
(vi) f(x) = tan x, f(x) = sec x
when x =(2n+1)  / 2 , nZ.
(vii) f(x) = cot x, f(x) = cosec x when x = n  ,
nZ.
(viii) f(x) = coth x, f(x) = cosech x at x = 0.

5. 7. PROPERTIES OF CONTINUOUS FUNCTION
The sum, difference, product, quotient (If Dr  0)
and composite of two continuous functions are
always continuous functions. Thus if f(x) and g(x)
are continuous functions then following are also
continuous functions:
(a) f(x) + g(x)
(b) f(x) – g(x)
(c) f(x) . g(x)
(d)  f(x) , where  is a constant
(e) f(x) /g(x), if g(x)  0
(f) f [g(x)]
For example -
(i) e2x + sin x is a continuous function be-
cause it is the sum of two continuous func-
tion e2x and sin x.
(ii) sin (x2 +2) is a continuous function because
it is the composite of two continuous func-
tions sin x and x2+2.
Note :
The product of one continuous and one discon-
tinuous function may or may not be continuous.
For example-
(i) f(x) = x is continuous and g(x) = cos 1/x is
discontinuous whereas their product
x cos 1/x is continuous.
(ii) f(x) = C is continuous and g(x) = sin 1/x is
discontinuous whereas their product
C sin 1/x is discontinuous.
DIFFERENTIABILITY
8. DIFFERENTIABILITY OF A FUNCTION
A function f(x) is said to be differentiable at a
point of its domain if it has a finite derivative at
that point. Thus f(x) is differentiable at x = a
 lim
x a

f x f a
x a
( ) ( )


exists finitely
 lim
h0
f a h f a
h
( ) ( )
 

= lim
h0
f a h f a
h
( ) ( )
 
 f' (a – 0) = f'(a+ 0)
 left- hand derivative = Right-hand derivative.
Generally derivative of f(x) at x = a is denoted by
f'(a) . So f' (a) = lim
x a

f x f a
x a
( ) ( )


Note : (i) Every differentiable function is necessarily
continuous but every continuous function
is not necessarily differentiable i.e.
Differentiability  continuity
but continuity 
 differentiability
8.1 Differentiability in an interval
(a) A function f(x) is said to be differentiable in
an open interval (a,b), if it is differentiable at
every point of the interval.
(b) A function f(x) is differentiable in a closed
interval [a,b] if it is –
(i) Differentiable at every point of interval (a,b)
(ii) Right derivative exists at x = a
(iii) Left derivative exists at x = b.
8.2 Differentiable function & their properties
A function is said to be a differentiable function
if it is differentiable at every point of its domain.
(a) Example of some differentiable functions:–
(i) Every polynomial function
(ii) Exponential function : ax, ex, e–x......
(iii) logarithmic functions : log ax, logex ,......
(iv) Trigonometrical functions : sin x, cos x,
(v) Hyperbolic functions : sinhx, coshx,......
(b) Examples of some non– differentiable
functions:
(i) |x | at x = 0
(ii) x  |x| at x = 0
(iii) [x], x  [x] at every n Z
(iv) x sin
1
x
F
H
GI
K
J, at x = 0
(v) cos
1
x
F
H
GI
K
J, at x = 0
(c) The sum, difference, product, quoteint
(Dr  0) and composite of two differentiable
functions is always a differentiable function.
Differentiability of function
Ex.10 The function f(x) = x2 – 2x is differentiable
at x = 2 because
Sol.  lim
x2
f x f
x
( ) ( )


2
2
= lim
x2
x x
x
2
2 0
2
 

 lim
x2
x = 2
Ex.11 Check the differentiability of the function
f(x) =
x x
x
x x
 

 
R
S
|
T
|
2 3
5 3
8 3
,
,
,
at x = 3.
Sol. For function to be differentiable
f'(3+h) = f' (3–h)
f' (3+h) = lim
h0
f h f
h
( ) ( )
3 3
 
 lim
h0
( )
3 2 5
  
h
h
= lim
h0
h
h
= 1

6. f' (3–h) = lim
h0
f h f
h
( ) ( )
3 3
 

= lim
h0
8 3 5
  

( )
h
h
=
h
h

= – 1
 f' (3+ h)  f' (3–h)
So function is not differentiable.
Ex.12 Check the differentiability of the function
f(x) =
x x x
x
sin ( / ),
,
1 0
0 0


R
S
T at x = 0
Sol. For function to be differentiable
f' (0+h) = f' (0–h)
f' (0+h) =
f h f
h
( ) ( )
0 0
 
 lim
h0
h
h
h
sin
1
0



lim
h 0
sin
1
h
F
H
GI
K
J
Which does not exist.
f' (0–h) = lim
h0
( )sin
 
F
H
G I
K
J

h
h
h
1
0
= lim
h0
sin 
F
H
G I
K
J
1
h
Which does not exist.
So function is not differentiable at x = 0
Here we can verify that
f (0+h) = f (0–h) = 0
So function is continuous at x = 0.
Ex.13 Check the differentiability of the function
f(x) =
1 0
1 0 2
2 2 2
2
,
sin , /
/ , /
x
x x
x x

  
   
R
S
|
T
|

  
b g
at
x =  / 2
Sol. f' (  / 2 + h) =
f h f
h
( / ) ( / )
 
2 2
 
= lim
h0
2 2 2 1 2
2
    
  
/ / sin /
h
h
b g b g
= lim
h0
2 1 1
2
  
h
h
= lim
h0
h= 0
= f'

2

F
H
G I
K
J
h =
f h f
h
 
2 2

F
H
G I
K
J
F
H
GI
K
J

= lim
h0
1
2
1
2
 
F
H
G I
K
J 
F
H
G I
K
J

sin sin
 
h
h
= lim
h0
1 2
 

cos h
h
=
lim
h0
cos h
h


1
= lim
h0
1 cosh
h
= 0
 function is differentiable at x =

2
Note : If a function f(x) is discontinuous at x = a
then it is not differentiable at that point.
Ex.14 Function f(x) =
sin ,
,
1
0
0 0
x
x
x
F
H
G
I
K
J 

R
S
|
T
|
is discontinuous at x = 0 , therefore it is not
differetiable at x = 0.

7. Ex.1 Function f(x) =
  
   

R
S
|
T
|
1 1
1 1
1 1
,
,
,
when x
x when x
when x
is con-
tinuous-
(A) Only at x = 1
(B) Only at x = – 1
(C) At both x = 1 and x = – 1
(D) Neither at x = 1 nor at x = – 1
Sol. f(–1–0) = –1, f(–1) = – (–1) = 1
 f( –1–0)  f(–1)
 f(x) is not continuous at x = –1
Further , f(1) = –1
f (1+0) = 1  f (1)  f(1+0)
 f(x) is not continuous at x = 1.
Ans.[D]
Ex.2 If f(x) =
x x x
x
k
cos( / ),
,
1 0
0 0


R
S
T
is continuous at x = 0, then-
(A) k < 0 (B) k > 0
(C) k = 0 (D) k 0
Sol. Since f(x) is continuous at x = 0
 x  0
lim
f(x) = f(0)
but f(0)= 0 ( given)
 x  0
lim
f(x) = x  0
lim
xk cos (1/x)
= 0, if k > 0. Ans.[B]
Ex.3 If f(x) =
1
2
0
1
2
0 0
1
2
1
2
3
2
1
2
1
1 1
  


  

R
S
|
|
|
|
T
|
|
|
|
x x
x
x
x x
x
,
,
,
,
,
then wrong statement is-
(A) f(x) is discontinuous at x = 0
(B) f(x) is continuous at x = 1/2
(C) f(x) is discontinuous at x= 1
(D) f(x) is continuous at x = 1/4
SOLVED EXAMPLES
Sol. Obviously function f(x) is discontinuous at x = 0
and x= 1 because the function is not defined,
when x< 0 and x> 1 , therefore f(0–0) and f(1+0)
do not exist. Again
f
1
2
0

F
H
G I
K
J=
Lim
x
1
2
3
2

F
H
G I
K
J
x = 1
f
1
2
0

F
H
G I
K
J=
Lim
x
1
2
1
2

F
H
G I
K
J
x = 0
 f
1
2
0

F
H
G I
K
J f
1
2
0

F
H
G I
K
J
 function f(x) is discontinuous at x=
1
2
.
Ans.[B]
Ex.4 If f(x) =
x x x
x
x
k x
3 2
2
16 20
2
2
2
  



R
S
|
T
| b g
,
,
is continu-
ous for all values of x, then the value of k is-
(A) 5 (B) 6
(C) 7 (D) 8
Sol.  f(x) is continuous at x =2
 f(2–0) = f(2+0) = f(2) = k
But f(2+0)
= h
Lim
 0
2 2 16 2 20
2 2
3 2
2
     
 
h h h
h
b g b g b g
b g
= h
Lim
 0
h h
h
3 2
2
7

= 7 Ans. [C]
Ex.5 If the function f(x) =
1 2
2 4
7 4
,
,
,
x
ax b x
x

  

R
S
|
T
|
is continuous at x= 2 and 4, then the values of
a and b are-
(A) 3,5 (B) 3,–5
(C) 0,3 (D) 0,5
Sol. Since f(x) is continuous at x= 2
 f(2) = Lim
x 
2
f(x)
 1 = Lim
x 
2
(ax+ b)
 1= 2a + b ...(1)

8. Again f(x) is continuous at x = 4,
 f(4) = Lim
x 
4
f(x)
 7 = Lim
x4
(ax+ b)
 7 = 4a + b ...(2)
Solving (1) and (2) , we get a= 3, b = –5.
Ans.[B]
Ex.6 If f(x) =
x when x Q
x when x Q
,
,

 
R
S
T , then f(x) is
continuous at-
(A) All rational numbers
(B) Zero only
(C) Zero and 1 only
(D) No where
Sol. Let us first examine continuity at x = 0.
f (0) = 0 ( 0 Q)
= f (0–0) = h
Lim
 0 f( 0- h) = h
Lim
 0 f(–h)
= h
Lim
 0 {–horhaccordingas–h Q or–h Q)
= 0
f( 0+0) = h
Lim
 0 f(0+h) = h
Lim
 0 f(h)
= h
Lim
 0 { h or –h} = 0
f(0) = f(0–0) = f(0+0)
 f(x) is continuous at x= 0.
Now let a  R, a  0, then
f(a–0) = h
Lim
 0 f( a–h)
= h
Lim
 0 {(a–h) or – (a–h) }
= a or –a, which is not unique.
 f(a–0) does not exist
 f(x) is not continuous at a  R0.
Hence f(x) is continuous only at x = 0.
Ans.[B]
Ex.7 f(x) = x –[x] is continuous at -
(A) x = 0 (B) x = –1
(C) x = 1 (D) x = 1/2
Sol. We know that [x] is discontinuous at every
integer. Therefore it is continuous only at
x = 1/2, while the function x is continuous at all
points x= 0, –1, 1, 1/2. Thus the given function
is continuous only at x = 1/2.
Ans.[D]
Ex.8 If f(x) =
1
3
2
2
1
2
2
3
2
2






R
S
|
|
|
T
|
|
|
sin
cos
, /
, /
sin
, /
x
x
x
a x
b x
x
x




b g
b g
is continuous at
x =  / 2, then value of a and b are-
(A) 1/2, 1/4 (B) 2,4
(C) 1/2,4 (D) 1/4,2
Sol. f

2
0

F
H
G I
K
J= h
Lim
 0
1
2
3
2
3
2
 
F
H
G I
K
J

F
H
G I
K
J
sin
cos


h
h
= h
Lim
 0
1
3
3
2
 cos
sin
h
h
= h
Lim
 0
1 1
3 1 1
2
  
 
cos cos cos
cos cos
h h h
h h
b g
e j
b g
b g
= 1/2
f

2
0

F
H
G I
K
J= h
Lim
 0
b h
h
1
2
2
2
2
 
F
H
G I
K
J
L
N
M O
Q
P
 
F
H
G I
K
J
L
N
M O
Q
P
sin



= h
Lim
 0
b h
h
1
4 2
 cos
b g
= h
Lim
 0
2 2
4
2
2
b h
h
sin /
=
b
8
Now f(x) is continuous at x=

2
 f

2
0

F
H
G I
K
J= f

2
0

F
H
G I
K
J= f

2
F
H
GI
K
J

1
2
=
b
8
= a
 a = 1/2, b = 4
Ans.[C]
Ex.9 If the function
f(x) =
1
2
1
1 3
6
12
3 6
    
  
 
R
S
|
|
T
|
|
sin
tan


x for x
ax b for x
x
for x
is continuous in the interval (– , 6), then the
value of a and b are respectively-
(A) 0,2 (B) 1,1
(C) 2,0 (D) 2,1

9. Sol. Obviously the function f(x) is continuous at
x= 1 and 3. Therefore Lim
x 
1
f(x) = f(1)
 a+ b = 2 ...(1)
and Lim
x 
3
f(x) = f(3)
 3a + b = 6 ...(2)
Solving (1) and (2) , we get a = 2, b = 0.
Ans.[C]
Ex.10 If f(x) =
1 4
0
0
16 4
0
2



 

R
S
|
|
|
T
|
|
|
cos
,
,
,
x
x
x
a x
x
x
x
then at x= 0 -
(A) f(x) is continuous, when a = 0
(B) f(x) is continuous, when a= 8
(C) f(x) is discontinuous for every value of a
(D) None of these
Sol. f(0–0) = Lim
x0
1 4
2
 cos x
x
=
2 2
2
2
sin x
x
= 8
f(0+0) =
Lim
x0
x
x
16 4
 
e j
×
16 4
16 4
 
 
x
x
= Lim
x0
x x
x
16 4
16 16
 
F
H
I
K
 
= 8
 f(0+0) = f(0–0)
 f(x) can be continuous at x = 0, if
f (0) = a = 8.
Ans.[B]
Ex.11 If f(x) =
sin
,
cos
,
,
x
x
x
x
x
x
x
k x




R
S
|
|
|
|
|
T
|
|
|
|
|
1
0
2
0
0
( Where [x] = greatest integer  x) is continu-
ous at x = 0, then k is equal to-
(A) 0 (B) 1 (C) –1 (D)Indeterminate
Sol. As given f(0–0) = f(0+0) = k
Now f(0–0) = h
Lim
 0
cos



h
h
h
b g
2
= h
Lim
 0
cos


F
H
G
I
K
J

h
2 1
1
bg= – 1
f (0+0) = h
Lim
 0
sin h
h 1 = h
Lim
 0
sin 0
0 1

= 0
 f(0–0)  f(0+0), so k is indeterminate.
Ans.[D]
Ex.12 If f(x) =
1 6 0
0
0 6
2 3
   

 
R
S
|
|
T
|
|
|sin | , /
,
, /
/|sin |
tan /tan
x x
b x
e x
a x
x x
b g 

is continuous at x= 0, then value of a,b are-
(A) 2/3, e2/3 (B) 1/3, e1/3
(C) 2/3, 1/3 (D) None of these
Sol. f (0–0) = h
Lim
 0 (1+ | sin (–h)|)a/|sin (–h)|
= h
Lim
 0 (1+ sinh)a/ sin h =ea
f(0+0) = h
Lim
 0 e
h
h
tan
tan
2
3 =
eh
h
h

F
H
G I
K
J
0
2
3
lim tan
tan
=
eh
h
h
0
2 2 2
3 2 3
lim sec
sec = e2/3
Now f(x) is continuous at x = 0
 f(0-0) = f(0+0) = f(0)
 ea = e2/3 = b
 a = 2/3, b= e2/3
Ans.[A]
Ex.13 f(x) = |x| is not differentiable at-
(A) x = –1 (B) x = 0
(C) x = 1 (D) None of these
Sol. at x = 0:
f'(0–0) = lim
h0
| |
0 0
 

h
h
= –1
f'(0+ 0) = lim
h0
| |
0 0
 
h
h
= 1
Now, since f' (0–0)  f'(0+0)
 f(x) is not differentiable at x= 0.
Ans.[B]

10. Ex.14 Function f(x) =
 
 
  
R
S
|
T
|
x if x
x if x
x x if x
,
,
,
0
0 1
1 1
2
3
, is
differentiable at -
(A) x = 0 but not at x = 1
(B) x = 1 but not at x = 0
(C) x = 0 and x = 1 both
(D) neither x = 0 nor x = 1
Sol. Differentiability at x = 0
R [f'(0)] = Lim
h0
f h f
h
( ) ( )
0 0
 
= Lim
h0
( )
0 0
2
 
h
h
= Lim
h0
h = 0
L [f'(0)] = Lim
h0
f h f
h
( ) ( )
0 0
 

= Lim
h0
  

( )
0 0
h
h
= – 1
 R [f' (0) ]  L [f'(0)]
 f(x) is not differentiable at x = 0
Differentiability at x = 1
R [f'(1)] =Lim
h0
f h f
h
( ) ( )
1 1
 
= Lim
h0
1 1 1 1
3
    
h h
h
b g b g
= Lim
h0
2 3 2 3
h h h
h
 
= 2
L [f'(1)] = Lim
h0
f h f
h
1 1
 

b g ( )
= Lim
h0
1 1
 

h
h
b g
= Lim
h0
 

2 2
h h
h
= 2
Thus R [f' (1) ] = L f'(1)]
 function f(x) is differentiable at x = 1
Ans.[B]
Ex.15 If f(x) =
3 1 1
4 1 4
x
x
x x
,
,
  
  
R
S
|
T
|
then at x = 1, f(x) is -
(A) Continuous but not differentiable
(B) Neither continuous nor differentiable
(C) Continuous and differentiable
(D) Differentiable but not continuous
Sol. Since f(1–0) = lim
x1
3x = 3
f(1+ 0) = lim
x1
(4–x) = 3
and f(1) = 31 = 3
 f(1–0) = f(1+0) = f(1)
 f(x) is continuous at x = 1
Again f' (1+ 0) = lim
x 
1
f x f
x
( ) ( )


1
1
= lim
x1
3 3
1
x
x


= lim
h0
3 3
1

h
h
= 3 lim
h0
3 1
h
h

= 3 log 3
and f'(1+ 0) = lim
x 
1
f x f
x
( ) ( )


1
1
= lim
x1
4 3
1
 

x
x
= – 1
 f' (1+0)  f'(1–0)
 f(x) is not differentiable at x = 1.
Ans.[A]
Ex.16 Function f(x) =
x
x
1  | |
is differentiable in
the set-
(A) (–  ,  ) (B) (–  ,0)
(C) (–  ,0)  (0,  ) (D) (0,  )
Sol. When x < 0, f(x) =
x
x
1
 f'(x) =
1
1 2
( )
 x
...(1)
which exists finitely for all x< 0
Also when x > 0, f(x) =
x
x
1
 f' (x) =
1
1 2
( )
 x
...(2)
which exists finitely for all x > 0. Also from
(1) and (2) we have
f
f
'( )
'( )
0 0 1
0 0 1
 
 
R
S
T  f'(0) = 1
Hence f(x) is differentiable  x R
Ans.[A]
Ex.17 If f(x) =
x
x
x
x
2 1
0
0 0
sin ,
,


R
S
|
T
|
, then
(A) f and f' are continuous at x = 0
(B) f is derivable at x = 0
(C) f and f' are derivable at x = 0
(D) f is derivable at x = 0 and f' is continuous
at x = 0

11. Sol. When x  0
f' (x) = 2x sin
1
x
+ x2 cos
1
x
. 
F
H
G I
K
J
1
2
x
= 2x sin
1
x
– cos
1
x
F
H
G
I
K
J
which exists finitely for all x  0
and f'(0) = lim
x0
f x f
x
( ) ( )


0
0
= lim
x0
x x
x
2
1
sin /
= 0
 f is also derivable at x = 0. Thus
f'(x) = 2
1 1
0
0 0
x
x x
x
x
sin cos ,
,
 

R
S
|
T
|
Also lim
x0
f'(x) = lim
x0
2
1 1
x
x x
sin cos

F
H
G I
K
J
=2– lim
x0
cos
1
x
But lim
x0
cos
1
x
does not exist, so lim
x0
f'(x)
does not exist. Hence f' is not continuous
(so not derivable) at x = 0.
Ans.[B]