UPSC NDA Math Model Paper 2

By using the definition of modulus function, the given function can be written as: f(x)={

x 1+x ,	x>0
0,	x=0
x 1−x ,	x<0

Since the expressions for f(x) change for x > 0 and x < 0, let us compare the limits of the derivatives as x → 0.
For x>0,f(x)=

x

1+x

⇒f′(x)=x[

d

dx

(

1

1+x

)] +(

d

dx

x)

1

1+x

⇒f′(x)=x

(−1)

(1+x)2

+

1

1+x

⇒f′(x)=

1

(1+x)2

⇒

lim

x→0

+f′(x)=1
Similarly, for x<0,f(x)=

x

1−x

⇒

lim

x→0

−f′(x)=

lim

x→0−

1

(1−x)2

=1
Since

lim

x→0+

f′(x)=

lim

x→0

−f′(x)=1, the function f(x) is differentiable at x=0, and f(0)=1
Also,

lim

x→∞

+f′(x) =

lim

x→∞

−f′(x)=0
∴ The function is differentiable in (-∞, ∞), i.e. it is differentiable everywhere.