### JEE Main On 08 April 2018 Question 23

##### Question: Let S={$t\in \mathbf{R}:f(x)=|x-\pi |\cdot ({e^{|x|}}-1)\sin |x| $ is not differentiable at $ \text{t }$} . Then the set S is equal to: [JEE Main Online 08-04-2018]

#### Options:

A) $ {\pi } $

B) $ {\text{ 0, }\pi } $

C) $ \phi $ (an empty set)

D) $ {0} $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

$ f(x)=|x-\pi |({e^{|x|}}-1)\sin |x| $ $ f(0)=0 $

$ f(0)=\underset{h\to 0}{\mathop{\lim }},\frac{f(h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }},\frac{(\pi -h)(e^{h}-1)(\sinh )}{h}=0 $

$ \Rightarrow $ Differentiable at $ x=0 $ $ f(\pi )=0 $

$ f(\pi )=\underset{h\to 0}{\mathop{\lim }},\frac{f(\pi +h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }},\frac{h({e^{\pi +h}}-1)sin(\pi +h)}{h}=0 $

$ \Rightarrow \text{differentiable at x=}\pi $

$ \Rightarrow S\text{ is an empty set}\text{.} $