NIMCET 2019 — Mathematics PYQ

A function is continuous If ${\lim }_x\to 0^{-}F(x)=F(0)={\lim }_{x\to 0^{+}}F(0)$

A function is differentiable if left hand derivative = right hand derrivative

CALCULATION:

{\mathrm{Given:}}f\left(x\right)=\left\{\begin{array}{cc}{x\sin\left(\frac{1}{x}\right)}&{if\:x>0}\\{0}&{x\leq0}\end{array}\right.

$\Rightarrow {\lim }_{x\to 0^{-}}f(x)=0,{\lim }_{x\to 0^{+}}xsin\frac{1}{x}=0$andf(0)=0

As we know that, a function is continuous If ${\lim }_{x\to 0^{-}}F(x)=F(0)={\lim }_{x\to 0^{+}}F(0)$

(2)

So the given function f(x) is continuous at x=0
Let's calculate the left hand derivative:
$$\underset{x\to 0^{-}}{\lim }\frac{d}{dx}f(0)=0$$

Similarly, let's calculate the right hand derivative:
$$\Rightarrow \underset{x\to 0^{+}}{\lim }\frac{d}{dx}[xsin\frac{1}{x}]=\underset{x\to 0^{+}}{\lim }[xsin\frac{1}{x}-\frac{1}{x}\cos \frac{1}{x}]=-\infty$$

As we can see that, left hand derivative $\ne$right hand derivative
So, the given function is not differentiable at x=0
Hence, option C is correct.