JEE Main 2014 — Continuity and Differentiability Question with Solution

We know that if $y\le k^2, \Rightarrow -k\le y\le k, k\in R.$

Given $\left|f\left(x\right)\right|\le x^2, \forall x\in R$

$\Rightarrow -x^2\le f\left(x\right)\le x^2 \forall x\in R$

Now, $\underset{x\to 0}{\lim }{\left(-x\right)}^2=\underset{x\to 0}{\lim }\left(x^2\right)=0,$

Hence, by using sandwich theorem, $\underset{x\to 0}{\lim }f\left(x\right)=0$

Further $\left|f\left(0\right)\right|\le 0 \Rightarrow f\left(0\right)=0$

Since, $\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)=0$

Hence, $f\left(x\right)$ is continuous at $x=0$

Now, if $x\ne 0,$ then $-\left(x\right)\le \frac{f\left(x\right)}{\left(x\right)}\le \left(x\right)$

Thus, by using sandwich theorem, we have $\underset{x\to 0}{\lim }-\left(x\right)\le \underset{x\to 0}{\lim }\frac{f\left(x\right)}{\left(x\right)}\le \underset{x\to 0}{\lim }\left(x\right)$

$\Rightarrow 0\le \underset{x\to 0}{\lim }\frac{f\left(x\right)}{\left(x\right)}\le 0$

$\Rightarrow \underset{x\to 0}{\lim }\frac{f\left(x\right)}{\left(x\right)}=0 ...\left(i\right)$

We know that a function $f\left(x\right)$ is differentiable at a point $x=a,$ if $\underset{h\to 0}{\lim }\frac{f\left(a-h\right)-\left(a\right)}{-h}=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-\left(a\right)}{h}$

Now, the left-hand derivative at $x=0$ is 

$\underset{h\to 0}{\lim }\frac{f\left(0-h\right)-f\left(0\right)}{-h}=\underset{h\to 0}{\lim }\frac{f\left(-h\right)}{-h}=0,$ ( by using equation $\left(i\right)$) 

And, the right-hand derivative at $x=0$ is

$\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0}{\lim }\frac{f\left(h\right)}{h}=0,$ ( by using equation $\left(i\right)$)

Hence, $f\left(x\right)$ is differentiable at $x=0.$