JEE Main 2014 — Mathematics Calculus

We know that if $y\leq {k}^{2},\Rightarrow -k\leq y\leq k,k\in R.$y≤k2,⇒−k≤y≤k,k∈R.

Given $|f(x)|\leq {x}^{2},\forall x\in R$∣f(x)∣≤x2,∀x∈R

$$\Rightarrow -{x}^{2}\leq f(x)\leq {x}^{2} \forall x\in R$$⇒−x2≤f(x)≤x2∀x∈R

Now, $\underset{x\rightarrow 0}{\mathrm{lim}}{(-x)}^{2}=\underset{x\rightarrow 0}{\mathrm{lim}}({x}^{2})=0,$x→0lim​(−x)2=x→0lim​(x2)=0,

Hence, by using sandwich theorem, $\underset{x\rightarrow 0}{\mathrm{lim}}f(x)=0$x→0lim​f(x)=0

Further $|f(0)|\leq 0 \Rightarrow f(0)=0$∣f(0)∣≤0⇒f(0)=0

Since, $\underset{x\rightarrow 0}{\mathrm{lim}}f(x)=f(0)=0$x→0lim​f(x)=f(0)=0

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

Now, if $x\neq 0,$x=0, then $-(x)\leq \frac{f(x)}{(x)}\leq (x)$−(x)≤(x)f(x)​≤(x)

Thus, by using sandwich theorem, we have $\underset{x\rightarrow 0}{\mathrm{lim}}-(x)\leq \underset{x\rightarrow 0}{\mathrm{lim}}\frac{f(x)}{(x)}\leq \underset{x\rightarrow 0}{\mathrm{lim}}(x)$x→0lim​−(x)≤x→0lim​(x)f(x)​≤x→0lim​(x)

$$\Rightarrow 0\leq \underset{x\rightarrow 0}{\mathrm{lim}}\frac{f(x)}{(x)}\leq 0$$⇒0≤x→0lim​(x)f(x)​≤0

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}\frac{f(x)}{(x)}=0...(i)$$⇒x→0lim​(x)f(x)​=0...(i)

We know that a function $f(x)$f(x) is differentiable at a point $x=a,$x=a, if $\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(a-h)-(a)}{-h}=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(a+h)-(a)}{h}$h→0lim​−hf(a−h)−(a)​=h→0lim​hf(a+h)−(a)​

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

$\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(0-h)-f(0)}{-h}=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(-h)}{-h}=0,$h→0lim​−hf(0−h)−f(0)​=h→0lim​−hf(−h)​=0, ( by using equation $(i)$(i))

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

$\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(0+h)-f(0)}{h}=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{f(h)}{h}=0,$h→0lim​hf(0+h)−f(0)​=h→0lim​hf(h)​=0, ( by using equation $(i)$(i))

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