Question

Let $f\left(x\right)=x|x|,g\left(x\right)=\sin x$ and $h\left(x\right)=\left(gof\right)\left(x\right)$. Then

• A. $h\left(x\right)$ is not differentiable at $x=0$.
• B. $h\left(x\right)$ is differentiable at $x=0$, but $h^{\prime }\left(x\right)$ is not continuous at $x=0$
• C. $h^{\prime }\left(x\right)$ is differentiable at $x=0$
• D. $h^{\prime }\left(x\right)$ is continuous at $x=0$ but it is not differentiable at $x=0$

Answer 1

The correct option is C $h^{\prime }\left(x\right)$ is differentiable at $x=0$

We have,
$f\left(x\right)=-x^2,x<0$
$f\left(x\right)=x^2,x\ge 0$
$g\left(x\right)=\sin x$
Hence,

$h\left(x\right)=g\left(f\right(x\left)\right)=\sin (-x^2),x<0$
$h\left(x\right)=g\left(f\right(x\left)\right)=\sin \left(x^2\right),x\ge 0$

As $x\to 0,h\left(x\right)\to 0$.
$h\left(0\right)=0$.

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

For $h\left(x\right)$, at $x=0$,

L.H.D $=-2x\times \cos (-x^2)=0$

R.H.D $=2x\times \cos \left(x^2\right)=0$

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

We have,
$h^{\prime }\left(x\right)=-2x\cos (-x^2),x<0$

$h^{\prime }\left(x\right)=2x\cos \left(x^2\right),x\ge 0$

Let us consider the derivative of $h^{\prime }\left(x\right)$ at $x=0$,
L.H.D $=-2\cos (-x^2)+4x^2\sin (-x^2)=-2$
R.H.D $=2\cos \left(x^2\right)-4x^2\sin \left(x^2\right)=2$

Hence, $h^{\prime }\left(x\right)$ is not differentiable at $x=0$.