Question

Let $f\left(x\right)=x|x|,g\left(x\right)=\sin \left(x\right)$ and $h\left(x\right)=(g\circ f)\left(x\right)$. 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 continuous but not differentiable at $x=0.$
  
• D. $h^{\prime }\left(x\right)$ is differentiable at $x=0.$

Answer 1

The correct option is C $h^{\prime }\left(x\right)$ is continuous but not differentiable at $x=0.$


$f\left(x\right)=x|x|$
$f\left(x\right)=\{\begin{array}{cc}{x}^2;& x\ge 0\\ -x^2;& x<0\end{array}$

$h\left(x\right)=(g\circ f)\left(x\right)$
$=\sin \left(x|x|\right)\forall x\in R$

$h\left(x\right)=\{\begin{array}{cc}\sin \left(x^2\right);& x\ge 0\\ -\sin \left(x^2\right);& x<0\end{array}$

$h^{\prime }\left(x\right)=\{\begin{array}{cc}2x\cos \left(x^2\right);& x\ge 0\\ -2x\cos \left(x^2\right);& x<0\end{array}$

$\text{L.H.L}=\underset{x\to 0^{-}}{lim}-2x\cos \left(x^2\right)=0$
$\text{R.H.L}=\underset{x\to 0^{+}}{lim}2x\cos \left(x^2\right)=0$
$\Rightarrow \text{L.H.L}=\text{R.H.L}$
$\therefore h^{\prime }\left(x\right)$ is continous at $x=0$

$\text{L.H.D}=\underset{x\to 0^{-}}{lim}\frac{h^{\prime }\left(x\right)-h^{\prime }\left(0\right)}{x-0}$
$=\underset{x\to 0^{-}}{lim}\frac{-2x\cos \left(x^2\right)-0}{x}$
$=\underset{x\to 0^{-}}{lim}-2\cos \left(x^2\right)$
$=-2$

$\text{R.H.D}=\underset{x\to 0^{+}}{lim}\frac{h^{\prime }\left(x\right)-h^{\prime }\left(0\right)}{x-0}$
$=\underset{x\to 0^{+}}{lim}\frac{2x\cos \left(x^2\right)-0}{x}$
$=\underset{x\to 0^{+}}{lim}2\cos \left(x^2\right)$
$=2$
$\Rightarrow \text{L.H.D}\ne \text{R.H.D}$
$\therefore h^{\prime }\left(x\right)$ is not differentiable at $x=0$