Question

Let $f\left(x\right)=x\left|x\right|$ and $g\left(x\right)=\sin x$.
Statement-1 : $gof$ is differentiable at $x=0$ and its derivative is continuous at that point
Statement-2 : $gof$ is twice differentiable at $x=0$.

• A. Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
• B. Statement-1 is true, Statement-2 is false
• C. Statement-1 is false, Statement-2 is true
• D. Statement-1 is true, Statement-2 is true; Statement-2 is correct explanation for Statement-1

Answer 1

The correct option is B Statement-1 is true, Statement-2 is false

$gof\left(x\right)=g\left(f\left(x\right)\right)=\sin \left(x\left|x\right|\right)$

$=\{\begin{array}{cc}-\sin x^2,& x<0\\ \sin x^2,& x\ge 0\end{array}$

Let the composite function $gof\left(x\right)$ be denoted by $H\left(x\right).$

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

${LH}^{\prime }\left(0\right)=\underset{h\to 0^{-}}{lim}\frac{H(0-h)-H\left(0\right)}{-h}$

$=\underset{h\to 0^{-}}{lim}\frac{-\sin h^2}{-h}$

$=\underset{h\to 0^{-}}{lim}\frac{\sin h^2}{h^2}.h=1.0=0$

${RH}^{\prime }\left(0\right)={lim}_{h\to 0^{+}}\frac{H(0+h)-H\left(0\right)}{h}$

$=\underset{h\to 0^{+}}{lim}\frac{\sin h^2-0}{h}=\underset{h\to 0^{+}}{lim}\left(\frac{\sin h^2-0}{h^2}\right)h$

$=1.0=0$
Thus $H\left(x\right)$ is differentiable at $x=0$
Also $H^{\prime }\left(x\right)=\{\begin{array}{cc}-2x\cos x^2,& x<0\\ 0,& x=0\\ 2x\cos x^2,& x>0\end{array}$

$H^{\prime }\left(x\right)$ is continuous at x=0 for $H^{\prime }\left(0\right)={LH}^{\prime }\left(0\right)={RH}^{\prime }\left(0\right)$

Again $H^{\prime \prime }\left(x\right)=\{\begin{array}{cc}-2\cos x^2+4x^2\sin x^2,& x<0\\ 2\cos x^2-4x^2\sin x^2,& x\ge 0\end{array}$

${LH}^{\prime \prime }\left(0\right)=-2$ and ${RH}^{\prime \prime }\left(0\right)=2$

Thus $H\left(x\right)$ is NOT twice differentiable at $x=0$

Hence both assertion and reason are true, but reason is not correct explanation of assertion.