Mathematics - Continuity - Differentiability Question with Solution | TestHub

$$\begin{array}{rl}& f(x)=x|x|\text{ and }g(x)=\sin x\\ & g\circ f(x)=\sin (x|x|)=\{\begin{array}{cc}-\sin x^2& ,x<0\\ \sin x^2& ,x\ge 0\end{array}\\ & \therefore \text{ (gof }{)}^{\prime }(x)=\{\begin{array}{cc}-2x\cos x^2& ,x<0\\ 2x\cos x^2& ,x\ge 0\end{array}\end{array}$$ Clearly, L(gof )' $(0)=0=\mathrm{R}(\text{ gof }{)}^{\prime }(0)$ $\therefore$ gof is differentiable at $x=0$ and also its derivative is continuous at $x=0$ $$\begin{array}{rl}& \text{ Now, (gof)" }(\mathrm{x})=\{\begin{array}{cc}-2\cos {\mathrm{x}}^2+4{\mathrm{x}}^2\sin {\mathrm{x}}^2& ,\mathrm{x}<0\\ 2\cos {\mathrm{x}}^2-4{\mathrm{x}}^2\sin {\mathrm{x}}^2& ,\mathrm{x}\ge 0\end{array}\\ & \therefore \mathrm{L}(\text{ gof }{)}^{\prime \prime }(0)=-2\text{ and }\mathrm{R}(\text{ gof }{)}^{\prime \prime }(0)=2\\ & \therefore \mathrm{L}(\text{ gof }{)}^{\prime \prime }(0)\ne \mathrm{R}(\text{ gof }{)}^{\prime \prime }(0)\end{array}$$ $\therefore gof(x)$ is not twice differentiable at $x=0$.