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

$$\begin{aligned} & f(x)=x|x| \text { and } g(x)=\sin x \\ & g \circ f(x)=\sin (x|x|)=\left\{\begin{array}{cc} -\sin x^2 & , x < 0 \\ \sin x^2 & , x \geq 0 \end{array}\right. \\ & \therefore \text { (gof })^{\prime}(x)=\left\{\begin{array}{cc} -2 x \cos x^2 & , x < 0 \\ 2 x \cos x^2 & , x \geq 0 \end{array}\right. \end{aligned}$$​f(x)=x∣x∣ and g(x)=sinxg∘f(x)=sin(x∣x∣)={−sinx2sinx2​,x<0,x≥0​∴ (gof )′(x)={−2xcosx22xcosx2​,x<0,x≥0​​ Clearly, L(gof )' $(0)=0=\mathrm{R}(\text { gof })^{\prime}(0)$(0)=0=R( gof )′(0) $\therefore$∴ gof is differentiable at $x=0$x=0 and also its derivative is continuous at $x=0$x=0 $$\begin{aligned} & \text { Now, (gof)" }(\mathrm{x})=\left\{\begin{array}{cc} -2 \cos \mathrm{x}^2+4 \mathrm{x}^2 \sin \mathrm{x}^2 & , \mathrm{x} < 0 \\ 2 \cos \mathrm{x}^2-4 \mathrm{x}^2 \sin \mathrm{x}^2 & , \mathrm{x} \geq 0 \end{array}\right. \\ & \therefore \mathrm{L}(\text { gof })^{\prime \prime}(0)=-2 \text { and } \mathrm{R}(\text { gof })^{\prime \prime}(0)=2 \\ & \therefore \mathrm{L}(\text { gof })^{\prime \prime}(0) \neq \mathrm{R}(\text { gof })^{\prime \prime}(0) \end{aligned}$$​ Now, (gof)" (x)={−2cosx2+4x2sinx22cosx2−4x2sinx2​,x<0,x≥0​∴L( gof )′′(0)=−2 and R( gof )′′(0)=2∴L( gof )′′(0)=R( gof )′′(0)​ $\therefore g o f(x)$∴gof(x) is not twice differentiable at $x=0$x=0.