$ \begin{aligned} & f: R \rightarrow R \quad f(x)=\left(x^2+\sin x\right)(x-1) \quad f\left(1^{+}\right)=f\left(1^{-}\right)=f(1)=0 \\ & f g(x): f(x) \cdot g(x) \quad f g: R \rightarrow R \\ & \text { let } f g(x)=h(x)=f(x) \cdot g(x) \quad h: R \rightarrow R \\ & \text { option }(c) \quad h^{\prime}(x)=f^{\prime}(x) g(x)+f(x) g^{\prime}(x) \\ & \qquad \begin{aligned} h^{\prime}(1) & =f^{\prime}(1) g(1)+0 \\ & \left(\text { as } f(1)=0, g^{\prime}(x) \text { exists }\right\} \end{aligned} \end{aligned} $
⇒ if $\mathrm{g}(\mathrm{x})$ is differentiable then $\mathrm{h}(\mathrm{x})$ is also differentiable (true) option (A) If $\mathrm{g}(\mathrm{x})$ is continuous at $\mathrm{x}=1$ then $\mathrm{g}\left(1^{+}\right)=\mathrm{g}\left(1^{-}\right)=\mathrm{g}(1)$
$ \begin{aligned} h^{\prime}\left(1^{+}\right)= & \lim _{h \rightarrow 0^{+}} \frac{h(1+h)-h(1)}{h} \\ & h^{\prime}\left(1^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{f(1+h) g(1+h)-0}{h}=f^{\prime}(1) g(1) \\ & h^{\prime}\left(1^{-}\right)=\lim _{h \rightarrow 0^{+}} \frac{f(1-h) g(1-h)-0}{-h}=f^{\prime}(1) g(1) \end{aligned} $
So $\quad h(x)=f(x) \cdot g(x)$ is differentiable
at $\mathrm{x}=1 \quad$ (True)
option (B) (D)
$ \begin{aligned} & h^{\prime}\left(1^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{\mathrm{h}(1+\mathrm{h})-\mathrm{h}(1)}{-\mathrm{h}} \\ & h^{\prime}\left(1^{+}\right)=\lim _{\mathrm{h} \rightarrow 0^{+}} \frac{f(1+\mathrm{h}) \mathrm{g}(1+\mathrm{h})}{\mathrm{h}}=\mathrm{f}^{\prime}(1) \mathrm{g}\left(1^{+}\right) \\ & \mathrm{h}^{\prime}\left(1^{-}\right)=\lim _{\mathrm{h} \rightarrow 0^{+}} \frac{f(1-\mathrm{h}) \mathrm{g}(1-\mathrm{h})}{-\mathrm{h}}=\mathrm{f}^{\prime}(1) \cdot \mathrm{g}\left(1^{-}\right) \\ & \Rightarrow \mathrm{g}\left(1^{+}\right)=\mathrm{g}\left(1^{-}\right) \end{aligned} $
So we cannot comment on the continuity and differentiability of the function.