$\text{ Let the function }\mathrm{f}:\mathrm{R}\to \mathrm{ℝ}\text{ be defined by }\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3-{\mathrm{x}}^2+(\mathrm{x}-1)\sin \mathrm{x}\text{ and let }\mathrm{g}:\mathrm{R}\to \mathrm{R}\text{ be an }$

$\text{ arbitrary function. Let fg }:\mathrm{ℝ}\to \mathrm{ℝ}\text{ be the product function defined by }\left(f\mathrm{g}\right)\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\text{ . Then }$

$\text{ which of the following statements is/are TRUE ? }$

$\begin{array}{r}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{f}\left(\mathrm{x}\right)=\left({\mathrm{x}}^2+\sin \mathrm{x}\right)(\mathrm{x}-1)\mathrm{f}\left(1^{+}\right)=\mathrm{f}\left(1^{-}\right)=\mathrm{f}\left(1\right)=0\\ \mathrm{fg}\left(\mathrm{x}\right):\mathrm{f}\left(\mathrm{x}\right)\cdot \mathrm{g}\left(\mathrm{x}\right)\mathrm{fg}:\mathrm{R}\to \mathrm{R}\\ \text{ let }\mathrm{fg}\left(\mathrm{x}\right)=\mathrm{h}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)\cdot \mathrm{g}\left(\mathrm{x}\right)\mathrm{h}:\mathrm{R}\to \mathrm{R}\\ \text{ option (c): }{h}^{\mathrm{\prime }}\left(x\right)=f^{\mathrm{\prime }}\left(x\right)g\left(x\right)+f\left(x\right)g^{\mathrm{\prime }}\left(x\right)\end{array}$

$\begin{array}{c} {\mathrm{h}}^{\mathrm{\prime }}\left(1\right)={\mathrm{f}}^{\mathrm{\prime }}\left(1\right)\mathrm{g}\left(1\right)+0\\ \left(\text{ as }\mathrm{f}\left(1\right)=0,{\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)\text{ exists }\right\}\end{array}$

$\begin{array}{l}\Rightarrow \text{ if }\mathrm{g}\left(\mathrm{x}\right)\text{ is differentiable then }\mathrm{h}\left(\mathrm{x}\right)\text{ is also differentiable (true) }\\ \text{ option }\left(\mathrm{A}\right) :\text{ If }\mathrm{g}\left(\mathrm{x}\right)\text{ is continuous at }\mathrm{x}=1\text{ then }\mathrm{g}\left(1^{+}\right)=\mathrm{g}\left(1^{-}\right)=\mathrm{g}\left(1\right)\\ {\mathrm{h}}^{\mathrm{\prime }}\left(1^{+}\right)=\underset{\mathrm{h}\to 0^{+}}{lim} \frac{\mathrm{h}(1+\mathrm{h})-\mathrm{h}\left(1\right)}{\mathrm{h}}\end{array}$

$\begin{array}{l}{\mathrm{h}}^{\mathrm{\prime }}\left(1^{+}\right)=\underset{\mathrm{h}\to 0^{+}}{lim} \frac{\mathrm{f}(1+\mathrm{h})\mathrm{g}(1+\mathrm{h})-0}{\mathrm{h}}={\mathrm{f}}^{\mathrm{\prime }}\left(1\right)\mathrm{g}\left(1\right)\\ {\mathrm{h}}^{\mathrm{\prime }}\left(1^{-}\right)=\underset{\mathrm{h}\to 0^{+}}{lim} \frac{\mathrm{f}(1-\mathrm{h})\mathrm{g}(1-\mathrm{h})-0}{-\mathrm{h}}={\mathrm{f}}^{\mathrm{\prime }}\left(1\right)\mathrm{g}\left(1\right)\\ \text{ So }\mathrm{h}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)\cdot \mathrm{g}\left(\mathrm{x}\right)\text{ is differentiable }\\ \text{ at }\mathrm{x}=1\text{ (True) }\end{array}$

$\text{ option (B) (D): }{\mathrm{h}}^{\mathrm{\prime }}\left(1^{+}\right)=\underset{\mathrm{h}\to 0^{+}}{lim} \frac{\mathrm{h}(1+\mathrm{h})-\mathrm{h}\left(1\right)}{\mathrm{h}}$

$\begin{array}{l}{\mathrm{h}}^{\mathrm{\prime }}\left(1^{+}\right)=\underset{\mathrm{h}\to 0^{+}}{lim} \frac{f(1+\mathrm{h})\mathrm{g}(1+\mathrm{h})}{\mathrm{h}}={\mathrm{f}}^{\mathrm{\prime }}\left(1\right)\mathrm{g}\left(1^{+}\right)\\ {\mathrm{h}}^{\mathrm{\prime }}\left(1^{-}\right)=\underset{\mathrm{h}\to 0^{+}}{lim} \frac{f(1-\mathrm{h})\mathrm{g}(1-\mathrm{h})}{-\mathrm{h}}={\mathrm{f}}^{\mathrm{\prime }}\left(1\right)\cdot \mathrm{g}\left(1^{-}\right)\\ \Rightarrow \mathrm{g}\left(1^{+}\right)=\mathrm{g}\left(1^{-}\right)\end{array}$

so we cannot comment on the continuity and differentiability of the function $g\left(x\right)$