since ƒ(x) = xg(x)

Now we check continuity of $f(\mathrm{x})$
at $x=a$
$ \begin{aligned} & \lim _{\mathrm{h} \rightarrow 0} f(\mathrm{a}+\mathrm{h})=f(\mathrm{a})+f(\mathrm{~b})+f(\mathrm{a})+f(\mathrm{~h}) \\ & \lim _{\mathrm{x} \rightarrow 0}(f(\mathrm{a})+f(\mathrm{~h})(1+f(\mathrm{a}))) \\ & \lim _{\mathrm{h} \rightarrow 0} f(\mathrm{a}+\mathrm{h})=f(\mathrm{a}) \end{aligned} $
$\therefore f(\mathrm{x})$ is continuous $\forall \mathrm{x} \in \mathrm{R}$
$ \begin{aligned} & \lim _{x \rightarrow 0} f(x)=f(0)=0 \quad\left(\lim _{x \rightarrow 0} f(x)=0\right) \\ & \therefore f(0)=0 \end{aligned} $
and $\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{1}=1$
$ \therefore f^{\prime}(0)=1 $
Now
$ f(\mathrm{x}+\mathrm{y})=f(\mathrm{x})+f(\mathrm{y})+f(\mathrm{x}) f(\mathrm{y}) $
using partial derivative (w.r.t. y)
$ \begin{aligned} & f^{\prime}(\mathrm{x}+\mathrm{y})+f^{\prime}(\mathrm{y})+f(\mathrm{x})+f^{\prime}(\mathrm{y}) \\ & \text { put } \mathrm{y}=0 \\ & f^{\prime}(\mathrm{x})=f^{\prime}(0)+f(\mathrm{x}) f^{\prime}(0) \\ & f^{\prime}(\mathrm{x})=1+f(\mathrm{x}) \\ & \int \frac{f^{\prime}(\mathrm{x})}{1+f(\mathrm{x})} \mathrm{d} \mathrm{x}=\int 1 \mathrm{dx} \\ & \ell \mathrm{n}|(1+f(\mathrm{x}))|=\mathrm{x}+\mathrm{C} \\ & f(0)=0 ; \mathrm{c}=0 \therefore|1+f(\mathrm{x})|=\mathrm{e}^{\mathrm{x}} \\ & 1+f(\mathrm{x})= \pm \mathrm{e}^{\mathrm{x}} \text { or } f(\mathrm{x})= \pm \mathrm{e}^{\mathrm{x}}-1 \end{aligned} $
Now $f(0)=0 \therefore f(\mathrm{x})=\mathrm{e}^{\mathrm{x}}-1$
$ \therefore f(\mathrm{x})=\mathrm{e}^{\mathrm{x}}-1 $
option (A) is correct and $f^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}} f^{\prime}(0)=1$ option(D) is correct
$ \begin{aligned} & \mathrm{g}(\mathrm{x})=\frac{f(\mathrm{x})}{\mathrm{x}}=\left\{\begin{array}{ccc} \frac{\mathrm{e}^{\mathrm{x}}-1}{\mathrm{x}} & ; & \mathrm{x} \neq 0 \\ 1 & ; & \mathrm{x}=0 \end{array}\right\} \\ & \mathrm{g}^{\prime}(0+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{~g}(0+\mathrm{h})-\mathrm{g}(0)}{\mathrm{h}} \\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{\frac{\mathrm{e}^{\mathrm{h}}-1}{\mathrm{~h}}-1}{\mathrm{~h}}=\frac{1}{2} \end{aligned} $
option B is correct