$\text{ Let }\mathit{f}:\mathrm{ℝ}\to \mathrm{ℝ}\text{ and }\mathit{g}:\mathrm{ℝ}\to \mathrm{ℝ}\text{ be functions satisfying }f(x+y)=f\left(x\right)+f\left(y\right)+f\left(x\right)f\left(y\right)\text{ and }f\left(x\right)=xg\left(x\right)$

$\text{ for all }x,y\in \mathrm{ℝ}\text{ . If }\underset{x\to 0}{lim} \mathrm{g}\left(\mathrm{x}\right)=1,\text{ then which of the following statements is/are TRUE? }$

$\begin{array}{l}\because \text{ Put }\mathrm{x}=\mathrm{y}=0\text{ is given relation }\\ \Rightarrow f\left(0\right)=f\left(0\right)+f\left(0\right)+f^2\left(0\right)\\ \Rightarrow f\left(0\right)=0\text{ or }-1\\ \because f(\mathrm{x}+\mathrm{y})=f\left(\mathrm{x}\right)+f\left(\mathrm{y}\right)+f\left(\mathrm{x}\right)\cdot f\left(\mathrm{y}\right)\\ \Rightarrow \frac{f(\mathrm{x}+\mathrm{y})-f\left(\mathrm{x}\right)}{\mathrm{y}}=\frac{f\left(\mathrm{y}\right)(1+f(\mathrm{x}\left)\right)}{\mathrm{y}}\\ \Rightarrow \underset{y\to 0}{lim} \left[\frac{f(\mathrm{x}+\mathrm{y})-f\left(\mathrm{x}\right)}{\mathrm{y}}\right]=\underset{y\to 0}{lim} (1+f(\mathrm{x}\left)\right)\cdot \frac{f\left(\mathrm{y}\right)}{\mathrm{y}} \left(\sin ce\underset{y\to 0}{lim}\frac{f\left(y\right)}{y}=\underset{y\to 0}{lim}g\left(y\right)=1 \right)\\ \Rightarrow f^{\mathrm{\prime }}\left(\mathrm{x}\right)=1+f\left(\mathrm{x}\right)\\ \Rightarrow f^{\mathrm{\prime }}\left(0\right)=1+f\left(0\right)\\ \Rightarrow f^{\mathrm{\prime }}\left(0\right)=1+0\\ \Rightarrow f^{\mathrm{\prime }}\left(0\right)=1\\ \\ \text{ Again }\frac{f^{\mathrm{\prime }}\left(x\right)}{1+f\left(x\right)}=1\Rightarrow \int \frac{f^{\mathrm{\prime }}\left(x\right)dx}{1+f\left(x\right)}dx=\int dx\\ \Rightarrow \ln (1+f(x\left)\right)=x+c\\ \Rightarrow \ln [1+f(\mathrm{x}\left)\right]=\mathrm{x} \left(\mathrm{since} \mathrm{f}\left(0\right)=0\right)\\ \Rightarrow 1+f\left(x\right))={\mathrm{e}}^{\mathrm{x}}\end{array}$

$\begin{array}{l}\Rightarrow f\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}-1\Rightarrow f^{\mathrm{\prime }}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}} \Rightarrow \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{differentiable} \mathrm{for} \mathrm{all} \mathrm{x}.\\ and f^{\mathrm{\prime }}\left(1\right)={\mathrm{e}}^{ }\\ \mathrm{g}\left(\mathrm{x}\right)=\frac{f\left(\mathrm{x}\right)}{\mathrm{x}}=\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}} \\ \text{ If }\mathrm{g}\left(0\right)=1\text{ then }\end{array}$

$\left[\mathrm{since} {\mathrm{g}}^{\mathrm{\prime }}\left(0\right)=\underset{\mathrm{h}\to 0}{lim} \frac{g(0+\mathrm{h})-\mathrm{g}\left(0\right)}{\mathrm{h}}\right]$

$\begin{array}{l}{g}^{\mathrm{\prime }}\left(0^{+}\right)=\underset{\mathrm{h}\to 0}{lim} \frac{\frac{{\mathrm{e}}^{\mathrm{h}}-1}{\mathrm{h}}-1}{\mathrm{h}}=\underset{\mathrm{h}\to 0}{lim} \frac{{\mathrm{e}}^{\mathrm{h}}-1-\mathrm{h}}{{\mathrm{h}}^2}=\frac{1}{2}\\ {\mathrm{g}}^{\mathrm{\prime }}\left(0^{-}\right)=\underset{\mathrm{h}\to 0}{lim} \frac{\mathrm{g}(0-\mathrm{h})-\mathrm{g}\left(0\right)}{-\mathrm{h}}=\underset{\mathrm{h}\to 0}{lim} \frac{{\displaystyle \frac{{\mathrm{e}}^{-\mathrm{h}}-1}{-h}}-1}{-h}\\ \underset{\mathrm{h}\to 0}{lim} \frac{{\mathrm{e}}^{-\mathrm{h}}-1+\mathrm{h}}{{\mathrm{h}}^2}=\frac{1}{2}\end{array}$

$\mathrm{g}\left(\mathrm{x}\right)\text{ is differentiable. }$