JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Continuity and Differentiability

From $f(x)=xg(x)$ and ${\lim }_{x\to 0}g(x)=1$, we get ${\lim }_{x\to 0}f(x)=0$, so $f(0)=0$. From the functional equation, $f$ is continuous everywhere. Differentiating $f(x+y)=f(x)+f(y)+f(x)f(y)$ with respect to $y$ and setting $y=0$: $f^{\prime }(x)=f^{\prime }(0)(1+f(x))$. Since $f^{\prime }(0)={\lim }_{x\to 0}\frac{f(x)}{x}={\lim }_{x\to 0}g(x)=1$, we have $f^{\prime }(x)=1+f(x)$. Solving this differential equation with $f(0)=0$ gives $f(x)=e^x-1$. Thus $f$ is differentiable everywhere (A is correct), $f^{\prime }(0)=1$ (D is correct), and $g(x)=\frac{e^x-1}{x}$ for $x\ne 0$, $g(0)=1$, which is differentiable everywhere (B is correct). But $f^{\prime }(1)=e\ne 1$ (C is false).