JEE Advanced 2020 — Paper 2 — Question 42

$\mathrm{f}(\mathrm{x})=\mathrm{x} \cdot \mathrm{g}(\mathrm{x}) \forall \mathrm{x} \in \mathrm{R}$

$\Rightarrow \quad \lim _{x \rightarrow 0} f(x)=\left(\lim _{x \rightarrow 0} x\right)\left(\lim _{x \rightarrow 0} g(x)\right)$

$\Rightarrow \quad \lim _{x \rightarrow 0} f(x)=0$

Also $\lim _{x \rightarrow 0}(f(x+y))=\lim _{x \rightarrow 0}(f(x)+f(y)+f(x) \cdot f(y))$

$\Rightarrow \quad \lim _{x \rightarrow 0} f(x+y)=f(y)$

$\Rightarrow \mathrm{f}(\mathrm{x})$ is continuous $\forall \mathrm{x} \in \mathrm{R}$

$\Rightarrow \quad \mathrm{f}(0)=0$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\Rightarrow \quad f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(h)+f(x) \cdot f(h)}{h}$

$\Rightarrow \quad f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(h)}{h} \cdot(1+f(x))$

$\Rightarrow \quad \int \frac{f^{\prime}(x) d x}{1+f(x)}=\int 1 d x$

$\Rightarrow \quad \ln |1+\mathrm{f}(\mathrm{x})|=\mathrm{x}+\mathrm{c}$

$\mathrm{f}(0)=0 \Rightarrow \mathrm{c}=0$

$\Rightarrow f(x)=e^{x}-1$

$\Rightarrow \quad \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}}$

$\Rightarrow \quad f(x)$ is differentiable and $f^{\prime}(0)=1$

Also $g(x)=\frac{f(x)}{x}, x \neq 0$ Now if $g(0)=1$

$\Rightarrow g(x)=\left\{\begin{array}{cc}\frac{e^{x}-1}{x}, & x \neq 0\\ 1, & x=0\end{array} \Rightarrow g(x)\right.$ is continuous at $x=0$

$\Rightarrow g^{\prime}(0)=\lim _{h \rightarrow 0} \frac{g(0+h)-g(0)}{h}=\lim _{h \rightarrow 0} \frac{\frac{e^{h}-1}{h}-1}{h}=\frac{1}{2}$

$\Rightarrow \mathrm{g}(\mathrm{x})$ is diff. $\forall \mathrm{x} \in \mathrm{R}$ if $\mathrm{g}(0)=1$