NIMCET 2024 Paper

Let us check the limit of the function at $\mathit{x}=0$,
$\underset{\mathit{x}⟶0}{\lim }\frac{\mathit{x}}{{\mathit{e}}^{\mathit{\pi }\mathit{x}}-1}=\frac{1}{\mathit{\pi }{\mathit{e}}^{\mathit{\pi }\mathit{x}}}=\frac{1}{\mathit{\pi }}$
Hence the function us continuous at $\mathit{x}=0$.
Let us check derivative of the function at $\mathit{x}=0$
${\mathit{f}}^{\prime }\left(0^{+}\right)=\underset{\mathit{h}⟶0}{\lim }\frac{\mathit{f}\left(0+\mathit{h}\right)-\mathit{f}\left(0\right)}{\mathit{h}}$
$=\underset{\mathit{h}⟶0}{\lim }\frac{\frac{\mathit{h}}{{\mathit{e}}^{\mathit{\pi }\mathit{h}}-1}-\frac{1}{\mathit{\pi }}}{\mathit{h}}$
$=\underset{\mathit{h}⟶0}{\lim }\frac{\mathit{h}\mathit{\pi }-\left({\mathit{e}}^{\mathit{h}\mathit{\pi }}-1\right)}{\mathit{h}\mathit{\pi }\left({\mathit{e}}^{\mathit{h}\mathit{\pi }}-1\right)}=\frac{\mathit{h}\mathit{\pi }-\left(\mathit{h}\mathit{\pi }+\frac{{\mathit{h}}^2{\mathit{\pi }}^2}{2!}+\right)}{\mathit{h}\mathit{\pi }\left(\mathit{h}\mathit{\pi }+\frac{{\mathit{h}}^2{\mathit{\pi }}^2}{2!}+\right)}$
$=\frac{-\frac{{\mathit{h}}^2{\mathit{\pi }}^2}{2!}+...}{{\mathit{h}}^2{\mathit{\pi }}^2+\frac{{\mathit{h}}^3{\mathit{\pi }}^3}{2!}+...}=-\frac{1}{2}$
Similarly, it can be shown that
${\mathit{f}}^{\prime }\left(0^{-}\right)=\frac{1}{2}$
Correct answer is (4).