NIMCET 2024 Paper

We know that for any value of $\mathit{x},\mathit{s}\mathit{i}\mathit{n}\mathit{x}$ is finite and its values remains between -1 and 1 .
${\mathit{f}}^{\prime }\left(0^{+}\right)=\frac{{\mathit{h}}^2\mathit{s}\mathit{i}\mathit{n}\left(\frac{1}{\mathit{h}}\right)-0}{\mathit{h}}=\mathit{h}\mathit{s}\mathit{i}\mathit{n}\left(\frac{1}{\mathit{h}}\right)=0$
${\mathit{f}}^{\prime }\left(0^{-}\right)=\frac{{\mathit{h}}^2\mathit{s}\mathit{i}\mathit{n}\left(-\frac{1}{\mathit{h}}\right)-0}{-\mathit{h}}=\mathit{h}\mathit{s}\mathit{i}\mathit{n}\left(\frac{1}{\mathit{h}}\right)=0$
Hence function is differentiable at $\mathit{x}=0$
${\mathit{f}}^{\prime }\left(\mathit{x}\right)=2\mathit{x}\mathit{s}\mathit{i}\mathit{n}\left(\frac{1}{\mathit{x}}\right)+{\mathit{x}}^2\cos \left(\frac{1}{\mathit{x}}\right)\left(-\frac{1}{{\mathit{x}}^2}\right)$
$\Rightarrow {\mathit{f}}^{\prime }\left(\mathit{x}\right)=2\mathit{x}\mathit{s}\mathit{i}\mathit{n}\left(\frac{1}{\mathit{x}}\right)-\cos \left(\frac{1}{\mathit{x}}\right)$
When $\mathit{x}⟶0,\cos \left(\frac{1}{\mathit{x}}\right)$ does not give a definite value, hence ${\mathit{f}}^{\prime }\left(\mathit{x}\right)$ is not continuous at $\mathit{x}=0$