NIMCET 2017 Question Paper with solutions

Now, (LHL at x = 0)
= $\underset{\mathit{x}\to 0^{-}}{\lim }$ ${\mathit{c}}^2\mathit{s}\mathit{i}\mathit{n}\left(\frac{1}{\mathit{x}}\right)$
$\underset{\mathit{h}\to 0}{\lim }{\left(0\mathit{h}\right)}^2\mathit{s}\mathit{i}\mathit{n}\left(\frac{1}{1-\mathit{h}}\right)$
= $\underset{\mathit{h}\to 0}{\lim }{\mathit{h}}^2$ sin $\left(\frac{-1}{\mathit{h}}\right)$ = 0
(RHL at x = 0) = $\underset{\mathit{x}\to 0^{+}}{\lim }$ ${\mathit{x}}^2\mathit{s}\mathit{i}\mathit{n}\left(\frac{1}{\mathit{x}}\right)$
= $\underset{\mathit{h}\to 0}{\lim }$ ${\left(0+\mathit{h}\right)}^2\mathit{s}\mathit{i}\mathit{n}\left(\frac{1}{0+\mathit{h}}\right)$
= $\underset{\mathit{h}\to 0}{\lim }{\mathit{h}}^2\mathit{s}\mathit{i}\mathit{n}\left(\frac{1}{\mathit{h}}\right)$ = 0
and f (0) = 0
∴ LHL = RHL = f (0)
∴ f (x) is continuous at x = 0,
again
(LHD at x = 0)
$\underset{\mathit{x}\to 0^{-}}{\lim }$ $\frac{\mathit{f}\left(\mathit{x}\right)-\mathit{f}\left(0\right)}{\mathit{x}-0}$
= $\underset{\mathit{x}\to 0^{-}}{\lim }$ $\frac{{\mathit{x}}^2\mathit{s}\mathit{i}\mathit{n}\left(\frac{1}{\mathit{x}}\right)-0}{\mathit{x}}$
= $\underset{\mathit{h}\to 0}{\lim }$ (0 - h) sin $\left(\frac{1}{0-\mathit{h}}\right)$
= $\underset{\mathit{h}\to 0}{\lim }$ h sin $\left(\frac{1}{\mathit{h}}\right)$ = 0
and RHD at x = 0 ,
$\underset{\mathit{x}\to 0^{+}}{\lim }$ $\frac{\mathit{f}\left(\mathit{x}\right)-\mathit{f}\left(0\right)}{\mathit{x}-0}$
= $\underset{\mathit{x}\to 0^{+}}{\lim }$ $\frac{{\mathit{x}}^2\mathit{s}\mathit{i}\mathit{n}\left(\frac{1}{\mathit{x}}\right)-0}{\mathit{x}}$
= $\underset{\mathit{x}\to 0^{+}}{\lim }$ x sin $\left(\frac{1}{\mathit{x}}\right)$
= $\underset{\mathit{h}\to 0}{\lim }$ (0 + h) sin $\left(\frac{1}{0+\mathit{h}}\right)$
= $\underset{\mathit{h}\to 0}{\lim }$ h sin $\frac{1}{\mathit{h}}$ = 0
∴ LHD = RHD = 0 - f'(x)
∴ Slope of f{x) = f'(x) = 0