NIMCET 2017 Question Paper with solutions

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