JEE Main 2014 — Continuity and Differentiability Question with Solution

Given $f\left(x\right)=\left\{\begin{array}{cc}\frac{\sqrt{2+cosx}-1}{{\left(\pi -x\right)}^2},& x\ne \pi \\ k,& x=\pi \end{array}\right.$ is continuous at $x=\pi$

Hence, $f\left(\mathrm{\pi }\right)=\underset{x\to \pi }{\lim }f\left(x\right)$

$\Rightarrow k=\underset{x\to \pi }{\lim }\frac{\sqrt{2+\cos x}-1}{{\left(\pi -x\right)}^2}$

Consider $L.H.L.=\underset{x\to {\pi }^{-}}{\lim }\frac{\sqrt{2+\cos \mathit{x}}-1}{{\left(\pi -x\right)}^2}$

$\underset{h\to 0}{\lim }\frac{\sqrt{2+\cos (\pi -h)}-1}{{\left(\pi -\left(\pi -h\right)\right)}^2}$

Also, $\cos \left(\pi -h\right)=-\cos h$

$\Rightarrow k=\underset{h\to 0}{\lim }\frac{\sqrt{2-\cos h}-1}{h^2}$

$\Rightarrow k=\underset{h\to 0}{\lim }\left(\frac{\sqrt{2-\cos h}-1}{h^2}\right)\times \left(\frac{\sqrt{2-\cos h}+1}{\sqrt{2-\cos h}+1}\right)$

$\Rightarrow \mathrm{k}\mathrm{=}\underset{h\to 0}{\lim }\frac{\left(2-\cos h\right)-1}{h^2\left\{\sqrt{2-\cos h}+1\right\}}$

$\Rightarrow \mathrm{k}\mathrm{=}\underset{h\to 0}{\lim }\left(\frac{1-\cos h}{h^2}\right)\underset{h\to 0}{\lim }\frac{1}{\left\{\sqrt{2-\cos h}+1\right\}}$

Now, using $\cos 2x=1-2{\sin }^{2}x, \Rightarrow 1-\cos 2x=2{\sin }^{2}x,$

$\Rightarrow k=\underset{h\to 0}{\lim }\frac{2{sin}^2\left(\frac{h}{2}\right)}{4{\left(\frac{h}{2}\right)}^2}\times \frac{1}{\left(\sqrt{2-1}+1\right)}$

$\Rightarrow k=\frac{1}{2}\times \frac{1}{2}$

$\Rightarrow k=\frac{1}{4}.$