JEE Main 2021 — Continuity and Differentiability Question with Solution

Given that 

$f\left(x\right)=\left\{\begin{array}{l}\frac{1}{x}{\log }_{\mathrm{e}}\left(\frac{1+\frac{x}{a}}{1-\frac{x}{ b}}\right) , x<0\\ k , x=0\\ \frac{{\cos }^{2}x-{\sin }^{2}x-1 }{\sqrt{x^2+1}-1} , x>0\end{array}\right.$

$f\left(x\right)$ is continuous at $x = 0$

$\mathrm{L}.\mathrm{H}.\mathrm{L} = \mathrm{R}.\mathrm{H}.\mathrm{L} = f\left(0\right)$

$\mathrm{LHL}=\underset{x\to 0^{-}}{\lim }\frac{\log \left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right)}{x}$

$=\underset{x\to 0^{-}}{\lim }\frac{\log \left(1+\frac{x}{a}\right)-\log \left(1-\frac{x}{b}\right)}{x}$

$=\underset{x\to 0^{-}}{\lim }\frac{\log \left(1+\frac{x}{a}\right)}{\frac{x}{a}}\times \frac{1}{a}-\frac{\log \left(1-\frac{x}{b}\right)}{-\frac{x}{b}}\times -\frac{1}{b}=\frac{1}{b}+\frac{1}{a}$

$\mathrm{RHL}=\underset{x\to 0^{+}}{\lim }\frac{{\cos }^{2}x-{\sin }^{2}x-1}{\sqrt{x^2+1}-1}\times \frac{\sqrt{x^2+1}+1}{\sqrt{x^2+1}+1}$

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

So, $\frac{1}{b}+\frac{1}{a}=k=-4 .$.

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

$\frac{1}{a} + \frac{1}{b} = -4.$

$\therefore \frac{1}{a} + \frac{1}{b} + \frac{4}{k} = -5.$