JEE Main 2021 — Continuity and Differentiability Question with Solution

A function $f\left(x\right)$ is continuous at a point $x=a,$ if $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right).$

Thus, for continuity of the given function, we have

$\underset{x\to 0}{\lim }f\left(x\right)=\alpha$

$\Rightarrow \underset{x\to 0}{\lim }\left[\frac{x^3}{(1-\cos 2x{)}^2}{\log }_e\left(\frac{1+2x{e}^{-2x}}{{\left(1-x{e}^{-x}\right)}^2}\right)\right]=\alpha$

Using, $\cos 2x=1-2{\sin }^{2}x, {\log }_e\left(\frac{m}{n}\right)={\log }_{e}m-{\log }_{e}n$ $\& {\log }_e{m}^n=n{\log }_{e}m,$

we get

$\underset{x\to 0}{\lim }\left[\frac{x^3}{{\left(2{\sin }^{2}x\right)}^2}\left({\log }_e\left(1+2x{e}^{-2x}\right)-{\log }_e{\left(1-x{e}^{-x}\right)}^2\right)\right]=\alpha$

$\Rightarrow \underset{x\to 0}{\lim }\left[\frac{x^3}{4{\sin }^{4}x}\left({\log }_e\left(1+2x{e}^{-2x}\right)-2{\log }_e\left(1-x{e}^{-x}\right)\right)\right]=\alpha$

$\Rightarrow \underset{x\to 0}{\lim }\left[\frac{x^4}{4x{\sin }^{4}x}\left(\frac{2x{e}^{-2x}·{\log }_e\left(1+2x{e}^{-2x}\right)}{2x{e}^{-2x}}+\frac{2x{e}^{-x}·{\log }_e\left(1-x{e}^{-x}\right)}{-x{e}^{-x}}\right)\right]=\alpha$

Now, using the standard limits $\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)=1 \& \underset{x\to 0}{\lim }\left(\frac{{\log }_e\left(1+x\right)}{x}\right)=1,$ we get

$\Rightarrow \underset{x\to 0}{\lim }\left[\frac{1}{4x}\left(2x{e}^{-2x}+2x{e}^{-x}\right)\right]=\alpha$$\Rightarrow \underset{x\to 0}{\lim }\left[\frac{2x}{4x}\left(e^{-2x}+e^{-x}\right)\right]=\alpha$

$\Rightarrow \underset{x\to 0}{\lim }\left[\frac{1}{2}\left(e^{-2x}+e^{-x}\right)\right]=\alpha$

$\Rightarrow \frac{1}{2}\left(2\right)=\alpha$

$\Rightarrow \alpha =1.$