JEE Main 2021 — Mathematics Calculus

A function $f(x)$f(x) is continuous at a point $x=a,$x=a, if $\underset{x\rightarrow a}{\mathrm{lim}}f(x)=f(a).$x→alim​f(x)=f(a).

Thus, for continuity of the given function, we have

$$\underset{x\rightarrow 0}{\mathrm{lim}}f(x)=\alpha$$x→0lim​f(x)=α

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}[\frac{{x}^{3}}{(1-\mathrm{cos}2x{)}^{2}}{\mathrm{log}}_{e}(\frac{1+2x{e}^{-2x}}{{(1-x{e}^{-x})}^{2}})]=\alpha$$⇒x→0lim​[(1−cos2x)2x3​loge​((1−xe−x)21+2xe−2x​)]=α

Using, $\mathrm{cos}2x=1-2{\mathrm{sin}}^{2}x,{\mathrm{log}}_{e}(\frac{m}{n})={\mathrm{log}}_{e}m-{\mathrm{log}}_{e}n$cos2x=1−2sin2x,loge​(nm​)=loge​m−loge​n &{\mathrm{log}}_{e}{m}^{n}=n{\mathrm{log}}_{e}m,

we get

$$\underset{x\rightarrow 0}{\mathrm{lim}}[\frac{{x}^{3}}{{(2{\mathrm{sin}}^{2}x)}^{2}}({\mathrm{log}}_{e}(1+2x{e}^{-2x})-{\mathrm{log}}_{e}{(1-x{e}^{-x})}^{2})]=\alpha$$x→0lim​[(2sin2x)2x3​(loge​(1+2xe−2x)−loge​(1−xe−x)2)]=α

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}[\frac{{x}^{3}}{4{\mathrm{sin}}^{4}x}({\mathrm{log}}_{e}(1+2x{e}^{-2x})-2{\mathrm{log}}_{e}(1-x{e}^{-x}))]=\alpha$$⇒x→0lim​[4sin4xx3​(loge​(1+2xe−2x)−2loge​(1−xe−x))]=α

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}[\frac{{x}^{4}}{4x{\mathrm{sin}}^{4}x}(\frac{2x{e}^{-2x}\cdot {\mathrm{log}}_{e}(1+2x{e}^{-2x})}{2x{e}^{-2x}}+\frac{2x{e}^{-x}\cdot {\mathrm{log}}_{e}(1-x{e}^{-x})}{-x{e}^{-x}})]=\alpha$$⇒x→0lim​[4xsin4xx4​(2xe−2x2xe−2x⋅loge​(1+2xe−2x)​+−xe−x2xe−x⋅loge​(1−xe−x)​)]=α

Now, using the standard limits \underset{x\rightarrow 0}{\mathrm{lim}}(\frac{\mathrm{sin}x}{x})=1&\underset{x\rightarrow 0}{\mathrm{lim}}(\frac{{\mathrm{log}}_{e}(1+x)}{x})=1, we get

$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}[\frac{1}{4x}(2x{e}^{-2x}+2x{e}^{-x})]=\alpha$⇒x→0lim​[4x1​(2xe−2x+2xe−x)]=α$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}[\frac{2x}{4x}({e}^{-2x}+{e}^{-x})]=\alpha$⇒x→0lim​[4x2x​(e−2x+e−x)]=α

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}[\frac{1}{2}({e}^{-2x}+{e}^{-x})]=\alpha$$⇒x→0lim​[21​(e−2x+e−x)]=α

$$\Rightarrow \frac{1}{2}(2)=\alpha$$⇒21​(2)=α

$$\Rightarrow \alpha =1.$$⇒α=1.