JEE Main 2021 — Mathematics Calculus

$$\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}f(x)=f(0)={Lim}_{x\rightarrow {0}^{-}}(x)$$x→0+lim​f(x)=f(0)=Limx→0−​(x)

$$=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}\frac{{\mathrm{cos}}^{-1}(1-{x}^{2})\cdot {\mathrm{sin}}^{-1}(1-x)}{x(1-x)(1+x)}$$=x→0+lim​x(1−x)(1+x)cos−1(1−x2)⋅sin−1(1−x)​

$$=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}\frac{{\mathrm{cos}}^{-1}(1-{x}^{2})}{x\cdot 1\cdot 1}\cdot \frac{\pi }{2}$$=x→0+lim​x⋅1⋅1cos−1(1−x2)​⋅2π​

Let $1-{x}^{2}=\mathrm{cos}\theta$1−x2=cosθ

$$=\frac{\pi }{2}\underset{\theta \rightarrow {0}^{+}}{\mathrm{lim}}\frac{\theta }{\sqrt{1-\mathrm{cos}\theta }}$$=2π​θ→0+lim​1−cosθ​θ​

$$=\frac{\pi }{2}\underset{\theta \rightarrow {0}^{+}}{\mathrm{lim}}\frac{\theta }{\sqrt{2}\mathrm{sin}\frac{\theta }{2}}=\frac{\pi }{\sqrt{2}}$$=2π​θ→0+lim​2​sin2θ​θ​=2​π​

Now, $\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}\frac{{\mathrm{cos}}^{-1}(1-{(1+x)}^{2}){\mathrm{sin}}^{-1}(-x)}{(1+x)-{(1+x)}^{3}}$x→0−lim​(1+x)−(1+x)3cos−1(1−(1+x)2)sin−1(−x)​

$$\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}\frac{\frac{\pi }{2}(-{\mathrm{sin}}^{-1}x)}{(1+x)(2+x)(-x)}$$x→0−lim​(1+x)(2+x)(−x)2π​(−sin−1x)​

$$\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}\frac{\frac{\pi }{2}}{1\cdot 2}\cdot \frac{{\mathrm{sin}}^{-1}x}{x}=\frac{\pi }{4}$$x→0−lim​1⋅22π​​⋅xsin−1x​=4π​

$$\Rightarrow \mathrm{RHL}\neq \mathrm{LHL}$$⇒RHL=LHL

Function can't be continuous

$\Rightarrow$⇒ No value of $\alpha$α exist