JEE Main 2024 — Limits Question with Solution

Given: $f\left(x\right)=\frac{{\cos }^{-1}\left(1-{\left\{x\right\}}^2\right){\sin }^{-1}\left(1-\left\{x\right\}\right)}{\left\{x\right\}-{\left\{x\right\}}^3}$

$\Rightarrow \mathrm{RHL}=\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0+h\right)$

$\Rightarrow \mathrm{RHL}=\underset{h\to 0}{\lim }f\left(h\right)$

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

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

Let ${\cos }^{-1}\left(1-h^2\right)=\theta \Rightarrow \cos \theta =1-h^2$

$\Rightarrow \mathrm{RHL}=\frac{\pi }{2}\underset{\theta \to 0}{\lim }\frac{\theta }{\sqrt{1-\cos \theta }}$

$\Rightarrow \mathrm{RHL}=\frac{\pi }{2}\underset{\theta \to 0}{\lim }\frac{1}{\sqrt{\frac{1-\cos \theta }{{\theta }^2}}}$

$\Rightarrow \mathrm{RHL}=\frac{\pi }{2}\frac{1}{\sqrt{{\displaystyle \frac{1}{2}}}}$

$\Rightarrow \mathrm{RHL}=\frac{\pi }{\sqrt{2}}$

$\Rightarrow R=\frac{\pi }{\sqrt{2}}$

Now finding left hand limit,

$\Rightarrow \mathrm{LHL}=\underset{x\to 0^{-}}{\lim }f\left(x\right)$

$\Rightarrow \mathrm{LHL}=\underset{h\to 0}{\lim }f\left(-h\right)$

$\Rightarrow \mathrm{LHL}=\underset{h\to 0}{\lim }\frac{{\cos }^{-1}\left(1-{\left\{-h\right\}}^2\right){\sin }^{-1}\left(1-\left\{-h\right\}\right)}{\left\{-h\right\}-{\left\{-h\right\}}^3}$

$\Rightarrow \mathrm{LHL}=\underset{h\to 0}{\lim }\frac{{\cos }^{-1}\left(1-{\left(-h+1\right)}^2\right){\sin }^{-1}\left(1-\left(-h+1\right)\right)}{\left(-h+1\right)-{\left(-h+1\right)}^3}$

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

$\Rightarrow \mathrm{LHL}=\underset{h\to 0}{\lim }\left(\frac{\pi }{2}\right)\frac{{\sin }^{-1}h}{\left(1-{\left(1-h\right)}^2\right)}$

$\Rightarrow \mathrm{LHL}=\frac{\pi }{2}\underset{h\to 0}{\lim }\left(\frac{{\sin }^{-1}h}{-h^2+2h}\right)$

$\Rightarrow \mathrm{LHL}=\frac{\pi }{2}\underset{h\to 0}{\lim }\left(\frac{{\sin }^{-1}h}{h}\right)\left(\frac{1}{-h+2}\right)$

$\Rightarrow \mathrm{LHL}=\frac{\pi }{4}$

$\Rightarrow L=\frac{\pi }{4}$

$\Rightarrow \frac{32}{{\pi }^2}\left(L^2+R^2\right)=\frac{32}{{\pi }^2}\left(\frac{{\pi }^2}{2}+\frac{{\pi }^2}{16}\right)$

$\Rightarrow \frac{32}{{\pi }^2}\left(L^2+R^2\right)=16+2$

$\Rightarrow \frac{32}{{\pi }^2}\left(L^2+R^2\right)=18$