JEE Main 2024 — 1 February, Shift 1 — Question 25

Finding right hand limit $$$$\begin{aligned}& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) & \quad=\lim _{h \rightarrow 0} f(h) & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^{2}\right) \sin ^{-1}(1-h)}{h\left(1-h^{2}\right)} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^{2}\right)}{h}\left(\frac{\sin ^{-1} 1}{1}\right)\end{aligned}$$$

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

$=\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{\theta}{\sqrt{1-\cos \theta}}$

$=\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{1}{\sqrt{\frac{1\cos \theta}{\theta^{2}}}}$

$=\frac{\pi}{2} \frac{1}{\sqrt{1 / 2}}$ $\mathrm{R}=\frac{\pi}{\sqrt{2}}$Now finding left hand limit$

$\begin{aligned}& L=\lim _{x \rightarrow 0^{-}} f(x) & =\lim _{h \rightarrow 0} f(-h)\end{aligned}$

$=\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-\{-h\}^{2}\right) \sin ^{-1}(1-\{-h\})}{\{-h\}-\{-h\}^{3}}$ $=\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-(-h+1)^{2}\right) \sin ^{-1}(1-(-h+1))}{(-h+1)-(-h+1)^{3}}$

$=\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(-h^{2}+2 h\right) \sin ^{-1} h}{(1-h)\left(1-(1-h)^{2}\right)}$

$=\lim _{h \rightarrow 0}\left(\frac{\pi}{2}\right) \frac{\sin ^{-1} \mathrm{~h}}{\left(1-(1-h)^{2}\right)}$

$=\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{-h^{2}+2 h}\right)$

$=\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{h}\right)\left(\frac{1}{-h+2}\right)$

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

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

$=18$