JEE Main 2024

Question

Let $\{x\}$ denote the fractional part of $x$ and $f(x) = \dfrac{\cos^{-1}(1-\{x\}^2)\sin^{-1}(1-\{x\})}{\{x\} - \{x\}^3}$, $x 
e 0$. If $L$ and $R$ respectively denote the left hand limit and the right hand limit of $f(x)$ at $x = 0$, then $\dfrac{32}{\pi^2}(L^2 + R^2)$ is equal to A). $18$ B). $17$ C). $16$ D). $19$

Edvaya · Accepted Answer

The correct answer is $18$. **Solution:** **Right-hand limit** ($x 	o 0^+$): $\{x\} = x 	o 0^+$, so $\{x\} 	o h$ as $h 	o 0^+$. $$R = \lim_{h 	o 0} \frac{\cos^{-1}(1-h^2)\sin^{-1}(1-h)}{h - h^3} = \lim_{h 	o 0} \frac{\cos^{-1}(1-h^2)}{h(1-h^2)} \cdot \sin^{-1}(1)$$ $$= \frac{\pi}{2} \lim_{h 	o 0} \frac{\cos^{-1}(1-h^2)}{h}$$ Let $\cos^{-1}(1-h^2) = t \implies \cos t = 1-h^2$. As $h 	o 0$, $t 	o 0$. $$R = \frac{\pi}{2}\lim_{t 	o 0} \frac{t}{\sqrt{1-\cos t}} = \frac{\pi}{2}\lim_{t 	o 0}\frac{t}{\sqrt{2}\sin(t/2)} = \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}/1} = \frac{\pi}{\sqrt{2}}$$ **Left-hand limit** ($x 	o 0^-$): $\{x\} = 1 + x 	o 1^-$ as $x 	o 0^-$. Let $\{x\} = 1-h$: $$L = \lim_{h 	o 0} \frac{\cos^{-1}(1-(1-h)^2)\sin^{-1}(1-(1-h))}{(1-h)-(1-h)^3} = \lim_{h 	o 0} \frac{\cos^{-1}(2h-h^2)\sin^{-1}h}{(1-h)(1-(1-h)^2)}$$ $$= \lim_{h 	o 0} \frac{\cos^{-1}(1-(h^2-2h+1)+1)\cdot\sin^{-1}h}{(1-h)\cdot h(2-h)} = \frac{\pi}{2}\lim_{h 	o 0}\frac{\sin^{-1}h}{h}\cdot\frac{1}{-h+2} = \frac{\pi}{2}\cdot\frac{1}{2} = \frac{\pi}{4}$$ $$\frac{32}{\pi^2}(L^2 + R^2) = \frac{32}{\pi^2}\!\left(\frac{\pi^2}{16} + \frac{\pi^2}{2}ight) = \frac{32}{\pi^2}\cdot\frac{9\pi^2}{16} = 18$$

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