Q33 / 90 · JEE Main 2024 (09 Apr Shift 2) · Mathematics
$\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$
is equal to
- A$\frac{5 \pi^2}{9}$
- Correct answerB$\frac{9 \pi^2}{8}$
- C$\frac{11 \pi^2}{10}$
- D$\frac{3 \pi^2}{2}$
Solution
$\begin{aligned} & \lim _{x \rightarrow \frac{\pi}{2}} \frac{0-\{\sin (2 x)+\cos (x)\} \cdot 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\ & =\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\{2 \sin x \cos x+\cos x\} 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\ & =\lim _{x \rightarrow \frac{\pi}{2}}\left\{\frac{2 \sin x \sin \left(\frac{\pi}{2}-x\right)}{2\left(x-\frac{\pi}{2}\right)}+\frac{\sin \left(\frac{\pi}{2}-x\right)}{2\left(\frac{\pi}{2}-x\right)}\right\} 3 x^2 \\ & =\left(1(1)+\frac{1}{2}\right) 3\left(\frac{\pi}{2}\right)^2 \\ & =\frac{9 \pi^2}{8}\end{aligned}$
Hence, the correct answer is option: 'B'.