JEE Main 2024 (Online) 9th April Evening ShiftQuestion 1
$\lim_{x\to\frac{\pi}{2}}\left(\frac{\int_{x^{3}}^{(\pi/2)^{3}}\left(\sin\left(2t^{1/3}\right)+\cos\left(t^{1/3}\right)\right)dt}{\left(x-\frac{\pi}{2}\right)^{2}}\right)$ is equal to
Limits & Continuity
Solution
Form $0/0$. Derivative of numerator: $-3x^2\left(\sin2x+\cos x\right)$ (zero at $\pi/2$ since $\sin\pi+\cos\frac\pi2=0$), so apply L'Hopital: $\frac{-3x^2(\sin2x+\cos x)}{2(x-\pi/2)}$, again $0/0$. Differentiate: $\frac{-6x(\sin2x+\cos x)-3x^2(2\cos2x-\sin x)}{2}$ at $\pi/2$: first term $0$, second $-3\frac{\pi^2}{4}(-2-1)=\frac{9\pi^2}{4}$. Limit $=\frac{9\pi^2}{8}$. Option (C).