JEE MAIN - Mathematics (2024 - 29th January Morning Shift - No. 18)
$$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{1 \over {{{\left( {x - {\pi \over 2}} \right)}^2}}}\int\limits_{{x^3}}^{{{\left( {{\pi \over 2}} \right)}^3}} {\cos \left( {{t^{{1 \over 3}}}} \right)dt} } \right)$$ is equal to
$$\frac{3 \pi^2}{4}$$
$$\frac{3 \pi^2}{8}$$
$$\frac{3 \pi}{4}$$
$$\frac{3 \pi}{8}$$
Explanation
Using L'hospital rule
$$\begin{aligned} & =\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} \frac{0-\cos x \times 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\ & =\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sin \left(x-\frac{\pi}{2}\right)}{2\left(x-\frac{\pi}{2}\right)} \times \frac{3 \pi^2}{4} \\ & =\frac{3 \pi^2}{8} \end{aligned}$$
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