Mathematics · Definite Integration

JEE Main 2024 — 29 January, Shift 1 — Question 6

\lim_{x \to \frac{\pi}{2}} \left( \frac{-1}{\left( x - \frac{\pi}{2} \right)^2} \int_{\left( \frac{\pi}{2} \right)}^{x^3} \cos\left(t^{1/3} \right) dt \right) \text{ is\; equal \; to}

Option A:
$\frac{3 \pi}{8}$
Option B:
$\frac{3 \pi^{2}}{4}$
Option C:
$\frac{3 \pi^{2}}{8}$
Correct
Option D:
$\frac{3 \pi}{4}$

Answer: C

Step-by-step solution

\text{Using L'Hôpital's Rule:}

\lim_{x \to \frac{\pi}{2}} \frac{0 - \cos x \cdot 3x^2}{2\left(x - \frac{\pi}{2}\right)}

= \lim_{x \to \frac{\pi}{2}} \frac{\sin\left( x - \frac{\pi}{2} \right)}{2\left( x - \frac{\pi}{2} \right)} \cdot \frac{3\pi^2}{4}

= 1 \cdot \frac{3\pi^2}{8} = \frac{3\pi^2}{8}

\boxed{\frac{3\pi^2}{8}}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Mathematics
Chapter: Definite Integration
Topic: Leibnitz rule & its application in limits

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