max 0leq xleq π (16sin (x/2 )cos 3 (x/2 ) ) is equal to:

Mathematics · Trigonometry Ratios and Identities

JEE Main 2026 — 4 April, Evening Shift — Question 36

$\max_{0\leq x\leq \pi}\left(16\sin \left(\frac{x}{2}\right)\cos^{3}\left(\frac{x}{2}\right)\right)$ is equal to:

Option A:
$\frac{3\sqrt{3}}{2}$
Option B:
$3\sqrt{3}$
Correct
Option C:
$4\sqrt{3}$
Option D:
$6\sqrt{3}$

Answer: B

Step-by-step solution

$E=16 \sin \frac{X}{2} \cos ^{3} \frac{X}{2}$ $\mathrm{E}=4 \sin \mathrm{x}[1+\cos \mathrm{x}]$ $\frac{\mathrm{dE}}{\mathrm{dx}}=4[\cos \mathrm{x}+\cos 2 \mathrm{x}]$ $=8 \cos \frac{3 x}{2} \cos \frac{x}{2}=0$ $\Rightarrow \cos \frac{3 x}{2}=0 \quad$ or $\cos \frac{x}{2}=0$ $\Rightarrow \mathrm{x}=\left\{\frac{\pi}{3}, \pi\right\}$ are critical points of the function $\mathrm{E}(0)=0$ $\mathrm{E}(\pi)=0$ E $\left(\frac{\pi}{3}\right)=3 \sqrt{3}$ ∴ maximum value of $\mathrm{E}=3 \sqrt{3}$

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Exam: JEE Main 2026
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: Trigonometry Ratios and Identities
Topic: Maximum and Minimum values of Trigonometric Expressions