Q37 / 75 · JEE Main 2026 (04 April Shift 2) · Mathematics
$\max_{0 \leq x \leq \pi}\left(16\sin\left(\dfrac{x}{2}\right)\cos^3\left(\dfrac{x}{2}\right)\right)$ is equal to:
- A$\dfrac{3\sqrt{3}}{2}$
- Correct answerB$3\sqrt{3}$
- C$4\sqrt{3}$
- D$6\sqrt{3}$
Solution
Let $y = \dfrac{x}{2}$. Since $0 \leq x \leq \pi$, we have $0 \leq y \leq \dfrac{\pi}{2}$.
The given expression becomes $f(y) = 16\sin y \cos^3 y$.
Differentiating with respect to $y$:
$f'(y) = 16(\cos y \cdot \cos^3 y + \sin y \cdot 3\cos^2 y(-\sin y))$
$f'(y) = 16\cos^2 y(\cos^2 y - 3\sin^2 y)$
For maximum value, $f'(y) = 0$:
$\cos^2 y - 3\sin^2 y = 0 \Rightarrow \tan^2 y = \dfrac{1}{3}$
Since $y \in \left[0, \dfrac{\pi}{2}\right]$, $\tan y = \dfrac{1}{\sqrt{3}} \Rightarrow y = \dfrac{\pi}{6}$.
Substituting $y = \dfrac{\pi}{6}$ in $f(y)$:
$f\left(\dfrac{\pi}{6}\right) = 16\sin\left(\dfrac{\pi}{6}\right)\cos^3\left(\dfrac{\pi}{6}\right)$
$f\left(\dfrac{\pi}{6}\right) = 16 \times \dfrac{1}{2} \times \left(\dfrac{\sqrt{3}}{2}\right)^3$
$f\left(\dfrac{\pi}{6}\right) = 8 \times \dfrac{3\sqrt{3}}{8} = 3\sqrt{3}$
The maximum value is $3\sqrt{3}$.
Answer: $3\sqrt{3}$
The given expression becomes $f(y) = 16\sin y \cos^3 y$.
Differentiating with respect to $y$:
$f'(y) = 16(\cos y \cdot \cos^3 y + \sin y \cdot 3\cos^2 y(-\sin y))$
$f'(y) = 16\cos^2 y(\cos^2 y - 3\sin^2 y)$
For maximum value, $f'(y) = 0$:
$\cos^2 y - 3\sin^2 y = 0 \Rightarrow \tan^2 y = \dfrac{1}{3}$
Since $y \in \left[0, \dfrac{\pi}{2}\right]$, $\tan y = \dfrac{1}{\sqrt{3}} \Rightarrow y = \dfrac{\pi}{6}$.
Substituting $y = \dfrac{\pi}{6}$ in $f(y)$:
$f\left(\dfrac{\pi}{6}\right) = 16\sin\left(\dfrac{\pi}{6}\right)\cos^3\left(\dfrac{\pi}{6}\right)$
$f\left(\dfrac{\pi}{6}\right) = 16 \times \dfrac{1}{2} \times \left(\dfrac{\sqrt{3}}{2}\right)^3$
$f\left(\dfrac{\pi}{6}\right) = 8 \times \dfrac{3\sqrt{3}}{8} = 3\sqrt{3}$
The maximum value is $3\sqrt{3}$.
Answer: $3\sqrt{3}$
Hence, the correct answer is option: 'B'.