JEE Main 2026 · 4 Apr · Shift 2 · Mathematics Q12

MathematicsSingle correct+4 / -1Question 12 of 75

$\displaystyle\max_{0\le x\le\pi}\left(16\sin\left(\frac x2\right)\cos^3\left(\frac x2\right)\right)$ is equal to:

\dfrac{3\sqrt3}2

3\sqrt3

4\sqrt3

6\sqrt3

Answer and worked solution

Official answer: B

Substitute $t=\frac x2\in[0,\frac\pi2]$ and maximise $\sin t\cos^3t$ .

Stationary point. The derivative is $\cos^2t(\cos^2t-3\sin^2t)$ , zero at $t=\frac\pi6$ , giving $\frac12\cdot\frac{3\sqrt3}8$ .

Maximum. $16\cdot\frac{3\sqrt3}{16}=3\sqrt3$ .

(B)

Topics:Maxima and minima Trigonometric ratios and identities

JEE Main 2026 · 4 Apr · Shift 2, Mathematics question 12