JEE Main 2026 (Online) 4th April Evening ShiftQuestion 1
$\max_{0\le x\le\pi}\left(16\sin\left(\frac{x}{2}\right)\cos^{3}\left(\frac{x}{2}\right)\right)$ is equal to:
Differentiation and its Application
Solution
Concept. Put $t=\frac x2\in\left[0,\frac\pi2\right]$ and maximise $h(t)=16\sin t\cos^3t$. Formulas used. (1) $\dfrac{d}{dt}\left(\sin t\cos^3t\right)=\cos^4t-3\sin^2t\cos^2t$Step 1 - Critical point. By (1), $$h'(t)=16\cos^2t\left(\cos^2t-3\sin^2t\right)=0.$$ Inside $\left(0,\frac\pi2\right)$, $\cos t\ne0$, so $\tan^2t=\frac13$ and $t=\frac\pi6$. Step 2 - Check. $h'>0$ for $t<\frac\pi6$ and $h'<0$ after it, so $t=\frac\pi6$ is the maximum. At the end points $h(0)=h\left(\frac\pi2\right)=0$. Step 3 - Value. $$h\left(\tfrac\pi6\right)=16\cdot\frac12\cdot\left(\frac{\sqrt3}{2}\right)^3=16\cdot\frac{3\sqrt3}{16}=3\sqrt3.$$ Answer: (B)