Application of DerivativesmediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Maximum of 16 sin(x/2)cos³(x/2) = 3√3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
max⁡0≤x≤π(16sin⁡(x2)cos⁡3(x2))\displaystyle\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
A332\dfrac{3\sqrt3}{2}
B333\sqrt3correct
C434\sqrt3
D636\sqrt3
Solution
Step 1: Simplify:
E=16sin⁡x2cos⁡3x2=8sin⁡xcos⁡2x2=8sin⁡x⋅1+cos⁡x2=4sin⁡x(1+cos⁡x).E=16\sin\frac x2\cos^3\frac x2=8\sin x\cos^2\frac x2=8\sin x\cdot\frac{1+\cos x}{2}=4\sin x(1+\cos x).
 Step 2: Differentiate:
dEdx=4[cos⁡x+cos⁡2x]=8cos⁡3x2cos⁡x2=0 ⇒ cos⁡3x2=0 or cos⁡x2=0.\frac{dE}{dx}=4[\cos x+\cos2x]=8\cos\frac{3x}{2}\cos\frac x2=0\ \Rightarrow\ \cos\frac{3x}{2}=0\ \text{or}\ \cos\frac x2=0.
 So the critical points in [0,π][0,\pi] are x=π3x=\dfrac{\pi}{3} and x=πx=\pi. Step 3: Evaluate: E(0)=0, E(π)=0E(0)=0,\ E(\pi)=0, and
E(π3)=4sin⁡π3(1+cos⁡π3)=4⋅32⋅32=33.E\left(\frac{\pi}{3}\right)=4\sin\frac{\pi}{3}\left(1+\cos\frac{\pi}{3}\right)=4\cdot\frac{\sqrt3}{2}\cdot\frac32=3\sqrt3.
 Step 4: Maximum value =33=3\sqrt3. Correct answer: (2)
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