Question

The maximum value of \[ E = 16\sin\frac{x}{2}\cos^3\frac{x}{2} \] where \(x\in[0,\pi]\), is

• A. \(3\)
• B. \(3\sqrt{3}\)
• C. \(6\sqrt{3}\)
• D. \(6\)

Hint:

When trigonometric powers appear, reduce them using identities before differentiating.

Answer 1

The Correct Option is B

Concept: Use trigonometric identities to simplify the expression and then find the critical points using differentiation.
Step 1: Simplify the expression \[ E = 16\sin\frac{x}{2}\cos^3\frac{x}{2} \] \[ E = 16\sin\frac{x}{2}\cos\frac{x}{2}\cos^2\frac{x}{2} \] Using identity: \[ 2\sin\frac{x}{2}\cos\frac{x}{2}=\sin x \] \[ E = 8\sin x \cos^2\frac{x}{2} \] Now, \[ \cos^2\frac{x}{2}=\frac{1+\cos x}{2} \] \[ E = 8\sin x \cdot \frac{1+\cos x}{2} \] \[ E = 4\sin x(1+\cos x) \]
Step 2: Differentiate \[ \frac{dE}{dx}=4[\cos x(1+\cos x)-\sin^2 x] \] \[ \frac{dE}{dx}=4[\cos x+\cos^2x-(1-\cos^2x)] \] \[ \frac{dE}{dx}=4[\cos x+2\cos^2x-1] \] This simplifies to \[ \frac{dE}{dx}=4(\cos x+\cos2x) \]
Step 3: Find critical points \[ \cos x+\cos2x=0 \] \[ 2\cos\frac{3x}{2}\cos\frac{x}{2}=0 \] Thus, \[ \cos\frac{3x}{2}=0 \quad \text{or} \quad \cos\frac{x}{2}=0 \] Hence, \[ x=\frac{\pi}{3},\;\pi \]
Step 4: Evaluate function \[ E(0)=0 \] \[ E(\pi)=0 \] \[ E\left(\frac{\pi}{3}\right)=4\sin\frac{\pi}{3}\left(1+\cos\frac{\pi}{3}\right) \] \[ =4\cdot\frac{\sqrt3}{2}\cdot\frac{3}{2} \] \[ =3\sqrt3 \] Thus maximum value: \[ \boxed{3\sqrt3} \]