JEE Main 2026 Apr 4 Shift 2, Mathematics Q12: Maxima & Minima

Use $2\sin (x/2)\cos (x/2)=\sin x$ and $2{\cos }^2(x/2)=1+\cos x$:

$E(x)=16\sin (x/2){\cos }^3(x/2)=8\cdot [2\sin (x/2)\cos (x/2)]\cdot {\cos }^2(x/2)=8\sin x\cdot \frac{1+\cos x}{2}=4\sin x(1+\cos x)$.

$\frac{dE}{dx}=4[\cos x(1+\cos x)-{\sin }^{2}x]=4[\cos x+{\cos }^{2}x-(1-{\cos }^{2}x)]=4(2{\cos }^{2}x+\cos x-1)$.

Setting this to zero: $(2\cos x-1)(\cos x+1)=0$, so $\cos x=1/2$ (i.e. $x=\pi /3$) or $\cos x=-1$ (i.e. $x=\pi$).

Evaluate: $E(0)=0$, $E(\pi )=0$, $E(\pi /3)=4\cdot \frac{\sqrt{3}}{2}\cdot \frac{3}{2}=3\sqrt{3}$.

Max $=3\sqrt{3}$.