Applications of Derivatives
Trigonometric Maxima/Minima on Closed Interval
nta_pyq_2023_apr
Grade 12
Question:
$\max_{0\leq x\leq\pi}\left\{x-2\sin x\cos x+\dfrac{1}{3}\sin 3x\right\}=$
$\dfrac{\pi+2-3\sqrt{3}}{6}$
$\pi$
$0$
$\dfrac{5\pi+2+3\sqrt{3}}{6}$
Step-by-Step Solution
Key Concept: $f'(x)=1-2\cos 2x+\cos 3x$. Expand using trig identities and factor to find critical points $x=0,\ \frac{\pi}{6},\ \frac{5\pi}{6}$.
$f'(x)=1-2\cos 2x+\cos 3x=(2\cos x+\sqrt{3})(2\cos x-\sqrt{3})(\cos x-1)=0$. Critical points: $x=0,\frac{\pi}{6},\frac{5\pi}{6}$. Maximum at $x=\frac{5\pi}{6}$: $f\!\left(\frac{5\pi}{6}\right)=\frac{5\pi+2+3\sqrt{3}}{6}$.
Correct Answer: 4