Q. If $P = \frac{1}{2}\sin^{2}{\theta + \frac{1}{3}\cos^{2}\theta}$, then
A) $\frac{1}{3} \leq P \leq \frac{1}{2}$
B) $P \geq \frac{1}{2}$
C) $2 \leq P \leq 3$
D) $- \frac{\sqrt{13}}{6} \leq P \leq \frac{\sqrt{13}}{6}$
Explanation
(a)
Given, $P = \frac{1}{2}\sin^{2}\theta + \frac{1}{3}\cos^{2}\theta$
$= \frac{1}{2}\ \left( 1 - \cos^{2}\theta \right) + \frac{1}{3}\cos^{2}\theta$
$= \frac{1}{2} - \frac{1}{6}\cos^{2}\theta$
Since, $0 \leq \cos^{2}\theta \leq 1$
$\Rightarrow \ \ - \frac{1}{6} \leq - \frac{1}{6}\cos^{2}\theta \leq 0$
$\Rightarrow \ \frac{1}{3} \leq \frac{1}{2} - \frac{1}{6}\cos^{2}\theta \leq \frac{1}{2}$
$\Rightarrow \ \frac{1}{3} \leq P \leq \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.