Question

If $P=\frac{1}{2}{\sin }^2\theta +\frac{1}{3}{\cos }^2\theta$, then

• A. $\frac{1}{3}\le P\le \frac{1}{2}$
• B. $P\ge \frac{1}{2}$
• C. $2\le P\le 3$
• D. $-\frac{\sqrt{13}}{6}\le P\le \frac{\sqrt{13}}{6}$

Answer 1

Answer: (A)

$\frac{1}{3}\le P\le \frac{1}{2}$

Given, $P=\frac{1}{2}{\sin }^2\theta +\frac{1}{3}{\cos }^2\theta$
$=\frac{1}{2}(1-{\cos }^2\theta )+\frac{1}{3}{\cos }^2\theta$
$=\frac{1}{2}-\frac{1}{6}{\cos }^2\theta$
Since, $0\le {\cos }^2\theta \le 1$
$\Rightarrow -\frac{1}{6}\le -\frac{1}{6}{\cos }^2\theta \le 0$
$\Rightarrow \frac{1}{3}\le \frac{1}{2}-\frac{1}{6}{\cos }^2\theta \le \frac{1}{2}$
$\Rightarrow \frac{1}{3}\le P\le \frac{1}{2}$