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. $p\le \frac{1}{2}$
• C. $2\le p\le 3$
• D. $-\frac{\sqrt{13}}{6}\le p\le \frac{\sqrt{13}}{6}$

Answer 1

The correct option is A

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

Explanation for the correct option:

Step 1: Apply this formula ${\cos }^2\theta =1-{\sin }^2\theta$:

We have; $p=\frac{1}{2}{\sin }^2\theta +\frac{1}{3}{\cos }^2\theta$

$\begin{array}{rcl}p& =& \frac{1}{2}{\sin }^2\theta +\frac{1}{3}(1-{\sin }^2\theta )\\ & =& \frac{1}{2}{\sin }^2\theta +\frac{1}{3}-\frac{{\sin }^2\theta }{3}\\ & =& \frac{1}{3}+\left[\frac{1}{2}-\frac{1}{3}\right]{\sin }^2\theta \\ & =& \frac{1}{3}+\frac{1}{6}{\sin }^2\theta \end{array}$

Which gives, $p\ge \frac{1}{3}$

Step 2: converting ${\sin }^2\theta$ in terms of ${\cos }^2\theta$ we have:

$$\begin{gathered} p=\frac{1}{2}(1-{\cos }^2\theta )+\frac{1}{3}{\cos }^2\theta \\ =\frac{1}{2}+[-\frac{1}{2}+\frac{1}{3}]{\cos }^2\theta \\ =\frac{1}{2}-\frac{1}{6}{\cos }^2\theta \end{gathered}$$

Hence, $p\le \frac{1}{2}$

Therefore, $\frac{1}{3}\le p\le \frac{1}{2}$

Hence, option (A) is the correct option.