Question

If $A={\cos }^2\theta +{\sin }^4\theta$, then for all values of $\theta$

• A. $1\le A\le 2$
• B. $\frac{13}{16}\le A\le 1$
• C. $\frac{3}{4}\le A\le \frac{13}{16}$
• D. $\frac{3}{4}\le A\le 1$

Answer 1

The correct option is D $\frac{3}{4}\le A\le 1$

We have, $A={\cos }^2\theta +{\sin }^2\theta {\sin }^2\theta$

$\Rightarrow A\le {\cos }^2\theta +{\sin }^2\theta .1(\because {\sin }^2\theta \le 1)$

$\Rightarrow A\le 1$

Again, $A=(1-{\sin }^2\theta )+{\sin }^4\theta \Rightarrow A={({\sin }^2\theta -\frac{1}{2})}^2+(1-\frac{1}{4})\Rightarrow A\ge \frac{3}{4}$

Hence $\frac{3}{4}\le A\le 1$.