Question

Given $A={\sin }^2\theta +{\cos }^4\theta ,$ then for all real $\theta$, which of the following is true?

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

Answer 1

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

$A={\sin }^2\theta +{\cos }^4\theta ={\sin }^2\theta +{\cos }^2\theta .{\cos }^2\theta \le {\sin }^2\theta +{\cos }^2\theta =1$ ....... $(\because {\cos }^2\theta \le 1)$

$\Rightarrow A\le 1$

And ${\sin }^2\theta +{\cos }^4\theta =1-{\cos }^2\theta +{\cos }^4\theta$

$=1+({\cos }^4\theta -{\cos }^2\theta )=1+{({\cos }^2\theta -\frac{1}{2})}^2-\frac{1}{4}$

$=\frac{3}{4}+{({\cos }^2\theta -\frac{1}{2})}^2\ge \frac{3}{4}$

$\Rightarrow \frac{3}{4}\le A\le 1$