Question

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

• A. $1\le A\le 2$
• B. $\frac{3}{4}\le A\le 1$
• C. $\frac{13}{16}\le A\le 1$
• D. None of these

Answer 1

Answer: (B)

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

$A={\cos }^2\theta +{\sin }^4\theta$

$A={\cos }^2\theta +{\sin }^2\theta {\sin }^2\theta \le {\cos }^2\theta +{\sin }^2\theta$

$A={\cos }^2\theta +{\sin }^2\theta {\sin }^2\theta \le 1$ $({\sin }^2\theta +{\cos }^2\theta =1)$. ----$\left(1\right)$

or $A={\cos }^2\theta +{\sin }^4\theta$

$A=(1-{\sin }^2\theta )+{\sin }^4\theta$

$A=({\sin }^2\theta {)}^2+(\frac{1}{2}{)}^2-2\times \frac{1}{2}\times {\sin }^2\theta +(1-\frac{1}{4})$

$A=({\sin }^2\theta -\frac{1}{2}{)}^2+\frac{3}{4}\ge \frac{3}{4}$ ----------$\left(2\right)$

From equation $1$ and $2$,

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

so option $B$ will be the correct answer