Question

If $p={\sin }^{2}x+{\cos }^{4}x$, then

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

Answer 1

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

Given, $p={\sin }^{2}x+{\cos }^{4}x$
$={\sin }^{2}x+{\cos }^{2}x(1-{\sin }^{2}x)$
$=({\sin }^{2}x+{\cos }^{2}x)-{\sin }^{2}x{\cos }^{2}x$
$=1-{\sin }^{2}x{\cos }^{2}x\le 1....\left(i\right)$
Again, $p=1-{\cos }^{2}x+{\cos }^{4}x$
$={({\cos }^{2}x-\frac{1}{2})}^2+\frac{3}{4}$
$\Rightarrow p\ge \frac{3}{4}....\left(ii\right)$
$\therefore$ From Eqs. (i) and (ii), we get
$\frac{3}{4}\le p\le 1$.