Question

If $y={\sin }^{2}x+{\cos }^{4}x\forall x\in R,$ then

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

Answer 1

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

$y={\sin }^{2}x+{\cos }^{4}x....\left(1\right)y={\sin }^{2}x+(1-{\sin }^{2}x{)}^2={\sin }^{2}x+1-2{\sin }^{2}x+{\sin }^{4}x=1-{\sin }^{2}x+{\sin }^{4}x....(2)$
Adding $\left(1\right)$ and $\left(2\right),$ we have
$2y=1+{\sin }^{4}x+{\cos }^{4}x=1+({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x=1+1-\frac{{\sin }^{2}2x}{2}2y=2-\frac{{\sin }^{2}2x}{2}y=1-\frac{{\sin }^{2}2x}{4}$
Max. value $=1$
Min. value $=\frac{3}{4}$

Alternate Solution:
$y={\sin }^{2}x+(1-{\sin }^{2}x{)}^2={\sin }^{2}x+1-2{\sin }^{2}x+{\sin }^{4}x=1-{\sin }^{2}x+{\sin }^{4}x$
$={({\sin }^{2}x-\frac{1}{2})}^2+\frac{3}{4}$

Now, $0\le {\sin }^{2}x\le 1$
$\Rightarrow -\frac{1}{2}\le {\sin }^{2}x-\frac{1}{2}\le \frac{1}{2}$
$\Rightarrow 0\le {({\sin }^{2}x-\frac{1}{2})}^2\le \frac{1}{4}$
$\Rightarrow \frac{3}{4}\le {({\sin }^{2}x-\frac{1}{2})}^2+\frac{3}{4}\le \frac{1}{4}+\frac{3}{4}$
$\Rightarrow \frac{3}{4}\le y\le 1$