Question

If $f\left(x\right)={\sin }^{6}x+{\cos }^{6}x,x\in R,$ then $f\left(x\right)$ lies in the interval

• A. $[\frac{1}{2},1]$
• B. $[\frac{1}{2},\frac{5}{8}]$
• C. $[\frac{1}{4},1]$
• D. $[\frac{7}{8},1]$

Answer 1

The correct option is C $[\frac{1}{4},1]$

$f\left(x\right)={\sin }^{6}x+{\cos }^{6}x$
$=({\sin }^{2}x{)}^3+({\cos }^{2}x{)}^3$
$=({\sin }^{2}x+{\cos }^{2}x{)}^3-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)$
$=1-3{\sin }^{2}x{\cos }^{2}x$
$=1-3{\sin }^{2}x(1-{\sin }^{2}x)$
$=3{\sin }^{4}x-3{\sin }^{2}x+1$
$=3[{({\sin }^{2}x-\frac{1}{2})}^2+\frac{1}{12}]$

For $x\in R,$
$0\le {\sin }^{2}x\le 1$
$\Rightarrow -\frac{1}{2}\le {\sin }^{2}x-\frac{1}{2}\le \frac{1}{2}$
$\Rightarrow 0\le {({\sin }^{2}x-\frac{1}{2})}^2\le \frac{1}{4}$
$\Rightarrow \frac{1}{12}\le {({\sin }^{2}x-\frac{1}{2})}^2+\frac{1}{12}\le \frac{1}{3}$
$\Rightarrow \frac{1}{4}\le 3[{({\sin }^{2}x-\frac{1}{2})}^2+\frac{1}{12}]\le 1$

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

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