Question

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

• A. $\frac{1}{2}\le f\left(x\right)\le 1$
• B. $\frac{1}{\sqrt{2}}\le f\left(x\right)\le 1$
• C. $\frac{1}{4}\le f\left(x\right)\le 1$
• D. $\frac{1}{2\sqrt{2}}\le f\left(x\right)\le 1$

Answer 1

The correct option is C $\frac{1}{4}\le f\left(x\right)\le 1$

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

We know that $0\le {\sin }^{2}2x\le 1$
$\Rightarrow -\frac{3}{4}\le -\frac{3}{4}{\sin }^{2}2x\le 0$
$\Rightarrow \frac{1}{4}\le f\left(x\right)\le 1$