Question

If $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x-\frac{1}{2}\sin 2x,$ then the range of $f\left(x\right)$ is

• A. $[0,\frac{3}{2}]$
• B. $[-\frac{1}{2},\frac{7}{2}]$
• C. $[0,\frac{9}{8}]$
• D. $[\frac{3}{4},\frac{7}{8}]$

Answer 1

The correct option is C $[0,\frac{9}{8}]$

Given : $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x-\frac{1}{2}\sin 2x$
$\Rightarrow f\left(x\right)=1-2{\sin }^{2}x{\cos }^{2}x-\frac{1}{2}\sin 2x\Rightarrow f\left(x\right)=1-\frac{{\sin }^{2}2x}{2}-\frac{\sin 2x}{2}\Rightarrow f\left(x\right)=-\frac{{\sin }^{2}2x+\sin 2x-2}{2}\Rightarrow f\left(x\right)=-\frac{{(\sin 2x+\frac{1}{2})}^2-\frac{9}{4}}{2}$

We know that,
$-1\le \sin 2x\le 1\Rightarrow -\frac{1}{2}\le \sin 2x+\frac{1}{2}\le \frac{3}{2}\Rightarrow 0\le {(\sin 2x+\frac{1}{2})}^2\le \frac{9}{4}\Rightarrow -\frac{9}{8}\le \frac{{(\sin 2x+\frac{1}{2})}^2-\frac{9}{4}}{2}\le 0\Rightarrow 0\le \frac{-{(\sin 2x+\frac{1}{2})}^2+\frac{9}{4}}{2}\le \frac{9}{8}\therefore f\left(x\right)\in [0,\frac{9}{8}]$