Question

The range of $f\left(x\right)={\sin }^{2}x+{\cos }^{4}x$ is

• A. $\left[\frac{1}{2},1\right]$
• B. $\left[\frac{3}{4},1\right]$
• C. $[0,1]$
• D. $[0,\frac{1}{4}]$

Answer 1

The correct option is B $\left[\frac{3}{4},1\right]$

$f\left(x\right)={\sin }^{2}x+{(1-{\sin }^{2}x)}^2$

$=1+{\sin }^{2}x-2{\sin }^{2}x+{\sin }^{4}x$

$=1-{\sin }^{2}x(1-{\sin }^{2}x)$

$=1-{\sin }^{2}xco{s}^{2}x$

$=1-{\left(\frac{\sin 2x}{2}\right)}^2$

$\sin 2x\in [-1,1]$

$\Rightarrow {\sin }^{2}2x\in [0,1]$

$\Rightarrow {\left[\frac{\sin 2x}{2}\right]}^2\in [0,\frac{1}{4}]$

$\Rightarrow 0\le {\left[\frac{\sin 2x}{2}\right]}^2\le \frac{1}{4}$

$\Rightarrow 0\ge -{\left[\frac{\sin 2x}{2}\right]}^2\ge -\frac{1}{4}$

$\Rightarrow 1\ge 1-{\left[\frac{\sin 2x}{2}\right]}^2\ge 1-\frac{1}{4}$

$\Rightarrow 1\ge f\left(x\right)\ge \frac{3}{4}$

$\Rightarrow f\left(x\right)$ lies in $[\frac{3}{4},1]$