Question

The function $f:R\to R$ is defined by $f\left(x\right)=co{s}^{2}x+sin64x$. Then, f(R) =

• A. $[\frac{3}{4},1)$
• B. $(\frac{3}{4},1]$
• C. $[\frac{3}{4},1]$
• D. $(\frac{3}{4},1)$

Answer 1

The correct option is C

$[\frac{3}{4},1]$

Given:
$\Rightarrow f\left(x\right)=co{s}^{2}x+si{n}^{4}x\Rightarrow f\left(x\right)=1-si{n}^{2}x+si{n}^{4}x\Rightarrow f\left(x\right)={(si{n}^{2}x-\frac{1}{2})}^2+\frac{3}{4}$
The minimum value of f(x) is $\frac{3}{4}$
Also,
$si{n}^{2}x\le 1\Rightarrow si{n}^{2}x-\frac{1}{2}\le \frac{1}{2}\Rightarrow {(si{n}^{2}x-\frac{1}{2})}^2\le \frac{1}{4}\Rightarrow {(si{n}^{2}x-\frac{1}{2})}^2+\frac{3}{4}\le \frac{1}{4}+\frac{3}{4}\Rightarrow f\left(x\right)\le 1$
The maximum value of f(x) is 1.