Question

If $f\left(x\right)={\sin }^{6}x+{\cos }^{6}x$ then range of$f\left(x\right)$ is

• A. $[\frac{1}{4},1]$
• B. $[\frac{1}{4},\frac{3}{4}]$
• C. $[\frac{3}{4},1]$
• D. none of these

Answer 1

Answer: (A)

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

$f\left(x\right)$ $=$ $si{n}^{6}x+co{s}^{6}x$ $=$ $(si{n}^{2}x{)}^3+(co{s}^{2}x{)}^3$

$(a+b{)}^3=a^3+3ab(a+b)+b^3$

$=$ $(si{n}^{2}x+co{s}^{2}x{)}^3-3si{n}^{2}xco{s}^{2}x(si{n}^{2}x+co{s}^{2}x)$

$=$ $1-3si{n}^{2}x.co{s}^{2}x$ $=$ $1-\frac{3}{4}(2sinx.cosx{)}^2$

$=$ $1-\frac{3}{4}si{n}^{2}2x$

now, $si{n}^{2}2x$ $ϵ$ $[0,1]$, so Range of $f\left(x\right)$ will be [$\frac{1}{4}$, 1]

Hence, option 'A' is correct.