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# Find the maximum and minimum value of $\sin^6x+\cos^6x$

$\begin{array}{1 1}(a)\;1,0&(b)\;1,\large\frac{1}{4}\\(c)\;0,1&(d)\;2,1\end{array}$

Let $Y=\sin^6x+\cos^6x$
$\qquad=(\sin^2x)^3+(\cos^2x)^3$
$\qquad=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)$
$\qquad=1-3\sin^2x\cos^2x$
$\qquad=1-\large\frac{3}{4}$$.4\sin^2x\cos^2x \qquad=1-\large\frac{3}{4}$$(\sin 2x)^2$
When $(\sin 2x)^2$ is minimum (i.e) 0 then y will be maximum and when $(\sin 2x)^2$ is maximum (i.e) 1 then $y$ will be minimum.
$Y_{max}=1-\large\frac{3}{4}$$\times 0=1 Y_{min}=1-\large\frac{3}{4}$$\times 1=\large\frac{1}{4}$
Hence (b) is the correct answer.
edited Mar 22, 2014