If K = sin^6x + cos^6x, then K belongs to the | vedclass

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Question

$\mathrm{K}=\left(\sin ^{2} \mathrm{x}+\cos ^{2} \mathrm{x}\right)^{3}-3 \sin ^{2} \mathrm{x} \cos ^{2} \mathrm{x}\left(\sin ^{2} \mathrm{x}+\cos ^{2}

Vedclass · Accepted Answer

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

Vedclass · Answer

$\mathrm{K}=\left(\sin ^{2} \mathrm{x}+\cos ^{2} \mathrm{x}\right)^{3}-3 \sin ^{2} \mathrm{x} \cos ^{2} \mathrm{x}\left(\sin ^{2} \mathrm{x}+\cos ^{2} \mathrm{x}\right)$

$=(1)^{3}-3 \sin ^{2} x \cos ^{2} x(1)$

$=1-\frac{3}{4} \sin ^{2} 2 x$

Now, $0 \leq \sin ^{2} 2 x \leq 1 \quad \Rightarrow \quad 0 \leq \frac{3}{4} \sin ^{2} 2 x \leq \frac{3}{4}$

$\Rightarrow \quad-\frac{3}{4} \leq-\frac{3}{4} \sin ^{2} 2 x \leq 0$

$\Rightarrow \quad 1-\frac{3}{4} \leq 1-\frac{3}{4} \sin ^{2} 2 x \leq 1$

$\Rightarrow \quad \frac{1}{4} \leq 1-\frac{3}{4} \sin ^{2} 2 x \leq 1$

$\Rightarrow \quad \frac{1}{4} \leq \mathrm{K} \leq 1 \Rightarrow \mathrm{K} \in\left[\frac{1}{4}, 1\right]$

If $K = sin^6x + cos^6x$, then $K$ belongs to the interval

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