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

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(C) We know that $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$.Let $a = \sin^2x$ and $b = \cos^2x$.Then $K = (\sin^2x + \cos^2x)^3 - 3\sin^2x\cos^2x(\sin^2x + \cos

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

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(C) We know that $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$.Let $a = \sin^2x$ and $b = \cos^2x$.Then $K = (\sin^2x + \cos^2x)^3 - 3\sin^2x\cos^2x(\sin^2x + \cos^2x)$.Since $\sin^2x + \cos^2x = 1$,we have $K = 1 - 3\sin^2x\cos^2x$.Multiplying and dividing by $4$,we get $K = 1 - \frac{3}{4}(4\sin^2x\cos^2x) = 1 - \frac{3}{4}(\sin 2x)^2$.Since $0 \leq \sin^2 2x \leq 1$,we have $0 \leq \frac{3}{4}\sin^2 2x \leq \frac{3}{4}$.Multiplying by $-1$,we get $-\frac{3}{4} \leq -\frac{3}{4}\sin^2 2x \leq 0$.Adding $1$ to all parts,we get $1 - \frac{3}{4} \leq 1 - \frac{3}{4}\sin^2 2x \leq 1 + 0$.Thus,$\frac{1}{4} \leq K \leq 1$.Therefore,$K \in [\frac{1}{4}, 1]$.

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

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