If f(x) = sin^6 x + cos^6 x for x in R,then f | vedclass

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(C) Given,$f(x) = \sin^6 x + \cos^6 x$.We can write this as $f(x) = (\sin^2 x)^3 + (\cos^2 x)^3$.Using the identity $a^3 + b^3 = (a+b)(a^2 + b^2 - ab)

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

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(C) Given,$f(x) = \sin^6 x + \cos^6 x$.We can write this as $f(x) = (\sin^2 x)^3 + (\cos^2 x)^3$.Using the identity $a^3 + b^3 = (a+b)(a^2 + b^2 - ab)$,we get:$f(x) = (\sin^2 x + \cos^2 x)(\sin^4 x + \cos^4 x - \sin^2 x \cos^2 x)$.Since $\sin^2 x + \cos^2 x = 1$,this simplifies to:$f(x) = \sin^4 x + \cos^4 x - \sin^2 x \cos^2 x$.Adding and subtracting $2\sin^2 x \cos^2 x$,we get:$f(x) = (\sin^2 x + \cos^2 x)^2 - 3\sin^2 x \cos^2 x = 1 - 3\sin^2 x \cos^2 x$.Multiplying and dividing by $4$,we get $f(x) = 1 - \frac{3}{4}(2\sin x \cos x)^2 = 1 - \frac{3}{4}\sin^2(2x)$.Since $0 \leq \sin^2(2x) \leq 1$,we have:$1 - \frac{3}{4}(1) \leq f(x) \leq 1 - \frac{3}{4}(0)$.$\frac{1}{4} \leq f(x) \leq 1$.Thus,$f(x) \in \left[\frac{1}{4}, 1\right]$.

If $f(x) = \sin^6 x + \cos^6 x$ for $x \in R$,then $f(x)$ lies in the interval

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