Let {f_k}(x) = frac{1}{k}(sin^k x + cos^k x) | vedclass

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(A) Given ${f_k}(x) = \frac{1}{k}(\sin^k x + \cos^k x)$.For $k=4$,${f_4}(x) = \frac{\sin^4 x + \cos^4 x}{4}$.Using the identity $\sin^4 x + \cos^4 x =

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$\frac{1}{12}$

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(A) Given ${f_k}(x) = \frac{1}{k}(\sin^k x + \cos^k x)$.For $k=4$,${f_4}(x) = \frac{\sin^4 x + \cos^4 x}{4}$.Using the identity $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$,we get:${f_4}(x) = \frac{1 - 2\sin^2 x \cos^2 x}{4} = \frac{1}{4} - \frac{1}{2}\sin^2 x \cos^2 x$.For $k=6$,${f_6}(x) = \frac{\sin^6 x + \cos^6 x}{6}$.Using the identity $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 - 3\sin^2 x \cos^2 x$,we get:${f_6}(x) = \frac{1 - 3\sin^2 x \cos^2 x}{6} = \frac{1}{6} - \frac{1}{2}\sin^2 x \cos^2 x$.Now,${f_4}(x) - {f_6}(x) = (\frac{1}{4} - \frac{1}{2}\sin^2 x \cos^2 x) - (\frac{1}{6} - \frac{1}{2}\sin^2 x \cos^2 x) = \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12}$.

Let ${f_k}(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ for $k = 1, 2, 3, ...$. Then for all $x \in R$,the value of $f_4(x) - f_6(x)$ is equal to

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