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

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Question

(A) $f_4(x) = \frac{\sin^4 x + \cos^4 x}{4} = \frac{(\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x}{4} = \frac{1 - 2\sin^2 x \cos^2 x}{4} = \frac{1}{4}

Vedclass · Accepted Answer

$\frac{1}{12}$

Vedclass · Answer

(A) $f_4(x) = \frac{\sin^4 x + \cos^4 x}{4} = \frac{(\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x}{4} = \frac{1 - 2\sin^2 x \cos^2 x}{4} = \frac{1}{4} - \frac{1}{2}\sin^2 x \cos^2 x$$f_6(x) = \frac{\sin^6 x + \cos^6 x}{6} = \frac{(\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)}{6} = \frac{1 - 3\sin^2 x \cos^2 x}{6} = \frac{1}{6} - \frac{1}{2}\sin^2 x \cos^2 x$$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)$$f_4(x) - f_6(x) = \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12}$

मान लीजिए $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ जहाँ $k = 1, 2, 3, ...$ है। तो सभी $x \in R$ के लिए,$f_4(x) - f_6(x)$ का मान किसके बराबर है?

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