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Let `f_k(x) = 1/k(sin^k x + cos^k x)` where `x in RR` and `k gt= 1.` Then `f_4(x) - f_6(x)` equalsA. `1/4`B. `1/12`C. `1/6`D. `1/3` |
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Answer» Correct Answer - B We have, `f_(k)(x)=1/k(sin^(k)x+cos^(k)x)` `thereforef_(4)(x)-f_(6)(x)` `=1/4(sin^(4)x+cos^(4)x)-1/6(sin^(6)x+cos^(6)x)` `=1/4{(sin^(2)x+cos^(2)x)^(2)-2sin^(2)xcos^(2)x}` `-1/6{(sin^(2)x+cos^(2)x)^(3)-3sin^(2)x cos^(2)(sin^(2)x+cos^(2)x)}` `=1/4(1-2sin^(2)xcos^(2)x)-1/6(1-3sin^(2)xcos^(2)x)=1/12` |
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