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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)/(12)`B. `(5)/(12)`C. `(-1)/(12)`D. `(1)/(4)` |
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Answer» Correct Answer - A We have, `f_(k)(x) = (1)/(k)(sin^(h)x+ cos^(k)x), k = 1, 2, 3, ...` `therefore f_(4)(x) = (1)/(4)(sin^(4)x + cos^(4) x)` `=(1)/(4)((sin^(2)x + cos^(2)x)^(2)-2sin^(2)xcos^(2)x)` `=(1)/(4)(1-(1)/(2)(sin2x)^(2))=(1)/(4)-(1)/(8)sin^(2)2x` and `f_(6)(x)=(1)/(6)(sin^(6)x + cos^(6)x)` `=(1)/(6){(sin^(2)x+cos^(2)x)^(3)-3sin^(2) x cos^(2)x (sin^(2)x + cos^(2)x)}` `=(1)/(6){1-(3)/(4)(2sin x cos x)^(2)}=(1)/(6)-(1)/(8)sin^(2)2x` Now, `f_(4)(x) - f_(6)(x)=(1)/(4)-(1)/(6)=(3-2)/(12)=(1)/(12)` |
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