Single Choice

Let $$ f_{k}(x)=\displaystyle \frac{1}{k}(\sin ^{k}x+\cos ^{k}x)$$ where $$ x\in R$$ and $$ k\geq 1$$. Then $$ f_{4}(x)-f_{6}(x)$$ equals

A$$ \displaystyle \frac{1}{6}$$
B$$ \displaystyle \frac{1}{3}$$
C$$ \displaystyle \frac{1}{4}$$
D$$ \displaystyle \frac{1}{12}$$
Correct Answer

Solution

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


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