Question:
Let $f_{k}(x)=\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right)$ for $k=1,2$,
3 ,.... Then for all $x \in R$, the value of $\mathrm{f}_{4}(\mathrm{x})-\mathrm{f}_{6}(\mathrm{x})$ is equal to :-
Correct Option: , 4
Solution:
$f_{4}(x)-f_{6}(x)$
$=\frac{1}{4}\left(\sin ^{4} x+\cos ^{4} x\right)-\frac{1}{6}\left(\sin ^{6} x+\cos ^{6} x\right)$
$=\frac{1}{4}\left(1-\frac{1}{2} \sin ^{2} 2 x\right)-\frac{1}{6}\left(1-\frac{3}{4} \sin ^{2} 2 x\right)=\frac{1}{12}$
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