Trigonometrical Ratios and Identities 1 Question 5
5. Let $f _k(x)=\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right)$ for $k=1,2,3 \ldots$. Then, for all $x \in R$, the value of $f _4(x)-f _6(x)$ is equal to
(a) $\frac{1}{12}$
(b) $\frac{5}{12}$
(c) $\frac{-1}{12}$
(d) $\frac{1}{4}$
(2019 Main, 11 Jan I)
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Answer:
Correct Answer: 5. (b)
Solution:
- We have,
$$ \begin{aligned} f _k(x) & =\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right), k=1,2,3, \ldots \\ \therefore f _4(x) & =\frac{1}{4}\left(\sin ^{4} x+\cos ^{4} x\right) \\ & =\frac{1}{4}\left(\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x\right) \\ & =\frac{1}{4} 1-\frac{1}{2}(\sin 2 x)^{2}=\frac{1}{4}-\frac{1}{8} \sin ^{2} 2 x \end{aligned} $$
and $f _6(x)=\frac{1}{6}\left(\sin ^{6} x+\cos ^{6} x\right)$
$$ =\frac{1}{6}{\left(\sin ^{2} x+\cos ^{2} x\right)^{3}-3 \sin ^{2} x \cos ^{2} x\right. $$
$\left.\left(\sin ^{2} x+\cos ^{2} x\right) }$
$=\frac{1}{6} 1-\frac{3}{4}(2 \sin x \cos x)^{2}=\frac{1}{6}-\frac{1}{8} \sin ^{2} 2 x$
Now, $f _4(x)-f _6(x)=\frac{1}{4}-\frac{1}{6}=\frac{3-2}{12}=\frac{1}{12}$
