Question

Let $f_k\left(x\right)=\frac{1}{x}({\sin }^{k}x+{\cos }^{k}x)$ where $x\in R\text{and}k\ge 1$, then $f_4\left(x\right)-f_6\left(x\right)$ equals

• A. $\frac{1}{4}$
• B. $\frac{1}{12}$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{1}{12}$

Given, $1/4(\sin 4x-\cos 4x)-1/6(\sin 6x-\cos 6x)$

$=\left[3\right(\sin 4x-\cos 4x)-2(\sin 6x-\cos 6x\left)\right]/12$

$=\left[3\right(1-2\sin 2x\cos 2x)-2(1-3\sin 2x\cos 2x\left)\right]/12$

Hence,

option $\left(B\right)$ is correct answer.