Question

Let $f_k\left(x\right)=\frac{1}{k}({\sin }^{k}x+{\cos }^{k}x)$ for $k=1,2,3,\dots .$ Then for all $x\in R,$ the value of $f_4\left(x\right)-f_6\left(x\right)$ is equal to:

• A. $\frac{1}{12}$
• B. $\frac{-1}{12}$
• C. $\frac{1}{4}$
• D. $\frac{5}{12}$

Answer 1

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

$f_k\left(x\right)=\frac{1}{k}({\sin }^{k}x+{\cos }^{k}x)$
$f_4\left(x\right)-f_6\left(x\right)=\frac{1}{4}({\sin }^{4}x+{\cos }^{4}x)-\frac{1}{6}({\sin }^{6}x+{\cos }^{6}x)$
$=\frac{1}{4}(1-\frac{1}{2}{\sin }^{2}2x)-\frac{1}{6}(1-\frac{3}{4}{\sin }^{2}2x)$
$=\frac{1}{4}-\frac{1}{6}-\frac{1}{8}{\sin }^{2}2x+\frac{1}{8}{\sin }^{2}2x$
$=\frac{1}{12}$