Question

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

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

Answer 1

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

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