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

Answer 1

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

$f_4\left(x\right)-f_6\left(x\right)=\frac{1}{4}({\sin }^{4}x+{\cos }^{4}x)-\frac{1}{6}({\sin }^{6}x+{\cos }^{6}x)f_4\left(x\right)-f_6\left(x\right)=\frac{1}{4}(1-2{\sin }^{2}x{\cos }^{2}x)-\frac{1}{6}(1-3{\sin }^{2}x{\cos }^{2}x)f_4\left(x\right)-f_6\left(x\right)=\frac{1}{4}-\frac{{\sin }^{2}x{\cos }^{2}x}{2}-\frac{1}{6}+\frac{{\sin }^{2}x{\cos }^{2}x}{2}{f}_4\left(x\right)-f_6\left(x\right)=\frac{1}{12}$