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 $f_4\left(x\right)-f_6\left(x\right)=?$

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

Answer 1

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

$\begin{array}{cc}{f}_4\left(x\right)& =\frac{1}{4}({\sin }^{4}x+{\cos }^{4}x)\\ & =\frac{1}{4}\{{({\sin }^{2}x+{\cos }^{2}x)}^2-2{\sin }^{2}x{\cos }^{2}x\}\\ & =\frac{1}{4}(1-2{\sin }^{2}x{\cos }^{2}x)\end{array}$

$\begin{array}{cc}{f}_6\left(x\right)& =\frac{1}{6}({\sin }^{6}x+{\cos }^{6}x)\\ & =\frac{1}{6}\left\{({\sin }^{2}x+{\cos }^{2}x)({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x)\right\}\\ & =\frac{1}{6}\left\{\right(1\left)(1-3{\sin }^{2}x{\cos }^{2}x)\right\}\\ & =\frac{1}{6}(1-3{\sin }^{2}x{\cos }^{2}x)\end{array}$

$\begin{array}{cc}{f}_4\left(x\right)-f_6\left(x\right)& =\frac{1}{4}-\frac{1}{2}{\sin }^{2}x{\cos }^{2}x-\frac{1}{6}+\frac{1}{2}{\sin }^{2}x{\cos }^{2}x\\ & =\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\end{array}$

Answer: option (B)