Question

Let $f_k\left(x\right)=\frac{1}{k}(si{n}^{k}x+co{s}^{k}x)$ where $xϵR$ and $k⩾1$
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 $\frac{1}{12}$

Let $f_k\left(x\right)=\frac{1}{k}(si{n}^{k}x+co{s}^{k}x)$
Consider
$f_4\left(x\right)-f_6\left(x\right)=\frac{1}{4}(si{n}^{4}x+co{s}^{4}x)-\frac{1}{6}(si{n}^{6}x+co{s}^{6}x)$
$=\frac{1}{4}[1-2si{n}^{2}xco{s}^{2}x]-\frac{1}{6}[1-3si{n}^{2}xco{s}^{2}x]$
$=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$