Question

If $f_n(x)=\frac{1}{n}({\cos }^{n}x+{\sin }^{n}x)$ , for $n=1,2,\mathrm{3.....},$ then $f_4(x)-f_6(x)$ is equal to

• A. 10
• B. $\frac{1}{12}$
• C. $\frac{1}{10}$
• D. 12

Answer 1

Answer: (B)

Given, $f_n\left(x\right)=\frac{1}{n}\left({\cos }^{n}x+{\sin }^{n}x\right)$,
where $n=1$, $2$, $3$, ...
Now, $f_4\left(x\right)-f_6\left(x\right)=\frac{1}{4}\left(co{s}^{4}x+{\sin }^{4}x\right)$
$-\frac{1}{6}\left(co{s}^{6}x+{\sin }^{6}x\right)$
$=\frac{1}{4}\left[{\left({\cos }^{2}x+{\sin }^{2}x\right)}^2-2{\cos }^{2}x{\sin }^{2}x\right]$
$-\frac{1}{6}\left[{\left({\cos }^{2}x\right)}^3+{\left({\sin }^{2}x\right)}^3\right]$
$=\frac{1}{4}\left[1-2{\sin }^2{\cos }^{2}x\right]-\frac{1}{6}{\left[({\cos }^{2}x+{\sin }^{2}x\right)}^3$
$-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)]$
$=\frac{1}{4}\left[1-2{\sin }^{2}x{\cos }^{2}x\right]-\frac{1}{6}\left[1-3{\sin }^{2}x{\cos }^{2}x\right]$
$=\frac{1}{4}-\frac{1}{2}{\sin }^{2}x{\cos }^{2}x-\frac{1}{6}+\frac{1}{2}{\sin }^{2}x{\cos }^{2}x$
$=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$