Question

Let $f_n\left(x\right)=\left(1/n\right)({\sin }^{n}x+{\cos }^{n}x)$ for $n=1,2,3,....$ then $f_4\left(3\pi /8\right)-f_6\left(3\pi /8\right)$ is equal to

• A. $2/(\sqrt{2}+1)$
• B. $1/12$
• C. $4/(\sqrt{3}-1)$
• D. $1/8$

Answer 1

The correct option is A $2/(\sqrt{2}+1)$

$f_n\left(x\right)=\left(\frac{1}{n}\right)({\sin }^{n}x+{\cos }^{n}x)$

$f_4\left(x\right)-f_6\left(x\right)=\left(\frac{1}{4}\right)({\cos }^{4}x+{\sin }^{4}x)-\frac{1}{6}({\cos }^{6}x+{\sin }^{6}x)$ ........$\left(1\right)$

using $a^3+b^3=(a+b)(a^2+b^2-ab)$

$({\cos }^{2}x{)}^3+({\sin }^{2}x{)}^3=({\sin }^{2}x+{\cos }^{2}x)({\sin }^4+{\cos }^{4}x-{\sin }^{2}x\cdot {\cos }^{2}x)$ ..........$\left(2\right)$

${\sin }^{2}x+{\cos }^{2}x=1$ ........$\left(3\right)$

put eq $\left(2\right)$ & eq $\left(1\right)$

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

$=\frac{2}{24}({\cos }^{4}x+{\sin }^{4}x)+\frac{1}{6}{\sin }^{2}x{\cos }^{2}x$

$=\frac{1}{12}({\cos }^{4}x+{\sin }^{4}x+2{\sin }^{2}x{\cos }^{2}x)$

$=\frac{1}{12}({\sin }^{2}x+{\cos }^{2}x{)}^2$

$=\frac{1}{12}$

Independent of x.