Question

For $x\in \left(0,\frac{\pi }{2}\right)$, let $f_n(x)=\int n\sin 2x\left({\sin }^{2n-2}x-{\cos }^{2n-2}x\right)dx-\frac{1}{2^{n-1}},n\in N$ and $f_n\left(\frac{\pi }{4}\right)=\frac{1}{2^{n-1}}$.
The derivative of $f_n(x)$ when $n=2$ and $x=\frac{\pi }{3}$, is

• A. 1
• B. $\frac{\sqrt{3}}{2}$
• C. 0
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

We have, $f_n(x)=\int n\sin 2x\left({\sin }^{2n-2}x-{\cos }^{2n-2}x\right)dx-\frac{1}{2^{n-1}}$
$=2n\int \left(\cos x\cdot {\sin }^{2n-1}x-\sin x{\cos }^{2n-1}x\right)dx-\frac{1}{2^{n-1}}$
$=2n\left(\frac{{\sin }^{2n}x}{2n}+\frac{{\cos }^{2n}x}{2n}\right)-\frac{1}{2^{n-1}}+C={\sin }^{2n}x+{\cos }^{2n}x-\frac{1}{2^{n-1}}+C$
$\text{ As, }{f}_n\left(\frac{\pi }{4}\right)=\frac{1}{2^{n-1}}\Rightarrow \frac{1}{2^n}+\frac{1}{2^n}-\frac{1}{2^{n-1}}+C=\frac{1}{2^{n-1}}\Rightarrow C=\frac{1}{2^{n-1}}$
$\therefore f_n(x)={\sin }^{2n}x+{\cos }^{2n}x$
We have, $f_2(x)={\sin }^{4}x+{\cos }^{4}x=1-\frac{1}{2}{\sin }^{2}2x$
$f_2^{\prime }(x)=-\sin 4x$
$\Rightarrow f_2^{\prime }\left(\frac{\pi }{3}\right)=-\sin \left(\frac{4\pi }{3}\right)=\frac{\sqrt{3}}{2}$