Question

For $x \in\left(0, \frac{\pi}{2}\right)$, let $f_n(x)=\int n \sin 2 x\left(\sin ^{2 n-2} x-\cos ^{2 n-2} x\right) d x-\frac{1}{2^{n-1}}, n \in N$ 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, $f_n(x)=\int n \sin 2 x\left(\sin ^{2 n-2} x-\cos ^{2 n-2} x\right) d x-\frac{1}{2^{n-1}}$
$=2 n \int\left(\cos x \cdot \sin ^{2 n-1} x-\sin x \cos ^{2 n-1} x\right) d x-\frac{1}{2^{n-1}}$
$=2 n\left(\frac{\sin ^{2 n} x}{2 n}+\frac{\cos ^{2 n} x}{2 n}\right)-\frac{1}{2^{n-1}}+C=\sin ^{2 n} x+\cos ^{2 n} x-\frac{1}{2^{n-1}}+C$
$\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}} $
$\therefore f _{ n }( x )=\sin ^{2 n } x +\cos ^{2 n } x$
We have, $f_2(x)=\sin ^4 x+\cos ^4 x=1-\frac{1}{2} \sin ^2 2 x$
$f _2^{\prime}( x )=-\sin 4 x $
$\Rightarrow f _2^{\prime}\left(\frac{\pi}{3}\right)=-\sin \left(\frac{4 \pi}{3}\right)=\frac{\sqrt{3}}{2}$