Question

Let $f\left(x\right)=\tan x-{\tan }^{3}x+{\tan }^{5}x-{\tan }^{7}x+...+\infty ,xϵ(0,\frac{\pi }{4})$, then

• A. $f^{\prime }\left(\frac{\pi }{12}\right)=\frac{1}{2}$
• B. ${lim}_{x\to 0}\frac{f\left(x\right)}{x}=1$
• C. ${\int }_0^{\frac{\pi }{6}}f\left(x\right)dx=\frac{1}{8}$
• D. $f\left(x\right)$ is an odd function

Answer 1

Answer: (B), (C), (D)

The correct options are
B

${lim}_{x\to 0}\frac{f\left(x\right)}{x}=1$

C

${\int }_0^{\frac{\pi }{6}}f\left(x\right)dx=\frac{1}{8}$

D

$f\left(x\right)$ is an odd function

Given series is a geometric progression with common ratio ${\tan }^{2}x$

So, $f\left(x\right)=\frac{\tan x}{1+{\tan }^{2}x}=\frac{\sin 2x}{2}$

$\to f^{\prime }\left(x\right)=\cos 2x$

$\Rightarrow f^{\prime }\left(\frac{\pi }{12}\right)=\frac{\sqrt{3}}{2}$

$\to {lim}_{x\to 0}\frac{f\left(x\right)}{x}=\frac{\sin 2x}{2x}=1$

$\to {\int }_0^{\frac{\pi }{6}}f\left(x\right)dx={\int }_0^{\frac{\pi }{6}}\frac{\sin 2x}{2}dx={\left[\frac{-\cos 2x}{4}\right]}_0^{\frac{\pi }{6}}=\frac{1}{8}$

Since, $f(-x)=-f\left(x\right),$ so, $f\left(x\right)$ is an odd function.