Question

Let $f\left(x\right)$ be differentiable function such that $f\left(\frac{x+y}{1-xy}\right)=f\left(x\right)+f\left(y\right)\forall x$ and $y$. lf $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=\frac{1}{3}$, then $f\left(1\right)$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{12}$
• C. $\frac{\pi }{6}$
• D. $\pi$

Answer 1

The correct option is B $\frac{\pi }{12}$

Given $f\left(\frac{x+y}{1-xy}\right)=f\left(x\right)+f\left(y\right)\forall x$ and $y$

As we know,

${\tan }^{-1}\left(\frac{x+y}{1-xy}\right)={\tan }^{-1}x+{\tan }^{-1}y$

Put $x=\tan \theta ,y=\tan \varphi$

$f\left(\tan \right(\theta +\varphi \left)\right)=f\left(\tan \theta \right)+f\left(\tan \varphi \right)$

So $f\left(x\right)=A{\tan }^{-1}x$

$\underset{x\to 0}{lim}\frac{A{\tan }^{-1}x}{x}=\frac{1}{3}$

$\Rightarrow A=\frac{1}{3}$

$f\left(x\right)=\frac{1}{3}{\tan }^{-1}x$

$f\left(1\right)=\frac{\pi }{12}$