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 $y$. If $\underset{x\to 0}{lt}\frac{f\left(x\right)}{x}=\frac{1}{3}$ then $f\left(1\right)$ equals

• A. $\frac{1}{4}$
• B. $\frac{1}{6}$
• C. $\frac{1}{12}$
• D. $\frac{1}{8}$

Answer 1

Answer: (C)

$\frac{1}{12}$

We have,

$f\left(\frac{x+y}{1-xy}\right)=f\left(x\right)+f\left(y\right)\forall x^{2}y$

$\Rightarrow f\left(x\right)=k{\tan }^{-1}\left(x\right)$

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

$k=\frac{1}{3}$

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

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

Hence, this is the answer.