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 $f\left(1\right)$ equals

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{12}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

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

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

$f\left(x\right)=ata{n}^{-1}x$

${lim}_{x\to 0}\frac{ata{n}^{-1}x}{x}=\frac{1}{3}$

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

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

$\Rightarrow f\left(1\right)=\frac{1}{3}ta{n}^{-1}\left(1\right)$

$=\frac{\pi }{12}$