Interesting Function

Calculus Level 4

Given a function $\large{f:\mathbb{R}\to\mathbb{R}}$ which satisfies:

$\large{f(x)+f(y)=f(\frac{x+y}{1-xy})}$ $\forall x,y \in\mathbb{R}$

also, $\large{\lim\limits_{x\to 0}\frac{f(x)}{x}=2}$

FIND the value of $\large{f(\frac{1}{\sqrt{3}})=a}$

ALSO FIND the value of $\large{f'(1)=b}$

ENTER YOUR ANSWER AS $\large{a+b}$

Answer upto $\text{5 decimal places}$ .

3 solutions

Tom Engelsman
Oct 2, 2015

Upon observation for x = y = 0, we have: f(0) + f(0) = f((0+0)/(1-0^2), or 2*f(0) = f(0) => f(0) = 0.

Differentiating this functional equation with respect to x and y gives:

f'(x) = [(1-xy) + y(x+y)]/(1-xy)^2 * f'((x+y)/(1-xy)) = (1+y^2)/(1-xy)^2 * f'((x+y)/(1-xy)) (i)

f'(y) = [(1-xy) + x(x+y)]/(1-xy)^2 * f'((x+y)/(1-xy)) = (1+x^2)/(1-xy)^2 * f'((x+y)/(1-xy)) (ii)

which yields the following relation:

f'(x)/(1+y^2) = f'(y)/(1+x^2), or f'(x) = A/(1+x^2) (A = real constant) (iii)

and integrating (iii) yields:

f(x) = A * arctan(x) + B, f(0) = 0, or f(x) = A * arctan(x).

Taking lim(x => 0) A * arctan(x) / x = 2 gives:

A*[1 / (1+x^2)] / 1 = 2, or A = 2 (by L'Hopital's Rule),

so f(x) = 2*arctan(x), which a = f(1/sqrt(3)) = pi/3 and b = f'(1) = 1. We ultimately obtain the result: a + b = pi/3 + 1 = 2.04719

Bob Kadylo
Jul 26, 2018

Rishabh Deep Singh
Feb 16, 2016

we can take f(x)=2.arctan(x) because it satisfy the above relation