Question

Let $f:R\to R$ be a continuous function satisfying $f\left(0\right)=1$ and $f\left(2x\right)-f\left(x\right)=x$, for all $x\in R.$ Then $f\left(2020\right)$ equals

Answer 1

$f\left(x\right)-f\left(\frac{x}{2}\right)=\frac{x}{2}...\left(1\right)$
$f\left(\frac{x}{2}\right)-f\left(\frac{x}{4}\right)=\frac{x}{4}...\left(2\right)$
$...$
$f\left(\frac{x}{2^{n-1}}\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2^n}...\left(n\right)$

Adding eqn$\left(1\right),\left(2\right),...,\left(n\right)$, we get
$f\left(x\right)-f\left(\frac{x}{2^n}\right)=\frac{x}{2}+\frac{x}{4}+...+\frac{x}{2^n}$
Applying limit both the sides
$f\left(x\right)-\underset{n\to \infty }{lim}f\left(\frac{x}{2^n}\right)=\underset{n\to \infty }{lim}\frac{x}{2}(1+\frac{1}{2}+\frac{1}{4}+...\infty )$
$\Rightarrow f\left(x\right)-f\left(0\right)=\frac{x}{2}\times \left(\frac{1}{1-\frac{1}{2}}\right)$
$\Rightarrow f\left(x\right)-1=x$
$\Rightarrow f\left(x\right)=x+1\forall x\in R$
$\therefore f\left(2020\right)=2020+1=2021$