Question

Let $f:R\to R$ be such that $f(2x-1)=f\left(x\right)$ for all $xϵR$. If $f$ is continuous at $x=1$ and $f\left(1\right)=1$, then

• A. $f\left(2\right)=1$
• B. $f\left(2\right)=2$
• C. $f$ is continuous only at $x=1$
• D. $f$ is continuous at all points

Answer 1

Answer: (A), (D)

The correct options are
A $f\left(2\right)=1$
D $f$ is continuous at all points

$x\to \left(\frac{x+1}{2}\right)$
$f\left(x\right)=f\left(\frac{x+1}{2}\right)=f\left(\frac{\frac{x+1}{2}+1}{2}\right)=f\left(\frac{x+1+2}{2^2}\right)$
$=f\left(\frac{x+1+2+2^2+2^3+..{.2}^{n-1}}{2^n}\right)$
$=f(\frac{x}{2^n}+\frac{2^n-1}{2^n})$
$=f(\frac{x}{2^n}+1-\frac{1}{2^n})$
Taking limit $n\to \infty$,
$f\left(x\right)=f\left(1\right)$, as $f\left(x\right)$ is continuous at $x=1$
$\Rightarrow f\left(x\right)$ is constant function