Question

Let $f:R\to R$ be such that $f(2x-1)=f\left(x\right)$ for all $x\in R$. 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

Replacing $x$ by $\left(\frac{x+1}{2}\right)$
$f\left(x\right)=f\left(\frac{x+1}{2}\right)$

Again, we put $\left(\frac{x+1}{2}\right)$ in place of $x$, we get
$f\left(\frac{x+1}{2}\right)=f\left(\frac{\frac{x+1}{2}+1}{2}\right)=f\left(\frac{x+1+2}{2^2}\right)$
Again, we put $\left(\frac{x+1}{2}\right)$ in place of $x$, we get
$f\left(\frac{x+1+2}{2^2}\right)=f\left(\frac{x+1+2+2^2}{2^3}\right)$

Repeating this process $n$ times , we get
$f\left(x\right)=f\left(\frac{x+1+2+2^2+2^3+...+2^{n-1}}{2^n}\right)$
$\Rightarrow f\left(x\right)=f(\frac{x}{2^n}+\frac{2^n-1}{2^n})$
$\Rightarrow f\left(x\right)=f(\frac{x}{2^n}+1-\frac{1}{2^n})$
Taking limit $n\to \infty$, we get
$f\left(x\right)=f\left(1\right)=1,\forall x\in R$.
$f\left(x\right)$ is continuous at $x=1$.
$\Rightarrow f\left(x\right)$ is a constant function.