Let f: R rightarrow R be such that f(2x-1) = | vedclass

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(A) Given $f(2x - 1) = f(x)$.For any $x$, we can write $f(x) = f(2x - 1) = f(2(2x - 1) - 1) = f(4x - 3) = f(2^n x - (2^n - 1))$.As $n \rightarrow \inf

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$f(2) = 1$

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(A) Given $f(2x - 1) = f(x)$.For any $x$, we can write $f(x) = f(2x - 1) = f(2(2x - 1) - 1) = f(4x - 3) = f(2^n x - (2^n - 1))$.As $n ightarrow \infty$, $2^n x - 2^n + 1 = 2^n(x - 1) + 1$.If $x 
eq 1$, then $2^n(x - 1) + 1 ightarrow \pm \infty$.Since $f$ is continuous at $x = 1$, we consider the limit as $x ightarrow 1$. Let $x_n$ be a sequence such that $x_n ightarrow 1$. Then $f(x_n) = f(2x_n - 1)$.By iterating the relation, $f(x) = f(1)$ for all $x$ in the domain because the sequence $x_{n+1} = \frac{x_n + 1}{2}$ converges to $1$.Since $f(1) = 1$, it follows that $f(x) = 1$ for all $x \in R$.Thus, $f(2) = 1$ and $f$ is a constant function, which is continuous everywhere.

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

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