Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a continuous function satisfying \( f(x) = f(x+1) \) for all \(x \in \mathbb{R}\). Then

• A. \(f\) is not necessarily bounded above.
• B. There exists a unique \(x_0 \in \mathbb{R}\) such that \(f(x_0+\pi) = f(x_0)\).
• C. There is no \(x_0 \in \mathbb{R}\) such that \(f(x_0+\pi) = f(x_0)\).
• D. There exist infinitely many \(x_0 \in \mathbb{R}\) such that \(f(x_0+\pi) = f(x_0)\).

Hint:

For a periodic continuous function with an irrational shift (like \(\pi\)), equality points occur infinitely often because the shift creates dense coverage over one full period.

Answer 1

The Correct Option is D

Step 1: Understanding the property.
Given \(f(x) = f(x+1)\), the function is periodic with period \(1\). We must determine how many \(x_0\) satisfy \(f(x_0+\pi) = f(x_0)\).
Step 2: Applying periodicity.
Since \(\pi\) is irrational with respect to the period \(1\), the sequence \(x_0 + n\pi\) (mod 1) is dense in \([0,1]\). Thus, by continuity, there are infinitely many \(x_0\) such that \(f(x_0+\pi) = f(x_0)\).
Step 3: Conclusion.
Hence, \(f(x_0+\pi) = f(x_0)\) has infinitely many solutions.