Q3. Let f : R → R be a continuous function satisfying f(x) = f(x + 1) for all x ∈ R.

Question

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function satisfying $f(x) = f(x + 1)$ for all $x \in \mathbb{R}$. Then which of the following statements is true?

(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 $f(x_0 + \pi) = f(x_0)$.

Answer 1

The correct answer is:

(D) There exist infinitely many $x_0 \in \mathbb{R}$ such that $f(x_0 + \pi) = f(x_0)$.

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Explanation:

Since the function $f$ is defined on the real numbers and satisfies $f(x) = f(x + 1)$ for all $x \in \mathbb{R}$, it means that the function has a period of 1. In other words, the function repeats its values every 1 unit along the x-axis.

Now, consider the function $g(x) = f(x + \pi)$. Since $f$ has a period of 1, $g(x)$ will have a period of $\pi$ because adding $\pi$ to $x$ is equivalent to shifting the function $f$ horizontally by $\pi$ units.

Since $f$ has a period of $\pi$, it means that for any value of $x_0$, $x_0 + \pi$ will be another point where $f$ takes the same value as $x_0$. Thus, there are infinitely many $x_0 \in \mathbb{R}$ such that $f(x_0 + \pi) = f(x_0)$. Therefore, option (D) is the correct statement.

This is AI generated answer with expert verified.