Let `f: R to R `, be a periodic function such that `{f(x):x in N}` is an infinite set then, the period of f(x) cannot be

Options

A. a rational
B. an irrational
C. e
D. `pi`

Explanation

To solve the problem, we need to analyze the properties of a periodic function \( f: \mathbb{R} \to \mathbb{R} \) under the given conditions. Let's break it down step by step.

### Step 1: Understanding Periodicity
A function \( f \) is periodic with period \( t \) if:
\[ f(x + t) = f(x) \quad \text{for all } x \in \mathbb{R} \]
This means that the function repeats its values every \( t \) units.

**Hint:** Recall the definition of periodic functions and what it means for a function to repeat its values.

### Step 2: Setting Up the Problem
Given that \( \{ f(x) : x \in \mathbb{N} \} \) is an infinite set, it implies that the function takes infinitely many distinct values for natural numbers \( x \).

**Hint:** Think about what it means for a set to be infinite in terms of the function's outputs.

### Step 3: Assume \( t \) is Rational
Let’s assume that \( t \) is a rational number. We can express \( t \) as:
\[ t = \frac{m}{n} \quad \text{where } m \text{ and } n \text{ are integers, and } n \neq 0 \]
This means that for every natural number \( x \), we have:
\[ f(x + t) = f(x) \implies f\left(x + \frac{m}{n}\right) = f(x) \]

**Hint:** Consider how the rational period \( t \) affects the values of \( f \) at natural numbers.

### Step 4: Analyzing the Implications
If we take \( x = n \) (where \( n \) is a natural number), then:
\[ f\left(n + \frac{m}{n}\right) = f(n) \]
This implies that \( f(n + k \cdot m) = f(n) \) for any integer \( k \). Hence, \( f \) would repeat its values at intervals of \( m \).

**Hint:** Think about how the periodicity leads to a finite number of distinct outputs if \( t \) is rational.

### Step 5: Conclusion
Since \( f \) must take infinitely many distinct values for natural numbers, the assumption that \( t \) is rational leads to a contradiction. Therefore, \( t \) cannot be a rational number.

Thus, we conclude that the period of \( f(x) \) cannot be a rational number.

**Final Answer:** The period of \( f(x) \) cannot be a rational number.