Understanding the Problem: Finding the Period of a Periodic Function

The question asks for the period of a periodic function \(f(x)\). We are given a specific condition that the function satisfies: \(f(x + 2\alpha) + f(x) = 0\) for all real numbers \(x\), where \(\alpha\) is a positive real number. The period of a function \(f(x)\) is the smallest positive value, let's call it \(T\), such that \(f(x+T) = f(x)\) for all \(x\) in the domain of the function.

Our goal is to use the given condition \(f(x + 2\alpha) + f(x) = 0\) to find this smallest positive value \(T\).

Step-by-Step Solution: Deriving the Period

Let's start with the given condition:

\(f(x + 2\alpha) + f(x) = 0\)

We can rewrite this condition as:

\(f(x + 2\alpha) = -f(x)\)    (Equation 1)

We are looking for a value \(T\) such that \(f(x+T) = f(x)\). Let's use Equation 1 to see if we can find such a relationship.

Replace \(x\) with \(x + 2\alpha\) in Equation 1:

\(f((x + 2\alpha) + 2\alpha) = -f(x + 2\alpha)\)

\(f(x + 4\alpha) = -f(x + 2\alpha)\)

Now, substitute \(f(x + 2\alpha)\) from Equation 1 into this new equation:

\(f(x + 4\alpha) = -(-f(x))\)

\(f(x + 4\alpha) = f(x)\)

This equation tells us that \(4\alpha\) is a value such that shifting the function by \(4\alpha\) brings it back to its original form. This means \(4\alpha\) is a period of the function \(f(x)\).

Now, we need to ensure that \(4\alpha\) is the smallest positive period. For a non-trivial periodic function (a function that is not constant), the smallest positive period is unique.

Let's consider if any smaller positive value, like \(2\alpha\), \(\alpha\), or \(\dfrac{\alpha}{2}\) could be the period. If \(T_0\) is the fundamental period (the smallest positive period), then any other period \(T\) must be an integer multiple of \(T_0\), i.e., \(T = n T_0\) for some positive integer \(n\).

We found that \(4\alpha\) is a period. If \(2\alpha\) were the period, then by definition, \(f(x+2\alpha) = f(x)\) for all \(x\). But the given condition is \(f(x+2\alpha) = -f(x)\). For both to be true, \(f(x) = -f(x)\), which implies \(2f(x) = 0\), or \(f(x) = 0\) for all \(x\). If \(f(x)\) is identically zero, it satisfies the condition, but the concept of a unique smallest positive period typically applies to non-zero functions. Assuming \(f(x)\) is not identically zero, \(2\alpha\) cannot be the period.

Similarly, if \(\alpha\) were the period, then \(f(x+\alpha) = f(x)\). Then \(f(x+2\alpha) = f((x+\alpha)+\alpha) = f(x+\alpha) = f(x)\). Again, this leads to \(f(x) = -f(x)\), implying \(f(x) = 0\).

If \(\dfrac{\alpha}{2}\) were the period, then \(f(x+\dfrac{\alpha}{2}) = f(x)\). Then \(f(x+\alpha) = f(x+\dfrac{\alpha}{2} + \dfrac{\alpha}{2}) = f(x+\dfrac{\alpha}{2}) = f(x)\), and \(f(x+2\alpha) = f(x+\alpha+\alpha) = f(x+\alpha) = f(x)\). This again implies \(f(x)=0\).

Since assuming \(2\alpha\), \(\alpha\), or \(\dfrac{\alpha}{2}\) as the period implies that \(f(x)\) must be the zero function (which has no unique smallest positive period in the usual sense), and given the structure of typical questions about periodic functions, it's highly probable that \(f(x)\) is a non-trivial function. In this case, \(4\alpha\) is the smallest positive value \(T\) such that \(f(x+T) = f(x)\) holds based on the given condition.

Why Other Options Are Incorrect

Let's quickly summarize why the other options for the period are incorrect for a non-trivial periodic function:

• 2\(\alpha\): The given condition is \(f(x + 2\alpha) = -f(x)\). If the period were \(2\alpha\), then \(f(x + 2\alpha) = f(x)\). This would mean \(f(x) = -f(x)\), or \(2f(x) = 0\), which implies \(f(x) = 0\). This only holds for the trivial zero function.
• \(\alpha\): If the period were \(\alpha\), then \(f(x + \alpha) = f(x)\). Repeated application would give \(f(x + 2\alpha) = f(x)\). As explained above, this implies \(f(x) = 0\).
• \(\dfrac{\alpha}{2}\): If the period were \(\dfrac{\alpha}{2}\), then \(f(x + \dfrac{\alpha}{2}) = f(x)\). Repeated application leads to \(f(x + \alpha) = f(x)\) and \(f(x + 2\alpha) = f(x)\). Again, this implies \(f(x) = 0\).

Thus, for any periodic function \(f(x)\) that is not identically zero and satisfies \(f(x + 2\alpha) = -f(x)\), the smallest positive period is \(4\alpha\).

Revision Table: Key Concepts

Additional Information: Properties of Anti-Periodic Functions

The condition \(f(x+a) = -f(x)\) for some constant \(a\) is sometimes referred to as a function being anti-periodic with anti-period \(a\). If a function is anti-periodic with anti-period \(a\), we can determine its regular period.

Given \(f(x+a) = -f(x)\).

Replace \(x\) with \(x+a\):

\(f((x+a)+a) = -f(x+a)\)

\(f(x+2a) = -f(x+a)\)

Now substitute \(f(x+a) = -f(x)\) into the equation:

\(f(x+2a) = -(-f(x))\)

\(f(x+2a) = f(x)\)

This shows that if a function is anti-periodic with anti-period \(a\), its regular period is \(2a\) (assuming \(f(x)\) is not the zero function). In our problem, the anti-period is \(2\alpha\). Following the same logic, the regular period is \(2 \times (2\alpha) = 4\alpha\).

This confirms our step-by-step derivation and provides a general concept related to the given functional equation.