Question

If \(f\) be a real valued function defined for all real numbers \(x\) such that for some fixed \(a>0\), it satisfies \(f(x+a)=\frac{1}{2}+\sqrt{f(x)-(f(x))^{2}}\ \forall x\), then \(f(x)\) is periodic with period:

• A. $a$
• B. $4a$
• C. $\frac{a}{2}$
• D. $2a$

Hint:

Equations structured like $f(x+a) = \frac{1}{2} + \sqrt{f(x) - f(x)^2}$ are built around the parametric properties of circle arcs or trigonometric squares. If you substitute $f(x) = \frac{1}{2} + \frac{1}{2}\sin(2\theta)$, the step relation turns into a neat phase shift that returns to its origin after exactly two iterations!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A function is periodic if there exists a \( T>0 \) such that \( f(x+T) = f(x) \). We will iterate the given functional relationship to find the cycle length.
Step 2: Key Formula or Approach:
1. Let \( f(x) - 1/2 = g(x) \).
2. Rewrite the equation in terms of \( g(x) \).
3. Find \( g(x+a) \) and \( g(x+2a) \).
Step 3: Detailed Explanation:
Given: \( f(x + a) - 1/2 = \sqrt{f(x) - f(x)^2} \).
Square both sides:
\[ (f(x+a) - 1/2)^2 = f(x) - f(x)^2 \]
Now, find the value at \( x + 2a \):
\[ (f(x+2a) - 1/2)^2 = f(x+a) - f(x+a)^2 \]
Note the algebraic identity: \( f - f^2 = 1/4 - (f - 1/2)^2 \).
Applying this to the original equation:
\[ (f(x+a) - 1/2)^2 = 1/4 - (f(x) - 1/2)^2 \quad \dots(1) \]
Now write the expression for \( x+2a \):
\[ (f(x+2a) - 1/2)^2 = 1/4 - (f(x+a) - 1/2)^2 \]
Substitute \( (f(x+a) - 1/2)^2 \) from (1):
\[ (f(x+2a) - 1/2)^2 = 1/4 - [1/4 - (f(x) - 1/2)^2] \]
\[ (f(x+2a) - 1/2)^2 = (f(x) - 1/2)^2 \]
Since \( f(x+a) \ge 1/2 \), the term \( f(x) - 1/2 \) must eventually be positive. Taking the root:
\[ f(x+2a) - 1/2 = f(x) - 1/2 \implies f(x+2a) = f(x) \].
The period is \( 2a \).
Step 4: Final Answer:
The function is periodic with period \( 2a \).