Let f be a real valued periodic function defined for all real numbers x such that for some fixed ` a gt 0` , `f(x+a)=(1)/(2)+sqrt(f(x)-{f(x)}^(2))` for all x .
Then , the period of f(x) is

Options

A. a
B. 2a
C. 3a
D. 4a

Explanation

To solve the problem, we need to find the period of the function \( f(x) \) defined by the equation:

\[ f(x + a) = \frac{1}{2} + \sqrt{f(x) - f(x)^2} \]

for some fixed \( a > 0 \).

### Step 1: Analyze the given functional equation

We start with the given equation:

\[ f(x + a) = \frac{1}{2} + \sqrt{f(x) - f(x)^2} \]

Let's denote this as **Equation (1)**.

### Step 2: Substitute \( x \) with \( x + a \)

Next, we substitute \( x \) with \( x + a \) in Equation (1):

\[ f((x + a) + a) = f(x + 2a) = \frac{1}{2} + \sqrt{f(x + a) - f(x + a)^2} \]

Now, we need to replace \( f(x + a) \) using Equation (1):

\[ f(x + 2a) = \frac{1}{2} + \sqrt{\left(\frac{1}{2} + \sqrt{f(x) - f(x)^2}\right) - \left(\frac{1}{2} + \sqrt{f(x) - f(x)^2}\right)^2} \]

### Step 3: Simplify the expression

Now, we simplify the expression for \( f(x + 2a) \):

Let \( y = f(x) \). Then, we have:

\[ f(x + a) = \frac{1}{2} + \sqrt{y - y^2} \]

Substituting this into the expression for \( f(x + 2a) \):

\[ f(x + 2a) = \frac{1}{2} + \sqrt{\left(\frac{1}{2} + \sqrt{y - y^2}\right) - \left(\frac{1}{2} + \sqrt{y - y^2}\right)^2} \]

### Step 4: Find \( f(x + 2a) \)

To find \( f(x + 2a) \), we recognize that we can express it in terms of \( f(x) \):

After simplification, we can find that:

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

This implies that the function \( f(x) \) is periodic with period \( 2a \).

### Conclusion

Thus, the period of the function \( f(x) \) is:

\[ \text{Period} = 2a \]