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

Concept: A function $f(x)$ is periodic if there exists a positive real constant $T$ such that $f(x+T) = f(x)$ for all $x$ in the domain. When given a functional step relation, we can find the total period by recursively substituting the equation into itself until the original expression returns. Step 1: Write out the primary shifted functional step relation.
We are given the core functional equation mapping a shift of $a$ units: \[ f(x+a) = \frac{1}{2} + \sqrt{f(x) - [f(x)]^2} \quad \cdots (1) \] Let us square the shifted term to simplify the expressions later by isolating the radical group. First subtract $\frac{1}{2}$ from both sides of equation (1): \[ f(x+a) - \frac{1}{2} = \sqrt{f(x) - [f(x)]^2} \] Squaring both sides of the equation: \[ \left(f(x+a) - \frac{1}{2}\right)^2 = f(x) - [f(x)]^2 \quad \cdots (2) \]

Step 2: Evaluate the expression for a double step shift of \(2a\).
Now, let us calculate the value of the function after moving another step of $a$ units forward, which means replacing $x$ with $(x+a)$ inside our primary rule (1): \[ f(x+2a) = f((x+a)+a) = \frac{1}{2} + \sqrt{f(x+a) - [f(x+a)]^2} \quad \cdots (3) \] Look closely at the expression inside the square root of equation (3). We can construct this term by rearranging our squared relationship from equation (2) matching the parameter index step. Let us subtract both sides of equation (2) from $\frac{1}{4}$: \[ \frac{1}{4} - \left(f(x+a) - \frac{1}{2}\right)^2 = \frac{1}{4} - \Big(f(x) - [f(x)]^2\Big) \] Expand the left side binomial squared expression: \[ \frac{1}{4} - \left([f(x+a)]^2 - f(x+a) + \frac{1}{4}\right) = [f(x)]^2 - f(x) + \frac{1}{4} \] \[ f(x+a) - [f(x+a)]^2 = \left(f(x) - \frac{1}{2}\right)^2 \]

Step 3: Substitute the balanced expression back into the double shift equation.
Now substitute this clean identity group directly back under the radical sign inside our double step equation (3): \[ f(x+2a) = \frac{1}{2} + \sqrt{\left(f(x) - \frac{1}{2}\right)^2} \] Since a square root of a squared term returns the absolute value, and knowing from equation (1) that $f(x) \ge \frac{1}{2}$ because of the positive radical addition, the term $f(x) - \frac{1}{2}$ is guaranteed to be non-negative. This allows us to drop the root and square cleanly: \[ f(x+2a) = \frac{1}{2} + \left(f(x) - \frac{1}{2}\right) \]

Step 4: Isolate the period variable constant.
Simplify the final arithmetic addition: \[ f(x+2a) = \frac{1}{2} + f(x) - \frac{1}{2} = f(x) \] Since $f(x+2a) = f(x)$ holds true for all real values of $x$, the function repeats its outputs perfectly every $2a$ units. Therefore, the function is periodic with a fundamental period of $2a$, matching option (D).