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
This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$2a$
Determine Function Periodicity
We are given the functional equation $f(x+a) = \frac{1}{2} + \sqrt{f(x) - (f(x))^2}$ for a real-valued function $f(x)$, where $a > 0$. We need to find the period of $f(x)$.
1. Analyze Domain Constraints
- The expression under the square root must be non-negative: $f(x) - (f(x))^2 \ge 0$.
- Let $y = f(x)$. Then $y - y^2 \ge 0$, which implies $y(1-y) \ge 0$.
- This inequality holds for $0 \le y \le 1$. Therefore, the range of the function is restricted: $0 \le f(x) \le 1$ for all $x$.
2. Transform the Functional Equation
- Let $f(x) = \frac{1}{2} + g(x)$. Since $0 \le f(x) \le 1$, we have $-\frac{1}{2} \le g(x) \le \frac{1}{2}$.
- Substitute this into the given equation: $ \frac{1}{2} + g(x+a) = \frac{1}{2} + \sqrt{\left(\frac{1}{2} + g(x)\right) - \left(\frac{1}{2} + g(x)\right)^2} $ $ g(x+a) = \sqrt{\left(\frac{1}{2} + g(x)\right) - \left(\frac{1}{4} + g(x) + g(x)^2\right)} $ $ g(x+a) = \sqrt{\frac{1}{4} - g(x)^2} $
- This implies $g(x+a) \ge 0$ for all $x$.
3. Derive Periodicity of $g(x)$
- Square both sides of $g(x+a) = \sqrt{\frac{1}{4} - g(x)^2}$: $ g(x+a)^2 = \frac{1}{4} - g(x)^2 $ $ g(x+a)^2 + g(x)^2 = \frac{1}{4} $
- Replace $x$ with $x+a$: $ g(x+2a)^2 + g(x+a)^2 = \frac{1}{4} $
- Comparing the two equations: $ g(x+2a)^2 + g(x+a)^2 = g(x+a)^2 + g(x)^2 $ $ g(x+2a)^2 = g(x)^2 $ $ g(x+2a) = \pm g(x) $
- Since $g(y) \ge 0$ for all $y$ (because $g(y) = g((y-a)+a) = \sqrt{\frac{1}{4}-g(y-a)^2} \ge 0$), we have $g(x+2a) = g(x)$.
- This shows that $g(x)$ is periodic with period $2a$.
4. Determine Period of $f(x)$
- Since $f(x) = \frac{1}{2} + g(x)$, and $g(x)$ is periodic with period $2a$, $f(x)$ is also periodic with period $2a$.
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