Question:

Step 1: Establish the fundamental constraint from the given relation. Given the equation $\sqrt{1 - f^2(x)} = g(x)$, the following facts are deduced: \begin{itemize} \item Since $g(x)$ is defined as a square root, $g(x) \geq 0$. \item Squaring both sides yields $f^2(x) + g^2(x) = 1$. \end{itemize} This constraint is used to analyze all statements. Step 2: Analyze statement (a) — Periodicity of $f(x)$ when $g(x)$ is periodic. If $g(x)$ has period 1, then $g(x+1) = g(x)$. From the constraint $f^2(x) + g^2(x) = 1$, it follows that: $$g^2(x+1) = g^2(x)$$ $$1 - f^2(x+1) = 1 - f^2(x)$$ $$f^2(x+1) = f^2(x)$$ This implies $f(x+1) = \pm f(x)$. This condition does not necessarily mean $f(x)$ has period $\frac{1}{2}$. For instance, $f(x)$ could have period 1 or 2. Therefore, statement (a) is false. Step 3: Analyze statement (b) — Derivative relationship when $f'(c) = -f(c) = 0.5$. Differentiate the constraint $f^2(x) + g^2(x) = 1$ with respect to $x$: $$2f(x)f'(x) + 2g(x)g'(x) = 0$$ Solving for $g'(x)$: $$g'(x) = -\frac{f(x)f'(x)}{g(x)}$$ Given $f'(c) = -f(c) = 0.5$, we have $f(c) = -0.5$ and $f'(c) = 0.5$. Calculate $g(c)$ using the constraint: $$g(c) = \sqrt{1 - f^2(c)} = \sqrt{1 - (-0.5)^2} = \sqrt{1 - 0.25} = \sqrt{0.75} = \frac{\sqrt{3}}{2}$$ Calculate $g'(c)$: $$g'(c) = -\frac{f(c)f'(c)}{g(c)} = -\frac{(-0.5)(0.5)}{\frac{\sqrt{3}}{2}} = \frac{0.25}{\frac{\sqrt{3}}{2}} = \frac{1/4}{\sqrt{3}/2} = \frac{1}{2\sqrt{3}}$$ Calculate the ratio $\frac{g'(c)}{g(c)}$: $$\frac{g'(c)}{g(c)} = \frac{\frac{1}{2\sqrt{3}}}{\frac{\sqrt{3}}{2}} = \frac{1}{2\sqrt{3}} \times \frac{2}{\sqrt{3}} = \frac{2}{2 \times 3} = \frac{1}{3}$$ Therefore, statement (b) is true. Step 4: Analyze statement (c) — Oddness of $f(x)$ when $g(x)$ is even. If $g(x)$ is an even function, then $g(-x) = g(x)$. From the constraint: $$g^2(-x) = g^2(x)$$ $$1 - f^2(-x) = 1 - f^2(x)$$ $$f^2(-x) = f^2(x)$$ This implies $f(-x) = \pm f(x)$. This condition means $f(x)$ could be an even function, an odd function, or neither. It is not necessarily odd. Therefore, statement (c) is false. Step 5: Analyze statement (d) — Continuity of $f(x)$ when $g(x)$ is continuous. From the constraint, $f^2(x) = 1 - g^2(x)$. If $g(x)$ is continuous, then $1 - g^2(x)$ is continuous. Let $h(x) = 1 - g^2(x)$. Then $f^2(x) = h(x)$. For $f(x)$ to be defined, $h(x) \ge 0$. While $h(x)$ is continuous, $f(x)$ is not necessarily continuous. For example, let $g(x) = 0$ for all $x$. Then $g(x)$ is continuous. From $f^2(x) = 1 - g^2(x)$, we have $f^2(x) = 1$, which implies $f(x) = \pm 1$. Consider the function $f(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ -1 & \text{if } x \notin \mathbb{Q} \end{cases}$. This function satisfies $f^2(x) = 1$ for all $x$, but it is discontinuous everywhere. Thus, $f(x)$ is not necessarily continuous. Therefore, statement (d) is false. Step 6: Conclusion. Based on the analysis, only statement (b) is necessarily true.