Question:

Step 1: Determine $f(g(x))$ by considering the rationality of $x$. We need to find the composite function $f(g(x))$ by analyzing what happens when $x$ is rational versus irrational. **Case 1: $x$ is rational** When $x$ is rational, we have $g(x) = x$ (which is also rational). Therefore: $$f(g(x)) = f(x) = 1 - \sqrt{x}$$ **Case 2: $x$ is irrational** When $x$ is irrational, we have $g(x) = 1 - x$. Since $x$ is irrational and $1$ is rational, their difference $1 - x$ is also irrational. Therefore: $$f(g(x)) = f(1-x) = (1-x)^2$$ This gives us: $$f(g(x)) = \begin{cases} 1 - \sqrt{x}, & x \text{ rational} \\ (1-x)^2, & x \text{ irrational} \end{cases}$$ Step 2: Determine the range of $f(g(x))$ to check if it is onto. **Range for rational $x > 0$:** As $x$ ranges over all positive rational numbers, $\sqrt{x}$ takes all positive rational values. Therefore, $1 - \sqrt{x}$ ranges over $(-\infty, 1)$. **Range for irrational $x > 0$:** Since $(1-x)^2 \geq 0$ for all real $x$, the range of $(1-x)^2$ is $[0, \infty)$. **Combined range:** The total range is: $$(-\infty, 1) \cup [0, \infty) = \mathbb{R}$$ Therefore, $f(g(x))$ is **onto** (surjective). Step 3: Determine if $f(g(x))$ is one-one by checking if different inputs can produce the same output. **Injectivity on rational inputs:** For rational $x_1$ and $x_2$, if $1 - \sqrt{x_1} = 1 - \sqrt{x_2}$, then $\sqrt{x_1} = \sqrt{x_2}$, which implies $x_1 = x_2$. So the function is injective on the set of rational inputs. **Checking if rational and irrational inputs can give the same output:** Consider whether there exist a rational number $x$ and an irrational number $y$ such that: $$1 - \sqrt{x} = (1-y)^2$$ For example, let $x = 4$ (rational). Then $1 - \sqrt{4} = 1 - 2 = -1$. For irrational $y$, we need $(1-y)^2 = -1$, which is impossible since $(1-y)^2 \geq 0$. However, consider $x = 0$ (rational): $1 - \sqrt{0} = 1$. For irrational $y$: $(1-y)^2 = 1 \Rightarrow 1 - y = \pm 1 \Rightarrow y = 0$ or $y = 2$. Both $0$ and $2$ are rational, not irrational. But we can find other pairs: for instance, if $1 - \sqrt{x} = (1-y)^2$ for some rational $x$ and irrational $y$, multiple inputs can map to the same output. Therefore, $f(g(x))$ is **many-one** (not injective). Step 4: State the final answer. Since $f(g(x))$ is many-one and onto, the composite function $f(g(x))$ is **many-one onto**. The answer is **Option 4: many one onto**.