<p>Let \(f(x)\) and \(g(x)\) are two functions defined from \(R^+ \to R\) such that</p><p>\[f(x) = \begin{cases} 1 - \sqrt{x}, & \text{if } x \text{ is rational} \\ x^2, & \text{if } x \text{ is irrational} \end{cases}\] and \[g(x) = \begin{cases} x, & \text{if } x \text{ is rational} \\ 1 - x, & \text{if } x \text{ is irrational} \end{cases}\]</p><p>The composite function \(f(g(x))\) is:</p>
Step-by-Step Solution
Key Concept: To find f(g(x)), you must first determine whether g(x) is rational or irrational based on whether x is rational or irrational, then apply the appropriate definition of f. The rationality of g(x) depends on the input type to g, not on the original x.
<p><strong>Step 1: Analyze f(g(x)) when x is rational</strong></p><p>If x ∈ ℚ⁺: g(x) = x, which is rational.</p><p>Since g(x) = x is rational, apply the first case of f: f(g(x)) = 1 - √x</p><p><strong>Step 2: Analyze f(g(x)) when x is irrational</strong></p><p>If x ∈ ℝ⁺ \ ℚ (irrational): g(x) = 1 - x, which is irrational (since 1 is rational and x is irrational).</p><p>Since g(x) = 1 - x is irrational, apply the second case of f: f(g(x)) = (1-x)²</p><p><strong>Step 3: Write the composite function</strong></p><p>$$f(g(x)) = \begin{cases} 1 - \sqrt{x}, & \text{if } x \text{ is rational} \\ (1-x)^2, & \text{if } x \text{ is irrational} \end{cases}$$</p><p>∴ Answer: D</p>
Correct Answer: D