If functions f(x) and g(x) are defined on \mathbb{R} \to \mathbb{R} such that f(x) = \begin{cases} x + 3, & x \text{ rational} \\ \frac{x + \sqrt{5}}{-x}, & x \text{ irrational} \end{cases} then (f - g)(x) is-
Step-by-Step Solution
<div class="solution"><p>Forced: f(2)=3, f(4)=5. Remaining: {1,3,5}→{1,2,4} with f(1)≠2, f(3)≠4.</p><p>f⁻¹(2)=3: possible via f(1)=4,f(3)=2,f(5)=1 ✓</p><p>f⁻¹(2)=5: possible via f(1)=4,f(3)=1,f(5)=2 ✓</p><p>f⁻¹(2)=1: impossible (f(1)≠2)</p><p><strong>Answer: (B),(C)</strong></p><div class="trap-box"><strong>Trap:</strong> The biconditional restricts both even AND odd inputs.</div><div class="key-concept"><strong>Key Concept:</strong> Finite bijection — lock forced values, distribute remainder under constraints</div></div>
Correct Answer: B