<p>Which of the following is inverse to itself?</p>
<p>(a) \(f(x) = \dfrac{1-x}{1+x}\)</p>
<p>(b) \(f(x) = e^{\log x}\)</p>
<p>(c) \(f(x) = 3^{x(x+1)}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: A function is inverse to itself (an involution) when f(f(x)) = x for all x in the domain. For relations, you must verify that if (a,b) is in the relation, then (b,a) is also in the relation, and applying the relation twice returns to the original element.
<p><strong>Step 1:</strong> Recall that a function f is inverse to itself (self-inverse/involution) if f⁻¹ = f, which means f(f(x)) = x for all x in domain, or equivalently, if (a,b) ∈ f then (b,a) ∈ f.</p><p><strong>Step 2:</strong> For a relation/function to be its own inverse, swapping the coordinates of every ordered pair must give the same relation back. This is the <strong>symmetric property about the line y = x</strong>.</p><p><strong>Step 3:</strong> Common self-inverse functions include:</p><ul><li>f(x) = x (identity)</li><li>f(x) = -x (negation)</li><li>f(x) = 1/x (reciprocal, x ≠ 0)</li><li>f(x) = c - x for constant c</li><li>Relations that are symmetric about y = x</li></ul><p><strong>Step 4:</strong> Check the given options: Evaluate which relation R satisfies the condition that (a,b) ∈ R ⟺ (b,a) ∈ R, or verify f(f(x)) = x algebraically.</p><p>∴ Answer: A</p>
Correct Answer: A