<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 its own inverse (involutive) if and only if f(f(x)) = x for all x in the domain, which geometrically means the function is symmetric about the line y = x. Check this by verifying that if (a,b) is in the relation, then (b,a) is also in the relation.
<p><strong>Step 1:</strong> Understand that f is inverse to itself means f⁻¹ = f, or equivalently f(f(x)) = x for all x in domain.</p><p><strong>Step 2:</strong> This requires the function to be symmetric about the line y = x. If (a, b) is in the function, then (b, a) must also be in the function.</p><p><strong>Step 3:</strong> Verify the given option by checking:</p><ul><li>The function is its own inverse (apply it twice, get identity)</li><li>OR check that swapping inputs and outputs gives the same relation</li><li>Common examples: f(x) = 1/x (for x ≠ 0), f(x) = -x, f(x) = (ax+b)/(cx+d) under specific conditions</li></ul><p><strong>Step 4:</strong> Eliminate other options by showing f(f(x)) ≠ x or that the inverse relation differs from the original.</p><p>∴ Answer: A</p>
Correct Answer: A