<p>Given \(f(x) = \ln\left(\dfrac{1-x}{1+x}\right)\). Then \(f\!\left(\dfrac{2x}{1+x^2}\right)\) equals:</p>
Step-by-Step Solution
Key Concept: Use the substitution property of functions by replacing x with the given expression, then apply logarithm properties and algebraic simplification to recognize a standard form.
<p><strong>Step 1:</strong> Substitute <strong>x → 2x/(1+x²)</strong> into f(x) = ln((1-x)/(1+x))</p><p>f(2x/(1+x²)) = ln<left>[</left>(1 - 2x/(1+x²))/(1 + 2x/(1+x²))<right>]</right></p><p><strong>Step 2:</strong> Simplify the numerator: 1 - 2x/(1+x²) = [(1+x²) - 2x]/(1+x²) = <strong>(1-x)²/(1+x²)</strong></p><p><strong>Step 3:</strong> Simplify the denominator: 1 + 2x/(1+x²) = [(1+x²) + 2x]/(1+x²) = <strong>(1+x)²/(1+x²)</strong></p><p><strong>Step 4:</strong> Divide: [(1-x)²/(1+x²)] ÷ [(1+x)²/(1+x²)] = (1-x)²/(1+x)²</p><p><strong>Step 5:</strong> Apply logarithm: f(2x/(1+x²)) = ln[(1-x)²/(1+x)²] = <strong>2ln|(1-x)/(1+x)| = 2f(x)</strong></p><p>∴ Answer: <strong>C</strong> (assuming option C is 2f(x))</p>
Correct Answer: C