<p>If \(f(x) = \log_e\left(\dfrac{1-x}{1+x}\right)\), \(|x| < 1\), then \(f\left(\dfrac{2x}{1+x^2}\right)\) is equal to:</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) is an odd function by verifying f(-x) = -f(x), which allows you to exploit symmetry properties for evaluating functional equations involving sums.
<p><strong>Step 1:</strong> Verify that f(x) is an odd function.</p><p>f(-x) = log₍ₑ₎((1-(-x))/(1+(-x))) = log₍ₑ₎((1+x)/(1-x))</p><p>= log₍ₑ₎((1+x)/(1-x)) = -log₍ₑ₎((1-x)/(1+x)) = -f(x) ✓</p><p><strong>Step 2:</strong> For any odd function, f(a) + f(-a) = 0. This is the key property to exploit in problems involving symmetric arguments or functional equations with sums of f at different points.</p><p><strong>Step 3:</strong> If the question asks for f(a) + f(b) where specific a and b are given (typically with symmetric properties), use the odd function property: f(a) + f(-a) = 0, or simplify using: f(a) + f(b) = log₍ₑ₎((1-a)/(1+a)) + log₍ₑ₎((1-b)/(1+b)) = log₍ₑ₎(((1-a)(1-b))/((1+a)(1+b)))</p><p><strong>Step 4:</strong> If b = -a, the numerator becomes (1-a)(1+a) = 1-a² and denominator becomes (1+a)(1-a) = 1-a², giving log₍ₑ₎(1) = 0.</p><p>∴ Answer: A</p>
Correct Answer: A