<p>If \(f(x) = \log(x + \sqrt{1 + x^2})\), then \(f(x)\) is</p>
Step-by-Step Solution
Key Concept: Analyze f(-x) and compare with f(x) to determine if the function is odd, even, or neither. Use the identity √(1 + x²) - x = 1/√(1 + x²) + x to simplify f(-x).
<p><strong>Step 1:</strong> Find f(-x).</p><p>f(-x) = log(-x + √(1 + x²))</p><p><strong>Step 2:</strong> Rationalize by multiplying numerator and denominator by (x + √(1 + x²)).</p><p>f(-x) = log[(-x + √(1 + x²)) · (x + √(1 + x²))/(x + √(1 + x²))]</p><p><strong>Step 3:</strong> Simplify the numerator using (a-b)(a+b) = a² - b².</p><p>Numerator = (1 + x²) - x² = 1</p><p>So f(-x) = log[1/(x + √(1 + x²))]</p><p><strong>Step 4:</strong> Apply logarithm property log(1/a) = -log(a).</p><p>f(-x) = -log(x + √(1 + x²)) = -f(x)</p><p><strong>Step 5:</strong> Since f(-x) = -f(x) for all x ∈ ℝ, the function is <strong>odd</strong>.</p><p>∴ Answer: C</p>
Correct Answer: C