Sets, Relations & Functions
Types of Functions
Grade 11
Question:
<p>The function of \(f(x) = \log\left(x + \sqrt{x^2 + 1}\right)\), is</p>
<p>an even function.</p>
<p>an odd function.</p>
<p>a periodic function.</p>
<p>neither an even nor an odd function.</p>
Step-by-Step Solution
Key Concept: To determine function properties, analyze the domain by checking when the argument of logarithm is positive, and examine f(-x) to test for odd/even behavior and monotonicity by taking derivatives.
<p><strong>Step 1: Find the Domain</strong></p><p>For f(x) = log(x + √(x² + 1)) to be defined, we need x + √(x² + 1) > 0</p><p>Since √(x² + 1) > |x| for all real x, we have:</p><p>• When x ≥ 0: x + √(x² + 1) > 0 ✓</p><p>• When x < 0: √(x² + 1) > |x| = -x, so x + √(x² + 1) > 0 ✓</p><p>Domain = ℝ (all real numbers)</p><p><strong>Step 2: Check if Function is Odd/Even</strong></p><p>f(-x) = log(-x + √((-x)² + 1)) = log(-x + √(x² + 1))</p><p>Rationalize: (-x + √(x² + 1)) = (x² + 1 - x²)/(x + √(x² + 1)) = 1/(x + √(x² + 1))</p><p>Therefore: f(-x) = log(1/(x + √(x² + 1))) = -log(x + √(x² + 1)) = -f(x)</p><p><strong>Step 3: Verify Monotonicity</strong></p><p>f'(x) = 1/(x + √(x² + 1)) · (1 + x/√(x² + 1)) = 1/√(x² + 1) > 0 for all x</p><p>Function is strictly increasing on ℝ</p><p>∴ <strong>Answer: B</strong> (Odd function with domain ℝ, strictly increasing)</p>
Correct Answer: B