Which of the following functions is an odd function?
(A) |x - 2| + (x + 2) \text{sgn}(x + 2)
(B) \frac{1}{x(e - 1)} + \frac{1}{2x}
(C) \log(\sin(x + \sqrt{1 + \sin^2 x}))
(D) e^{4x} (2^{x - 1})^4
Step-by-Step Solution
Key Concept: A function f(x) is odd if f(-x) = -f(x) for all x in its domain. We need to check this property for each option by substituting -x and verifying the relationship.
<p><strong>Step 1: Test Option (A):</strong> f(x) = |x - 2| + (x + 2)sgn(x + 2)</p><p>f(-x) = |-x - 2| + (-x + 2)sgn(-x + 2) = |x + 2| + (-x + 2)sgn(-x + 2)</p><p>This does not equal -f(x). Not odd.</p><p><strong>Step 2: Test Option (B):</strong> f(x) = 1/[x(e - 1)] + 1/(2x)</p><p>f(-x) = 1/[-x(e - 1)] + 1/(-2x) = -1/[x(e - 1)] - 1/(2x) = -f(x) ✓</p><p>This appears to be odd, BUT we must check the domain. The term e - 1 ≈ 1.718, so the function is defined for all x ≠ 0. However, let's verify Option C first.</p><p><strong>Step 3: Test Option (C):</strong> f(x) = log(sin(x + √(1 + sin²x)))</p><p>Let g(x) = sin(x + √(1 + sin²x)). We need to show g(-x) = 1/g(x), which would make log(g(-x)) = -log(g(x)).</p><p>Note: √(1 + sin²(-x)) = √(1 + sin²x) (since sin²(-x) = sin²x)</p><p>g(-x) = sin(-x + √(1 + sin²x))</p><p>Using the identity: sin(A + B)·sin(A - B) = sin²A - sin²B</p><p>Let A = √(1 + sin²x) and observe that sin(x + √(1 + sin²x))·sin(-x + √(1 + sin²x)) = sin²(√(1 + sin²x)) - sin²x = 1 + sin²x - sin²x = 1</p><p>Therefore: sin(-x + √(1 + sin²x)) = 1/sin(x + √(1 + sin²x))</p><p>Thus: f(-x) = log(1/sin(x + √(1 + sin²x))) = -log(sin(x + √(1 + sin²x))) = -f(x) ✓</p><p>This is ODD.</p><p><strong>Step 4: Test Option (D):</strong> f(x) = e^(4x)·(2^(x-1))^4 = e^(4x)·2^(4x-4)</p><p>f(-x) = e^(-4x)·2^(-4x-4) ≠ -f(x). Not odd.</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C