Relations & Functions
Extension of Functions
Grade 12
Question:
<p>Let the function <strong>f(x) = x² + x + sin x - cos x</strong> be defined on the interval [0, 1]. Find the odd and even extensions of f(x) in the interval [-1, 1].</p>
<p>(a) Odd: {x² + x + sin x - cos x, 0 ≤ x ≤ 1; -x² + x + sin x + cos x, -1 < x < 0} and Even: {x² + x + sin x - cos x, 0 ≤ x ≤ 1; x² - x - sin x - cos x, -1 ≤ x < 0}</p>
<p>(b) Odd: {x² + x + sin x - cos x, 0 ≤ x ≤ 1; x² - x - sin x - cos x, -1 ≤ x < 0} and Even: {x² + x + sin x - cos x, 0 ≤ x ≤ 1; -x² + x + sin x + cos x, -1 < x < 0}</p>
<p>(c) does not exist</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Odd extension uses g(x) = -f(-x) for negative x values; even extension uses g(x) = f(-x). Apply this to each term of the function.
<p><strong>For Odd Extension:</strong></p><p>An odd function satisfies g(-x) = -g(x). For x ∈ [-1, 0], we need g(x) = -f(-x).</p><p>g(x) = -[(-x)² + (-x) + sin(-x) - cos(-x)] = -[x² - x - sin x - cos x] = -x² + x + sin x + cos x</p><p><strong>For Even Extension:</strong></p><p>An even function satisfies g(-x) = g(x). For x ∈ [-1, 0], we need g(x) = f(-x).</p><p>g(x) = (-x)² + (-x) + sin(-x) - cos(-x) = x² - x - sin x - cos x</p><p>∴ Answer is (a).</p>
Correct Answer: a