Relations & Functions
Types of functions
Grade 12
Question:
<p><strong>Paragraph for Question nos. 658 and 659</strong><br>Let \(f(x) = x^{2010} + x^{1010} - x^{510} + x^{210} + x^2\). If \(f(x)\) is divided by \(x^2(x^2 - 1)\), then we get remainder as \(g(x)\), function of \(x\).</p><p>If \(g(x)\) is defined from \(R\) to \(R\), then:</p>
<p>(a) \(g(x)\) is injective but not surjective</p>
<p>(b) \(g(x)\) is surjective but not injective</p>
<p>(c) \(g(x)\) is neither injective nor surjective</p>
<p>(d) \(g(x)\) is injective and surjective</p>
Step-by-Step Solution
Key Concept: When f(x) is divided by x²(x² - 1), the remainder g(x) must have degree less than 4. Use the division algorithm: f(x) = q(x)·x²(x² - 1) + g(x), where g(x) = ax³ + bx² + cx + d. Find coefficients by evaluating f at roots of x²(x² - 1) and using derivatives.
<p><strong>Step 1:</strong> Apply division algorithm: f(x) = q(x)·x²(x² - 1) + g(x), where deg(g) < 4.</p><p>So g(x) = ax³ + bx² + cx + d</p><p><strong>Step 2:</strong> At x = 1: f(1) = 1 + 1 - 1 + 1 + 1 = 3, and g(1) = a + b + c + d = 3</p><p><strong>Step 3:</strong> At x = -1: f(-1) = 1 + 1 - 1 + 1 + 1 = 3, and g(-1) = -a + b - c + d = 3</p><p><strong>Step 4:</strong> At x = 0: f(0) = 0, and g(0) = d = 0</p><p><strong>Step 5:</strong> Since x = 0 is a root of multiplicity 2 in divisor, f'(0) = g'(0). Calculate f'(x) = 2010x^{2009} + 1010x^{1009} - 510x^{509} + 210x^{209} + 2x, so f'(0) = 0, giving g'(0) = c = 0</p><p><strong>Step 6:</strong> From steps 2, 3, 4, 5: a + b = 3 and -a + b = 3, yielding a = 0, b = 3</p><p>∴ g(x) = 3x² and g: ℝ → ℝ is well-defined</p><p>Answer: B</p>
Correct Answer: B