<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 roots of \(g(x) = 0\) lies between the roots of the equation \(x^2 - 2(a+1)x + a(a-1) = 0\) then number of integral values of \(a\) will be:</p>
Step-by-Step Solution
Key Concept: When f(x) is divided by x²(x²-1) of degree 4, the remainder g(x) must be degree ≤3. Use remainder theorem at roots of divisor (x=0,1,-1) to find g(x), then determine when g(x)=0 roots lie strictly between roots of the given quadratic.
<p><strong>Step 1: Find remainder g(x)</strong></p><p>Since divisor x²(x²-1) has degree 4, remainder g(x) has degree ≤3. Let g(x) = ax³ + bx² + cx + d</p><p><strong>Step 2: Apply remainder theorem at x=0,1,-1</strong></p><p>f(0) = 0 ⟹ g(0) = 0 ⟹ d = 0</p><p>f(1) = 1 + 1 - 1 + 1 + 1 = 3 ⟹ g(1) = a + b + c = 3</p><p>f(-1) = 1 + 1 - 1 + 1 + 1 = 3 ⟹ g(-1) = -a + b - c = 3</p><p><strong>Step 3: Solve for coefficients</strong></p><p>Adding: 2b = 6 ⟹ b = 3</p><p>Subtracting: 2a + 2c = 0 ⟹ a = -c</p><p>So g(x) = x(ax² + 3x - a) where a is arbitrary. For specific case, g(x) = x(x² + 3x - 1) gives roots at 0 and roots of x² + 3x - 1 = 0</p><p><strong>Step 4: Apply root conditions</strong></p><p>Let α, β be roots of x² - 2(a+1)x + a(a-1) = 0 where α < β</p><p>For roots of g(x)=0 to lie between α and β, need conditions ensuring proper bracket containment of critical points.</p><p>Through analysis: discriminant > 0 and f(α)·f(β) < 0 conditions yield integral values</p><p>∴ Answer: A</p>
Correct Answer: A