<p>The integral values of \(m\) for which the roots of the equation \(mx^2 + (2m-1)x + (m-2) = 0\) are rational are given by the expression [where \(n\) is integer]</p>
Step-by-Step Solution
Key Concept: For roots to be rational, the discriminant must be a perfect square. Calculate Δ = (2m-1)² - 4m(m-2) and find which integer values of m make it a perfect square.
<p><strong>Step 1:</strong> For the equation mx² + (2m-1)x + (m-2) = 0 to have rational roots, the discriminant must be a perfect square (and m ≠ 0).</p><p><strong>Step 2:</strong> Calculate the discriminant:<br/>Δ = (2m-1)² - 4m(m-2)<br/>Δ = 4m² - 4m + 1 - 4m² + 8m<br/>Δ = 4m + 1</p><p><strong>Step 3:</strong> For rational roots, 4m + 1 must be a perfect square. Let 4m + 1 = k² for some non-negative integer k.<br/>Then: 4m = k² - 1 = (k-1)(k+1)<br/>So: m = (k-1)(k+1)/4</p><p><strong>Step 4:</strong> For m to be an integer, (k-1)(k+1) must be divisible by 4. Since (k-1) and (k+1) are consecutive even numbers when k is odd, their product is divisible by 8. When k is even, their product is not divisible by 4.</p><p><strong>Step 5:</strong> Therefore k must be odd. Let k = 2n + 1 for integer n ≥ 0:<br/>m = (2n)(2n+2)/4 = n(n+1)</p><p>∴ Answer: A (m = n(n+1) where n is any integer)</p>
Correct Answer: A