<p>The least non-negative integral value of \(\lambda\) for which the equation \(2x^2 - 2(2\lambda+1)x + \lambda(\lambda+1) = 0\) has one root less than \(\lambda\) and other root greater than \(\lambda\), is equal to:</p>
Step-by-Step Solution
Key Concept: For a quadratic f(x) = ax² + bx + c with a > 0, one root lies below α and the other above α if and only if f(α) < 0. This converts the root separation problem into a single inequality evaluation.
<p><strong>Step 1: Apply the root separation condition</strong></p><p>For quadratic f(x) = 2x² - 2(2λ+1)x + λ(λ+1) with positive leading coefficient, one root is less than λ and the other is greater than λ if and only if f(λ) < 0.</p><p><strong>Step 2: Calculate f(λ)</strong></p><p>f(λ) = 2λ² - 2(2λ+1)·λ + λ(λ+1)</p><p>= 2λ² - 2λ(2λ+1) + λ(λ+1)</p><p>= 2λ² - 4λ² - 2λ + λ² + λ</p><p>= (2 - 4 + 1)λ² + (-2 + 1)λ</p><p>= -λ² - λ</p><p>= -λ(λ+1)</p><p><strong>Step 3: Apply the inequality</strong></p><p>We need: f(λ) < 0</p><p>-λ(λ+1) < 0</p><p>λ(λ+1) > 0</p><p><strong>Step 4: Solve the inequality</strong></p><p>λ(λ+1) > 0 when λ < -1 or λ > 0</p><p><strong>Step 5: Find the least non-negative integral value</strong></p><p>Since λ must be non-negative (λ ≥ 0), we need λ > 0.</p><p>The least non-negative integer satisfying λ > 0 is λ = 1.</p><p><strong>Verification:</strong> For λ = 1: f(1) = -1(2) = -2 < 0 ✓</p><p>For λ = 0: f(0) = 0 (not < 0) ✗</p><p>∴ Answer: D (or 1, depending on options provided)</p>
Correct Answer: D