<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 ax² + bx + c = 0 with a > 0, one root lies below λ and the other above λ if and only if f(λ) < 0. This is because the parabola opens upward, so λ must lie between the two roots.
<p><strong>Step 1:</strong> Let f(x) = 2x² - 2(2λ + 1)x + λ(λ + 1). Since a = 2 > 0, for one root to be less than λ and the other greater than λ, we need f(λ) < 0.</p><p><strong>Step 2:</strong> Calculate f(λ):</p><p>f(λ) = 2λ² - 2(2λ + 1)λ + λ(λ + 1)</p><p>= 2λ² - 4λ² - 2λ + λ² + λ</p><p>= 2λ² - 4λ² + λ² - 2λ + λ</p><p>= -λ² - λ</p><p>= -λ(λ + 1)</p><p><strong>Step 3:</strong> For the required condition: f(λ) < 0</p><p>-λ(λ + 1) < 0</p><p>λ(λ + 1) > 0</p><p>This inequality holds when λ < -1 or λ > 0.</p><p><strong>Step 4:</strong> Since we need the least non-negative integral value of λ, we check λ ≥ 0:</p><p>From λ > 0, the least non-negative integer is λ = 1.</p><p>∴ Answer: D (which is 1)</p>
Correct Answer: D