<p><strong>32.</strong> The range of value of \(\lambda\) for which the expression \(\dfrac{2x^2 - 5x + 3}{4x - \lambda}\) can take all real values for \(x \in R - \left\{\dfrac{\lambda}{4}\right\}\), is:</p>
Step-by-Step Solution
Key Concept: For the rational expression to take all real values, rearrange as a quadratic in x: 2x² - (5+4λ)x + (3+λy) = 0 must have real solutions for all real y. This requires the discriminant in x to be non-negative for all y, which is only possible if the coefficient of y² in the discriminant equals zero.
<p><strong>Step 1:</strong> Let y = (2x² - 5x + 3)/(4x - λ). Rearrange to get: 2x² - 5x + 3 = y(4x - λ)</p><p><strong>Step 2:</strong> Rearrange as a quadratic in x: 2x² - (5 + 4y)x + (3 + λy) = 0</p><p><strong>Step 3:</strong> For the expression to take ALL real values of y, this quadratic must have real solutions for every real y. This requires Δ ≥ 0 for all y.</p><p><strong>Step 4:</strong> Calculate discriminant: Δ = (5 + 4y)² - 4(2)(3 + λy) = 25 + 40y + 16y² - 24 - 8λy = 16y² + (40 - 8λ)y + 1</p><p><strong>Step 5:</strong> For Δ ≥ 0 for ALL y ∈ ℝ, the quadratic in y must have no real roots or be always positive. This requires: (40 - 8λ)² - 4(16)(1) ≤ 0</p><p><strong>Step 6:</strong> Solve: (40 - 8λ)² ≤ 64 → |40 - 8λ| ≤ 8 → 32 ≤ 8λ ≤ 48 → 4 ≤ λ ≤ 6</p><p>∴ Answer: B (4 ≤ λ ≤ 6)</p>
Correct Answer: B