<p>Find the number of integral values of \(k\) for which \(e^{\lambda^2 - 2\lambda + 1 + \ln 3}\) and \(e^{-(\lambda^2 - 2\lambda + 1) + \ln 2}\), where \(\lambda \in R - \{1\}\) are the roots of the equation \(x^2 - (3k+1)x + 3k^2 - k + 2 = 0\).</p>
Step-by-Step Solution
Key Concept: Recognize that the exponents simplify to (λ²-2λ+1) = (λ-1)², and use Vieta's formulas by setting the sum and product of roots equal to the symmetric functions of the given exponential expressions. The constraint that both expressions must be real roots forces a relationship between k and the quadratic's parameters.
<p><strong>Step 1:</strong> Simplify the exponents. Note that λ² - 2λ + 1 = (λ-1)². Let t = (λ-1)² where t > 0 (since λ ∈ ℝ - {1}).</p><p>The roots become: α = e^(t + ln 3) = 3e^t and β = e^(-t + ln 2) = 2e^(-t)</p><p><strong>Step 2:</strong> Apply Vieta's formulas. For the equation x² - (3k+1)x + 3k² - k + 2 = 0:</p><p>• Sum of roots: α + β = 3e^t + 2e^(-t) = 3k + 1</p><p>• Product of roots: α·β = (3e^t)(2e^(-t)) = 6 = 3k² - k + 2</p><p><strong>Step 3:</strong> From the product equation: 6 = 3k² - k + 2</p><p>3k² - k - 4 = 0</p><p>(3k - 4)(k + 1) = 0</p><p>k = 4/3 or k = -1</p><p><strong>Step 4:</strong> Verify with the sum condition. For each k value, check if there exists t > 0 satisfying 3e^t + 2e^(-t) = 3k + 1:</p><p>• For k = -1: 3e^t + 2e^(-t) = -2. Since e^t > 0 and e^(-t) > 0, LHS > 0, but RHS < 0. ✗</p><p>• For k = 4/3: 3e^t + 2e^(-t) = 5. By AM-GM inequality, 3e^t + 2e^(-t) ≥ 2√(6) ≈ 4.899 < 5. Solution exists. ✓</p><p><strong>Step 5:</strong> Among {-1, 4/3}, only k = 4/3 is invalid as an integral value.</p><p>∴ Answer: <strong>0</strong> integral values of k</p>
Correct Answer: 0