Quadratic Equations
Roots of quadratic / exponential expressions
Grade 11
Question:
<p><strong>878.</strong> Find the number of integral values of <em>k</em> 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 λ² - 2λ + 1 = (λ-1)² and use Vieta's formulas to relate the sum and product of the given exponential roots to the quadratic coefficients. The constraint λ ∈ ℝ - {1} ensures the exponents are well-defined and non-zero.
<p><strong>Step 1:</strong> Simplify the exponents. Let (λ-1)² = t where t > 0 (since λ ≠ 1).</p><p>Root 1: e^(t + ln 3) = 3e^t</p><p>Root 2: e^(-t + ln 2) = 2e^(-t)</p><p><strong>Step 2:</strong> Apply Vieta's formulas for x² - (3k+1)x + 3k² - k + 2 = 0.</p><p>Sum of roots: 3e^t + 2e^(-t) = 3k + 1</p><p>Product of roots: (3e^t)(2e^(-t)) = 6 = 3k² - k + 2</p><p><strong>Step 3:</strong> Solve 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> Check the sum equation for each value.</p><p>For k = -1: 3e^t + 2e^(-t) = -2 (impossible, LHS > 0)</p><p>For k = 4/3: 3e^t + 2e^(-t) = 5</p><p>Let y = e^t (y > 1 since t > 0): 3y + 2/y = 5</p><p>3y² - 5y + 2 = 0 → (3y - 2)(y - 1) = 0</p><p>y = 2/3 (rejected, need y > 1) or y = 1 (rejected, need y > 1)</p><p><strong>Step 5:</strong> No valid integral values of k satisfy both conditions.</p><p>∴ Answer: 0</p>
Correct Answer: 0