<p><strong>168.</strong> Let \(x_1\) and \(x_2\) (\(x_1 > x_2\)) are the roots of the equation \(9^{\log_9(x^2 - 4x + 5)} = x - 1\), then the value of \(\tan(x_1)\pi + \sec(x_2)\pi\) is:</p>
Step-by-Step Solution
Key Concept: Recognize that 9^(log₉(u)) = u for u > 0, which transforms the equation to x² - 4x + 5 = x - 1. Then verify that the quadratic's argument in the logarithm must remain positive for valid roots.
<p><strong>Step 1:</strong> Apply the property a^(log_a(u)) = u. Since 9^(log₉(x² - 4x + 5)) = x² - 4x + 5, the equation becomes:</p><p>x² - 4x + 5 = x - 1</p><p><strong>Step 2:</strong> Simplify to get x² - 5x + 6 = 0</p><p><strong>Step 3:</strong> Factor: (x - 2)(x - 3) = 0, giving x₁ = 3 and x₂ = 2</p><p><strong>Step 4:</strong> Verify validity: x² - 4x + 5 = (x-2)² + 1 > 0 for all x ✓, and both x - 1 = 2, 1 > 0 ✓</p><p><strong>Step 5:</strong> Calculate tan(x₁π) + sec(x₂π):<br/>• tan(3π) = tan(π) = 0 (since tan has period π)<br/>• sec(2π) = 1/cos(2π) = 1/1 = 1</p><p>∴ Answer: 0 + 1 = <strong>1</strong></p>
Correct Answer: D