<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(A)) = A, so the equation simplifies to x² - 4x + 5 = x - 1. The constraint that the logarithm argument must be positive (x² - 4x + 5 > 0) is always satisfied, but the domain restriction x - 1 > 0 is crucial for validity.
<p><strong>Step 1:</strong> Simplify using the property 9^(log₉(A)) = A (when A > 0):<br>x² - 4x + 5 = x - 1</p><p><strong>Step 2:</strong> Rearrange to standard form:<br>x² - 5x + 6 = 0<br>(x - 2)(x - 3) = 0<br>Roots: x = 2 or x = 3</p><p><strong>Step 3:</strong> Apply domain restrictions. Since 9^(log₉(...)) must equal x - 1, we need x - 1 > 0, so x > 1.<br>Both x = 2 and x = 3 satisfy this initially. However, checking x² - 4x + 5 > 0:<br>• At x = 2: 4 - 8 + 5 = 1 > 0 ✓<br>• At x = 3: 9 - 12 + 5 = 2 > 0 ✓<br>Both are valid algebraically.</p><p><strong>Step 4:</strong> Re-examine the original equation more carefully. The standard interpretation yields x₁ = 3 and x₂ = 2 (with x₁ > x₂).</p><p><strong>Step 5:</strong> Calculate tan(x₁π) + sec(x₂π):<br>tan(3π) = 0 (since tan has period π and tan(0) = 0)<br>sec(2π) = 1/cos(2π) = 1/1 = 1<br>Therefore: 0 + 1 = 1</p><p>∴ Answer: D</p>
Correct Answer: D