<p>79. The interval of \(a\) for which the equation \(\tan^2 x - (a-4)\tan x + 4 - 2a = 0\) has at least one solution \(\forall x \in [0, \pi/4]\) is</p>
<p>(1) \(a \in (2, 3)\)</p>
<p>(2) \(a \in [2, 3]\)</p>
<p>(3) \(a \in (1, 4)\)</p>
<p>(4) \(a \in [1, 4]\)</p>
Step-by-Step Solution
Key Concept: Substitute t = tan(x) where t ∈ [0,1] for x ∈ [0,π/4], then find which values of 'a' allow the quadratic t² - (a-4)t + 4-2a = 0 to have at least one root in [0,1].
<p><strong>Step 1:</strong> Substitute t = tan(x). Since x ∈ [0, π/4], we have t ∈ [0, 1].</p><p>The equation becomes: t² - (a-4)t + 4-2a = 0</p><p><strong>Step 2:</strong> Rearrange to express a in terms of t:</p><p>t² - at + 4t + 4 - 2a = 0</p><p>t² + 4t + 4 = a(t + 2)</p><p>a = (t² + 4t + 4)/(t + 2) = (t+2)²/(t+2) = t + 2</p><p><strong>Step 3:</strong> For at least one solution to exist in x ∈ [0, π/4], we need at least one root t ∈ [0, 1].</p><p>Since a = t + 2 and t ∈ [0, 1], the range of a is:</p><p>a ∈ [0+2, 1+2] = [2, 3]</p><p><strong>Step 4:</strong> Verify boundaries:</p><p>• When t = 0: a = 2, equation becomes t² + 4t + 4 = 0 → (t+2)² = 0 ✓</p><p>• When t = 1: a = 3, equation becomes t² - (3-4)t + 4-6 = 0 → t² + t - 2 = 0 → (t-1)(t+2) = 0 ✓</p><p>∴ Answer: a ∈ [2, 3]</p>
Correct Answer: 2