<p>The equation \(x^3 - 6x^2 + 9x + \lambda = 0\) have exactly one root in (1, 3) then find the number of integral values of \([\lambda + 1]\) (where [.] denotes the greatest integer function)</p>
Step-by-Step Solution
Key Concept: Factor the cubic as x(x-3)² + λ = 0, then analyze f(x) = -x(x-3)² on (1,3). For exactly one root in (1,3), λ must lie strictly between the maximum and minimum values of -x(x-3)² on this interval.
<p><strong>Step 1:</strong> Rewrite the equation as x³ - 6x² + 9x + λ = 0, or equivalently: -x(x-3)² = λ</p><p><strong>Step 2:</strong> Let f(x) = -x(x-3)². Find critical points: f'(x) = -(x-3)² - 2x(x-3) = -(x-3)[(x-3) + 2x] = -(x-3)(3x-3) = -3(x-3)(x-1)</p><p>Critical points: x = 1 and x = 3 (boundaries of our interval)</p><p><strong>Step 3:</strong> Evaluate f(x) at critical point and check behavior on (1,3):</p><p>• f(1) = -1(1-3)² = -1(4) = -4</p><p>• f(3) = -3(3-3)² = 0</p><p>• f(2) = -2(2-3)² = -2(1) = -2</p><p><strong>Step 4:</strong> On interval (1,3), f(x) is continuous. f'(x) = -3(x-3)(x-1):</p><p>• For x ∈ (1,3): (x-3) < 0 and (x-1) > 0, so f'(x) > 0 (f is increasing)</p><p>Thus f increases from f(1) = -4 to f(3) = 0 on [1,3]</p><p><strong>Step 5:</strong> For exactly one root in (1,3), the horizontal line y = λ must intersect the curve y = f(x) exactly once in the open interval (1,3).</p><p>This requires: -4 < λ < 0 (strict inequalities for open interval)</p><p><strong>Step 6:</strong> Find integral values of [λ+1]:</p><p>• When -4 < λ < 0, we have -3 < λ+1 < 1</p><p>• Since λ+1 ∈ (-3, 1), the possible values of [λ+1] are: -3, -2, -1, 0</p><p>• That's 4 integral values</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: 4