<p>Find the number of positive integral values of <i>k</i> for which \(kx^2 + (k - 3)x + 1 < 0\) for at least one positive <i>x</i>.</p>
Step-by-Step Solution
Key Concept: For a quadratic expression to be always positive for all real x, the discriminant must be negative and the leading coefficient must be positive. We need to find when kx² + (k-3)x + 1 > 0 for all x ∈ ℝ.
<p><strong>Step 1:</strong> For kx² + (k-3)x + 1 > 0 for all x ∈ ℝ, we need:</p><p>(i) k > 0 (parabola opens upward)</p><p>(ii) Discriminant < 0 (no real roots)</p><p><strong>Step 2:</strong> Calculate the discriminant:</p><p>Δ = (k-3)² - 4(k)(1) < 0</p><p>Δ = k² - 6k + 9 - 4k < 0</p><p>Δ = k² - 10k + 9 < 0</p><p><strong>Step 3:</strong> Factorize the quadratic inequality:</p><p>k² - 10k + 9 < 0</p><p>(k - 1)(k - 9) < 0</p><p><strong>Step 4:</strong> Solve the inequality:</p><p>(k - 1)(k - 9) < 0 implies 1 < k < 9</p><p><strong>Step 5:</strong> Find positive integral values of k in the range 1 < k < 9:</p><p>k ∈ {2, 3, 4, 5, 6, 7, 8}</p><p><strong>Step 6:</strong> Count the positive integers:</p><p>Number of values = 7</p><p><strong>∴ Answer: 7</strong></p>
Correct Answer: 7