<p>Let \(f(x) = (k-3)x^2 - 2kx + 3k - 6\) where \(x \in R\). If the range of \(f(x)\) is \([0, \infty)\), then the value of \(k\) can be:</p>
Step-by-Step Solution
Key Concept: For range to be [0, ∞), the function must be a perfect square (discriminant = 0) with positive leading coefficient, or a linear function with minimum value 0. This requires both the discriminant condition and coefficient conditions to be satisfied simultaneously.
<p><strong>Step 1:</strong> For range [0, ∞), the function must touch x-axis at exactly one point (minimum value = 0) and open upward.</p><p><strong>Step 2:</strong> If k ≠ 3, we have a quadratic. For minimum value 0: discriminant Δ = 0</p><p>Δ = (-2k)² - 4(k-3)(3k-6) = 0</p><p>4k² - 4(k-3)·3(k-2) = 0</p><p>4k² - 12(k-3)(k-2) = 0</p><p>k² - 3(k² - 5k + 6) = 0</p><p>k² - 3k² + 15k - 18 = 0</p><p>-2k² + 15k - 18 = 0</p><p>2k² - 15k + 18 = 0</p><p>(2k - 3)(k - 6) = 0</p><p>So k = 3/2 or k = 6</p><p><strong>Step 3:</strong> Check leading coefficient condition (k - 3 > 0):</p><p>• k = 3/2: k - 3 = -3/2 < 0 ✗ (parabola opens downward, range would be (-∞, 0])</p><p>• k = 6: k - 3 = 3 > 0 ✓ (parabola opens upward) ✓</p><p><strong>Step 4:</strong> Check k = 3: f(x) = -6x + 3, a linear function with range ℝ ✗</p><p>∴ Answer: k = 6 (Option C)
Correct Answer: C