<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 the range to be [0, ∞), the function must be a non-negative quadratic (k ≠ 3) with vertex touching the x-axis exactly once, meaning discriminant = 0 and leading coefficient > 0.
<p><strong>Step 1:</strong> For range [0, ∞), we need f(x) ≥ 0 for all x ∈ ℝ. This requires:</p><ul><li>f(x) is a quadratic (k ≠ 3)</li><li>Leading coefficient > 0, so k - 3 > 0 ⟹ k > 3</li><li>Discriminant Δ = 0 (parabola touches x-axis once)</li></ul><p><strong>Step 2:</strong> Calculate discriminant:</p><p>Δ = (-2k)² - 4(k-3)(3k-6)</p><p>= 4k² - 4(k-3)·3(k-2)</p><p>= 4k² - 12(k-3)(k-2)</p><p>= 4k² - 12(k² - 5k + 6)</p><p>= 4k² - 12k² + 60k - 72</p><p>= -8k² + 60k - 72</p><p><strong>Step 3:</strong> Set Δ = 0:</p><p>-8k² + 60k - 72 = 0</p><p>k² - 7.5k + 9 = 0</p><p>2k² - 15k + 18 = 0</p><p>(2k - 3)(k - 6) = 0</p><p>⟹ k = 3/2 or k = 6</p><p><strong>Step 4:</strong> Apply condition k > 3:</p><p>k = 3/2 < 3 ✗ (rejected)</p><p>k = 6 > 3 ✓ (accepted)</p><p>∴ Answer: k = 6 (Option C)</p>
Correct Answer: C