Let k be the largest integer such that the equation (x−1)² + 2kx + 11 = 0 has no real roots. If y is a positive real number, then the least possible value of k/(4y) + 9y is:

Options

1. 9
2. 2
3. 4
4. 6

Correct Answer

6

Explanation

Solution: Step 1: Expand the quadratic (x − 1)² + 2kx + 11 = x² − 2x + 1 + 2kx + 11 = x² + (2k − 2)x + 12 a = 1, b = 2k − 2, c = 12 Step 2: No real roots → discriminant < 0 Δ = b² − 4ac < 0 → (2k − 2)² − 48 < 0 → (2k − 2)² < 48 |2k − 2| < √48 = 4√3 → 2 − 4√3 < 2k < 2 + 4√3 → 1 − 2√3 < k < 1 + 2√3 Approximate: 1 − 3.464 < k < 1 + 3.464 → −2.464 < k < 4.464 Largest integer k = 4 Step 3: Minimize f(y) = k/(4y) + 9y Substitute k = 4 → f(y) = 1/y + 9y Derivative: f'(y) = −1/y² + 9 = 0 → y² = 1/9 → y = 1/3 Minimum value: f(1/3) = 1/(1/3) + 9 × (1/3) = 3 + 3 = 6 Answer: 6

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