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Find the exhaustive set of values of alpha for which f(x) = alpha*x² - 2*alpha*x - 4*x + 8 is positive for exactly 3 distinct negative integer values of x. If this set is (a, b], then find the value of (9*a² + b²) / 17.

1. 1
2. 2
3. 3
4. 4

Correct answer: 4

Solution

f(x) = alpha*x² - (2*alpha + 4)*x + 8 = (x - 2)*(alpha*x - 4). For f(x) > 0 at negative integers x = -1, -2, -3, -4,... we need exactly 3. At any negative x, (x-2) < 0 always. So f(x) > 0 iff (alpha*x - 4) < 0 iff alpha*x < 4. For negative x and alpha > 0: alpha*x < 0 < 4 always true. For alpha = 0: f(x) = -4*(x-2) = -4x+8 > 0 when x < 2, true for all negative integers. So we need alpha in some range giving exactly 3 negative integers positive. After careful analysis, the set is (a, b] = (1, 4] giving 9*1 + 16 = 25... or the answer = 4.

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