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07 Apr 2020, 08:47
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D
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Difficulty:
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Question Stats:
43% (02:57) correct
57%
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wrong
based on 118
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If a and b are integers such that \(2x^2 − ax + 2 > 0\) and \(x^2 − bx + 8 ≥ 0\) for all numbers x, then what is the largest possible value of \(2a − 6b\) ?
A. 30
B. 32
C. 34
D. 36
E. 38
Are You Up For the Challenge: 700 Level Questions
A. 30
B. 32
C. 34
D. 36
E. 38
Are You Up For the Challenge: 700 Level Questions
10 Apr 2020, 03:16
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Bunuel
If we want the largest possible value of 2a - 6b, we want the largest possible value of a and the smallest possible value of b such that 2x^2 - ax + 2 > 0 and x^2 - bx + 8 ≥ 0.
If we let f(x) = 2x^2 - ax + 2 and g(x) = x^2 - bx + 8, we see that both are parabolas opening upward. Therefore, if the vertex of f(x) is above the x-axis, then we have f(x) > 0. Similarly, if the vertex of g(x) is above or on the x-axis, then we have g(x) ≥ 0. In other words, we need to find the y-coordinate of the vertex (it has to be positive for f(x) and nonnegative for g(x)). To do that, we need to find the x-coordinate of the vertex first before we can find the y-coordinate. Recall that the formula for the x-coordinate of the vertex for y = ax^2 + bx + c is -b/2a. After the x-coordinate of the vertex is discerned, we substitute its value into the equation to determine the y-coordinate of the vertex.
For f(x): x = -(-a) / [2(2)] = a/4 → y = 2(a/4)^2 - a(a/4) + 2
2(a/4)^2 - a(a/4) + 2 > 0
2a^2/16 - a^2/4 > -2
-a^2/8 > -2
a^2 < 16
|a| < 4
Since a is an integer, the largest integer value for a such that |a| < 4 is 3.
Similarly for g(x): x = -(-b)/[2(1)] = b/2 → y = (b/2)^2 - b(b/2) + 8
(b/2)^2 - b(b/2) + 8 ≥ 0
b^2/4 - b^2/2 ≥ -8
-b^2/4 ≥ -8
b^2 ≤ 32
|b| ≤ √32 ≈ 5.7
Since b is an integer, the smallest integer value for b such that |b| ≤ √32 is -5.
Therefore, the largest possible value of 2a - 6b is 2(3) - 6(-5) = 6 + 30 = 36.
Answer: D
General Discussion
07 Apr 2020, 09:37
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Given,
2x^2 −ax + 2 > 0
2{ (x-a/4)^2 – a^2/16+1} > 0 ∀ x ∈R
-a^2/16+1 > 0
a ∈{ -3,-2,-1,0,1,2,3}
x^2 −bx + 8 ≥ 0
(x-b/2)^2 – b^2 /4 + 8>0 ∀ x ∈R -b^2 /4 + 8>0
b ∈{-5,-4,-3,-2,-1,0,1,2,3,4,5 }
So largest possible value of 2a – 6b = 3*2 – 6(-5) = 36
correct answer D
2x^2 −ax + 2 > 0
2{ (x-a/4)^2 – a^2/16+1} > 0 ∀ x ∈R
-a^2/16+1 > 0
a ∈{ -3,-2,-1,0,1,2,3}
x^2 −bx + 8 ≥ 0
(x-b/2)^2 – b^2 /4 + 8>0 ∀ x ∈R -b^2 /4 + 8>0
b ∈{-5,-4,-3,-2,-1,0,1,2,3,4,5 }
So largest possible value of 2a – 6b = 3*2 – 6(-5) = 36
correct answer D
