If a and b are integers such that 2x^2 − ax + 2 0 and x^2 − bx + 8 ≥

Question

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

ScottTargetTestPrep · Accepted Answer

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) / = 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)/ = 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

preetamsaha · Answer

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

samarpan.g28 · Answer

We have to use the concept of maxima-minima here. for any quadratic eqn ax^2 + bx + c = 0, maxima or minima is (-b)/(2a) (depending on sign of a).
For getting the largest value of 2a-6b, a shall be the maximum and b shall be the minimum.
\(2x^2 − ax + 2 > 0\) for this eqn, x is maximum at -(-a)/(2*2)=a/4. Put a/4 in the same eqn in place of x and we will get a^2<16 i.e. a<4 and a>-4 or a has a value range of {-3,-2,-1,0,1,2,3}. We will take the maximum value i.e. a=3.

\(x^2 − bx + 8 ≥ 0\) for this eqn, x is minimum at -(-b)/(2)=b/2. Put b/2 in the same eqn in place of x and we will get b^2<=32 i.e. b<=5.something or b>=-5.something. Since a and b are integers, the minimum value of b will be b=-5.

Therefore, 2*3-6*-5=36(D).

SKP220292 · Answer

That's not correct. At a/4, you get the minima (not maxima) of the quadratic expression (not of x itself). This is an upward facing parabola, which means there is no maxima, there is only a minima. Maxima is infinite for an upward facing parabola. You got the answer correct, but the reasoning is incorrect and math is conceptually wrong. Conceptual mistakes could expose gaps in other questions, so beware of that.

For the first expression, you find the minima which is x = a/4. Replace this value of x in the equation and use the logic - the minima of the quadratic must be greater than 0 for the quadratic inequality to hold true. That gives you |a|<4 (not a<4). Apply similar principle to the second equation and go from there.

If a and b are integers such that 2x^2 − ax + 2 > 0 and x^2 − bx + 8 ≥

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions