Question

If a and b are integers such that 2x2 −ax + 2 > 0 and x2 −bx + 8 ≥ 0 for all real numbers x, then the largest possible value of 2a−6b is

Answer 1

Correct Answer: 36

Step 1: Analyze the Inequality 

We are given the inequality:

\[ 2x^2 - ax + 2 > 0 \quad \forall x \in \mathbb{R} \]

This quadratic is always positive for all real \( x \), so its discriminant must be negative:

\[ \Delta = a^2 - 4 \cdot 2 \cdot 2 < 0 \Rightarrow a^2 < 16 \Rightarrow -4 < a < 4 \]

Step 2: Consider \( a \) as an Integer

The largest possible integer value of \( a \) satisfying \( a^2 < 16 \) is \( a = 3 \).

Step 3: Analyze Second Inequality

Next inequality:

\[ -4 < ax^2 - bx + 8 \geq 0 \quad \forall x \in \mathbb{R} \]

This quadratic lies entirely within the interval \( (-4, 0] \) for all real \( x \), implying that its maximum value is at most 0 and minimum value is more than -4. For this to happen, it must have real roots (or just touch the x-axis), and the discriminant condition must hold:

\[ \Delta = b^2 - 4ac \leq 0 \quad \text{where } a = a, c = 8 \]

We assume coefficient of \( x^2 \) is 1 (for normalization), so:

\[ b^2 \leq 4 \cdot 8 = 32 \Rightarrow -\sqrt{32} \leq b \leq \sqrt{32} \Rightarrow -4\sqrt{2} \leq b \leq 4\sqrt{2} \]

Since \( \sqrt{2} \approx 1.414 \), then \( 4\sqrt{2} \approx 5.656 \), and so \( b \in [-5, 5] \) (integer values).

Step 4: Maximize \( 2a - 6b \)

We want to maximize:

\[ 2a - 6b \]

Given \( a = 3 \) (max possible), and \( b = -5 \) (min possible), we get:

\[ 2a - 6b = 2(3) - 6(-5) = 6 + 30 = \boxed{36} \]

Final Answer:

\[ \boxed{36} \]