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
Explanation:
We have to maximize the value of 2a - 6b. Therefore let us look for the largest possible value of a and the smallest possible value of b.
If 2x2 – ax + 2 > 0 for all value of x, the graph of 2x2 – ax + 2 is above the x-axis or the roots of the quadratic equations 2x2 – ax + 2 = 0 are imaginary or its discriminant is less than 0. ∴ We have a2 – 16 < 0 or a2 < 16 or –4 < a < 4. ⇒ The largest possible value of a is 3.
If x2 – bx + 8 ≥ 0, the discriminant of the equation is less than or equal to zero. ⇒ b2 - 32 ≤ 0 ⇒ -4√2 ≤ b ≤ 4√2 ⇒ -5.64 ≤ b ≤ 5.64 ∴ The smallest possible value of b is –5.
Therefore the maximum possible value of 2a - 6b = 2(3) – 6(–5) = 36.
Hence, 36.
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