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

2x2-ax+2>0∀x∈R
⇒ Δ<0
⇒ a2-4×2×2<0
⇒ a2<16
⇒ -4<ax2-bx+8≥0∀x∈R
⇒ b2-4(8)≤0
⇒ -4√2≤b≤4√2
As b is integer -5≤b≤5
Therefore, maximum possible value of 2a – 6b is 2(3) – 6(–5) = 36