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
Correct Answer: 36
$2 x^{2}-a x+2>0 \forall x \in R$
$\Rightarrow \Delta<0$
$\Rightarrow$ $a^{2}-4 \times 2 \times 2<0$
$\Rightarrow a^{2}<16$
$\Rightarrow-4$x^{2}-b x+8 \geq 0 \forall x \in R$
$\Rightarrow b^{2}-4(8) \leq 0$
$\Rightarrow-4 \sqrt{2} \leq b \leq 4 \sqrt{2}$
As b is integer $-5 \leq b \leq 5$
Therefore, maximum possible value of 2a – 6b is 2(3) – 6(–5) = 36
$2 x^{2}-a x+2>0 \forall x \in R$
$\Rightarrow \Delta<0$
$\Rightarrow$ $a^{2}-4 \times 2 \times 2<0$
$\Rightarrow a^{2}<16$
$\Rightarrow-4
$\Rightarrow b^{2}-4(8) \leq 0$
$\Rightarrow-4 \sqrt{2} \leq b \leq 4 \sqrt{2}$
As b is integer $-5 \leq b \leq 5$
Therefore, maximum possible value of 2a – 6b is 2(3) – 6(–5) = 36
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