Question

If ${\log }_2(a{x}^2+x+a)\ge 1\forall x\in R$, then exhaustive set of value of $a$ is

Answer 1

$10g_2(a{x}^2+x+a)\ge 1$

$\Rightarrow a{x}^2+x+a\ge 2$

$\Rightarrow a{x}^2+x+a-2\ge 0$

$\therefore a>0-\left(i\right)$

$\Delta \le 0$

$\Rightarrow 1-4a(a-2)\le 2$

$\Rightarrow 1-4a^2+8a\le 0$

$\Rightarrow 4a^2-8a-1\ge 0$

$\therefore a+(-\infty ,\frac{2-\sqrt{5}}{2})\cup (\frac{2+\sqrt{5}}{2},\infty )-\left(ii\right)$

From $\left(i\right)$ & $\left(ii\right)$

$aϵ(\frac{2+\sqrt{5}}{2},\infty )$