Question

Find the number of integral value(s) of $a\in (8,\infty )$ for which the expression ${\log }_{10}\left(4x^2-ax+3\right)$ is always non-negative for all $x\in [1,3]$.

Answer 1

Answer: 0

Let $y={\log }_{10}\left(4x^2-ax+3\right)$
Since $y$ is non-negative for all $x\in [1,3]$
$\therefore y\ge 0\text{ for all }x\in [1,3]$
$\Rightarrow \left(4x^2-ax+3\right)\ge 1\text{ for all }x\in [1,3]$
$4x^2-ax+2\ge 0\text{ for all }x\in [1,3]$
Case-I : When $\frac{a}{8}\le 1$
This case is not possible.
Case-II : $1<\frac{a}{8}<3\Rightarrow a\in (8,24)$
now, $f\left(\frac{a}{8}\right)\ge 0\Rightarrow a\in [-4\sqrt{2},4\sqrt{2}]$
This is not possible.
Case-III : $\frac{a}{8}\ge 3\Rightarrow a\ge 24$
$f(3)\ge 0$
$a\le \frac{38}{3}$, which is not possible.
Hence, from the above cases $a\in \{\varphi \}$.