Question

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

Answer 1

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