Question

If the range of values of $'a'$ for which $f\left(\right.x\left.\right)=\left(log\right)_{a}\left(4 a x - x^{2}\right)$ is strictly increasing $\forall x\in \left[\frac{3}{2} , 2\right]$ is $\left(\right.p,q\left]\right.\cup\left(\right.r,\infty \left.\right)$ then the value of $\left(\right.2p+4q+r\left.\right)$ equals

Answer 1

$f^{'}\left(\right.x\left.\right)=\frac{\left(\right. 4 a - 2 x \left.\right)}{\left(4 ax - x^{2}\right) loga}\geq 0$
$f^{'}\left(\right.x\left.\right)=\frac{\left(\right. x - 2 a \left.\right)}{x \left(\right. x - 4 a \left.\right) loga}\geq 0$
$C-1:a\in \left(\right.1,\infty \left.\right)\Rightarrow \frac{\left(\right. x - 2 a \left.\right)}{x \left(\right. x - 4 a \left.\right)}\geq 0$

$x\in \left(\right.0,2a\left.\right)$ so always $\uparrow$
$C-2:a\in \left(\right.0,1\left.\right)$
$\frac{\left(\right. x - 2 a \left.\right)}{x \left(\right. x - 4 a \left.\right)} < 0$

$x\in \left(\right.2a,4a\left.\right)$
$2a\leq \frac{3}{2}\Rightarrow a\leq \frac{3}{4}$ and $4a>2\Rightarrow a>\frac{1}{2}$
So $a\in (\frac{1}{2} , \frac{3}{4}]\cup [1,\infty )$