Question

The smallest value of $\left(\frac{-8p}{7}\right)$ for which $|x^2-5x+7-p|=6+|x^2-5x+1-p|$ for all $x\in [-1,3]$ is

Answer 1

$|x^2-5x+7-p|=6+|x^2-5x+1-p|$
or, $\left|\right(x^2-5x+1-p)+6|=|x^2-5x+1-p|+|6|$
We must have
$6(x^2-5x+1-p)\ge 0\forall x\in [-1,3]$
$\Rightarrow x^2-5x+(1-p)\ge 0\forall x\in [-1,3]$

$y=x^2-5x+(1-p)$ is an upward parabola with vertex at $(\frac{5}{2},-p-\frac{21}{4})$
$\frac{5}{2}\in [-1,3]$
$\therefore -p-\frac{21}{4}\ge 0$
$\Rightarrow p+\frac{21}{4}\le 0$
$\Rightarrow p\le \frac{-21}{4}$
$\therefore p\in (-\infty ,\frac{-21}{4}]$
$\Rightarrow$ The smallest value of $\left(\frac{-8p}{7}\right)=\frac{-8}{7}\times \frac{-21}{4}=6$