Question

Let $f\left(x\right)$ be a real valued continuous function satisfying
$f^3\left(x\right)-5f^2\left(x\right)+10f\left(x\right)-12\le 0$
and $f^2\left(x\right)-7f\left(x\right)+12\le 0.$
A ray of light coming along the curve $y=f\left(x\right)$ from positive direction of $x-$axis and strikes the inner surface of mirror $y=2\sqrt{3x}$. If $l$ is the length of the reflected ray contained within the parabola $y^2=12x,$ then the value of $12l$ is

Answer 1

$f^3\left(x\right)-5f^2\left(x\right)+10f\left(x\right)-12\le 0$
at $f\left(x\right)=3,27-45+30-12=57-57=0$
$\Rightarrow (f\left(x\right)-3)(f^2\left(x\right)-2f\left(x\right)+4)\le 0$
$\Rightarrow (f\left(x\right)-3)({\left(f\right(x)-1)}^2+3)\le 0$
$\Rightarrow f\left(x\right)\le 3\dots \left(1\right)$

and $f^2\left(x\right)-7f\left(x\right)+12\le 0$
$\Rightarrow (f\left(x\right)-3)(f\left(x\right)-4)\le 0$
$\Rightarrow 3\le f\left(x\right)\le 4\dots \left(2\right)$

from (1) and (2), $f\left(x\right)=3$

Ray coming along the line $y=3$ strikes at $P(\frac{3}{4},3)$ . Then, after reflection it passes through the focus (3, 0).
$3=2a{t}_1\Rightarrow 3=2\times 3t_1\Rightarrow t_1=\frac{1}{2}$
$t_2=-2\{\because t_1{t}_2=-1\}\therefore Q\equiv (12,-12)$
Length PQ, $l=\sqrt{{(12-\frac{3}{4})}^2+{(-12-3)}^2}=\sqrt{\frac{{\left(45\right)}^2}{16}+225}=\sqrt{\frac{2025+3600}{16}}=\frac{75}{4}\therefore 12l=75\times 3=225$