(i) On the same axes, sketch the graphs of $y=x+2+\frac{1}{x-1}$ and $y=|2 x+2|$, stating the coordinates of any points of intersections with the axes, turning points and the equations of any asymptotes.
[5]
(ii) Hence solve the inequality
$$
x+2+\frac{1}{x-1}>|2 x+2|
$$
giving your answers in exact form.
[5]
(i)

(ii)$\frac{-1-\sqrt{37}}{6}<x<\frac{1-\sqrt{5}}{2}$ or $1<x<\frac{1+\sqrt{5}}{2}$
(i)

(ii) Intersection between
$y=x+2+\frac{1}{x-1}$ and $y=2 x+2$
$2x+2 =x+2+\frac{1}{x-1} $
$x(x-1) =1 $
$x^2-x-1 =0 $
$x =\frac{1 \pm \sqrt{5}}{2}$
Intersection between
$y=x+2+\frac{1}{x-1}$ and $y=-2 x-2$
$-2 x-2 =x+2+\frac{1}{x-1} $
$-3 x-4 =\frac{1}{x-1} $
$(3 x+4)(1-x) =1 $
$-3 x^2-x+3 =0 $
$x =\frac{-1 \pm \sqrt{37}}{6}$
From the graph, the intersection occurs at $x<-1$. So, $x=\frac{-1-\sqrt{37}}{6}$
From the graphs in (i), the solution is $\frac{-1-\sqrt{37}}{6}<x<\frac{1-\sqrt{5}}{2}$ or $1<x<\frac{1+\sqrt{5}}{2}$