Let $f\left(x\right)=\left\{\begin{array}{l}a\frac{\left|x^2-x-2\right|}{2+x-x^2},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x<2\\ b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em},x=2\\ \frac{x-\left[x\right]}{x-2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em},x>2\end{array}\right.$

If $f$ is continuous at $x = 2$ then

$\frac{x-\left[x\right]}{x-2}=1$for $2<x<3$

So, $b=\underset{x\to 2+}{lim} f\left(x\right)=1.$ For $1<x<2,\frac{a\left|x^2-x-2\right|}{-\left(x^2-x-2\right)}$

$=\frac{a|x-2||x+1|}{-(x-2)(x+1)}=\frac{-a(x-2)(x+1)}{-(x-2)(x+1)}=a$

For$f$ to be continuous$a = b = 1$.