If $\mathrm{a}$ and $\mathrm{b} (\mathrm{a}>\mathrm{b})$ are points of discountinuity of the function

$\text{⨍ (x)=}\left\{\begin{array}{ccc}3-2{\mathrm{x}}^2& ,& \text{⨍ or x≤0}\\ 2\mathrm{x}+3& ,& \text{⨍ or 0<x≤1}\\ 2{\mathrm{x}}^2-3\mathrm{x}& ,& \text{⨍ or 1<x<2}\\ 2\mathrm{x}-3& ,& \text{⨍ or 2≤x<3}\\ \left|\mathrm{x}\right|& ,& \text{⨍ or x≥3}\end{array}\right.$

then $3\mathrm{a}-\mathrm{b}=$

$\begin{array}{l}\underset{x\to 0}{\mathrm{Lt}}f\left(x\right)={\mathrm{Lt}}_{x\to 0-}\left(3-2x^2\right)=3,\underset{x\to 0+}{\mathrm{Lt}}f\left(x\right)=\underset{x\to 0+}{\mathrm{Lt}}2x+3=3\\ {\mathrm{Lt}}_{x\to 1-}f\left(x\right)={\mathrm{Lt}}_{x\to 1-}2x+3=5,\underset{x\to 1+}{Lt}f\left(x\right)=\underset{x\to 1+}{Lt}2x^2-3x=-1\text{. }\\ \underset{x\to 2-}{\mathrm{Lt}}f\left(x\right)=\underset{x\to 2-}{\mathrm{Lt}}2x^2-3x=0,\underset{x\to 2+}{\mathrm{Lt}}f\left(x\right)=\underset{x\to 2+}{Lt}2x-3=1\Rightarrow a=2,b=1\Rightarrow 3a-b=5\text{. }\\ \end{array}$