If $f : \mathrm{ℝ}\to \mathrm{ℝ}$ is defined by $f\left(x\right)=\left\{\begin{array}{ccc}x-1,& \text{ for }& x\le 1\\ 2-x^2,& \text{ for }& 1<x\le 3\\ x-10,& \text{ for }& 3<x<5\\ 2x,& \text{ for }& x\ge 5\end{array}\right.$then  set of points of discontinuty of f is

$\underset{x\to 1^{+}}{Lt}f\left(x\right)=\underset{x\to 1^{+}}{Lt}(x-1)=0,\underset{x\to 1^{-}}{Lt}f\left(x\right)=\underset{x\to 1^{-}}{Lt}\left(2-x^2\right)=2-1=1$

$\therefore f\left(x\right)$ is discontinuous at 1.

$\underset{x\to 3^{-}}{Lt}f\left(x\right)=\underset{x\to 3^{-}}{Lt}\left(2-x^2\right)=2-9=-7,\underset{x\to 3^{+}}{Lt}f\left(x\right)=\underset{x\to 3^{+}}{Lt}(x-10)=3-10=-7$

$\therefore f\left(x\right)$ is discontinuous at 3.

$\underset{x\to 5^{-}}{Lt}f\left(x\right)=\underset{x\to 5^{-}}{Lt}(x-10)=5-10=-5,\underset{x\to 5^{+}}{Lt}f\left(x\right)=\underset{x\to 5^{+}}{Lt}2x=10$

$\therefore f\left(x\right)$ is discontinuous at 5.