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Given:

$f(x)=\left\{\begin{array}{c}1+x^{2}, \text { if } 0 \leq \mathrm{x} \leq 1 \\ 2-x, \text { if } x>1\end{array}\right.$

We observe

$(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}\left(1+(1-h)^{2}\right)=\lim _{h \rightarrow 0}\left(2+h^{2}-2 h\right)=2$

(RHL at $x=1$ ) $=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)==\lim _{h \rightarrow 0}(2-(1+h))=\lim _{h \rightarrow 0}(1-h)=1$

$\therefore \lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)$

Thus, f(x) is discontinuous at x = 1.