If f(x) = begin{cases} 1 + x^2, & text{when } | vedclass

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(C) To check the continuity at $x = 1$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$.$1$. Left-hand limit $(LHL)$: $\lim_{x

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$f(x)$ is discontinuous at $x = 1$

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(C) To check the continuity at $x = 1$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$.$1$. Left-hand limit $(LHL)$: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} (1 + x^2) = 1 + (1)^2 = 2$.$2$. Right-hand limit $(RHL)$: $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} (1 - x) = 1 - 1 = 0$.Since $\lim_{x 	o 1^-} f(x) 
e \lim_{x 	o 1^+} f(x)$,the limit does not exist at $x = 1$.Therefore,$f(x)$ is discontinuous at $x = 1$.

If $f(x) = \begin{cases} 1 + x^2, & \text{when } 0 \le x \le 1 \\ 1 - x, & \text{when } x > 1 \end{cases}$,then

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