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Question: Answered & Verified by Expert
If $f(x)=\left\{\begin{array}{l}1+x^2, \text { when } 0 \leq x \leq 1 \\ 1-x, \text { when } x\gt1\end{array}\right.$, then
MathematicsContinuity and DifferentiabilityJEE Main
Options:
  • A $\lim _{x \rightarrow 1^{+}} f(x) \neq 0$
  • B $\lim _{x \rightarrow 1^{-}} f(x) \neq 2$
  • C $f(x)$ is discontinuous at $x=1$
  • D None of these
Solution:
2888 Upvotes Verified Answer
The correct answer is: $f(x)$ is discontinuous at $x=1$
$\lim _{x \rightarrow 1^+} f(x)=0$ and $\lim _{x \rightarrow 1^-} f(x)=1+1=2$.
Hence $f(x)$ is discontinuous at $x=1$.

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