MCQ
If $f(x)=\left\{\begin{array}{cl}x: & 0 \leq x<\frac{1}{2} \\ 1-x: & \frac{1}{2} \leq x<1\end{array}\right.$, then
  • $f ( x )$ is continuous at $x=\frac{1}{2}$
  • B
    $f ( x )$ is discontinuous at $x=\frac{1}{2}$
  • C
    $\lim _{x \rightarrow \frac{1}{2}^{-}} f(x)=1$
  • D
    $\lim _{x \rightarrow \frac{1}{2}^{+}} f(x)=1$

Answer

Correct option: A.
$f ( x )$ is continuous at $x=\frac{1}{2}$
(A)
$f \left(\frac{1}{2}\right)=1-\frac{1}{2}=\frac{1}{2}$
$\lim _{x \rightarrow \frac{1}{2}^{-}} f (x)=\lim _{x \rightarrow \frac{1}{2}^{-}}(x)=\frac{1}{2}$
$\lim _{x \rightarrow \frac{1}{2}^{+}} f(x)=\lim _{x \rightarrow \frac{1^{+}}{2}}(1-x)=1-\frac{1}{2}=\frac{1}{2}$
$\therefore \lim _{x \rightarrow \frac{1}{2}^{-}} f (x)=\lim _{x \rightarrow \frac{1}{2}^{+}} f (x)= f \left(\frac{1}{2}\right)$
$\therefore f (x)$ is continuous at $x=\frac{1}{2}$.

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