MCQ
The points of discontinuity of the function $\text{f(x)}=\begin{cases}\frac{1}{5}(2\text{x}^2+3),&\text{x}\leq1\\6-5\text{x},&1<\text{x}<3\\\text{x}-3,&\text{x}\geq3\end{cases}$ is $($are$) :$
  • A
    $x = 1$
  • $x = 3$
  • C
    $x = 1, 3$
  • D
    None of these

Answer

Correct option: B.
$x = 3$
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}\frac{1}{5}(2\text{x}^2+3)=1$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}6-5\text{x}=1$
Function is continuou at $x = 1$
$\lim\limits_{\text{x}\rightarrow3^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}6-5\text{x}=-9$
$\lim\limits_{\text{x}\rightarrow3^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}\text{x}-3=0$
Function is discontinuous at $x = 3$

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