The points of discontinuity of the function f(x)= 1 5 (2x^2+3),&x… - Vidyadip

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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