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Continuity — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsContinuity1 Mark
MCQ
The points of discontinuity of the function $\text{f(x)}=\begin{cases}2\sqrt{\text{x}},&0\leq\text{x}\leq1\\4-2\text{x},&1<\text{x}<\frac{5}{2}\\2\text{x}-7,&\frac{5}{2}\leq\text{x}\leq4\end{cases}$ is $($are$)$:
  • A
    $\text{x}=1,\text{x}=\frac{5}{2}$
  • ✓
    $\text{x}=\frac{5}{2}$
  • C
    $\text{x}=1,\frac{5}{2},4$
  • D
    $\text{x}=0,4$
✓

Answer

Correct option: B.
$\text{x}=\frac{5}{2}$
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}2\sqrt{\text{x}}=2$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1}4-2\text{x}=2$
Function is continuous at $x = 1$
$\lim\limits_{\text{x}\rightarrow\frac{5}{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{5}{2}}4-2\text{x}=-1$
$\lim\limits_{\text{x}\rightarrow\frac{5}{2}^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{5}{2}}2\text{x}-7=-2$
Function is discontinuous at $\text{x}=\frac{5}{2}$

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