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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
If f(x) defind by $\text{f(x)}=\begin{cases}\frac{|\text{x}^2-\text{x}|}{\text{x}^2-\text{x}},&\text{x}\neq0,1\\1,&\text{x}=0\\-1,&\text{x}=1\end{cases}$ then $f(x)$ is continuse for all:
  • A
    $x$
  • B
    $x$ except at $x = 0$
  • C
    $x$ except at $x = 1$
  • ✓
    $x$ except at $x = 0$ and $x = 1$
✓

Answer

Correct option: D.
$x$ except at $x = 0$ and $x = 1$
Given function $\text{f(x)}=\frac{|\text{x}^2-\text{x}|}{\text{x}^2-\text{x}}$
Consider,
$\text{f}(0^+)=\lim\limits_{\text{x}\rightarrow0}\frac{|\text{x}^2-\text{x}|}{\text{x}^2-\text{x}}=\lim\limits_{\text{x}\rightarrow0}\frac{|\text{x(x}-1)|}{\text{x(x}-1)}=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x(x}-1)}{\text{x(x}-1)}=1$
$\text{f}(0^-)=\lim\limits_{\text{x}\rightarrow0}\frac{|\text{x}^2-\text{x}|}{\text{x}^2-\text{x}}=\lim\limits_{\text{x}\rightarrow0}\frac{|\text{x}(\text{x}-1)|}{\text{x}(\text{x}-1)}=\lim\limits_{\text{x}\rightarrow0}\frac{-\text{x}(\text{x}-1)}{\text{x}(\text{x}-1)}=-1$
Also, for $f(1^+)$ and $f(1^-)$ you can check.

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