If f(x) defind by f(x)= |x^2-x| x^2-x ,&x 0,1 1,&x=0 -1,&x=1 then f(x) is continuse for all:

If f(x) defind by f(x)= |x^2-x| x^2-x ,&x 0,1 1,&x=0 -1,&x=1 then…

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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