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  •  If f(x) defined by <img alt="\mathrm{f(x)=\left\{\begin{array}{cl}\frac{\left|x^2-x\right|}{x^2-x} & , x \neq 0,1 \\ 1 & , x=0 \\ -1 & , x=1\end{array}\right.}" src="https://

 If f(x) defined by \mathrm{f(x)=\left\{\begin{array}{cl}\frac{\left|x^2-x\right|}{x^2-x} & , x \neq 0,1 \\ 1 & , x=0 \\ -1 & , x=1\end{array}\right.} 

then f(x) is continuous for all

Option: 1

x


Option: 2

x  except at x=0


Option: 3

x except at x=1
 


Option: 4

x except at x=0 and x=1.


Answers (1)

best_answer

 We have:
\mathrm{f(x) =\left\{\begin{array}{cc} \frac{x^2-x}{x^2-x}=1, & \text { if } x<0 \text { or } x>1 \\ -\frac{\left(x^2-x\right)}{x^2-x}=-1, & \text { if } 0<x<1 \\ 1, & , \text { if } x=0 \\ -1 & , \text { if } x=1 \end{array}\right. }
\mathrm{ =\left\{\begin{array}{cc} 1, \text { if } x \leq 0 \text { or } x>1 \\ -1, \text { if } 0<x \leq 1 \end{array}\right. }

Now, \mathrm{ \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0} 1=1~ and, \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}-1=-1 }
Clearly,   \mathrm{\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x). So, f(x) }  is not continuous at x=0.
It can be easily seen that it is not continuous at x=1 also.

Posted by

manish painkra

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