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  • <img alt="\mathrm{f(x)=\frac{\left|x^3-3 x^2+2 x\right|}{x^3-3 x^2+2 x}.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%5Cfrac%7B%5Cleft%7Cx%5E3-3%20x%5E2&plus;2

\mathrm{f(x)=\frac{\left|x^3-3 x^2+2 x\right|}{x^3-3 x^2+2 x}.} The number of points \mathrm{a}, where \mathrm{ \operatorname{Lt}_{x \rightarrow a} f(x)} doesn't exist, is
 

Option: 1

2


Option: 2

1


Option: 3

3


Option: 4

4


Answers (1)

best_answer

\mathrm{f(x)=\frac{|x(x-1)(x-2)|}{x(x-1)(x-2)}=\frac{|x|}{x} \frac{|x-1||| x-2 \mid}{x-1} \frac{x-2}{x-2}}

Consider the points \mathrm{x=0,1,2;} we redefine the function

\mathrm{ f(x)=\left\{\begin{array}{ccc} (-1)(-1)(-1)=-1 & , & x<0 \\ (1)(-1)(-1)=1 & , & 0<x<1 \\ (1)(1)(-1)=-1 & , & 1<x<2 \\ (1)(1)(1)=1 & , & x>2 \end{array}\right. }

\mathrm{ Lt f(x) \, as \: x \rightarrow 0,1,2 } does not exist as both R.H.L. and L.H.L. will be different. Hence \mathrm{ a=0,1,2. }

Hence option 3 is correct.

Posted by

avinash.dongre

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