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  • A function  is defined by <img alt="\mathrm{f(x)=\left\{\begin{array}{ll}\frac{\le

A function \mathrm{f(x)} is defined by \mathrm{f(x)=\left\{\begin{array}{ll}\frac{\left[x^2\right]-1}{x^2-1}, & \text { for } x^2 \neq 1 \\ 0, & \text { for } x^2=1\end{array}\right.. At\, \, x=1,} which of the following is correct?

 

 

Option: 1

 L.H.L. doesn't exist
 


Option: 2

 R.H.L. doesn't exist
 


Option: 3

 L.H.L. exists
 


Option: 4

 None of these


Answers (1)

best_answer

 Because of \mathrm{\left[x^2\right] } let us consider the case when \mathrm{x^2<1 }and when \mathrm{1<x^2<2 } for continuity at x=1.
\mathrm{\begin{aligned} \therefore \quad\left[x^2\right] & =0 \text { when } x^2<1 \text { i.e., }-1<x<1 \\ & {\left[x^2\right]=1 \text { when } 1<x^2<2 \text { i.e., } x>1 } \end{aligned} }
Hence, we redefine the function as
\mathrm{f(x)=\left\{\begin{array}{cc} \frac{-1}{x^2-1}, & x<1 \\ 0, & x=1 \\ \frac{1-1}{x^2-1}=0, & x>1 \end{array}\right. }

Now, L.H.L.\mathrm{ =\lim _{h \rightarrow 0} \frac{-1}{(1-h)^2-1} }
\mathrm{=\lim _{h \rightarrow 0} \frac{-1}{-2 h+h^2}=\infty \therefore \text { L.H.L. doesn't exist. } }
 

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Gunjita

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