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\mathrm{f(x)=\frac{x}{1+[x]}} ,  is discontinuous in set \mathrm{s} , where \mathrm{s} is

Option: 1

\mathrm{\{[-1,0) \cup integers \}}


Option: 2

\left \{\text{ all real numbers }\right \}


Option: 3

\left \{\text{ all positive rational }\right \}


Option: 4

none of these


Answers (1)

\mathrm{f(x)=\frac{x}{1+n}, \quad n \leq x<n+1}
 

(i) clearly, if \mathrm{n=-1} then \mathrm{f(x)} is not defined

i.e. \mathrm{f(x)} is not defined \mathrm{\forall x \in[-1,0)}

(ii) if \mathrm{n \neq-1}
\begin{array}{rll} f(x)=-\frac{x}{2} ; & -3 \leq x<-2 & (n=-3) \\ =-x ; & -2 \leq x<-1 \quad & (n=-2) \\ = & x ; \quad 0 \leq x<+1 \quad(n=0) \\ = & \ldots \ldots \ldots \end{array}

Clearly, the function is discontinuous at all integral values of \mathrm{x}.

From (i) and (ii) \mathrm{f(x)} is not continuous in set \mathrm{S}

\mathrm{\mathrm{S}=\{[-1,0) \cup \text { integers }\}}.

Posted by

Sumit Saini

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