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If \mathrm{f(x)}=\left\{\begin{array}{r}\mathrm{x \text {, when } x \in Q} \\\mathrm{ -x \text {, when } x \notin Q}\end{array}\right.  then discuss the continuity of \mathrm{f(x)}

Option: 1

  f(x) is continuous of x=0  and limit does not exist ist.


Option: 2

f(x) is not continues at os x=0 and limit exist.
 


Option: 3

 f(x) is continuous only at x=0 and limit exist 


Option: 4

 None 


Answers (1)

best_answer

\mathrm{ f(x)}=\left\{\begin{array}{l}\mathrm{x \text {, when } x \in Q }\\\mathrm{ -x \text {, when } x \notin Q}\end{array}\right.


Let us first examine continuity at x=0
\begin{aligned} &\mathrm{ \therefore \quad \text { L.H.L. }=f(0-0)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h) }\\ &\mathrm{ =\lim _{h \rightarrow 0}[-h \text { or } h \text { according as }-h \in Q }\\ & \text { or }\mathrm{-h \notin Q]=0} \text {. } \\ & \text { and R.H.L. }\mathrm{=f(0+0)=\lim _{h \rightarrow 0} f(0+h)} \\ & \mathrm{=\lim _{h \rightarrow 0}[h \text { or }-h]=0} \\ & \Rightarrow \mathrm{f(0)=f(0-0)=f(0+0)} \\ & \Rightarrow \text {f(x) is continuous at x=0} \\ \end{aligned}

\mathrm{Now,~ Icl a \in R, a=0~ then}\\ \begin{aligned} \mathrm{f(a-0)} & =\mathrm{\lim _{a \rightarrow 0} f(a-h) }\\ & \mathrm{=\lim _{a \rightarrow 0}((a-h) \operatorname{cor}-(a-h)\}} \\ & \mathrm{=a \text { or }-a \text { which is not unique. }}\\ \end{aligned}\\ \Rightarrow \text{f(a-0) docs not cust.}\\ \Rightarrow \text{f(x) is not continuous at a}\mathrm{ \in R-p p_4.}\\ \text{Hence, f(x) is continuous only at r=0 }

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jitender.kumar

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