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  • Let   be defined as <img alt="\mathrm {f(x)=\left\{\begin{array}{cl}0, & x

Let \mathrm{f : R \rightarrow R }  be defined as \mathrm {f(x)=\left\{\begin{array}{cl}0, & x \text { is irrational } \\ \sin |x|, & x \text { is rational }\end{array}\right.}Then which of the following is true?

 

 

Option: 1

 f is discontinuous for all x
 


Option: 2

 f is continuous for all x
 


Option: 3

f is discontinuous at x=k \pi, where k is an integer
 


Option: 4

 f is continuous at  x=k \pi , where k is an integer.


Answers (1)

\mathrm{: f(x) is\, \, continuous\, \, at\, \, x=a\, \, if \, \lim _{h \rightarrow 0} f(a+h)=f(a) }
We consider the continuity at \mathrm{x=k \pi(k \in I) }
\mathrm{\therefore \quad \lim _{h \rightarrow 0} f(k \pi+h)=f(k \pi) \text { for continuity at } x=k \pi(k \in I) }
Now, \mathrm{ \lim _{h \rightarrow 0} f(k \pi+h) }
\mathrm{ \begin{aligned} & =\left\{\begin{array}{cc} \lim _{h \rightarrow 0} 0, & \text { if } k \pi+h \text { is irrational } \\ \lim _{h \rightarrow 0} \sin |k \pi+h|, & \text { if } k \pi+h \text { is rational } \end{array}\right. \\ & =0 \\ & {[\because \sin |k \pi|=0 \text { as } k \in I]} \\ & \text { And } f(k \pi)=0 \\ & {[\because k \pi \text { is irrational as } k \in I]} \\ & \end{aligned} }
 

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Kshitij

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