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  • Let <img alt="\mathrm{f(x)=\left\{\begin{array}{lll} \sin ^2 x, & x \text { rational } \\ -\sin ^2 x, & x \text { irrational } \end{array}\right.}" src="https://entrancecorner.oncodeco

Let \mathrm{f(x)=\left\{\begin{array}{lll} \sin ^2 x, & x \text { rational } \\ -\sin ^2 x, & x \text { irrational } \end{array}\right.} Then set of points, where f(x) is continuous is 

Option: 1

\mathrm{\left\{(2 n+1) \frac{\pi}{2}, \quad n \in I\right\}}


Option: 2

null set 


Option: 3

\mathrm{\{\mathrm{n} \pi, \mathrm{n} \in \mathrm{I}\}}


Option: 4

set of all rational numbers 


Answers (1)

best_answer

If a is any number for which \mathrm{\sin a \neq 0}, then we don't get any fix tendency of f(x) to approaches where \mathrm{x \rightarrow a}

so  \mathrm{\lim _{x \rightarrow a} f(x)} does not exist

but if  \mathrm{\sin a=0, \lim _{x \rightarrow a} f(x)=0}

Posted by

vishal kumar

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