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If \mathrm{f(x)=[x](\sin k x)^\rho} is continuous for real x, then (where [.] represents the greatest integer function) 

Option: 1

\mathrm{k \in\{n \pi, n \in I\}, p>0}


Option: 2

\mathrm{k \in\{2 n \pi, n \in I\}, p>0}


Option: 3

\mathrm{k \in\{n \pi, n \in I\}, p \in R-\{0\}}


Option: 4

\mathrm{k \in\{n \pi, n \in I, n \neq 0\}, p \in R-\{0\}}


Answers (1)

best_answer

\mathrm{f(x)=[x](\sin k x)^\rho} 

\mathrm{(\sin k x)^p} is continuous function \mathrm{\forall x \in R, k \in R} and \mathrm{p>0,[x]} is discontinuous at \mathrm{x \in I}

For \mathrm{\begin{aligned} & k=n \pi, n \in I \\ &\: \: \: \: \: \: \: \quad f(x)=[x](\sin (n \pi x))^p \\ & \quad \lim _{x \rightarrow a} f(x)=0, a \in I \\ &\: \: \text { and } f(a)=0 \end{aligned}}

So, f(x) becomes continuous for all \mathrm{x\: \epsilon \: R}

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shivangi.bhatnagar

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