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If \mathrm{f(x)=[x] \sin \frac{\pi}{[x+1]}}, then the points of discontinuity of \mathrm{f(x)} in domain of the function are

Option: 1

\mathrm{\forall x \in \boldsymbol{I}}


Option: 2

\mathrm{\forall x \in \boldsymbol{R}}


Option: 3

\mathrm{all x \in \boldsymbol{I}-\{0\}}


Option: 4

\forall x \in R \sim\{0\}


Answers (1)

best_answer

\mathrm{f(x)}  is defined

if \mathrm{[x+1] \neq 0 \Rightarrow 0 \leq x+1<1}
i.e. \mathrm{[x+1]=0\, if \, -1 \leq x<0}. Thus dom \mathrm{f=R \sim[-1,0)}.

We have \mathrm{\sin \left(\frac{\pi}{[x+1]}\right)} continuous at all points of \mathrm{R \sim[-1,0)}. and \mathrm{[x]}  continuous on \mathrm{R \sim I}, where \mathrm{ I} denote the set of integers. Thus the points where \mathrm{ f} can possibly be discontinuous are,-3,-2,-1,0,1,2, \ldots \ldots .$. But for $0 \leq x<1,[x]=0$ and $\sin \left(\frac{\pi}{[x+1]}\right)  is defined. Therefore, f(x)=0$ for $0 \leq x<1.

Also, f is not defined on [-1,0), so the continuity of f at 0 means continuity of f from right at 0 . Since f is continuous from right at 0, f is continuous at 0 . Hence the set of discontinuities of f is I \sim\{0\}.
 

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Gaurav

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