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If  \mathrm{f(x)=[x \sin \pi x]} , then f(x) is

Option: 1

continuous at x=0


Option: 2

continuous in (-1,1)
 


Option: 3

differentiable at x= -2


Option: 4

differentiable in (-1,1)


Answers (1)

best_answer

If -1 \leq x \leq 1 , then \mathrm{0 \leq x \sin \pi x \leq 1 / 2~ and ~so ~f(x)=[x \sin \pi x]=0} (by definition of the greatest integer function).
\mathrm{ \therefore f(x)=[x \sin \pi x]=0 \text { when }-1 \leq x \leq 1 \text {. } }
If 1<x<1+h, where h is a small positive real number, then
\mathrm{ \pi<\pi x<\pi+\pi h \Rightarrow-1<\sin \pi x<0 \Rightarrow-1<x \sin \pi x<0 }
So, \mathrm{ f(x)=[x \sin \pi x]=-1 }  in the right neighbourhood of x=1. Thus, f(x) is constant and equal to zero in [-1,1] and so f(x) is differentiable and hence continuous on $(-1,1)$. At x=1, f(x) is discontinuous because \mathrm{ \lim f(x)=0~and ~\lim f(x)=-1~. Hence ~\lim _{x \rightarrow 1^{-}} f(x)=0 } and it is not differentiable at x=1.

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