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				   F(x)=integrate [f(t)]dt from 0 to x where x>0 and [.] denotes greatest int. func. .Then,
is f(x) continuous but not diff. for x=1,2,3,........ If yes,how?


6 years ago

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										Ans:f(t) should be given. f(t) = t$F(x) = \int_{0}^{x}[t]dt$Apply the newton leibniz’s theorm here, we have,$F'(x) = [x].1 - [0].0 = [x]$$F'(x) = f(x)$$f(x) = [x]$To check the continuity at integer ‘a’LHL = RHL$\lim_{x\rightarrow a^{-}}[x] = \lim_{x\rightarrow a^{+}}[x]$$LHL = \lim_{x\rightarrow a^{-}}[x] = a-1$$RHL = \lim_{x\rightarrow a^{+}}[x] = a$So f(x) is not continuous at integer ‘a’. So it will definitely will not be differentiable at integer ‘a’.Thanks & RegardsJitender SinghIIT DelhiaskIITians Faculty

2 years ago

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