Let [x] denote the greatest integer less than or equal to x. If f(x)=[xsinπx] , then f(x) is :
A
continuous at x=0
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B
continuous in (−1,0)
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C
differentiable at x=1
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D
differentiable in (−1,1)
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Solution
The correct options are B continuous at x=0 C continuous in (−1,0) D differentiable in (−1,1) We have −1<x<1⇒0≤xsinπx≤12
Also xsinπx becomes negative and numerically less than 1 when x is slightly greater than 1 and so by definition of [x]
f(x)=[xsinπx]=1 when 1<x<1+h Thus f(x) is continuous and equal to 0 in the closed interval [−1,1] and so f(x) is continuous in the open interval (−1,1)
At x=1, f(x) is discontinuous since limh→0(1−h)=0 and limh→0(1+h)=−1 ∴f(x) is not differentiable at x=1