Let f(x)=2x3−5. Then, number of points of discontinuity of f(x) in (1,2)

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answer is D.

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Detailed Solution

f(x)=2x3−5 is an increasing function of x on (1,2) and f(1)=−3,f(2)=11. Between −3 and 11 there are thirteen points where f(x) is discontinuous.

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