Let f(x)=2x3−5. Then, number of points of discontinuity of f(x) in (1,2)
0
3
10
13
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.