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Question

Let f:RR be defined as f(x)=(|x1|+|4x11|)[x22x2], where [.] denotes the greatest integer function. Then the number of points of discontinuity of f(x) in (12,52) is

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Solution

f(x)=(|x1|+|4x11|)[x22x2]
or, f(x)=(|x1|+|4x11|)[(x1)23]
or, f(x)=(|x1|+|4x11|)g(x) ([(x1)2]3)h(x)


g(x) is continuous for all real values of x.
Possible points of discontinuity of h(x) are the points where (x1)2 takes integral values.
At x=1,
h(1)=h(1+)=h(0)=3
So, f(x) is continuous at x=1

At x=2,
h(2)=03=3
h(2+)=13=2
h is discontinuous at x=2, so f(x) is discontinuous at x=2

At x=2+1,
h((2+1))=13=2
h((2+1)+)=23=1
h is discontinuous at x=2+1, so f(x) is discontinuous at x=2+1

In (12,52), there are two points of discontinuity.

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