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The function f(x)=[x]^2-[x^2] is discont...

The function `f(x)=[x]^2-[x^2]` is discontinuous at (where `[gamma]` is the greatest integer less than or equal to `gamma`), is discontinuous at

A

all integers

B

all integers except 0 and 1

C

all integers except 0

D

all integers except 1

Text Solution

Verified by Experts

The correct Answer is:
D
We know that the function [x] is discontinuous at all integers points. Therefore. `[x]^(2)` is discontinuous at all integers points.
Also, `[x]^(2)` is discontinuous at `x=pm sqrt(|n|)," where n"in Z`
We observe that
`underset(x to 1^(-))lim f(x)= underset(h to 0)lim f(1-h)=underset(h to 0)lim [1-h]^(2)-[(1-h)^(2)]`
`Rightarrow underset(x to 1^(-))lim f(x)=0-0=0`
`underset(x to 1^(+))lim f(x)=underset(H to 0)lim f(1+h)=underset(h to 0)lim [1+h]^(2)-[(1+h)^(2)]`
`and f(1)=[1]^(2)-[1]^(2)=0`
`therefore underset(x to 1^(-))lim f(x)=underset(x to 1^(+))lim f(x)=f(1)`
So, f(x) is continuous at x=1
f(x) is discontinuous at all other integer points.
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