The function f(x) = [x]^2 - [x^2],(where [y] | vedclass

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(D) Given $f(x) = [x]^2 - [x^2]$.For $-1 < x < 0$,$[x] = -1$ and $[x^2] = 0$,so $f(x) = (-1)^2 - 0 = 1$.For $x = 0$,$f(0) = 0^2 - 0 = 0$.For $0 < x <

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All integers except $1$

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(D) Given $f(x) = [x]^2 - [x^2]$.For $-1 For $x = 0$,$f(0) = 0^2 - 0 = 0$.For $0 For $x = 1$,$f(1) = 1^2 - 1 = 0$.For $1 For $x = \sqrt{2}$,$f(\sqrt{2}) = 1^2 - 2 = -1$.For $\sqrt{2} For $x = \sqrt{3}$,$f(\sqrt{3}) = 1^2 - 3 = -2$.For $\sqrt{3} For $x = 2$,$f(2) = 2^2 - 4 = 0$.For $2 For $x = \sqrt{5}$,$f(\sqrt{5}) = 2^2 - 5 = -1$.By analyzing the behavior at integers $n$,we find that the function is discontinuous at all integers except $x = 1$.

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

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