If [.] denotes the greatest integer function, | vedclass

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(C) Given $f(x) = [x]^2 - [x^2]$.For any integer $n$,let $x = n + h$,where $0 \le h < 1$.Then $f(n+h) = [n+h]^2 - [(n+h)^2] = n^2 - [n^2 + 2nh + h^2]

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all integers except $1$

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(C) Given $f(x) = [x]^2 - [x^2]$.For any integer $n$,let $x = n + h$,where $0 \le h Then $f(n+h) = [n+h]^2 - [(n+h)^2] = n^2 - [n^2 + 2nh + h^2] = n^2 - n^2 - [2nh + h^2] = -[2nh + h^2]$.At $x = n$,$f(n) = [n]^2 - [n^2] = n^2 - n^2 = 0$.For $x 	o n^-$,let $x = n - h$ where $h 	o 0^+$. Then $f(n-h) = [n-h]^2 - [(n-h)^2] = (n-1)^2 - [n^2 - 2nh + h^2]$.For $n=0$,$f(0)=0$. $\lim_{x 	o 0^-} f(x) = [-h]^2 - [h^2] = (-1)^2 - 0 = 1$. Since $1 
eq 0$,$f(x)$ is discontinuous at $x=0$.For $n=1$,$f(1)=0$. $\lim_{x 	o 1^-} f(x) = [1-h]^2 - [(1-h)^2] = 0^2 - 0 = 0$. $\lim_{x 	o 1^+} f(x) = [1+h]^2 - [(1+h)^2] = 1^2 - 1 = 0$. Since $0=0$,$f(x)$ is continuous at $x=1$.For any other integer $n 
eq 0, 1$,the function is discontinuous because the limit from the left and right will not equal $f(n)=0$.Thus,$f(x)$ is discontinuous at all integers except $1$.

If $[.]$ denotes the greatest integer function,then $f(x) = [x]^2 - [x^2]$ is discontinuous at

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