Let f(x) = [x^2 - x] + |-x + [x]|,where x in | vedclass

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(D) Given $f(x) = [x^2 - x] + |-x + [x]|$.We know that $-x + [x] = -\{x\}$,where $\{x\}$ is the fractional part of $x$.Thus,$f(x) = [x^2 - x] + |-\{x\

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continuous at $x = 1$,but not continuous at $x = 0$

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(D) Given $f(x) = [x^2 - x] + |-x + [x]|$.We know that $-x + [x] = -\{x\}$,where $\{x\}$ is the fractional part of $x$.Thus,$f(x) = [x^2 - x] + |-\{x\}| = [x^2 - x] + \{x\}$.Check continuity at $x = 0$:$f(0) = [0^2 - 0] + \{0\} = 0 + 0 = 0$.$f(0^+) = \lim_{h 	o 0^+} [h^2 - h] + \{h\} = [-0.0001] + 0 = -1 + 0 = -1$.Since $f(0) 
eq f(0^+)$,$f$ is discontinuous at $x = 0$.Check continuity at $x = 1$:$f(1) = [1^2 - 1] + \{1\} = 0 + 0 = 0$.$f(1^+) = \lim_{h 	o 0^+} [(1+h)^2 - (1+h)] + \{1+h\} = [1 + 2h + h^2 - 1 - h] + h = [h + h^2] + h = 0 + 0 = 0$.$f(1^-) = \lim_{h 	o 0^+} [(1-h)^2 - (1-h)] + \{1-h\} = [1 - 2h + h^2 - 1 + h] + (1-h) = [-h + h^2] + 1 - h = -1 + 1 - 0 = 0$.Since $f(1) = f(1^+) = f(1^-) = 0$,$f$ is continuous at $x = 1$.Therefore,$f$ is continuous at $x = 1$ but not at $x = 0$.

Let $f(x) = [x^2 - x] + |-x + [x]|$,where $x \in R$ and $[t]$ denotes the greatest integer less than or equal to $t$. Then,$f$ is

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