Let f(x) = [ x ^ 2 - x] + |- x + [x]| , where x ∈ ℝ and [t] denotes...

Given,

$f\left(x\right) = \left[{x}^{2} - x\right] + \left|- x + \left[x\right]\right|$

$⇒ f\left(x\right) = \left[{x}^{2} - x\right] + \left|x - \left[x\right]\right|, \left\{\text{as}\left|- A\right| = \left|A\right|\right\}$

$⇒ f\left(x\right) = \left[{x}^{2} - x\right] + \left|\left\{x\right\}\right|, \left\{\text{as}\left[x\right] + \left\{x\right\} = x\right\}$

$⇒ f\left(x\right) = \left[{x}^{2} - x\right] + \left\{x\right\} ( ∵ \left\{x\right\} \ge 0 )$

Now at $x = 0$, $f\left(0\right) = 0$

And $f\left({0}^{+}\right) = - 1, \left\{\text{as}{x}^{2} - x < 0\text{for}x \to {0}^{+}\right\}$

So, function is discontinuous at $x = 0$

Now at $x = 1$, $f\left(1\right) = 0$,

$f\left({1}^{+}\right) = 0 + 0 = 0$ and $f\left({1}^{-}\right) = - 1 + 1 = 0$

Hence, the function is continuous at $x = 1$.