Let f: (- 2, 2) → ℝ be defined by f(x) = \ cases x[x] & , & - 2 < x...

Given,

$f: \left(- 2, 2\right) \to \mathbb{R}$ be defined by f\left(x\right) = \left\{\begin{cases}x\left[x\right] & , & - 2 < x < 0 \\ \left(x - 1\right)\left[x\right] & , & 0 \le x < 2\end{cases}\right)</p><p>⇒ f\left(x\right) = \left\{\begin{array}{ll}- 2x, - 2 \begin{array}{ll}0, 0\right)

Now plotting the diagram of the above function we get,

Now from above diagram we can say that, $y = f$\left(x\right)$is same as$y = \left|f\left(x\right)\right|$</p><p>Hence, the function is not continuous at one point and non differentiable at three points, so$m = 1, n = 3$</p><p>Hence,$m + n = 4