Let [x] denote the greatest integer function and f(x) = max\ 1 + x...

Given,

$f\left(x\right) = max\left\{1 + x + \left[x\right], 2 + x, x + 2\left[x\right]\right\}, 0 \le x \le 2$

⇒ f\left(x\right) = \left\{\begin{cases}max\left\{x + 1, x + 2, x\right\} & 0 \le x < 1 \\ max\left\{x + 2, x + 2, x + 2\right\} & 1 \le x < 2 \\ max\left\{5, 4, 6\right\} & x = 2\end{cases}\right)</p><p>⇒ f\left(x\right) = \left\{$\begin{cases}x + 2, & 0 \le x < 1 \\ x + 2, & 1 \le x < 2 \\ 6, & x = 2\end{cases}$\right)

⇒ f ( x ) = \left\{\begin{cases}x + 2 & 0 \le x < 2 \\ 6 & x = 2\end{cases}\right)</p><p>So,f$is not continuous at$x = 2$as$f\left({2}^{-}\right) = 4 \text{$and$} f\left(2\right) = 6$</p><p>And$f$is differentiable in \left(0, 2\right)$ as $x + 2$ is a linear function,

$∴ m = 1, n = 0$

⇒ {\left(m + n\right)}^{2} + 2 = 1 + 2 = 3

Hence this is the required option.