Let f(x) = \ cases ( (x - [x]))/(x - [x]), & x ∈ (- 2, - 1) max(2x,...

Given,

f\left(x\right) = \left\{\begin{cases}\dfrac{\sin \left(x - \left[x\right]\right)}{x - \left[x\right]}, & x \in \left(- 2, - 1\right) \\ max\left(2x, 3\left[\left|x\right|\right]\right), & \left|x\right| < 1 \\ 1, & otherwise\end{cases}\right)</p><p>On simplifying we get,</p><p>f\left(x\right) = \left\{$\begin{cases}\dfrac{\sin \left(x + 2\right)}{x + 2}, & x \in \left(- 2, - 1\right) \\ max\left\{2x, 0\right\} & , x \in \left(- 1, 1\right) \\ 1 & otherwise\end{cases}$\right)

$f\left(- {2}^{+}\right) = \lim_{h \to 0}f\left(- 2 + h\right) = \lim_{h \to 0}\dfrac{\sin \left(- 2 + h + 2\right)}{- 2 + h + 2} = \lim_{h \to 0}\dfrac{sinh}{h} = 1$

$f$ is continuous at $x = - 2$

$f\left(- {1}^{-}\right) = \lim_{h \to 0}\dfrac{\sin \left(- 1 - h + 2\right)}{\left(- 1 - h + 2\right)} = \sin 1$

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

$f\left({1}^{+}\right) = 1 \text{ and } f\left({1}^{-}\right) = 0 ⇒ f$ is not continuous at $x = 1$

$f$ is continuous but not diff. at $x = 0$

\left$$\begin{array}{ll}⇒ fisdiscontinuousatx = - 1 \text{ and } 1 \\ \text{ and } fisnotdiff . atx = - 1, 0 \text{ and } 1\end{array}$$\right\} ⇒ $$\begin{array}{ll}m = 2 \\ n = 3\end{array}$$$