f(x) = begin{cases} frac{sin(x-[x])}{x-[x]} & | vedclass

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(C) For $x \in (-2, -1)$,$[x] = -2$,so $f(x) = \frac{\sin(x+2)}{x+2}$.For $x \in (-1, 0)$,$[|x|] = 0$,so $f(x) = \max\{2x, 0\} = 0$. For $x \in [0, 1)

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$(2, 3)$

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(C) For $x \in (-2, -1)$,$[x] = -2$,so $f(x) = \frac{\sin(x+2)}{x+2}$.For $x \in (-1, 0)$,$[|x|] = 0$,so $f(x) = \max\{2x, 0\} = 0$. For $x \in [0, 1)$,$[|x|] = 0$,so $f(x) = \max\{2x, 0\} = 2x$.Thus,$f(x) = \begin{cases} \frac{\sin(x+2)}{x+2} & , x \in (-2, -1) \ 0 & , x \in (-1, 0) \ 2x & , x \in [0, 1) \ 1 & , 	ext{otherwise} \end{cases}$.Checking continuity:At $x = -1$: $f(-1^+) = 0$,$f(-1^-) = \lim_{x 	o -1^-} \frac{\sin(x+2)}{x+2} = \sin(1) 
eq 0$. Discontinuous.At $x = 0$: $f(0^-) = 0$,$f(0^+) = 0$,$f(0) = 0$. Continuous.At $x = 1$: $f(1^-) = 2(1) = 2$,$f(1^+) = 1$. Discontinuous.So,$m = 2$ (points $x = -1, 1$).Checking differentiability:At $x = -1$: Discontinuous,so not differentiable.At $x = 0$: $f'(0^-) = 0$,$f'(0^+) = 2$. Not differentiable.At $x = 1$: Discontinuous,so not differentiable.So,$n = 3$ (points $x = -1, 0, 1$).The ordered pair is $(2, 3)$.

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