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(C) The function $f(t)$ is a piecewise linear function. For $t \in [2n-1, 2n+1]$,$f(t)$ is the line segment connecting $(2n-1, f(2n-1))$ and $(2n+1, f

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$5$

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(C) The function $f(t)$ is a piecewise linear function. For $t \in [2n-1, 2n+1]$,$f(t)$ is the line segment connecting $(2n-1, f(2n-1))$ and $(2n+1, f(2n+1))$.Given $f(2n-1) = (-1)^{n+1} 2$,we have:$f(1) = 2, f(3) = -2, f(5) = 2, f(7) = -2, f(9) = 2$.For $1 For $3 For $5 For $7 $g(x) = \int_1^x f(t) dt$. $g(1) = 0$. $g(x) = 0$ when the net signed area from $1$ to $x$ is zero.From the graph,the area from $1$ to $3$ is $\frac{1}{2}(2)(2) + \frac{1}{2}(2)(-2) = 0$. So $g(3) = 0$.The area from $3$ to $5$ is $\frac{1}{2}(2)(-2) + \frac{1}{2}(2)(2) = 0$. So $g(5) = 0$.The area from $5$ to $7$ is $\frac{1}{2}(2)(2) + \frac{1}{2}(2)(-2) = 0$. So $g(7) = 0$.In $(1, 8]$,$g(x) = 0$ at $x = 3, 5, 7$. Thus $\alpha = 3$.$\beta = \lim_{x \rightarrow 1^+} \frac{g(x)}{x-1} = g'(1^+) = f(1^+) = 2$.Therefore,$\alpha + \beta = 3 + 2 = 5$.

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