Piecewise Function Integration and Limit Evaluation

To find the value of $\alpha + \beta$, we break the problem down into two parts: determining the limit $\beta$ and counting the number of solutions $\alpha$ for the equation $g(x) = 0$ in the interval $(1, 8]$.

Step 1: Definition and Analysis of $f(t)$

The given piecewise definition for $f(t)$ shows that $f(t)$ is a continuous piecewise linear function connecting the points $(2n-1, (-1)^{n+1}2)$ for $n \in \mathbb{N}$.

Evaluating $f(2n-1)$ for initial values of $n$:

• For $n = 1$: $t = 1 \implies f(1) = (-1)^2 \times 2 = 2$
• For $n = 2$: $t = 3 \implies f(3) = (-1)^3 \times 2 = -2$
• For $n = 3$: $t = 5 \implies f(5) = (-1)^4 \times 2 = 2$
• For $n = 4$: $t = 7 \implies f(7) = (-1)^5 \times 2 = -2$
• For $n = 5$: $t = 9 \implies f(9) = (-1)^6 \times 2 = 2$

For $2n-1 < t < 2n+1$, $f(t)$ is a linear interpolation between $(2n-1, f(2n-1))$ and $(2n+1, f(2n+1))$. Specifically:

• On $[1, 3]$, $f(t)$ is the line segment from $(1, 2)$ to $(3, -2)$: $f(t) = 4 - 2t$
• On $[3, 5]$, $f(t)$ is the line segment from $(3, -2)$ to $(5, 2)$: $f(t) = 2t - 8$
• On $[5, 7]$, $f(t)$ is the line segment from $(5, 2)$ to $(7, -2)$: $f(t) = 12 - 2t$
• On $[7, 9]$, $f(t)$ is the line segment from $(7, -2)$ to $(9, 2)$: $f(t) = 2t - 16$

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Step 2: Evaluation of $\beta$

The function $g(x)$ is defined as: $g(x) = \int_{1}^{x} f(t) \, dt$

We need to evaluate: $\beta = \lim_{x \to 1^+} \frac{g(x)}{x - 1}$

For $x \in (1, 3)$, using $f(t) = 4 - 2t$: $g(x) = \int_{1}^{x} (4 - 2t) \, dt = \left[ 4t - t^2 \right]_1^x = (4x - x^2) - (4 - 1) = -x^2 + 4x - 3 = -(x - 1)(x - 3)$

Now, substituting $g(x)$ into the limit expression: $\beta = \lim_{x \to 1^+} \frac{-(x - 1)(x - 3)}{x - 1} = \lim_{x \to 1^+} (3 - x) = 2$

(Alternatively, by Leibniz's Integral Rule / definition of derivative, $\beta = g'(1^+) = f(1) = 2$.)

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Step 3: Evaluation of $\alpha$ (Number of solutions of $g(x) = 0$ in $(1, 8]$)

By the Fundamental Theorem of Calculus, $g'(x) = f(x)$.

Let's evaluate $g(x)$ at key integer points $x \in [1, 8]$ by computing the area under $f(t)$:

• $g(1) = 0$
• $g(2) = \int_{1}^{2} (4 - 2t) \, dt = \left[ 4t - t^2 \right]_1^2 = 4 - 3 = 1$
• $g(3) = \int_{1}^{3} (4 - 2t) \, dt = \left[ 4t - t^2 \right]_1^3 = 3 - 3 = 0$
• $g(4) = g(3) + \int_{3}^{4} (2t - 8) \, dt = 0 + \left[ t^2 - 8t \right]_3^4 = -1$
• $g(5) = g(3) + \int_{3}^{5} (2t - 8) \, dt = 0 + \left[ t^2 - 8t \right]_3^5 = 0$
• $g(6) = g(5) + \int_{5}^{6} (12 - 2t) \, dt = 0 + \left[ 12t - t^2 \right]_5^6 = 1$
• $g(7) = g(5) + \int_{5}^{7} (12 - 2t) \, dt = 0 + \left[ 12t - t^2 \right]_5^7 = 0$
• $g(8) = g(7) + \int_{7}^{8} (2t - 16) \, dt = 0 + \left[ t^2 - 16t \right]_7^8 = -1$

Now, analyzing monotonicity using $g'(x) = f(x)$:

1. On $(1, 2)$: $f(x) > 0 \implies g(x)$ is strictly increasing from $g(1) = 0$ to $g(2) = 1$. Thus, $g(x) > 0$ for all $x \in (1, 2]$.
2. On $[2, 4]$: $f(x) < 0$ for $x \in (2, 4) \implies g(x)$ is strictly decreasing from $g(2) = 1$ to $g(4) = -1$. By the Intermediate Value Theorem, $g(x) = 0$ has exactly one solution in $(2, 4)$, which is $x = 3$.
3. On $[4, 6]$: $f(x) > 0$ for $x \in (4, 6) \implies g(x)$ is strictly increasing from $g(4) = -1$ to $g(6) = 1$. Thus, $g(x) = 0$ has exactly one solution in $(4, 6)$, which is $x = 5$.
4. On $[6, 8]$: $f(x) < 0$ for $x \in (6, 8) \implies g(x)$ is strictly decreasing from $g(6) = 1$ to $g(8) = -1$. Thus, $g(x) = 0$ has exactly one solution in $(6, 8)$, which is $x = 7$.

Therefore, the solutions of $g(x) = 0$ in the interval $(1, 8]$ are $x = 3, 5, 7$. Hence, the number of solutions is: $\alpha = 3$

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Step 4: Final Calculation

$\alpha + \beta = 3 + 2 = 5$