Step 1: Analyze the function f(t)

The function $f(t)$ is defined on $[1, \infty)$. Let's determine its values at odd integers $t=2n-1$ for $n \in \mathbb{N}$.

• For $n=1$, $t=1$: $f(1) = (-1)^{1+1} \cdot 2 = 2$.
• For $n=2$, $t=3$: $f(3) = (-1)^{2+1} \cdot 2 = -2$.
• For $n=3$, $t=5$: $f(5) = (-1)^{3+1} \cdot 2 = 2$.
• For $n=4$, $t=7$: $f(7) = (-1)^{4+1} \cdot 2 = -2$.

For $t$ in an interval $(2n-1, 2n+1)$, the function is defined as a linear interpolation between $f(2n-1)$ and $f(2n+1)$: $f(t) = \dfrac{(2n+1-t)}{2} f(2n-1)+\dfrac{(t-(2n-1))}{2} f(2n+1)$ This is the equation of a line segment connecting the points $(2n-1, f(2n-1))$ and $(2n+1, f(2n+1))$. Therefore, the graph of $f(t)$ is a continuous piecewise linear function, forming a triangular wave. The points on the graph are $(1,2), (3,-2), (5,2), (7,-2), \dots$. The function $f(t)$ crosses the t-axis at even integers, i.e., $f(2n)=0$ for $n \in \mathbb{N}$.

Step 2: Calculate $\alpha$, the number of solutions to $g(x)=0$

The function $g(x)$ is defined as $g(x) = \int_1^x f(t) \, dt$. We need to find the number of roots of $g(x)=0$ in the interval $(1, 8]$.

By the Fundamental Theorem of Calculus, $g'(x) = f(x)$. The extrema of $g(x)$ occur when $g'(x)=f(x)=0$, which happens at $x=2, 4, 6, 8, \dots$.

We can calculate the values of $g(x)$ at integer points by finding the net area under the curve of $f(t)$. The graph of $f(t)$ consists of triangles with base 1 above and below the t-axis.

• $g(1) = \int_1^1 f(t) \, dt = 0$.
• $g(2) = \int_1^2 f(t) \, dt$. This is the area of a triangle with base $2-1=1$ and height $f(1)=2$. So, $g(2) = \frac{1}{2} \cdot 1 \cdot 2 = 1$.
• $g(3) = \int_1^3 f(t) \, dt = g(2) + \int_2^3 f(t) \, dt$. The second integral is the area of a triangle with base $3-2=1$ and height $f(3)=-2$. So, $g(3) = 1 + \frac{1}{2} \cdot 1 \cdot (-2) = 1 - 1 = 0$. Thus, $x=3$ is a solution.
• $g(4) = g(3) + \int_3^4 f(t) \, dt$. The integral is the area of a triangle with base $4-3=1$ and height $f(3)=-2$. So, $g(4) = 0 + \frac{1}{2} \cdot 1 \cdot (-2) = -1$.
• $g(5) = g(4) + \int_4^5 f(t) \, dt$. The integral is the area of a triangle with base $5-4=1$ and height $f(5)=2$. So, $g(5) = -1 + \frac{1}{2} \cdot 1 \cdot 2 = -1 + 1 = 0$. Thus, $x=5$ is a solution.
• $g(6) = g(5) + \int_5^6 f(t) \, dt$. The integral is the area of a triangle with base $6-5=1$ and height $f(5)=2$. So, $g(6) = 0 + \frac{1}{2} \cdot 1 \cdot 2 = 1$.
• $g(7) = g(6) + \int_6^7 f(t) \, dt$. The integral is the area of a triangle with base $7-6=1$ and height $f(7)=-2$. So, $g(7) = 1 + \frac{1}{2} \cdot 1 \cdot (-2) = 1 - 1 = 0$. Thus, $x=7$ is a solution.
• $g(8) = g(7) + \int_7^8 f(t) \, dt$. The integral is the area of a triangle with base $8-7=1$ and height $f(7)=-2$. So, $g(8) = 0 + \frac{1}{2} \cdot 1 \cdot (-2) = -1$.

Now, let's analyze the behavior of $g(x)$:

• In $(1,3]$, $g(x)$ increases from $g(1)=0$ to a maximum of $g(2)=1$, then decreases to $g(3)=0$. The only root in $(1,3]$ is $x=3$.
• In $(3,5]$, $g(x)$ decreases from $g(3)=0$ to a minimum of $g(4)=-1$, then increases to $g(5)=0$. The only root in $(3,5]$ is $x=5$.
• In $(5,7]$, $g(x)$ increases from $g(5)=0$ to a maximum of $g(6)=1$, then decreases to $g(7)=0$. The only root in $(5,7]$ is $x=7$.
• In $(7,8]$, $g(x)$ decreases from $g(7)=0$ to $g(8)=-1$. There are no roots in $(7,8]$.

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

Step 3: Calculate $\beta$

We need to find the limit $\beta=\lim\limits_{x \rightarrow 1^+} \dfrac{g(x)}{x-1}$. Since $g(1) = \int_1^1 f(t) \, dt = 0$, this limit is of the indeterminate form $\frac{0}{0}$. We can use L'Hôpital's Rule. $\beta = \lim_{x \rightarrow 1^+} \dfrac{g'(x)}{\frac{d}{dx}(x-1)} = \lim_{x \rightarrow 1^+} \dfrac{g'(x)}{1}$ By the Fundamental Theorem of Calculus, $g'(x) = f(x)$. So, $\beta = \lim_{x \rightarrow 1^+} f(x)$ Since $f(t)$ is continuous, this limit is equal to $f(1)$. From Step 1, we calculated $f(1) = 2$. Therefore, $\beta=2$.

Alternatively, the expression for $\beta$ is the definition of the right-hand derivative of $g(x)$ at $x=1$. $\beta = g'(1^+) = f(1^+) = f(1) = 2$

Step 4: Calculate the final value

We are asked to find the value of $\alpha + \beta$. $\alpha + \beta = 3 + 2 = 5$

Thus, the value of $\alpha+\beta$ is 5.