$\text{Let the function}\mathrm{ f}:[1,\infty )\to R\text{ be defined by}$

$f\left(t\right)=\left\{\begin{array}{ll}2(-1){}^{n+1}& \text{; if}t=2n-1,n\in N\\ \frac{2n+1-t}{2}f(2n-1)+\frac{t-(2n-1)}{2}f(2n+1)& \text{; if}\mathrm{ 2}n-1<t<2n+1,n\in N\end{array}\right..$

$\text{Define}g\left(x\right)={\int }_1^{x}f\left(t\right)dt,x\in (1,\infty ).$

$\text{Let}\mathrm{ \alpha }\text{ denote the number of solutions of the equation}\mathrm{ g}\left(x\right)=\mathrm{0 }\text{in the interval}(1,8]\text{and }\beta =\underset{x\to 1^{+}}{lim}\frac{g\left(x\right)}{x-1}.$

$\text{Then the value of }\alpha +\mathrm{\beta }\text{is equal to \_\_\_\_\_}$

The correct answer is 5.

Step 1: Analyze the function $f\left(t\right)$

The given function $f:[1,\infty )\to R$ is defined at odd integer points $t=2n-1$ (for $n\in N$) as:

$f(2n-1)=2(-1{)}^{n+1}$

Evaluating this at specific values of $n$:

For $n=1$: $t=1\Rightarrow f\left(1\right)=2(-1{)}^2=2$
For $n=2$: $t=3\Rightarrow f\left(3\right)=2(-1{)}^3=-2$
For $n=3$: $t=5\Rightarrow f\left(5\right)=2(-1{)}^4=2$
For $n=4$: $t=7\Rightarrow f\left(7\right)=2(-1{)}^5=-2$

For any interval $2n-1<t<2n+1$, the expression:

$f\left(t\right)=\frac{2n+1-t}{2}f(2n-1)+\frac{t-(2n-1)}{2}f(2n+1)$

represents a linear interpolation between $f(2n-1)$ and $f(2n+1)$. Thus, $f\left(t\right)$ is a continuous piecewise linear function connecting the points $(1,2),(3,-2),(5,2),(7,-2)$, and so on.

Step 2: Find the value of $\beta$

We are given:

$\beta =\underset{x\to 1^{+}}{lim}\frac{g\left(x\right)}{x-1}$

Since $g\left(1\right)={\int }_1^{1}f\left(t\right)dt=0$, this limit represents the right-hand derivative of $g\left(\mathrm{x}\right)$ at $x=1$:

$\beta =\underset{x\to 1^{+}}{lim}\frac{g\left(x\right)-g\left(1\right)}{x-1}=g^{\text{'}}\left(1^{+}\right)$

By the Fundamental Theorem of Calculus, $g^{\text{'}}\left(x\right)=f\left(x\right)$. Therefore:

$\beta =f\left(1\right)=2$

Step 3: Find the value of $\alpha$

$\alpha$ is the number of solutions to $g\left(x\right)=0$ in the interval $(1,8]$.g(x)=∫1xf(t)dt across sub-intervals:

1. For $x\in [1,3]$:
The straight line from $(1,2)$ to $(3,-2)$ is given by $f\left(\mathrm{t}\right)=4-2t$.
$g\left(x\right)={\int }_1^x(4-2t)dt=[4t-t^2{]}_1^x=-(x-1\left)\right(x-3)$.
For $x\in (1,3)$, $g\left(x\right)>0$.
At $x=3$, $g\left(3\right)=0$. This is the 1st solution.

2. For $x\in [3,5]$:
The function $f\left(t\right)$ goes linearly from $-2$ at $t=3$ to $2$ at $t=5$.
By symmetry, the net area under $f\left(t\right)$ over $[3,5]$ is $0$.
Thus, $g\left(5\right)=g\left(3\right)+{\int }_3^{5}f\left(\mathrm{t}\right)dt=0+0=0$.
For $x\in (3,5)$, $g\left(x\right)<0$.
At $x=5$, $g\left(5\right)=0$. This is the 2nd solution.

3. For $x\in [5,7]$:
Similarly, $f\left(t\right)$ goes linearly from $2$ at $t=5$ to $-2$ at $t=7$.
The net area over $[5,7]$ is $0$.
Thus, $g\left(7\right)=g\left(5\right)+0=0$.
For $x\in (5,7)$, $g\left(x\right)>0$.
At $x=7$, $g\left(7\right)=0$. This is the 3rd solution.

4. For $x\in (7,8]$:
In this region, $f\left(t\right)$ is strictly negative (increasing from $-2$ at $t=7$ to $-0.5$ at $t=8$).
Hence, $g\left(x\right)$ decreases from $0$ to negative values, so there are no roots in $(7,8]$.

Therefore, the solutions of $g\left(x\right)=0$ in $(1,8]$ are $x=3,5,7$, which gives:

$\alpha =3$

Step 4: Compute $\alpha +\beta$

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