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Mathematics Question on Integral Calculus

Let the function f:[1,∞)→Rf:[1,\infin)→\R be defined by
f(t)={(−1)n+12,if t=2n−1,n∈N, (2n+1−t)2f(2n−1)+(t−(2n−1))2f(2n+1)if 2n−1<t<2n+1,n∈N.f(t) = \begin{cases} (-1)^{n+1}2, & \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 } 2n-1<t<2n+1,n\in\N. \end{cases}
Define g(x)=∫1xf(t)dt,x∈(1,∞).g(x)=\int\limits_{1}^{x}f(t)dt,x\in(1,\infin). Let α denote the number of solutions of the equation g(x) = 0 in the interval (1, 8] and β=lim⁡x→1+g(x)x−1β=\lim\limits_{x→1+}\frac{g(x)}{x-1}. Then the value of α + β is equal to _____.

Answer

The correct answer is: 5.