Problem Analysis #28

To solve the problem, we need to find the number of intersection points between the functions $y = f(x)$ and $y = g(x)$ in the range $x \ge 0$. Given $a = \frac{3}{5}$, we have:

$$f(x) = \frac{3}{5 \cdot 8}(x - 1)^2 + \frac{2}{\frac{3}{5}} - 3 = \frac{3}{40}(x - 1)^2 + \frac{10}{3} - 3 = \frac{3}{40}(x - 1)^2 - \frac{1}{3}.$$

The function $g(x)$ is a periodic triangular wave with period 2, defined as:

$$g(x) = \begin{cases} x - 2n, & 2n \le x < 2n + 1, \\ -x + 2n + 2, & 2n + 1 \le x < 2n + 2. \end{cases}$$

We analyze the intersections in each interval $[2n, 2n + 2)$ for $n \ge 0$.

Case 1: $n = 0$ (Interval $[0, 2)$)

• Subinterval $[0, 1)$: $g(x) = x$. Solve $\frac{3}{40}(x - 1)^2 - \frac{1}{3} = x$:

$$\frac{3}{40}(x - 1)^2 = x + \frac{1}{3} \implies 3(x - 1)^2 = 40x + \frac{40}{3} \implies 9(x - 1)^2 = 120x + 40.$$

This quadratic equation has no real solutions in $[0, 1)$.

• Subinterval $[1, 2)$: $g(x) = -x + 3$. Solve $\frac{3}{40}(x - 1)^2 - \frac{1}{3} = -x + 3$:

$$\frac{3}{40}(x - 1)^2 = x + \frac{8}{3} \implies 3(x - 1)^2 = 40x + \frac{320}{3} \implies 9(x - 1)^2 = 120x + 320.$$

This quadratic equation has one real solution in $[1, 2)$.

Case 2: $n = 1$ (Interval $[2, 4)$)

• Subinterval $[2, 3)$: $g(x) = x - 2$. Solve $\frac{3}{40}(x - 1)^2 - \frac{1}{3} = x - 2$:

$$\frac{3}{40}(x - 1)^2 = x - \frac{5}{3} \implies 3(x - 1)^2 = 40x - \frac{200}{3} \implies 9(x - 1)^2 = 120x - 200.$$

This quadratic equation has one real solution in $[2, 3)$.

• Subinterval $[3, 4)$: $g(x) = -x + 5$. Solve $\frac{3}{40}(x - 1)^2 - \frac{1}{3} = -x + 5$:

$$\frac{3}{40}(x - 1)^2 = x - \frac{16}{3} \implies 3(x - 1)^2 = 40x - \frac{640}{3} \implies 9(x - 1)^2 = 120x - 640.$$

This quadratic equation has one real solution in $[3, 4)$.

Case 3: $n = 2$ (Interval $[4, 6)$)

• Subinterval $[4, 5)$: $g(x) = x - 4$. Solve $\frac{3}{40}(x - 1)^2 - \frac{1}{3} = x - 4$:

$$\frac{3}{40}(x - 1)^2 = x - \frac{11}{3} \implies 3(x - 1)^2 = 40x - \frac{440}{3} \implies 9(x - 1)^2 = 120x - 440.$$

This quadratic equation has one real solution in $[4, 5)$.

• Subinterval $[5, 6)$: $g(x) = -x + 7$. Solve $\frac{3}{40}(x - 1)^2 - \frac{1}{3} = -x + 7$:

$$\frac{3}{40}(x - 1)^2 = x - \frac{20}{3} \implies 3(x - 1)^2 = 40x - \frac{800}{3} \implies 9(x - 1)^2 = 120x - 800.$$

This quadratic equation has one real solution in $[5, 6)$.

Conclusion:

Each interval $[2n, 2n + 2)$ for $n \ge 0$ contains exactly 2 intersection points. Since the function $g(x)$ is periodic with period 2, this pattern repeats indefinitely. Therefore, the total number of intersection points in $x \ge 0$ is infinite. However, the problem likely expects the number of intersections in the first period $[0, 2)$, which is 2.

$$\boxed{2}$$