Problem 1590

1. Define the functions and conditions:
Let $f(x) = x^2 - x$ and $g(x) = |x| + |x-1| + 1$. We need to find the range of $(p, q)$ such that the line $y = px + q$ intersects $y = x^2 - x$ but does not intersect $y = |x| + |x-1| + 1$.

2. **Intersection with $y = x^2 - x$:**
The line $y = px + q$ intersects the parabola $y = x^2 - x$ if the quadratic equation $x^2 - (p+1)x + q = 0$ has real roots. This requires the discriminant to be non-negative:
$$\Delta = (p+1)^2 - 4q \geq 0$$
Solving for $q$, we get:
$$q \leq \frac{(p+1)^2}{4}$$

3. **Non-intersection with $y = |x| + |x-1| + 1$:**
The function $g(x) = |x| + |x-1| + 1$ can be broken into three linear segments:
- For $x \leq 0$: $g(x) = 1 - 2x$
- For $0 \leq x \leq 1$: $g(x) = 2$
- For $x \geq 1$: $g(x) = 2x - 1$

The line $y = px + q$ must lie below $g(x)$ for all $x$:
- For $x \leq 0$: $px + q < 1 - 2x$
- For $0 \leq x \leq 1$: $px + q < 2$
- For $x \geq 1$: $px + q < 2x - 1$

4. Analyzing the inequalities:
- For $x \leq 0$: $px + q < 1 - 2x$
$$q < 1 - (2 + p)x \quad \text{(since \( x \leq 0 \))}$$
The most restrictive condition is when $x = 0$:
$$q < 1$$

- For $0 \leq x \leq 1$: $px + q < 2$
$$q < 2 \quad \text{(since \( 0 \leq x \leq 1 \))}$$

- For $x \geq 1$: $px + q < 2x - 1$
$$q < 2x - (p+1)x \quad \text{(since \( x \geq 1 \))}$$
The most restrictive condition is when $x = 1$:
$$q < 2 - p$$

5. Combining the conditions:
- For $p \leq 0$:
$$q < 2$$
- For $p > 0$:
$$q < 2 - p$$

6. **Range of $(p, q)$:**
Combining the conditions from the discriminant and the non-intersection:
$$q \leq \frac{(p+1)^2}{4} \quad \text{and} \quad q < \begin{cases} 2 & \text{if } p \leq 0 \\ 2 - p & \text{if } p > 0 \end{cases}$$

7. Finding the area:
The area of the region in the $(p, q)$-plane is bounded by these inequalities. We need to integrate over the valid region:
$$\text{Area} = \int_{-2}^{0} \left(2 - \frac{(p+1)^2}{4}\right) dp + \int_{0}^{2} \left(2 - p - \frac{(p+1)^2}{4}\right) dp$$

- For $-2 \leq p \leq 0$:
$$\int_{-2}^{0} \left(2 - \frac{(p+1)^2}{4}\right) dp = \int_{-2}^{0} \left(2 - \frac{p^2 + 2p + 1}{4}\right) dp = \int_{-2}^{0} \left(2 - \frac{p^2}{4} - \frac{p}{2} - \frac{1}{4}\right) dp$$
$$= \int_{-2}^{0} \left(\frac{7}{4} - \frac{p^2}{4} - \frac{p}{2}\right) dp$$
$$= \left[\frac{7p}{4} - \frac{p^3}{12} - \frac{p^2}{4}\right]_{-2}^{0}$$
$$= \left(0 - 0 - 0\right) - \left(\frac{7(-2)}{4} - \frac{(-2)^3}{12} - \frac{(-2)^2}{4}\right)$$
$$= 0 - \left(-\frac{14}{4} + \frac{8}{12} + \frac{4}{4}\right)$$
$$= \frac{14}{4} - \frac{2}{3} - 1$$
$$= \frac{14}{4} - \frac{2}{3} - \frac{4}{4}$$
$$= \frac{14 - 4}{4} - \frac{2}{3}$$
$$= \frac{10}{4} - \frac{2}{3}$$
$$= \frac{5}{2} - \frac{2}{3}$$
$$= \frac{15}{6} - \frac{4}{6}$$
$$= \frac{11}{6}$$

- For $0 \leq p \leq 2$:
$$\int_{0}^{2} \left(2 - p - \frac{(p+1)^2}{4}\right) dp = \int_{0}^{2} \left(2 - p - \frac{p^2 + 2p + 1}{4}\right) dp = \int_{0}^{2} \left(2 - p - \frac{p^2}{4} - \frac{p}{2} - \frac{1}{4}\right) dp$$
$$= \int_{0}^{2} \left(\frac{7}{4} - p - \frac{p^2}{4} - \frac{p}{2}\right) dp$$
$$= \int_{0}^{2} \left(\frac{7}{4} - \frac{3p}{2} - \frac{p^2}{4}\right) dp$$
$$= \left[\frac{7p}{4} - \frac{3p^2}{4} - \frac{p^3}{12}\right]_{0}^{2}$$
$$= \left(\frac{7(2)}{4} - \frac{3(2)^2}{4} - \frac{(2)^3}{12}\right) - \left(0 - 0 - 0\right)$$
$$= \left(\frac{14}{4} - \frac{12}{4} - \frac{8}{12}\right)$$
$$= \left(\frac{14}{4} - \frac{12}{4} - \frac{2}{3}\right)$$
$$= \left(\frac{14 - 12}{4} - \frac{2}{3}\right)$$
$$= \left(\frac{2}{4} - \frac{2}{3}\right)$$
$$= \left(\frac{1}{2} - \frac{2}{3}\right)$$
$$= \left(\frac{3}{6} - \frac{4}{6}\right)$$
$$= -\frac{1}{6}$$

Adding the two areas:
$$\text{Total Area} = \frac{11}{6} + \left(-\frac{1}{6}\right) = \frac{10}{6} = \frac{5}{3}$$

The final answer is $\boxed{\frac{5}{3}}$.