Key Concepts and Formulas

• Inequalities: Understanding how to solve and combine inequalities, including those involving square roots.
• Geometric Interpretation of Equations: Recognizing the geometric shapes represented by equations (lines and circles).
• Quadratic Formula: Solving quadratic equations of the form $ax^2 + bx + c = 0$ using $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

Step 1: Understand the Region E and the Point (p, p+1)

The region $E$ is defined by $3-x \leq y \leq \sqrt{9-x^2}$ and $0 \leq x \leq 3$. This region is bounded below by the line $y = 3-x$ and above by the semicircle $y = \sqrt{9-x^2}$ (or $x^2 + y^2 = 9, y \geq 0$). We are given the point $(p, p+1)$ lies inside this region. This means it must satisfy the inequalities defining $E$, with strict inequalities. However, we will use non-strict inequalities initially to find the boundaries, and then interpret the final result accordingly.

Step 2: Set up the Inequalities

Since $(p, p+1)$ lies inside $E$, it must satisfy:

1. $3 - p \leq p + 1$ (The point must lie above or on the line $y = 3-x$)
2. $p + 1 \leq \sqrt{9 - p^2}$ (The point must lie below or on the semicircle $y = \sqrt{9 - p^2}$)
3. $0 \leq p \leq 3$ (The x-coordinate must be within the given bounds)

Step 3: Solve the First Inequality

$3 - p \leq p + 1$ $2 \leq 2p$ $1 \leq p$ So, $p \geq 1$.

Step 4: Solve the Second Inequality

$p + 1 \leq \sqrt{9 - p^2}$

First, consider the domain of the square root: $9 - p^2 \geq 0$, which means $p^2 \leq 9$, so $-3 \leq p \leq 3$.

Since we want the point to lie inside the region, we can assume $p+1 > 0$ because $y = 3-x$ and $y = \sqrt{9-x^2}$ intersect when $3-x = \sqrt{9-x^2}$ or $(3-x)^2 = 9-x^2$ or $9 - 6x + x^2 = 9 - x^2$ or $2x^2 - 6x = 0$ or $2x(x-3) = 0$. Then $x = 0, 3$ are the intersection points. Thus, we require $p > 0$. Since $p+1$ must be strictly less than $\sqrt{9-p^2}$, we know that $p+1$ is positive. Therefore, we can square both sides:

$(p + 1)^2 \leq 9 - p^2$ $p^2 + 2p + 1 \leq 9 - p^2$ $2p^2 + 2p - 8 \leq 0$ $p^2 + p - 4 \leq 0$

Find the roots of $p^2 + p - 4 = 0$ using the quadratic formula: $p = \frac{-1 \pm \sqrt{1^2 - 4(1)(-4)}}{2(1)} = \frac{-1 \pm \sqrt{17}}{2}$

Since the parabola opens upwards, the inequality $p^2 + p - 4 \leq 0$ is satisfied between the roots: $\frac{-1 - \sqrt{17}}{2} \leq p \leq \frac{-1 + \sqrt{17}}{2}$

Since $p+1>0$, $p>-1$, and $p \geq 1$, we only need to consider the positive root: $p \leq \frac{-1 + \sqrt{17}}{2}$

Also, we need to consider the fact that the point is inside the region. Thus, we have the inequality $p < \frac{-1 + \sqrt{17}}{2}$.

Step 5: Combine All Inequalities

We have $p \geq 1$, $p \leq \frac{-1 + \sqrt{17}}{2}$, and $0 \leq p \leq 3$. Combining these, we get:

$1 \leq p \leq \frac{-1 + \sqrt{17}}{2}$.

Since we want the point to lie strictly inside the region, we must have strict inequalities: $1 < p < \frac{-1 + \sqrt{17}}{2}$.

Thus, $a = 1$ and $b = \frac{-1 + \sqrt{17}}{2}$.

Step 6: Calculate b^2 + b - a^2

$b^2 + b - a^2 = \left(\frac{-1 + \sqrt{17}}{2}\right)^2 + \frac{-1 + \sqrt{17}}{2} - 1^2$ $= \frac{1 - 2\sqrt{17} + 17}{4} + \frac{-1 + \sqrt{17}}{2} - 1$ $= \frac{18 - 2\sqrt{17}}{4} + \frac{-2 + 2\sqrt{17}}{4} - \frac{4}{4}$ $= \frac{18 - 2\sqrt{17} - 2 + 2\sqrt{17} - 4}{4}$ $= \frac{12}{4} = 3$

Common Mistakes & Tips

• Remember to consider the domain of the square root when solving inequalities.
• Be careful when squaring inequalities; make sure both sides are non-negative.
• Don't forget the original constraints on the variables (in this case, $0 \leq x \leq 3$).
• Pay attention to whether the problem asks for the point to be inside or on the region. This determines whether you use strict or non-strict inequalities.

Summary

The problem involves finding the range of values for $p$ such that the point $(p, p+1)$ lies strictly inside the region $E$ defined by $3-x \leq y \leq \sqrt{9-x^2}$ and $0 \leq x \leq 3$. We set up inequalities based on the boundaries of the region and solved them to find the interval $(a, b)$ for $p$. Finally, we calculated $b^2 + b - a^2$ and found it to be 3.

Final Answer

The final answer is \boxed{3}.