Key Concepts and Formulas

• Rational Roots of a Quadratic Equation: A quadratic equation $ax^2 + bx + c = 0$, where $a, b, c$ are rational, has rational roots if and only if its discriminant, $D = b^2 - 4ac$, is a perfect square.
• Area Under a Curve: The area bounded by the curve $y = f(x)$, the x-axis ($y=0$), and the lines $x=a$ and $x=b$ is given by $\int_a^b f(x) \, dx$, provided $f(x) \geq 0$ on $[a, b]$.

Step-by-Step Solution

Step 1: Analyze the quadratic equation and the condition for rational roots

The given quadratic equation is $x^2 - px + \frac{5}{4}p = 0$. We need to find the maximum integer value of $p$ in the interval $[0, 10]$ such that the roots of this equation are rational. The condition for rational roots involves the discriminant being a perfect square.

Step 2: Calculate the discriminant

The discriminant $D$ of the quadratic equation $ax^2 + bx + c = 0$ is given by $D = b^2 - 4ac$. In our case, $a = 1$, $b = -p$, and $c = \frac{5}{4}p$. Therefore, $D = (-p)^2 - 4(1)\left(\frac{5}{4}p\right) = p^2 - 5p = p(p-5).$

Step 3: Apply the condition for rational roots

For the roots to be rational, the discriminant $D = p(p-5)$ must be a perfect square. Let $p(p-5) = k^2$ for some non-negative integer $k$.

Step 4: Find the maximum integral value of $p$ in $[0,10]$

We need to find the largest integer $p$ in the interval $[0, 10]$ such that $p(p-5)$ is a perfect square. We test integer values:

• $p = 0$: $0(0-5) = 0$, which is a perfect square.
• $p = 1$: $1(1-5) = -4$, which is not a perfect square.
• $p = 2$: $2(2-5) = -6$, which is not a perfect square.
• $p = 3$: $3(3-5) = -6$, which is not a perfect square.
• $p = 4$: $4(4-5) = -4$, which is not a perfect square.
• $p = 5$: $5(5-5) = 0$, which is a perfect square.
• $p = 6$: $6(6-5) = 6$, which is not a perfect square.
• $p = 7$: $7(7-5) = 14$, which is not a perfect square.
• $p = 8$: $8(8-5) = 24$, which is not a perfect square.
• $p = 9$: $9(9-5) = 9(4) = 36 = 6^2$, which is a perfect square.
• $p = 10$: $10(10-5) = 50$, which is not a perfect square.

The integer values of $p$ in $[0, 10]$ that make $p(p-5)$ a perfect square are $0, 5,$ and $9$. The maximum of these is $9$. Thus, $q = 9$.

Step 5: Define the region using the value of $q$

The region is given by $\{(x, y): 0 \leq y \leq (x-q)^2, 0 \leq x \leq q\}$. Substituting $q = 9$, we have $\{(x, y): 0 \leq y \leq (x-9)^2, 0 \leq x \leq 9\}$.

Step 6: Set up the definite integral for the area

The area of the region is given by the integral $\text{Area} = \int_0^9 (x-9)^2 \, dx.$

Step 7: Evaluate the integral

$\text{Area} = \int_0^9 (x-9)^2 \, dx = \left[ \frac{(x-9)^3}{3} \right]_0^9 = \frac{(9-9)^3}{3} - \frac{(0-9)^3}{3} = 0 - \frac{(-9)^3}{3} = \frac{729}{3} = 243.$

Common Mistakes & Tips

• Remember that the discriminant must be a perfect square, not just non-negative, for rational roots.
• When testing values of $p$, make sure to check all integers in the given interval.
• Be careful with the limits of integration when setting up the area integral.

Summary

We first found the maximum integral value of $p$ in $[0, 10]$ for which the roots of the given quadratic equation are rational. This involved calculating the discriminant and testing integer values. We found that $q=9$. Then, we calculated the area of the region defined by the given inequalities using the value of $q$, which resulted in an area of 243.

Final Answer

The final answer is \boxed{243}, which corresponds to option (B).