Question:

Step 1: Determine the condition for rational roots of the given quadratic equation. For the quadratic equation $ax^2+bx+c=0$ to have rational roots, its discriminant $D$ must be a perfect square. The given equation is $x^2-px+\dfrac{5}{4}p=0$. The coefficients are $a=1$, $b=-p$, and $c=\dfrac{5}{4}p$. The discriminant is: $$D = b^2 - 4ac = (-p)^2 - 4(1)\left(\dfrac{5}{4}p\right)$$ $$D = p^2 - 5p$$ Step 2: Find the integral values of $p$ in the given range for which the discriminant is a perfect square. For rational roots, $D = p^2 - 5p$ must be a perfect square. Let $p^2 - 5p = k^2$ for some non-negative integer $k$. We need to find integral values of $p$ in the range $[0,10]$ that satisfy this condition. Let's test each integer value of $p$ from $0$ to $10$: \begin{itemize} \item If $p=0$, $D = 0^2 - 5(0) = 0 = 0^2$. (Perfect square) \item If $p=1$, $D = 1^2 - 5(1) = 1 - 5 = -4$. (Not a perfect square) \item If $p=2$, $D = 2^2 - 5(2) = 4 - 10 = -6$. (Not a perfect square) \item If $p=3$, $D = 3^2 - 5(3) = 9 - 15 = -6$. (Not a perfect square) \item If $p=4$, $D = 4^2 - 5(4) = 16 - 20 = -4$. (Not a perfect square) \item If $p=5$, $D = 5^2 - 5(5) = 25 - 25 = 0 = 0^2$. (Perfect square) \item If $p=6$, $D = 6^2 - 5(6) = 36 - 30 = 6$. (Not a perfect square) \item If $p=7$, $D = 7^2 - 5(7) = 49 - 35 = 14$. (Not a perfect square) \item If $p=8$, $D = 8^2 - 5(8) = 64 - 40 = 24$. (Not a perfect square) \item If $p=9$, $D = 9^2 - 5(9) = 81 - 45 = 36 = 6^2$. (Perfect square) \item If $p=10$, $D = 10^2 - 5(10) = 100 - 50 = 50$. (Not a perfect square) \end{itemize} The integral values of $p$ in $[0,10]$ for which the roots are rational are $0, 5, 9$. Step 3: Identify the maximum integral value of $p$. The problem states that $q$ is the maximum integral value of $p$ in $[0,10]$ for which the roots are rational. From Step 2, the possible values are $0, 5, 9$. The maximum among these values is $9$. Therefore, $q=9$. Step 4: Set up the integral to calculate the area of the given region. The region is defined by $\{(x,y):0\leq y\leq(x-q)^2,\,0\leq x\leq q\}$. Substitute $q=9$ into the definition of the region: $\{(x,y):0\leq y\leq(x-9)^2,\,0\leq x\leq 9\}$ The area $A$ of this region can be calculated by integrating the function $y=(x-9)^2$ from $x=0$ to $x=9$: $$A = \int_{0}^{9} (x-9)^2 \, dx$$ Step 5: Evaluate the definite integral. To evaluate the integral, we can use a substitution or expand the integrand. Using substitution, let $u = x-9$. Then $du = dx$. When $x=0$, $u = 0-9 = -9$. When $x=9$, $u = 9-9 = 0$. The integral becomes: $$A = \int_{-9}^{0} u^2 \, du$$ Now, integrate $u^2$ with respect to $u$: $$A = \left[\frac{u^3}{3}\right]_{-9}^{0}$$ Substitute the limits of integration: $$A = \frac{(0)^3}{3} - \frac{(-9)^3}{3}$$ $$A = 0 - \frac{-729}{3}$$ $$A = \frac{729}{3}$$ $$A = 243$$ Step 6: State the final answer. The maximum integral value of $p$ is $q=9$, and the area of the region is $243$. Comparing this with the given options, the area matches Option 1. The final answer is $\boxed{\text{243}}$.