Let $$q$$ be the maximum integral value of $$p$$ in $$[0, 10]$$ for which the roots of the equation $$x^2 - px + \frac{5}{4}p = 0$$ are rational. Then the area of the region $$\{(x,y) : 0 \leq y \leq (x-q)^2, 0 \leq x \leq q\}$$ is

We need to find the maximum integral value of $$p$$ in $$[0, 10]$$ for which the equation $$x^2 - px + \frac{5}{4}p = 0$$ has rational roots, and then compute the area of the given region.

Condition for rational roots.

The discriminant of $$x^2 - px + \frac{5p}{4} = 0$$ is:

$$D = p^2 - 4 \cdot \frac{5p}{4} = p^2 - 5p = p(p - 5)$$

For rational roots, $$D$$ must be a non-negative perfect square.

Find the maximum integral $$p$$ in $$[0, 10]$$.

We need $$p(p-5) = k^2$$ for some non-negative integer $$k$$. Testing integer values:

$$p = 0$$: $$0 \cdot (-5) = 0 = 0^2$$ (valid)

$$p = 5$$: $$5 \cdot 0 = 0 = 0^2$$ (valid)

$$p = 9$$: $$9 \cdot 4 = 36 = 6^2$$ (valid)

$$p = 10$$: $$10 \cdot 5 = 50$$ (not a perfect square)

Check $$p = 6, 7, 8$$: $$6, 14, 24$$ — none are perfect squares.

The maximum integral value is $$q = 9$$.

Compute the area.

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

$$\text{Area} = \int_0^9 (x - 9)^2\, dx$$

Let $$u = x - 9$$, so $$du = dx$$. When $$x = 0$$, $$u = -9$$; when $$x = 9$$, $$u = 0$$.

$$\text{Area} = \int_{-9}^{0} u^2\, du = \left[\frac{u^3}{3}\right]_{-9}^{0} = 0 - \frac{(-9)^3}{3} = 0 - \frac{-729}{3} = 243$$

The area is $$243$$.

The correct answer is Option A: $$243$$.