Let q be the maximum integral value of p in [ | vedclass

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(A) For the roots of the quadratic equation $x^2 - px + \frac{5}{4}p = 0$ to be rational,the discriminant $D$ must be a perfect square.$D = (-p)^2 - 4

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$243$

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(A) For the roots of the quadratic equation $x^2 - px + \frac{5}{4}p = 0$ to be rational,the discriminant $D$ must be a perfect square.$D = (-p)^2 - 4(1)(\frac{5}{4}p) = p^2 - 5p$.We are given $p \in [0, 10]$ and $p$ is an integer.Let $p^2 - 5p = k^2$ for some non-negative integer $k$.Testing integer values of $p$ in $[0, 10]$:If $p=0, D=0$ (perfect square).If $p=1, D=-4$.If $p=2, D=-6$.If $p=3, D=-6$.If $p=4, D=-4$.If $p=5, D=0$ (perfect square).If $p=6, D=6$.If $p=7, D=14$.If $p=8, D=24$.If $p=9, D=81 - 45 = 36 = 6^2$ (perfect square).If $p=10, D=100 - 50 = 50$.The maximum integral value of $p$ is $q = 9$.The area of the region is given by $\int_{0}^{9} (x - 9)^2 dx$.Let $u = x - 9$,then $du = dx$. When $x=0, u=-9$; when $x=9, u=0$.Area $= \int_{-9}^{0} u^2 du = \left[ \frac{u^3}{3} \right]_{-9}^{0} = 0 - (\frac{-729}{3}) = 243$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

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

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