JEE MAIN - Mathematics (2023 - 30th January Evening Shift - No. 7)
Let $q$ be the maximum integral value of $p$ in $[0,10]$ for which the roots of the equation $x^2-p x+\frac{5}{4} p=0$ are rational. Then the area of the region $\left\{(x, y): 0 \leq y \leq(x-q)^2, 0 \leq x \leq q\right\}$ is :
$\frac{125}{3}$
243
164
25
Explanation
Given equation : $$4{x^2} - 4px + 5p = 0$$
_30th_January_Evening_Shift_en_7_1.png)
for rational roots, D must be perfect square
$$D = 16{p^2} - 4 \times 4 \times 5p = 16p(p - 5)$$
So, max. Integral value of $$p = 9$$ for making D is perfect square
$$\therefore$$ $$q = 9$$
Area of shared region
$$ = \int\limits_0^9 {{{(x - 9)}^2}dx} $$
$$ = \left. {{{{{(x - 9)}^3}} \over 3}} \right|_0^9 = {{{9^3}} \over 3} = 243$$ sq. units
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