Let q be the maximum integral value of p in [0,10] for which the roots of the equation x2-p x+(5/4) p=0 are rational. Then the area of the region (x, y): 0 ≤ y ≤(x-q)2, 0 ≤ x ≤ q is

Question

Let q be the maximum integral value of p in [0,10] for which the roots of the equation x2-p x+(5/4) p=0 are rational. Then the area of the region (x, y): 0 ≤ y ≤(x-q)2, 0 ≤ x ≤ q is (A) 164 (B) 243 (C) (125/3) (D) 25

Tardigrade Admin · Accepted Answer

x2-p x+(5 p/4)=0 D=p2-5 p=p(p-5) ∴ q=9 0 ≤ y ≤(x-9)2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/ed24acf61fa606c643dbd882c02c06f9-.png" /> Area =∫ limits09(x-9)2 dx =243

Q. 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 ${(x, y) : 0 \leq y \leq (x - q)^{2}, 0 \leq x \leq q}$ is

164

243

$\frac{125}{3}$

25

$x^{2} - p x + \frac{5 p}{4} = 0$
$D = p^{2} - 5 p = p (p - 5)$
$∴ q = 9$
$0 \leq y \leq (x - 9)^{2}$

Area $= 0 \int 9 (x - 9)^{2} d x = 243$

Q. Let q be the maximum integral value of p in [0,10] for which the roots of the equation x2−px+45​p=0 are rational. Then the area of the region {(x,y):0≤y≤(x−q)2,0≤x≤q} is

164

243

3125​

25

x2−px+45p​=0 D=p2−5p=p(p−5) ∴q=9 0≤y≤(x−9)2 Area =0∫9​(x−9)2dx=243

Q. 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 ${(x, y) : 0 \leq y \leq (x - q)^{2}, 0 \leq x \leq q}$ is

$\frac{125}{3}$

$x^{2} - p x + \frac{5 p}{4} = 0$
$D = p^{2} - 5 p = p (p - 5)$
$∴ q = 9$
$0 \leq y \leq (x - 9)^{2}$

Area $= 0 \int 9 (x - 9)^{2} d x = 243$