Let q be the maximum integral value of p in [0,10] for which the roots of the equation x2−px+45p=0 are rational. Then the area of the region {(x,y):0≤y≤(x−q)2,0≤x≤q} is
(A)243✓ Correct
(B)25
(C)3125
(D)164
Answer: (A)
Solution
Rational roots of x2−px+45p=0 require the discriminant D=p2−5p=p(p−5) to be a perfect square. Testing integers in [0,10], this holds for p=0,5,9 (giving D=0,0,36), so the largest such value is q=9. The region 0≤y≤(x−9)2,0≤x≤9 has area ∫09(x−9)2dx=[3(x−9)3]09=0−3(−9)3=243.
JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option