KVPY SA Exam 2011 Question Paper

f(x) = ${\mathrm{ax}}^2$ + bx + c
given f(1) = 0
⇒ a + b + c = 0
and 40 < f(6) < 50
⇒ 40 < 36a + 6b + c < 50
⇒ 40 < 35a + 5b < 50
⇒ 8 < 7a + b < 10
7a + b = integer = 9 ....(1)
and 60 < f(7) < 70
⇒ 60 < 49a + 7b + c < 70
⇒ 60 < 48a + 6b < 70
⇒ 10 < 8a + b < 11.6
8a + b = integer = 11 .....(2)
Solving (1) & (2)
a = 2, b = –5, c = 3
$\therefore \mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^2-5\mathrm{x}+3$
f(50) = 4753
1000 t < f(50) < 1000(t + 1)
(1000 × 4) < 4753 < 1000(4 + 1)
∴ t = 4