UPSEE Solved Model Paper 1

Let (α,β) be the point with integral coordinates and lying in the interior of the region common to the circle x2+y2 = 16 and the parabola y2 = 4x. Then
𝛼2+𝛽2 − 16 < 0 𝑎𝑛𝑑 𝛽2 − 4𝛼 < 0
It is evident from the figure that 0 < 𝛼 < 4
⇒ 𝛼 = 1,2,3 [∵ 𝛼 ∈ 𝑍]
When a = 1:
𝛽2 < 4𝛼 ⇒ 𝛽2 < 4 ⇒ 𝛽 = 0,1 ,-1,-2
So, the points are (1,0), (1, 1) and (1,-1)
When a = 2:
𝛽2 < 4 ⇒ 𝛽2 < 8 ⇒ 𝛽 = 0,1,2
So, the points are (2,0), (2,1), (2, 2),(2,-1) and(2,-2)
When a =3:
𝛽2 < 4𝛼 ⇒ 𝛽2 < 12 ⇒ 𝛽 = 1,2,3
So, the points are (3,0), (3, 1), (3, 2), (3, 3),(3,-1),(3,-2) and (3,-3)
Out of these points (3, 3) and (3-3) does not satisfy
𝛼2+𝛽2 − 16 < 0
Thus, the points lying in the region are
(1, 0), (1, 1), (1,-1),(2, 0), (2, 1),(2, 2),(2,-1), (2,-2),(3, 0), (3, 1) , (3, 2),(3,-1) and (3,-2).
So total 13 Points.