KCET 2019 Math Solved Paper

(B)

y2 = x → (1)
x2 + y2 = 2x → (2)
Equation (2) is a circle with centre (1, 0) and radius 1.
Solving (1) and (2), we get the points of intersection (0, 0) and (1, 1)
(x–1)2+y2 = 1
y2 = x
(x–1)2 + x = 1
x2– x = 0
x(x – 1) = 0
x = 0, x = 1
area =

1

∫

0

{√1−(x−1)2−√x}dx
=−[

x 3 2

3 2

]01 +
=−

2

3

+{0+

π

4

} = −

2

3

+

π

4