Tardigrade
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Tardigrade
Question
Mathematics
Let A= (x, y) ∈ R 2: y ≥ 0,2 x ≤ y ≤ √4-(x-1)2 and B= (x, y) ∈ R × R: 0 ≤ y ≤ min 2 x, √4-(x-1)2 . Then the ratio of the area of A to the area of B is
Q. Let
A
=
{
(
x
,
y
)
∈
R
2
:
y
≥
0
,
2
x
≤
y
≤
4
−
(
x
−
1
)
2
}
and
B
=
{
(
x
,
y
)
∈
R
×
R
:
0
≤
y
≤
min
{
2
x
,
4
−
(
x
−
1
)
2
}
}
.
Then the ratio of the area of
A
to the area of
B
is
2144
151
JEE Main
JEE Main 2023
Application of Integrals
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A
π
+
1
π
B
π
+
1
π
−
1
C
π
−
1
π
+
1
D
π
−
1
π
Solution:
y
2
+
(
x
−
1
)
2
=
4
shaded portion
=
circular
(
O
A
BC
)
−
Ar
(
△
O
A
B
)
=
4
π
(
4
)
−
2
1
(
2
)
(
1
)
A
=
(
π
−
1
)
Area
B
=
Ar
(
△
A
OB
)
+
Area of arc of circle
(
A
BC
)
=
2
1
(
1
)
(
2
)
+
4
π
(
2
)
2
=
π
+
1
B
A
=
π
+
1
π
−
1