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JEE Main Math Class 11 Coordinate Geometry Part 6 Questions
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© examsnet.com
Question : 1
Total: 33
One of the points of intersection of the curves
y
=
1
+
3
x
−
2
x
2
and
y
=
1
x
is
(
1
2
,
2
)
. Let the area of the region enclosed by these curves be
1
24
(
ℓ
√
5
+
m
)
−
nlog
e
(
1
+
√
5
)
, where
ℓ
,
m
,
n
∈
N
. Then
ℓ
+
m
+
n
is equal to
[4 Apr 2024 Shift 1]
32
30
29
31
Validate
Solution:
👈: Video Solution
A
=
1
+
√
5
2
∫
1
2
(
1
+
3
x
−
2
x
2
−
1
x
)
dx
A
=
[
x
+
3
x
2
2
−
2
x
3
3
−
ln
x
]
1
2
1
+
√
5
2
A
=
1
+
√
5
2
+
3
2
(
1
+
√
5
2
)
2
−
2
3
(
1
+
√
5
2
)
3
−
ln
(
1
+
√
5
2
)
−
1
2
−
3
2
(
1
4
)
+
2
3
(
1
8
)
+
ln
(
1
2
)
A
=
1
2
+
√
5
2
+
3
8
+
3
4
√
5
+
15
8
−
4
3
−
2
3
√
5
−
1
2
−
3
8
+
1
12
−
ℓ
n
(
1
+
√
5
)
=
√
5
(
1
2
+
3
4
−
2
3
)
+
15
8
−
4
3
+
1
12
−
ln
(
1
+
√
5
)
=
14
24
√
5
+
15
24
−
ln
(
1
+
√
5
)
© examsnet.com
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