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JEE Main Math Class 11 Coordinate Geometry Part 5 Questions
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© examsnet.com
Question : 9
Total: 100
Let
Y
=
Y
(
X
)
be a curve lying in the first quadrant such that the area enclosed by the line
Y
−
y
=
Y
′
(
x
)
(
X
−
x
)
and the co-ordinate axes, where
(
x
,
y
)
is any point on the curve, is always
−
y
2
2
Y
′
(
x
)
+
1
,
Y
′
(
x
)
≠
0
. If
Y
(
1
)
=
1
, then
12
Y
(
2
)
equals_____
[30-Jan-2024 Shift 2]
Your Answer:
Validate
Solution:
👈: Video Solution
A
=
1
2
(
−
y
Y
′
(
x
)
+
x
)
(
y
−
x
Y
∕
x
)
=
−
y
2
2
Y
′
(
x
)
+
1
⇒
(
−
y
+
x
Y
′
(
x
)
)
(
y
−
x
Y
′
(
x
)
)
=
−
y
2
+
2
Y
′
(
x
)
−
y
2
+
x
y
Y
′
(
x
)
+
x
y
Y
′
(
x
)
−
x
2
[
Y
′
(
x
)
]
2
=
−
y
2
+
2
Y
′
(
x
)
2
x
y
−
x
2
Y
′
(
x
)
=
2
d
y
d
x
=
2
x
y
−
2
x
2
d
y
d
x
−
2
x
y
=
−
2
x
2
I.F.
=
e
−
2
ln
x
=
1
x
2
y
⋅
1
x
2
=
2
3
x
−
3
+
c
Put
x
=
1
,
y
=
1
1
=
2
3
+
c
⇒
c
=
1
3
Y
=
2
3
⋅
1
X
+
1
3
X
2
⇒
12
Y
(
2
)
=
5
3
×
12
=
20
© examsnet.com
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