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KEAM 2015 Math Paper
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© examsnet.com
Question : 18
Total: 120
If
x
′
+
y
′
=
1
,
then
y
y
′
′
+
(
y
′
)
2
+
1
=
0
y
y
′
′
+
2
(
y
′
)
2
+
1
=
0
y
y
′
′
−
2
(
y
′
)
2
+
1
=
0
y
y
′
′
+
(
y
′
)
2
−
1
=
0
y
y
′
′
−
2
(
y
′
)
2
−
1
=
0
Validate
Solution:
It is given that,
x
2
+
y
2
=
1
On differentiating both sides w.r.t.
x
, we get
⇒
x
+
y
y
′
=
0
Again, on differentiating, we get
1
+
y
′
⋅
y
′
+
y
⋅
y
′
′
=
0
⇒
y
y
′
′
+
(
y
′
)
2
+
1
=
0
© examsnet.com
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