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Exams
Joint Entrance Examination
Mathematics
Differential Equations
Question 267
Question 267 of 490
Medium
MCQ
If a curve
y
=
f
(
x
)
y=f(x)
y
=
f
(
x
)
, through
(
1
,
2
)
(1,2)
(
1
,
2
)
, solves
2
x
2
d
y
=
(
2
x
y
+
y
2
)
d
x
2x^2\,dy=(2xy+y^2)\,dx
2
x
2
d
y
=
(
2
x
y
+
y
2
)
d
x
, then
f
(
1
2
)
f\left(\dfrac12\right)
f
(
2
1
)
is:
A
1
1
+
log
e
2
\dfrac{1}{1+\log_e2}
1
+
lo
g
e
2
1
B
1
1
−
log
e
2
\dfrac{1}{1-\log_e2}
1
−
lo
g
e
2
1
C
1
+
log
e
2
1+\log_e2
1
+
lo
g
e
2
D
−
1
1
+
log
e
2
\dfrac{-1}{1+\log_e2}
1
+
lo
g
e
2
−
1
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Solutions · 1
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Official solution
With
y
=
v
x
y=vx
y
=
v
x
:
−
2
v
=
ln
x
+
c
\dfrac{-2}{v}=\ln x+c
v
−
2
=
ln
x
+
c
, i.e.
−
2
x
y
=
ln
x
+
c
\dfrac{-2x}{y}=\ln x+c
y
−
2
x
=
ln
x
+
c
. Using
(
1
,
2
)
(1,2)
(
1
,
2
)
:
c
=
−
1
c=-1
c
=
−
1
. At
x
=
1
/
2
x=1/2
x
=
1/2
:
y
=
1
1
+
log
e
2
y=\dfrac{1}{1+\log_e2}
y
=
1
+
lo
g
e
2
1
.
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