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Question
Mathematics
If f(x)= loge ((1-x/1+x)), |x|<1 , then f((2x/1+x2)) is equal to :
Q. If
f
(
x
)
=
lo
g
e
(
1
+
x
1
−
x
)
,
∣
x
∣
<
1
,
then
f
(
1
+
x
2
2
x
)
is equal to :
3509
219
JEE Main
JEE Main 2019
Relations and Functions
Report Error
A
2
f
(
x
)
52%
B
2
f
(
x
2
)
19%
C
(
f
(
x
)
)
2
17%
D
−
2
f
(
x
)
12%
Solution:
f
(
x
)
=
lo
g
e
(
1
+
x
1
−
x
)
,
∣
x
∣
<
1
f
(
1
+
x
2
2
x
)
=
ℓ
n
(
1
+
1
+
x
2
2
x
1
−
1
+
2
x
2
2
x
)
=
ℓ
n
(
(
x
+
1
)
2
(
x
−
1
)
2
)
=
2
ℓ
n
∣
∣
1
+
x
1
−
x
∣
∣
=
2
f
(
x
)