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JEE Main 25 June 2022 Shift 1 Solved Paper
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© examsnet.com
Question : 13
Total: 90
Let
f
:
R
→
R
and
g
:
R
→
R
be two functions defined by
f
(
x
)
=
log
e
(
x
2
+
1
)
−
e
−
x
+
1
and
g
(
x
)
=
1
−
2
e
2
x
e
x
. Then, for which of the following range of
α
, the inequality
f
(
g
(
(
α
−
1
)
2
3
)
)
>
f
(
g
(
α
−
5
3
)
)
holds ?
[25-Jun-2022-Shift-1]
(
2
,
3
)
(
−
2
,
−
1
)
(
1
,
2
)
(
−
1
,
1
)
Validate
Solution:
👈: Video Solution
f
(
x
)
=
log
e
(
x
2
+
1
)
−
e
−
x
+
1
f
′
(
x
)
=
2
x
x
2
+
1
+
e
−
x
=
2
x
+
1
x
+
e
−
x
>
0
∀
x
∈
R
g
(
x
)
=
e
−
x
−
2
e
x
g
′
(
x
)
−
−
e
−
x
−
2
e
x
<
0
∀
x
∈
R
⇒
f
(
x
)
is increasing and g
(
x
)
is decreasing function.
f
(
g
(
(
α
−
1
)
2
3
)
)
>
f
(
g
(
α
−
5
3
)
)
⇒
(
α
−
1
)
2
3
<
α
−
5
3
=
α
2
−
5
α
+
6
<
0
=
(
α
−
2
)
(
α
−
3
)
<
0
=
α
∈
(
2
,
3
)
© examsnet.com
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