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JEE Main Math Class 12 Application of Derivatives Part 3 Questions
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© examsnet.com
Question : 25
Total: 40
Let
g
(
x
)
=
f
(
x
)
+
f
(
1
−
x
)
and
f
n
(
x
)
>
0
,
x
∈
(
0
,
1
)
. If
g
is decreasing in the interval
(
0
,
α
)
and increasing in the interval
(
α
,
1
)
, then
tan
−
1
(
2
α
)
+
tan
−
1
(
α
+
1
α
)
is equal to
[10-Apr-2023 shift 2]
5
π
4
π
3
π
4
3
π
2
Validate
Solution:
g
(
x
)
=
f
(
x
)
+
f
(
1
−
x
)
&
f
′
′
(
x
)
>
0
,
x
∈
(
0
,
1
)
g
′
(
x
)
=
f
′
(
x
)
−
f
′
(
1
−
x
)
=
0
⇒
f
′
(
x
)
=
f
′
(
1
−
x
)
x
=
1
−
x
x
=
1
2
g
′
(
x
)
=
0
at
x
=
1
2
g
′
′
(
x
)
=
f
′
′
(
x
)
+
f
′
′
(
1
−
x
)
>
0
g
is concave up
hence
α
=
1
2
tan
−
1
2
α
+
tan
−
1
1
α
+
tan
−
1
α
+
1
α
tan
−
1
1
+
tan
−
1
2
+
tan
−
1
3
=
π
© examsnet.com
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