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Mathematics
Mathematics
Problem
Let
g
(
x
)
=
f
(
x
)
+
f
(
1
−
x
)
g(x) = f(x) + f(1-x)
g
(
x
)
=
f
(
x
)
+
f
(
1
−
x
)
and
f
′
′
(
x
)
>
0
f''(x) > 0
f
′′
(
x
)
>
0
,
x
∈
(
0
,
1
)
x \in (0, 1)
x
∈
(
0
,
1
)
. If
g
g
g
is decreasing in the interval
(
0
,
α
)
(0, \alpha)
(
0
,
α
)
and increasing in the interval
(
α
,
1
)
(\alpha, 1)
(
α
,
1
)
, then
tan
−
1
(
2
α
)
+
tan
−
1
(
1
α
)
+
tan
−
1
(
α
+
1
α
)
\tan^{-1}(2\alpha) + \tan^{-1}\left(\frac{1}{\alpha}\right) + \tan^{-1}\left(\frac{\alpha+1}{\alpha}\right)
tan
−
1
(
2
α
)
+
tan
−
1
(
α
1
)
+
tan
−
1
(
α
α
+
1
)
is equal to: (a)
3
π
2
\frac{3\pi}{2}
2
3
π
(b)
π
\pi
π
(c)
5
π
4
\frac{5\pi}{4}
4
5
π
(d)
3
π
4
\frac{3\pi}{4}
4
3
π
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