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Byju's Answer
Standard XII
Mathematics
Derivative from First Principle
limx →∞∑r = 1...
Question
lim
x
→
∞
n
∑
r
=
1
tan
−
1
(
2
r
1
−
r
2
+
r
4
)
is equal to
A
π
/
4
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B
π
/
2
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C
3
π
4
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D
None of these
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Solution
The correct option is
B
π
/
2
Given,
lim
n
→
∞
n
∑
r
=
1
t
a
n
−
1
(
2
r
1
−
r
2
+
r
4
)
NOw,
t
a
n
−
1
(
(
r
2
+
r
)
−
(
r
2
−
r
)
1
+
(
r
2
+
r
)
(
r
2
−
r
)
)
=
t
a
n
−
1
(
r
2
+
r
)
−
t
a
n
−
1
(
r
2
−
r
)
summing over
r
=
1
to
r
=
n
,the terms will get cancel and we get
=
>
t
a
n
−
1
(
n
2
+
n
)
−
t
a
n
−
1
(
0
)
=
>
lim
n
→
∞
t
a
n
−
1
(
n
2
+
n
)
=
>
t
a
n
−
1
(
∞
)
=
>
π
2
Suggest Corrections
0
Similar questions
Q.
lim
n
→
∞
n
∑
r
=
1
tan
−
1
(
2
r
1
−
r
2
+
r
4
)
is equal to
Q.
∑
n
r
=
1
t
a
n
−
1
(
2
r
−
1
1
+
2
2
r
−
1
)
is equal to
Q.
The value of
∞
∑
r
=
1
tan
−
1
2
r
2
+
r
2
+
r
4
is equal to
Q.
If
S
=
90
∑
r
=
1
tan
−
1
(
2
r
2
+
r
2
+
r
4
)
,
then
8190
(
cot
S
)
is equal to
Q.
If
S
=
90
∑
r
=
1
t
a
n
−
1
(
2
r
2
+
r
2
+
r
4
)
, then
8190
c
o
t
S
is equal to
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