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TS EAMCET 14-Sep-2020 Shift 1 Solved Paper
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© examsnet.com
Question : 9
Total: 160
The roots of the equation
(
x
−
1
)
5
=
32
(
x
+
1
)
5
are
1
+
2
e
2
k
π
i
5
1
−
2
e
2
k
π
i
5
,
k
=
1
,
2
,
3
,
4
,
5
1
−
2
e
2
k
π
i
5
1
+
2
e
2
k
π
i
5
,
k
=
0
,
1
,
2
,
3
,
4
1
,
2
ω
.
3
ω
2
,
2
ω
+
3
ω
2
,
5
ω
2
+
7
3
+
2
e
2
(
k
+
1
)
π
i
5
3
−
2
e
2
(
k
+
1
)
π
i
5
,
k
=
0
,
1
,
2
,
3
,
4
Validate
Solution:
We have,
(
x
−
1
)
5
=
32
(
x
+
1
)
5
⇒
(
x
−
1
x
+
1
)
5
=
32
⇒
x
−
1
x
+
1
=
(
32
)
1
/
5
⇒
x
−
1
x
+
1
=
2
e
2
k
π
i
5
Apply componendo and dividendo
2
2
x
=
1
−
2
e
2
k
π
t
5
1
+
2
e
2
k
π
d
5
⇒
x
=
1
+
2
e
2
k
π
t
5
1
−
2
e
2
k
π
i
5
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