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TS EAMCET 14-Sep-2020 Shift 2 Solved Paper
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© examsnet.com
Question : 4
Total: 160
If a and b are any two real numbers, then
|
2
a
−
2
b
−
4
4
a
4
a
4
2
−
b
−
a
4
2
b
2
b
b
−
a
−
2
|
=
4
[
(
a
+
b
)
3
+
8
(
a
+
b
)
2
+
16
(
a
+
b
)
+
8
]
1
2
(
a
+
b
+
2
)
3
2
[
(
a
+
b
)
3
+
6
(
a
+
b
)
2
+
12
(
a
+
b
)
+
8
]
(
a
+
b
+
2
)
3
Validate
Solution:
We have,
=
2
|
2
a
−
2
b
−
4
4
a
4
a
4
2
−
b
−
a
4
2
b
2
b
b
−
a
−
2
|
=
2
|
a
−
b
−
2
2
a
2
a
4
2
−
b
−
a
4
2
b
2
b
b
−
a
−
2
|
=
2
|
−
(
a
+
b
+
2
)
0
2
a
a
+
b
+
2
−
(
a
+
b
+
2
)
4
0
a
+
b
+
2
b
−
a
−
2
|
C
1
→
C
1
−
C
2
and
C
2
→
C
2
−
C
3
=
2
(
a
+
b
+
2
)
2
|
−
1
0
2
a
1
−
1
4
0
1
b
−
a
−
2
|
=
2
(
a
+
b
+
2
)
2
(
−
1
(
−
b
+
a
+
2
−
4
)
+
2
a
(
1
)
=
2
(
a
+
b
+
2
)
2
(
a
+
b
+
2
)
=
2
(
a
+
b
+
2
)
3
=
2
[
(
a
+
b
)
3
+
6
(
a
+
b
)
2
+
12
(
a
+
b
)
+
8
]
© examsnet.com
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