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AP EAMCET Engineering 22 Sep 2020 Shift 1 Paper
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© examsnet.com
Question : 58
Total: 160
Assuming
x
to be so small that
x
2
and higher powers of
x
can be neglected, the coefficient of
x
in
(
1
−
x
)
1
∕
3
+
(
1
−
5
x
)
2
(
16
−
x
)
1
∕
4
is equal to
989
96
989
192
−
989
96
−
989
192
Validate
Solution:
Given expression
(
1
−
x
)
1
/
3
+
(
1
−
5
x
)
2
(
16
−
x
)
1
/
4
=
1
2
[
(
1
−
x
)
1
/
3
(
1
−
x
16
)
−
1
/
4
+
(
1
−
5
x
)
2
(
1
−
x
16
)
−
1
/
4
]
=
1
2
[
(
1
−
1
3
x
)
(
1
+
x
64
)
+
(
1
−
10
x
)
(
1
+
x
64
)
]
(On ignoring the higher degree terms)
=
1
2
(
1
−
x
3
+
x
64
+
1
−
10
x
+
x
64
)
(On ignoring the higher degree terms)
=
1
2
(
2
−
x
(
1
3
+
10
−
1
32
)
)
∴
Coefficient of
x
=
−
32
+
960
−
3
192
=
−
989
192
© examsnet.com
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