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KEAM 2016 Math Paper
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© examsnet.com
Question : 34
Total: 120
The coefficient of
x
2
in the expansion of the determinant
|
x
2
x
3
+
1
x
5
+
2
x
2
+
3
x
2
+
x
x
3
+
x
4
x
+
4
x
3
+
x
5
2
3
|
is
-10
-8
-2
-6
8
Validate
Solution:
For the coefficient of
x
2
, on expanding along
R
1
, we get
∆
=
x
2
[
8
x
2
+
8
x
−
x
6
−
2
x
7
−
x
8
]
−
(
x
3
+
1
)
[
8
x
3
+
24
−
x
4
−
x
5
−
4
x
3
−
4
x
4
]
+
(
x
5
+
2
)
[
x
6
+
x
7
+
3
x
3
+
3
x
4
−
x
3
−
x
2
−
4
x
2
−
4
x
]
=
8
x
4
+
8
x
3
−
x
8
−
2
x
9
−
x
10
−
8
x
6
−
24
x
3
+
x
7
+
x
8
+
4
x
6
+
4
x
7
−
8
x
3
−
24
+
x
4
+
x
5
+
4
x
3
+
4
x
4
+
x
11
+
x
12
+
3
x
8
+
3
x
9
−
x
8
−
x
7
−
4
x
7
−
4
x
6
+
2
x
6
+
2
x
7
+
6
x
3
+
6
x
4
−
2
x
3
−
2
x
2
−
8
x
2
−
8
x
Coefficient of
x
2
=
−
2
−
8
=
−
10
© examsnet.com
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