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KEAM 2018 Math Paper
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© examsnet.com
Question : 17
Total: 120
(
√
3
−
√
2
)
4
−
(
√
3
−
√
2
)
4
=
20
√
6
30
√
6
5
√
10
40
√
6
10
√
6
Validate
Solution:
Take
(
a
+
b
)
4
=
4
C
0
a
4
+
4
C
1
a
3
b
+
4
C
2
a
2
b
2
+
4
C
3
a
b
3
+
4
C
4
b
4
=
4
C
0
a
4
+
4
C
1
a
3
b
+
4
C
2
a
2
b
2
+
4
C
1
a
b
3
+
4
C
0
b
4
(
∵
n
C
r
=
n
C
n
−
r
)
=
1
×
a
4
+
4
a
3
b
+
4
×
3
2
a
2
b
2
+
4
a
b
3
+
1
×
b
4
⇒
(
a
+
b
)
4
=
a
4
+
4
a
3
b
+
6
a
2
b
2
+
4
a
b
3
+
b
4
.....(i)
Similarly,
(
a
−
b
)
4
=
a
4
−
4
a
3
b
+
6
a
2
b
2
−
4
a
b
3
+
b
4
....(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(
a
+
b
)
4
−
(
a
−
b
)
4
=
8
a
3
b
+
8
a
b
3
=
8
a
b
(
a
2
+
b
2
)
Now, putting
a
=
√
3
and
b
=
√
2
(
√
3
+
√
2
)
4
−
(
√
3
−
√
2
)
4
=
8
√
3
√
2
[
(
√
3
)
2
+
(
√
2
)
2
]
=
8
√
6
(
3
+
2
)
=
8
√
6
×
5
=
40
√
6
© examsnet.com
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