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JEE Advanced 2019 Paper 1
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© examsnet.com
Question : 54
Total: 54
Let
ω
≠
1
be a cube root of unity. Then the minimum of the set
{
|
a
+
b
ω
+
c
ω
2
|
2
:
a
,
b
,
c
distinct non-zero integers" } equals _________
Your Answer:
Validate
Solution:
|
a
+
b
ω
+
c
ω
2
|
2
=
|
a
+
b
(
−
1
2
+
i
√
3
2
)
+
C
(
−
1
2
−
i
√
3
2
)
|
2
=
|
(
a
−
b
2
−
c
2
)
+
i
√
3
2
(
b
−
c
)
|
2
=
(
a
−
b
2
−
c
2
)
2
+
3
4
(
b
−
c
)
2
=
1
4
[
(
2
a
−
b
−
c
)
2
+
3
(
b
−
c
)
2
]
a
,
b
,
c
{
−
1
,
0
,
1
}
Min value:
a
=
−
1
,
b
=
1
,
c
=
0
=
1
4
[
(
2
−
1
)
2
+
3
(
1
)
2
]
=
12
4
=
3
© examsnet.com
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