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Test Index
JEE Main Math Class 11 Complex Numbers and Quadratic Equations Part 2 Questions
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© examsnet.com
Question : 16
Total: 100
Let
C
be the set of all complex numbers.
Let
S
1
=
{
z
∈
C
:
|
z
−
2
|
≤
1
}
and
S
2
=
{
z
∈
C
:
z
(
1
+
i
)
+
z
(
1
−
i
)
≥
4
}
.
Then, the maximum value of
z
−
5
2
|
2
for
z
∈
S
1
∩
S
2
is equal to
[2021, 27 July Shift-II]
3
+
2
√
2
4
5
+
2
√
2
2
3
+
2
√
2
2
5
+
2
√
2
4
Validate
Solution:
Let
S
1
=
{
z
∈
C
:
|
z
−
2
|
≤
1
}
and
S
2
=
{
z
∈
C
:
z
(
1
+
i
)
+
z
(
1
−
i
)
≥
4
}
Now
|
z
−
2
|
≤
1
Let
z
=
x
+
i
y
⇒
|
x
+
i
y
−
2
|
≤
1
⇒
(
x
−
2
)
2
+
y
2
≤
1
Also,
z
(
1
+
i
)
+
z
(
1
−
i
)
≥
4
⇒
(
x
+
i
y
)
(
1
+
i
)
+
(
x
−
i
y
)
(
1
−
i
)
≥
4
⇒
2
x
−
2
y
≥
4
⇒
x
−
y
≥
2
Let point on circle be
A
(
2
+
cos
θ
,
sin
θ
)
,
l
θ
∈
[
−
3
π
4
,
π
4
]
(
A
P
)
2
=
(
2
+
cos
θ
−
5
2
)
2
+
sin
2
θ
⇒
(
A
P
)
2
=
cos
2
θ
+
1
4
−
cos
θ
+
sin
2
θ
⇒
(
A
P
)
2
=
5
4
−
cos
θ
For
(
A
P
)
2
to be maximum,
θ
=
−
3
π
4
⇒
(
A
P
)
2
=
5
4
+
1
√
2
⇒
(
A
P
)
2
=
5
+
2
√
2
4
© examsnet.com
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