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Tardigrade
Question
Mathematics
Let S 1, S 2 and S 3 be three sets defined as S 1= z ∈ C :|z-1| ≤ √2 S 2= z ∈ C: operatornameRe((1- i ) z ) ≥ 1 S 3= z ∈ C: operatornameIm( z ) ≤ 1 Then the set S1 ∩ S2 ∩ S3
Q. Let
S
1
,
S
2
and
S
3
be three sets defined as
S
1
=
{
z
∈
C
:
∣
z
−
1∣
≤
2
}
S
2
=
{
z
∈
C
:
Re
((
1
−
i
)
z
)
≥
1
}
S
3
=
{
z
∈
C
:
Im
(
z
)
≤
1
}
Then the set
S
1
∩
S
2
∩
S
3
2404
170
JEE Main
JEE Main 2021
Sets
Report Error
A
is a singleton
0%
B
has exactly two elements
50%
C
has infinitely many elements
25%
D
has exactly three elements
25%
Solution:
For
∣
z
−
1∣
≤
2
,
z
lies on and inside the circle of radius
2
units and centre
(
1
,
0
)
.
For
S
2
Let
z
=
x
+
i
y
Now,
(
1
−
i
)
(
z
)
=
(
1
−
i
)
(
x
+
i
y
)
R
e
((
1
−
i
)
z
)
=
x
+
y
⇒
x
+
y
≥
1
⇒
S
1
∩
S
2
∩
S
3
has infinity many elements