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Byju's Answer
Standard XII
Mathematics
nth Term of HP
lf log1/2|z...
Question
lf
log
1
2
|
z
−
2
|
>
log
1
2
|
z
|
then
A
x
>
1
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B
x
<
1
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C
x
<
2
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D
x
>
2
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Solution
The correct option is
A
x
>
1
L
e
t
z
=
x
+
i
y
log
1
2
|
z
−
2
|
>
log
1
2
|
z
|
⇒
|
z
−
2
|
<
|
z
|
⇒
|
x
−
2
+
i
y
|
<
|
x
+
i
y
|
⇒
√
x
2
−
4
x
+
4
+
y
2
<
√
x
2
+
y
2
⇒
−
4
x
+
4
<
0
⇒
x
>
1.
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0
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Standard XII Mathematics
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