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Byju's Answer
Standard XII
Mathematics
Range of Quadratic Expression
Find the set ...
Question
Find the set of all solutions of the equation
2
|
y
|
−
|
2
y
−
1
−
1
|
=
2
y
−
1
+
1
.
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Solution
When
2
y
−
1
−
1
<
0
,
2
y
−
1
<
1
⇒
2
y
−
1
<
2
0
⇒
y
−
1
<
0
⇒
y
<
1
(
∵
2
x
is a strictly increasing function)
Case 1: (
y
<
0
)
⇒
2
(
−
y
)
−
(
1
−
2
y
−
1
)
=
2
y
−
1
+
1
⇒
2
(
−
y
)
=
2
⇒
y
=
(
−
1
)
Case 2: (
0
≤
y
<
1
)
⇒
2
y
−
(
1
−
2
y
−
1
)
=
2
y
−
1
+
1
⇒
2
y
=
2
⇒
y
=
1
Case 3: (
y
>
1
)
⇒
2
y
−
(
2
y
−
1
−
1
)
=
2
y
−
1
+
1
⇒
2
y
=
2
y
−
1
+
2
y
−
1
=
2
(
2
y
−
1
)
=
2
y
So, all of
y
>
1
,
y
∈
Real Numbers
ie, This equation is satisfied
∀
y
∈
{
(
−
1
)
}
∪
[
1
,
∞
)
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0
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Q.
Find the set of all solutions of the equation
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|
−
∣
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, the solution includes
Q.
From the set of all solution of the equation
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Q.
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Q.
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Q.
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