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Answered by
8
Hi ,
| x - 5 | < -1
| x - 5 | never be negative
positive number < -1 is not
possible .
option K is correct ,
I hope this helps you.
:)
| x - 5 | < -1
| x - 5 | never be negative
positive number < -1 is not
possible .
option K is correct ,
I hope this helps you.
:)
Answered by
6
it's based on Modulus concept .
Let's start to break Modulus .
|x - 5 | < -1
case 1 : when x ≥ 5
then, ( x - 5) > 0 hence, mode break with positive sign .
e.g (x - 5) < -1
x - 5 + 1 < 0
x - 4 < 0 => x < 4
now,
there is no any set which is common in x > 5 and x < 4
hence, x∈ Φ
case 2 : - when x < 5
then, (x - 5) < 0, hence mode break with negative sign .
e.g., -(x - 5) < -1
-x + 5 < -1
-x + 5 + 1 < 0
-x + 6 < 0 => x > 6
now, take common in x < 5 and x > 6 ,
we get , x ∈ Φ
hence, there are no any set in which
|x - 5 | < -1
so, Option K is correct .
you can understand it with graph too .
see attached graph ,
above graph doesn't intersects or overlaps the below graph .
hence, x ∈ Φ
Let's start to break Modulus .
|x - 5 | < -1
case 1 : when x ≥ 5
then, ( x - 5) > 0 hence, mode break with positive sign .
e.g (x - 5) < -1
x - 5 + 1 < 0
x - 4 < 0 => x < 4
now,
there is no any set which is common in x > 5 and x < 4
hence, x∈ Φ
case 2 : - when x < 5
then, (x - 5) < 0, hence mode break with negative sign .
e.g., -(x - 5) < -1
-x + 5 < -1
-x + 5 + 1 < 0
-x + 6 < 0 => x > 6
now, take common in x < 5 and x > 6 ,
we get , x ∈ Φ
hence, there are no any set in which
|x - 5 | < -1
so, Option K is correct .
you can understand it with graph too .
see attached graph ,
above graph doesn't intersects or overlaps the below graph .
hence, x ∈ Φ
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