the solution set of the following inequality modulus x minus 1 greater than or equal to modulus x minus 3
Answers
I think this is what you mean:
| x-1 | ≥ | x-3 |
you can solve this question in more than one way.
One of these is,
| x-1 | - | x-3 | ≥ 0
Find the critical points, here they are 1 and 3.
solve for x<1,
-(x-1)-(-(x-3))≥0
-x+1+x-3≥0
-2≥0 (not possible)
Thus, x should be greater than 1
For 1 ≤ x ≤ 3
(x-1)-(-(x-3))≥0
2x-4≥0
x≥2
So one solution set is 2≤x≤3
For x>3
(x-1)-(x-3)≥0
2≥0
Therefore from above two results,
x≥2
Hope you got it
Given : | x - 1 | ≥ | x - 3 |
To find : Find x
Solution:
|x| =x if x≥ 0
|x| = - x if x < 0
| x - 1 | ≥ | x - 3 |
case 1
x ≥ 3
=> x - 1 ≥ x - 3
=> - 1 ≥ - 3
Which is true
Hence x ≥ 3
x ∈ [3 , ∞)
Case 2
x < 1
-(x - 1) ≥ -(x - 3)
=> -x + 1 ≥ -x + 3
=> 1 ≥ 3
Not possible hence
x can not be < 1
case 3
1 ≤ x < 3
x - 1 ≥ -(x - 3)
=> x - 1 ≥ -x + 3
=> 2x ≥ 4
=> x ≥ 2
=> x ∈ [ 2 , 3)
x ∈ [ 2 , 3) ∪ x ∈ [3 , ∞)
=> x ∈ [2 , ∞) or x ≥ 2
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