CBSE Class 11-science Answered
Find values of x for:-
||x-1|-2|=|x-3|
![question image](http://images.topperlearning.com/topper/new-ate/008422Screenshot20200809164315896com.google.android.apps.docs.jpg)
Asked by 9377777388ts | 09 Aug, 2020, 17:36: PM
TO solve: ||x-1|-2|=|x-3|
![If space straight x greater or equal than 1 rightwards double arrow straight x minus 1 greater or equal than 0
rightwards double arrow open vertical bar straight x minus 1 close vertical bar equals straight x minus 1
rightwards double arrow open vertical bar open vertical bar straight x minus 1 close vertical bar minus 2 close vertical bar equals open vertical bar straight x minus 1 minus 2 close vertical bar equals open vertical bar straight x minus 3 close vertical bar
rightwards double arrow open vertical bar straight x minus 3 close vertical bar equals open vertical bar straight x minus 3 close vertical bar
rightwards double arrow open vertical bar open vertical bar straight x minus 1 close vertical bar minus 2 close vertical bar equals open vertical bar straight x minus 3 close vertical bar space is space true space for space every space straight x greater or equal than 1
If space 0 less than straight x less than 1 space rightwards double arrow space straight x minus 1 less than 0
rightwards double arrow open vertical bar open vertical bar straight x minus 1 close vertical bar minus 2 close vertical bar equals open vertical bar negative straight x plus 1 minus 2 close vertical bar equals open vertical bar negative straight x minus 1 close vertical bar
rightwards double arrow open vertical bar negative straight x minus 1 close vertical bar equals open vertical bar straight x minus 3 close vertical bar
As space space 0 less than straight x rightwards double arrow negative straight x less than 0 rightwards double arrow negative straight x minus 1 less than negative 1 less than 0
rightwards double arrow open vertical bar negative straight x minus 1 close vertical bar equals negative left parenthesis negative straight x minus 1 right parenthesis equals straight x plus 1
rightwards double arrow straight x plus 1 equals negative left parenthesis straight x minus 3 right parenthesis space space... space since space straight x less than 1 less than 3
rightwards double arrow straight x plus 1 equals negative straight x plus 3
rightwards double arrow 2 straight x equals 2 rightwards double arrow straight x equals 1
Thus comma space open vertical bar open vertical bar straight x minus 1 close vertical bar minus 2 close vertical bar equals open vertical bar straight x minus 3 close vertical bar space is space true space for space every space straight x greater or equal than 1
Hence comma space the space solution space is space straight x element of left square bracket 1 comma space infinity right parenthesis.](https://images.topperlearning.com/topper/tinymce/cache/0d380bdfe580096cf3c6e1c98051ffce.png)
Answered by Renu Varma | 13 Aug, 2020, 12:49: PM
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