CBSE Class 11-science Answered
Dear Student,
The critical points are where the quantity inside the modulus (in the given expression) becomes zero. Hence the critical points here are:-2 and -1.
Case1:
x -2:
2 -(x+2) - [-(2 x+1 - 1)] = 2 x+1 +1................(because for X -2, |x+2| and |2 x+1 - 1| are both negative)
2 -(x+2) +(2 x+1 - 1) = 2 x+1 +1
2 -(x+2)=21
-x-2 = 1
x=-3
which is in the domain considered, i.e.,X -2, Hence x=-3 is a solution.
Similarly;
Case 2:
-2<x-1:
2 (x+2) - [-(2 x+1 - 1)] = 2 x+1 +1
2 (x+2) +(2 x+1 - 1) = 2 x+1 +1
2 (x+2)=21
x+2 = 1
x=-1
which is in the considered domain; -2<x-1. Hence x=-1 is a solution.
Case 3:
x>-1:
2 (x+2) - (2 x+1 - 1) = 2 x+1 +1
2 (x+2) - 2 x+1 + 1 = 2 x+1 +1
2 (x+2) = 2.2 x+1
2 (x+2) = 2 x+2
LHS=RHS (for all) values of x. Hence, the entire considered interval is the solution.
So, x-1; x=-3.
Regards Topperlearning.