1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days
8104911739

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

022-62211530

Mon to Sat - 11 AM to 8 PM

Asked by varunsinghal2011 9th May 2010, 1:55 PM

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:

-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:

-2-1:

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; -2-1. 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.

Answered by Expert 29th May 2010, 11:43 AM
• 1
• 2
• 3
• 4
• 5
• 6
• 7
• 8
• 9
• 10

You have rated this answer /10

### Free related questions

13th November 2016, 11:25 PM
20th November 2016, 2:00 PM
23rd November 2016, 4:49 PM
24th November 2016, 2:46 PM
RELATED STUDY RESOURCES :