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.