Question

If, $2^{|x+1|}-2^x=|2^x-1|+1$ then

• A. $x\in [0,\infty )$
• B. $x\in [0,\infty )\cup \{-2\}$
• C. $x\in (0,\infty )\cup \{-2\}$
• D. $x\in [-2,\infty )$

Answer 1

The correct option is B $x\in [0,\infty )\cup \{-2\}$

$2^{|x+1|}-2^x=|2^x-1|+1$
Critical points $x+1=0\Rightarrow x=-1$
and
$2^x-1=0\Rightarrow x=0$

There are three regions
$x<-1,x\in [-1,0],x>0$

$\text{case-1},x<-1$
For $x<-1,|x+1|=-(x+1),|2^x-1|=-(2^x-1)$
Now, equation will be :
$2^{-(x+1)}-2^x=-(2^x-1)+1$
$2^{-(x+1)}=2$
$-x-1=1$
$x=-2$ Also $-2<-1$
So, $x=-2\cdots \left(1\right)$

$\text{case-2},x\in [-1,0]$
For $x\in [-1,0],|x+1|=(x+1),|2^x-1|=-(2^x-1)$
Now, equation will be :
$2^{(x+1)}-2^x=-(2^x-1)+1$
$2^{(x+1)}=2$
$x+1=1$
$x=0$ Also $0\in [-1,0]$
So, $x=0\cdots \left(2\right)$

$\text{case-3},x>0$
For $x>0,|x+1|=(x+1),|2^x-1|=(2^x-1)$
Now, equation will be :
$2^{(x+1)}-2^x=-(2^x-1)+1$
$2^{(x+1)}=2^{x+1}$ always true
$x\in R$ But $x>0$
So, $x>0\cdots \left(3\right)$

By set $\left(1\right)\cup \left(2\right)\cup \left(3\right)$
$x\in [0,\infty )\cup \{-2\}$