Question

The solution of the equation $2^{|x+2|}-|2^{x+1}-1|=2^{x+1}+1$ is

• A. $\{-3,-1\}$
• B. $\{-3\}\cup [1,\infty )$
• C. $\{-3\}\cup [-1,\infty )$
• D. $\{-\infty ,-3]\cup \{-1\}$

Answer 1

The correct option is C $\{-3\}\cup [-1,\infty )$

|x + 2| vanishes at x = - 2 and $|2^{x+1}-1|$ vanishes at x = - 1, hence we divide the problem into three intervals:
(i) If x < - 2, then |x + 2| = - (x + 2) also $x+1\le -1$
$\Rightarrow 2^{x+1}<2^{-1}=\frac{1}{2}\Rightarrow 2^{x+1}<1\Rightarrow |2^{x+1}-1|=-(2^{x+1}-1)\therefore Equationis{2}^{-x-2}+2^{x+1}-1=2^{x+1}+1\Rightarrow 2^{-x-2}=2\Rightarrow x=-3$
(ii) If $-2\le x<-1$, the |x + 2| = x + 2 also,
$x+1<0\Rightarrow 2^{x+1}<1\Rightarrow |2^{x+1}-1|=-(2^{x-1}-1)$
$\therefore$ Equation is $2^{x+2}+2^{x+1}-1=2^{x+1}+1$
$\Rightarrow 2^{x+2}=2\Rightarrow x=-1\text{/}ϵ[-2,-1)$
(iii) If $x\ge -1,then|x+2|=x+2and|2^{x+1}-1|=2^{x+1}-1$
$\therefore$ Equation is $2^{x+2}-2^{x+1}+1=2^{x+1}+1,$
$\Rightarrow 2^{x+2}=2^{x+2}$ which is identity
$\therefore$ All x such that $x\ge -1$ satisfy the equation
Hence, the solution set is $xϵ\{-3\}\cup [-1,\infty )$