Question

The set of solutions of $2^{|x|}-|2^{x-1}-1|=2^{x-1}+1$ is

• A. $\{-1\}\cup [1,\infty )$
• B. $[1,\infty )$
• C. $[-1,\infty )$
• D. $\left\{0\right\}\cup [1,\infty )$

Answer 1

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

$2^{|x|}-|2^{x-1}-1|=2^{x-1}+1$
For the given function,
Case $I:x<0$
$2^{x-1}<1\Rightarrow |2^{x-1}-1|=1-2^{x-1}{2}^{-x}+2^{x-1}-1=2^{x-1}+1\Rightarrow 2^{-x}=2\Rightarrow x=-1$

Case $II:0\le x<1$
$|2^{x-1}-1|=-2^{x-1}+12^x+2^{x-1}-1=2^{x-1}+1\Rightarrow 2^x=2\Rightarrow x=1$
Therefore, no solution.

Case $III:x\ge 1$
$2^x-2^{x-1}+1=2^{x-1}+1\Rightarrow 2^x=2\cdot 2^{x-1}\Rightarrow 2^x=2^x\therefore x\in [1,\infty )$
So the solution set is $\{-1\}\cup [1,\infty )$