Question

The number of real solutions of $1+|e^x-1|=e^x(e^x-2)$ is

• A. \N
• B. 1
• C. 2
• D. 4

Answer 1

The correct option is B 1

$1+1+|e^x-1|=e^{2x}-2e^x+1\Rightarrow 2+|e^x-1|=|e^x-1{|}^2\Rightarrow |e^x-1{|}^2-|e^x-1|-2=0\therefore |e^x-1|=2,-1(but|e^x-1|\ne -1)\Rightarrow |e^x-1|=2\Rightarrow e^x-1=\pm 2\therefore e^x=1\pm 2=3,-1(but{e}^x\ne -1)\Rightarrow e^x=3\Rightarrow x=lo{g}_{e}3.$
$\therefore$ No. of solutions is one.