Question

The number of real solutions of the equation $1+\left|e^x-1\right|=e^x\left(e^x-2\right)$ is

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

Answer: (A)

We have, $1+\left|e^x-1\right|=e^x\left(e^x-2\right)$
$1+1+\left|e^x-1\right|=e^{2x}-2e^x+1={\left(e^x-1\right)}^2$
$1+1+\left|e^x-1\right|={\left|e^x-1\right|}^2$
Let $\left(e^x-1\right)=y$, then
$y^2-y-2=0\Rightarrow y=2,-1$
$\left|e^x-1\right|=2\Rightarrow e^x-1=\pm 2$
$\Rightarrow e^x=1\pm 2\Rightarrow e^x=3,-1$
Or $e^x=3\Rightarrow x={\log }_{e}3$
$\therefore$ There is one real solution of the equation.