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, $1+\left|e^{x}-1\right|=e^{x}\left(e^{x}-2\right)$
$1+1+\left|e^{x}-1\right|=e^{2 x}-2 e^{x}+1=\left(e^{x}-1\right)^{2}$
$1+1+\left|e^{x}-1\right|=\left|e^{x}-1\right|^{2}$
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.