Question

The equation $e^{x-1}$ + x - 2 = 0 has

• A. one real root
• B. two real roots
• C. three real roots
• D. four real roots.

Answer 1

The correct option is A one real root

Clearly, x = 1 satisfies the given equation.
Assume that $f\left(x\right)=e^{x-1}+x-2=0$ has a real root $\alpha$ other than x = 1. We may suppose
that $\alpha >1$ (the case $\alpha <1$ is exactly similar).
Applying Rolle's theorem on [1, $\alpha$] (if $\alpha <1$ apply the theorem on[$\alpha$, 1]), we get $\beta$
$ϵ(1,\alpha )$ such that f'($\beta$) = 0.
But $f^{\prime }\left(\beta \right)=e^{\beta -1}+1,$ so that $e^{\beta -1}=-1,$ which is not possible.
Hence there is no real root other than 1.