The number of real roots of the equation e $\mathrm{e}^{(x-1)}+\mathrm{x}$ $-2=0$ is

Let $f(x)=e^{x-1}+x-2$ check for $x=1$ Then, $\mathrm{f}(1)=\mathrm{e}^{0}+1-2=0$ So, $x=1$ is a real root of the equation $f(x)=0$ Let $x=\alpha$ be the other root such that $\alpha>1$ or $\alpha<1$. Consider the interval $[1, \alpha]$ or $[\alpha, 1]$ Clearly $\mathrm{f}(1)=\mathrm{f}(\alpha)=0$ By Rolle's theorem $\mathrm{f}(\mathrm{x})=0$ has a root in $(1, \alpha)$ or in $(\alpha, 1)$. But $f^{\prime}(x)=e^{x-1}+1>0,$ for all $x$. Thus $\mathrm{f}^{\prime}(\mathrm{x}) \neq 0,$ for any $\mathrm{x} \in(1, \alpha)$ or $\mathrm{x} \in(\alpha, 1)$ which is a contradiction. Hence, $\mathrm{f}(\mathrm{x})=0$ has no real root other than 1 .