Question

The number of real roots of the equation $x^{x-1}+x-2=0$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Let $f\left(x\right)=e^{x-1}+x-2\Rightarrow f^{\prime }\left(x\right)=e^{x-1}+1>0\forall x\in R$
Also when $x\to$ $\infty ,f\left(x\right)\to \infty$ and when $x\to -\infty$
Further $f\left(x\right)$ is continuous, hence its graph cuts $x$- axis only at one point.
Hence, equation $f\left(x\right)=0$ has only one root.
Alternative method:
Also $e^{x-1}=2-x$
As shown in the figure, graphs of $y=e^{x-1}$ and $y=2-x$ cuts at only one point. Hence, there is only one root.