Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$1$

Let $f\left(x\right)=e^{x-1}+x-2$

Check for $x=1$

Then, $f\left(1\right)=e^0+1-2=0$

So, $x=1$ is a real root of the equation $f\left(x\right)=0$

Let $x=\alpha$ be the other root such that $\alpha >1$ or $\alpha <1$

consider the interval $[1,\alpha ]$ or $[\alpha ,1]$

Clearly $f\left(1\right)=f\left(\alpha \right)=0$

By Rolle's theorem $f^{\prime }\left(x\right)=0$ has a root in $(1,\alpha )$ or in $(\alpha ,1)$

But $f^{\prime }\left(x\right)=e^{x-1}+1>0$ for all $x$. Thus, $f^{\prime }\left(x\right)\ne 0$, for
any $x\in (1,\alpha )$ or $x\in (\alpha ,1)$, which is a contradiction.

Hence, $f\left(x\right)=0$ has no real root other than $1$.