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

Answer 2

Let $f(x)=e^{x-1}+x-2$
Check for $x =1$
Then, $f(1)=e^{0}+1-2=0$
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 $f(1)=f(\alpha)=0$
By Rolle's theorem $f'(x)=0$ has a root in $(1, \alpha)$ or in $(\alpha, 1)$
But $f'(x)=e^{x-1}+1>0$ for all $x$. Thus, $f'(x) = 0$,
for any $x \in(1, \alpha)$ or $x \in(\alpha, 1)$, which is a contradiction.
Hence, $f(x)=0$ has no real root other than $1$.