Question

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

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

Answer 1

Answer: (A)

Clearing $f\left(1\right)=e^{1-1}+1-2$
$=1+1-2=0$
$\therefore 1$ is a root of the equation
$f\left(x\right)=e^{x-1}+x-2=0$
Let $\alpha > 1$ be also a root
Then $f\left(\alpha\right)=0$
$\therefore $ by Rolle’s Theorem
$f'\left(x\right)=0$ for at least one $x$ between $1$ and $\alpha$.
Now $f'\left(x\right)=e^{x-1}+1=0$
$\Rightarrow e^{x-1}=-1$
which is not possible $\quad\left[\because x \in\left(1, \alpha\right) \text{i.e}., x > 1\right]
\therefore $ there exists no root $> 1$
Similarly we can show that no root $< 1$, exists.
Thus there is only one root.