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
$f^{\prime }\left(x\right)=0$ for at least one $x$ between $1$ and $\alpha$.
Now $f^{\prime }\left(x\right)=e^{x-1}+1=0$
$\Rightarrow e^{x-1}=-1$
which is not possible $\left[\because x\in \left(1,\alpha \right)\text{i.e}.,x>1\right]<br>\therefore$ there exists no root $>1$
Similarly we can show that no root $<1$, exists.
Thus there is only one root.