Explanation:
(A) : Given, equation is \(\mathrm{e}^{\mathrm{x}-1}+\mathrm{x}-2=0\)
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 \(\mathrm{x}=\alpha\) be the other root such that \(\alpha>1\) or \(\alpha\lt 1\).
Consider the interval \([1, \alpha]\) or \([\alpha, 1]\).
Clearly \(\mathrm{f}(1)=\mathrm{f}(\alpha)=0\)
By Rolle's Theorem \(\mathrm{f}^{\prime}(\mathrm{x})=0\) has a root in \((1, \alpha)\) or in \((\alpha, 1)\)
But \(f^{\prime}(x)=e^{x-1}+1>0\), for all \(x\)
Thus, \(\mathrm{f}^{\prime}(\mathrm{x}) \neq 0\), for any \(\mathrm{x} \in(1, \alpha)\) or \(\mathrm{x} \in(\alpha, 1)\), which is a contradiction.
Hence, \(\mathrm{f}(\mathrm{x})=0\) has no real root other than 1 .