Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Let $f(x)=e^{x-1}+x-2$ check for $x=1$ Then, $\mathrm{f}(1)={\mathrm{e}}^0+1-2=0$ 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 $\mathrm{f}(1)=\mathrm{f}(\alpha )=0$ By Rolle's theorem $\mathrm{f}(\mathrm{x})=0$ has a root in $(1,\alpha )$ or in $(\alpha ,1)$. But $f^{\prime }(x)=e^{x-1}+1>0,$ for all $x$. Thus ${\mathrm{f}}^{\prime }(\mathrm{x})\ne 0,$ 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 .