Mathematics - Quadratic Equation Question with Solution | TestHub

$f(x)=e^{x-k}$ then inverse of $f(x);f^{-1}(x)=(\ln x)+k$

and also both functions are increasing, therefore

$\mathrm{f}(\mathrm{x})={\mathrm{f}}^{-1}(\mathrm{x})$ is equivalent to $\mathrm{f}(\mathrm{x})={\mathrm{f}}^{-1}(\mathrm{x})=\mathrm{x}$

$\Rightarrow \ln \mathrm{x}+\mathrm{k}=\mathrm{x}$ should have a solution

$\Rightarrow \mathrm{k}=\mathrm{x}-\ln \mathrm{x}$

Now, let $\mathrm{g}(\mathrm{x})=\mathrm{x}-\ln \mathrm{x}$

has least value 1 as ${\mathrm{g}}^{\prime }(\mathrm{x})=1-\frac{1}{\mathrm{x}}$ has a minimum at $\mathrm{x}=1$

and ${\lim }_{\mathrm{x}\to 0^{+}}\mathrm{g}(\mathrm{x}),{\lim }_{\mathrm{x}\to \infty }\mathrm{g}(\mathrm{x})$ both approach to $\infty$.