Find the least value of `k` such that the equation `(ln x) + k = e^(x-k)` has a solution.

Options

A. `e`
B. `1//e`
C. `1`
D. None of these

Answer

C

Explanation

Then inverse of `f(x), f^(-1)f(lnx)+k,`
`f(x)=e^(x-k)` and also both function are increasing therefore
`f(x)=f^(-1)(x)` is equivalent to
`f(x)=f^(-1)(x)=x`
`impliesInx+k =x` should have a solution
`impliesk=x-z Inx`
Now,let `g(x)=x-lnx`
has least value 1 as `g'(x)=1-(1)/(x)` has a minimum at x=1 and `underset(xrarrO^(+))lim g(x), underset(xrarroo)lim g(x)` both approach to oo.`