Daily questions _191105

Known function \ (F (X) = {\ LN} X + (\ -a mathrm {E}) + X B \) , where \ (\ mathrm {e} \ ) is the natural logarithm in base, if the inequality \ (f (x) \ leqslant 0 \) constant is established, \ (\ dfrac ba \) maximum value of \ (\ underline {\ qquad \ qquad} \) .
Analysis:
a title, apparently \ (a> \ mathrm {E} \) , otherwise, there is always \ (X = \ mathrm {E} ^ {. 1-B} \) , such that \ [f (x) = 1 -b + (\ mathrm {e} -a) x + b> 0. \] does not match the title set, and therefore there must be \ (A> \ mathrm {E} \) . At this time constancy \ [\ forall x> 0, f (x) \ leqslant f \ left (\ dfrac { 1} {a- \ mathrm {e }} \ right) = - {\ ln} (a- \ mathrm {e}) - 1 + b \ leqslant 0. \] so \ (b \ leqslant {\ ln } ( A- \ mathrm {E}) +. 1 \) , then\ [\ Dfrac {b} { a} \ leqslant \ dfrac {{\ ln} (a- \ mathrm {e}) + 1} {a} \ leqslant \ dfrac {\ dfrac {a- \ mathrm {e}} {\ mathrm {e}} +
1} {a} = \ dfrac {1} {\ mathrm {e}}. \] Thus if and only if \ ((a, b) = (2 \ mathrm {e}, \ 2)) when, \ (\ dfrac BA \) has its maximum value \ (\ dfrac. 1 {{} \ mathrm {E}} \) .