Track / Stage 5 / 77 of 400 #677 of 2000

Problem 677

AIME late

Algebra Difficulty 5.2 Find the answer

9 If for any real number $x$ , there is $f(x)=\log _{a}\left(2+\mathrm{e}^{x-1}\right) \leqslant-1$ , then the range of real number $a$ is $\qquad$ . (where $e$ is an irrational number, $e=2.71828 \cdots$ )

Your answer

Official solution

(9) $\left[\frac{1}{2}, 1\right)$ Hint: When $01$ , from the condition we get $2+\mathrm{e}^{x-1} \leqslant \frac{1}{a}$ , which means $a \leqslant \frac{1}{2+\mathrm{e}^{x-1}}$ always holds, but this is impossible. Therefore, the range of real number $a$ is $\left[\frac{1}{2}, 1\right)$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.