
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
If for any $x \in D$, the inequality $f_{1}(x) \leqslant f(x) \leqslant f_{2}(x)$ holds, then the function $f(x)$ is called a "compromise function" of $f_{1}(x)$ to $f_{2}(x)$ over the interval $D$. Given that the function $f(x)=(k-1)x-1$, $g(x)=0$, $h(x)=(x+1)\ln x$, and $f(x)$ is a "compromise function" of $g(x)$ to $h(x)$ over the interval $[1,2e]$, then the set of values of the real number $k$ is \_\_\_\_\_\_.

Present the answer in LaTex format: \boxed{Your answer}
user