9. Which of the following statistics can measure the dispersion of the sample $x _ { 1 } , x _ { 2 } , \cdots , x _ { n }$? ( )\\
A. The standard deviation of the sample $x _ { 1 } , x _ { 2 } , \cdots , x _ { n }$\\
B. The median of the sample $x _ { 1 } , x _ { 2 } , \cdots , x _ { n }$\\
C. The range of the sample $x _ { 1 } , x _ { 2 } , \cdots , x _ { n }$\\
D. The mean of the sample $x _ { 1 } , x _ { 2 } , \cdots , x _ { n }$

【Answer】AC\\
【Solution】\\
【Analysis】Determine which of the given options measure data dispersion and which measure central tendency.

【Detailed Solution】By the definition of standard deviation, standard deviation measures data dispersion. \\
By the definition of median, median measures central tendency. \\
By the definition of range, range measures data dispersion. \\
By the definition of mean, mean measures central tendency. \\
Therefore, the answer is: AC.\\
[Detailed Solution] From the given conditions, $f ( x ) = \left| e ^ { x } - 1 \right| = \left\{ \begin{array} { l } 1 - e ^ { x } , x < 0 \\ e ^ { x } - 1 , x \geq 0 \end{array} \right.$ , then $f ^ { \prime } ( x ) = \left\{ \begin{array} { l } - e ^ { x } , x < 0 \\ e ^ { x } , x > 0 \end{array} \right.$ ,\\
Thus point $A \left( x _ { 1 } , 1 - e ^ { x _ { 1 } } \right)$ and point $B \left( x _ { 2 } , e ^ { x _ { 2 } } - 1 \right)$ , $k _ { A M } = - e ^ { x _ { 1 } } , k _ { B N } = e ^ { x _ { 2 } }$ ,\\
Therefore $- e ^ { x _ { 1 } } \cdot e ^ { x _ { 2 } } = - 1 , x _ { 1 } + x _ { 2 } = 0$ ,\\
So $AM: y - 1 + e ^ { x _ { 1 } } = - e ^ { x _ { 1 } } \left( x - x _ { 1 } \right) , M \left( 0 , e ^ { x _ { 1 } } x _ { 1 } - e ^ { x _ { 1 } } + 1 \right)$ ,\\
Thus $| A M | = \sqrt { x _ { 1 } ^ { 2 } + \left( e ^ { x _ { 1 } } x _ { 1 } \right) ^ { 2 } } = \sqrt { 1 + e ^ { 2 x _ { 1 } } } \cdot \left| x _ { 1 } \right|$ ,\\
Similarly $| B N | = \sqrt { 1 + e ^ { 2 x _ { 2 } } } \cdot \left| x _ { 2 } \right|$ ,\\
Therefore $\frac { | A M | } { | B N | } = \frac { \sqrt { 1 + e ^ { 2 x _ { 1 } } } \cdot \left| x _ { 1 } \right| } { \sqrt { 1 + e ^ { 2 x _ { 2 } } } \cdot \left| x _ { 2 } \right| } = \sqrt { \frac { 1 + e ^ { 2 x _ { 1 } } } { 1 + e ^ { 2 x _ { 2 } } } } = \sqrt { \frac { 1 + e ^ { 2 x _ { 1 } } } { 1 + e ^ { - 2 x _ { 1 } } } } = e ^ { x _ { 1 } } \in ( 0,1 )$ .\\
Thus the answer is: $( 0,1 )$\\
[Key Point Explanation]\\
The key to solving this problem is to use the geometric meaning of the derivative to transform the condition $x _ { 1 } + x _ { 2 } = 0$ , and after eliminating one variable, the calculation yields the solution.

\section*{IV. Answer Questions: This section contains 6 questions totaling 70 points. Solutions should include written explanations, proofs, or calculation steps.}