Let's analyze the problem step-by-step:

Given:
- List $L$ consists of integers from 1 to 99 .
- Two integers $c$ and $d$ such that $\(c+d=100\)$ and $\(c \times d<0\)$.

Explanation of $c$ and $d$ :
- Since $\(c \times d<0, c\)$ and $d$ have opposite signs (one positive and one negative).
- Their sum is 100 .
- Example possible values: $\(c=101, d=-1\)$, or $\(c=150, d=-50\)$, etc.

Constructing the full list:
- The complete list is the integers from 1 to 99 , plus $c$ and $d$.
- So the list has 101 elements in total.

Evaluate each statement:
A. The average (mean) equals the median.
- Average of numbers 1 to 99 is:

$$
\(\frac{1+99}{2}=50\)
$$

- Median of integers 1 to 99 is the middle number: 50 (the 50 th number).
- Adding $c$ and $d$ :

$$
\(c+d=100\)
$$

- Sum of 1 to 99 is:

$$
\(\frac{99 \times 100}{2}=4950\)
$$

- Total sum of list $L$ :

$$
\(4950+100=5050\)
$$

- Number of elements $=101$
- New average:

$$
\(\frac{5050}{101} \approx 50\)
$$

- Median value:

With 101 numbers (odd length), the median is the 51st element when sorted.
- Since $d<0, d$ is less than 1 , so the lowest number in the list is $d$ (negative or zero or less than 1).
- The first number is negative (or less than 1 ), then 1 through 99 , then $c$ (possibly larger than 99).
- The 51 st number will be the 50 th integer (counting past the negative number); since $d$ adds one element less than 1 , the 51 st element will be 50 .
- So median is 50 .
- The mean is approximately 50 .

B. The range of $L$ is greater than 100.
- Range = difference between maximum and minimum values.
- Minimum value: since $\(c d<0\)$, one is negative; let's say $d$ is negative (less than 1 ).
- Maximum value: $\(c=100-d\)$, which is greater than 99 .
- The range:

$$
\(\text { Range }=c-d=c+|d|=100-d+|d|=100+|d-d|\)
$$


Since $d$ is negative:

$$
\(\text { Range }=c-d=c+|d|=100+2|d|\)
$$


Because $\(d \neq 0\)$, the range is definitely larger than 100 .
Therefore, statement $B$ is true.
C. The range of $L$ is greater than 200 .
- Is the range necessarily greater than 200 ?
- For example, if $\(c=101\)$ and $\(d=-1\)$, then range is:

$$
\(101-(-1)=102\)
$$

- 102 is not greater than 200 .
- So the range is not necessarily greater than 200.

Therefore, statement C is false.

Final answer:
- Statements that must be true:
$A$ and $B$
- Statement that need not be true:

C