The mode is the value that appears most frequently in a data set. For a list to have "undoubtedly one mode," one element must appear more frequently than any other element.

Since $n$ is an even number, let's consider the worst-case scenario for having multiple modes:

Worst-Case Scenario (Multiple Modes)
The most difficult way to guarantee a single mode is if we have two or more numbers that appear with the maximum possible equal frequency.

If $n$ is even, we can split the $n$ elements evenly between two different numbers, creating two modes.
- Example (n=10): A list like $\(\{1,1,1,1,1,2,2,2,2,2\}\)$ has two modes: 1 and 2 , each appearing $\(\frac{10}{2}=5\)$ times.
- General Case: A list of $n$ elements can have two modes, where each mode appears $\(\frac{n}{2}\)$ times.


The Minimum Number to Guarantee One Mode
To ensure that one element is the unique mode, its frequency must be greater than the frequency of any other element.
1. Maximum possible frequency for multiple modes: $\(\frac{n}{2}\)$.
2. Minimum frequency for a unique mode: If one element appears $\(\frac{n}{2}+1\)$ times, it guarantees that no other element can appear that frequently or more frequently.

If one element (say, ' $X$ ') appears $\(\frac{n}{2}+1\)$ times:
- The count for ' $X$ ' is $\(\frac{n}{2}+1\)$.
- The remaining elements available are $\(n-\left(\frac{n}{2}+1\right)=\frac{n}{2}-1\)$.
- The remaining $\(\frac{n}{2}-1\)$ elements must be distributed among all other numbers (including the one that isn't ' $X$ '). No single number (other than ' $X$ ') can have a frequency greater than $\(\frac{n}{2}-\)$ 1.
- Since $\(\frac{n}{2}+1>\frac{n}{2}-1\)$, the element ' X ' is guaranteed to be the unique mode.

Therefore, the minimum number of elements that must be the same to guarantee one mode is $\(\frac{n}{2}+1\)$.

The correct option is d) $\(\frac{n}{2}+1\)$.