If `k in I` such that `lim_(nrarroo) (cos.(kpi)/(4))^(2n)-(cos.(kpi)/(6))^(2n)=0,` then

Options

A. k must not be divisible by 24
B. k is divisible by 24 or k is divisible neither by 4 nor by 6
C. k must be divisible by 12 but not necessarity by 24
D. none of these

Explanation

To solve the problem, we need to analyze the limit given in the question:

\[ \lim_{n \to \infty} \left( \cos\left(\frac{k \pi}{4}\right)^{2n} - \cos\left(\frac{k \pi}{6}\right)^{2n} \right) = 0 \]

### Step 1: Understanding the Limit
In order for the limit to equal zero, both terms must approach the same value as \( n \) approaches infinity. This means that the two cosine terms must be equal or both must approach zero.

### Step 2: Case Analysis
We will analyze the cases based on the values of \( \cos\left(\frac{k \pi}{4}\right) \) and \( \cos\left(\frac{k \pi}{6}\right) \).

#### Case 1: Both Cosine Values Equal to 1
If both \( \cos\left(\frac{k \pi}{4}\right) \) and \( \cos\left(\frac{k \pi}{6}\right) \) equal 1, we have:

\[ \cos\left(\frac{k \pi}{4}\right) = 1 \implies \frac{k \pi}{4} = 2m\pi \implies k = 8m \]

\[ \cos\left(\frac{k \pi}{6}\right) = 1 \implies \frac{k \pi}{6} = 2p\pi \implies k = 12p \]

From these equations, we find that \( k \) must be divisible by both 8 and 12. The least common multiple of 8 and 12 is 24, so:

\[ k \text{ is divisible by } 24 \]

#### Case 2: Both Cosine Values Between -1 and 1
If both cosine values are between -1 and 1, we need to consider the condition that neither cosine can equal 1 or -1. This means \( k \) should not be divisible by 4 or 6.

#### Case 3: Both Cosine Values Equal to -1
If both cosine values equal -1, we have:

\[ \cos\left(\frac{k \pi}{4}\right) = -1 \implies \frac{k \pi}{4} = (2m + 1)\pi \implies k = 4(2m + 1) \]

\[ \cos\left(\frac{k \pi}{6}\right) = -1 \implies \frac{k \pi}{6} = (2p + 1)\pi \implies k = 6(2p + 1) \]

In this case, \( k \) must also satisfy both conditions, which leads to contradictions when trying to satisfy both equations simultaneously.

### Conclusion
From the analysis, we find that:
- If \( k \) is divisible by 24, the limit holds.
- If \( k \) is not divisible by 4 or 6, the limit also holds.
- The case where both cosines equal -1 is not possible.

Thus, the correct answer is that \( k \) must be divisible by 24 or neither by 4 nor by 6.

### Final Answer
The correct option is:
**Option 2: k is divisible by 24 or k is divisible neither by 4 nor by 6.**