If `f(x)=tan^2((pix)/(n^2-5n+8))+cot(n+m)pix;(n in N, m in Q)` is a periodic function with 2 as its fundamental period, then m can't belong to

Options

A. `(-infty,-2) cup (-1, infty)`
B. `(-infty,-3) cup (-2, infty)`
C. `(-2,-1) cup (-3, -2)`
D. `(-3,-5/2) cup (-5/2, -2)`

Answer

To solve the problem step by step, we need to analyze the function \( f(x) = \tan^2\left(\frac{\pi x}{n^2 - 5n + 8}\right) + \cot((n + m) \pi x) \) and determine the conditions under which it is periodic with a fundamental period of 2.

### Step 1: Identify the periods of the components of the function
The periodicity of the function is determined by the individual components:
- The period of \( \tan^2(x) \) is \( \pi \).
- The period of \( \cot(x) \) is \( \pi \).

### Step 2: Determine the period of \( \tan^2\left(\frac{\pi x}{n^2 - 5n + 8}\right) \)
The period of \( \tan^2\left(\frac{\pi x}{n^2 - 5n + 8}\right) \) can be calculated as follows:
\[ \text{Period} = \pi \cdot (n^2 - 5n + 8) \]

### Step 3: Determine the period of \( \cot((n + m) \pi x) \)
The period of \( \cot((n + m) \pi x) \) is given by:
\[ \text{Period} = \frac{\pi}{(n + m)} \]

### Step 4: Set the least common multiple (LCM) of the periods equal to 2
Since the function \( f(x) \) is periodic with a fundamental period of 2, we can set up the following equation:
\[ \text{LCM}\left(n^2 - 5n + 8, \frac{1}{n + m}\right) = 2 \]

### Step 5: Analyze the conditions for the LCM
The LCM of two numbers \( a \) and \( b \) can be expressed as:
\[ \text{LCM}(a, b) = \frac{a \cdot b}{\text{GCD}(a, b)} \]
For our case, we need to find conditions under which:
\[ \frac{(n^2 - 5n + 8)(n + m)}{\text{GCD}(n^2 - 5n + 8, n + m)} = 2 \]

### Step 6: Solve for \( n^2 - 5n + 8 \) and \( n + m \)
We will consider two cases:
1. \( n^2 - 5n + 8 = 1 \)
2. \( n^2 - 5n + 8 = 2 \)

#### Case 1: \( n^2 - 5n + 8 = 1 \)
This simplifies to:
\[ n^2 - 5n + 7 = 0 \]
The discriminant \( D = (-5)^2 - 4 \cdot 1 \cdot 7 = 25 - 28 = -3 \) is negative, so there are no real solutions.

#### Case 2: \( n^2 - 5n + 8 = 2 \)
This simplifies to:
\[ n^2 - 5n + 6 = 0 \]
Factoring gives:
\[ (n - 2)(n - 3) = 0 \implies n = 2 \text{ or } n = 3 \]

### Step 7: Determine the values of \( m \)
Now, substituting \( n = 2 \) and \( n = 3 \) into \( n + m \):
- For \( n = 2 \): \( n + m = 2 + m \)
- For \( n = 3 \): \( n + m = 3 + m \)

We need to ensure that:
\[ \frac{1}{2 + m} \leq 1 \quad \text{and} \quad \frac{1}{3 + m} \leq 1 \]

This leads to:
1. \( 2 + m \geq 1 \implies m \geq -1 \)
2. \( 3 + m \geq 1 \implies m \geq -2 \)

### Step 8: Conclusion on the values of \( m \)
Thus, \( m \) cannot belong to the intervals:
- \( (-2, -1) \)
- \( (-3, -2) \)

The final answer is that \( m \) cannot belong to the interval \( (-2, -1) \cup (-3, -2) \).