If `(x^(2) + 5)/(2) = x - 2` cos (m + n)` has at least one real root, the

Options

A. number of possible values of x is two
B. number of possible values of x is one
C. the value of ` m + n is (2n + 1 )pi`
D. the value of m +n is ` 2npi`

Explanation

To solve the equation \( \frac{x^2 + 5}{2} = x - 2 \cos(m + n) \) and determine the conditions for it to have at least one real root, we can follow these steps:

### Step 1: Rearranging the Equation
Start by multiplying both sides by 2 to eliminate the fraction:

\[ x^2 + 5 = 2(x - 2 \cos(m + n)) \]

### Step 2: Simplifying the Equation
Distributing on the right side gives:

\[ x^2 + 5 = 2x - 4 \cos(m + n) \]

Now, rearranging this equation to one side:

\[ x^2 - 2x + 5 + 4 \cos(m + n) = 0 \]

### Step 3: Identifying the Quadratic Form
This is a quadratic equation in the standard form \( ax^2 + bx + c = 0 \), where:

- \( a = 1 \)
- \( b = -2 \)
- \( c = 5 + 4 \cos(m + n) \)

### Step 4: Determining the Condition for Real Roots
For the quadratic equation to have at least one real root, the discriminant must be non-negative:

\[ D = b^2 - 4ac \geq 0 \]

Calculating the discriminant:

\[ D = (-2)^2 - 4(1)(5 + 4 \cos(m + n)) = 4 - 4(5 + 4 \cos(m + n)) \]

Simplifying this:

\[ D = 4 - 20 - 16 \cos(m + n) = -16 - 16 \cos(m + n) \]

### Step 5: Setting the Discriminant Greater than or Equal to Zero
To ensure there is at least one real root:

\[ -16 - 16 \cos(m + n) \geq 0 \]

This simplifies to:

\[ -16 \cos(m + n) \geq 16 \]

Dividing both sides by -16 (which reverses the inequality):

\[ \cos(m + n) \leq -1 \]

### Step 6: Finding Values of \( m + n \)
The cosine function achieves a value of -1 at specific angles:

\[ m + n = (2n + 1)\pi \quad \text{for } n \in \mathbb{Z} \]

### Conclusion
Thus, for the equation \( \frac{x^2 + 5}{2} = x - 2 \cos(m + n) \) to have at least one real root, the value of \( m + n \) must be of the form:

\[ m + n = 2n\pi + \pi \quad \text{for } n \in \mathbb{Z} \]