If `0lt alt 5, 0 lt blt 5` and `(x^2+5)/2 = x -2 cos(a+bx)` is satisfied for at least one real x then least value of `(a+b)/pi`is ?

Options

A. `pi//2`
B. `pi`
C. `3 pi`
D. `4 pi`

Explanation

To solve the problem, we need to analyze the equation given and find the least value of \((a + b)/\pi\) under the constraints \(0 < a < 5\) and \(0 < b < 5\).

### Step-by-step Solution:

1. **Start with the given equation:**
\[ \frac{x^2 + 5}{2} = x - 2 \cos(a + bx) \]

2. **Rearranging the equation:**
\[ 2 \cos(a + bx) = x - \frac{x^2 + 5}{2} \]
This simplifies to:
\[ 2 \cos(a + bx) = 2x - x^2 - 5 \]

3. **Dividing both sides by 2:**
\[ \cos(a + bx) = x - \frac{x^2}{2} - \frac{5}{2} \]

4. **Rearranging further:**
\[ \cos(a + bx) = -\frac{1}{2}x^2 + x - \frac{5}{2} \]

5. **Setting the range of \(\cos(a + bx)\):**
Since \(\cos(a + bx)\) must satisfy \(-1 \leq \cos(a + bx) \leq 1\), we can set up the inequalities:
\[ -1 \leq -\frac{1}{2}x^2 + x - \frac{5}{2} \leq 1 \]

6. **Solving the left inequality:**
\[ -1 \leq -\frac{1}{2}x^2 + x - \frac{5}{2} \]
Rearranging gives:
\[ 0 \leq -\frac{1}{2}x^2 + x - \frac{3}{2} \]
Multiplying through by -2 (reversing the inequality):
\[ 0 \geq x^2 - 2x + 3 \]
The quadratic \(x^2 - 2x + 3\) has no real roots (discriminant \(< 0\)), hence it is always positive. Thus, this inequality does not provide any restrictions.

7. **Solving the right inequality:**
\[ -\frac{1}{2}x^2 + x - \frac{5}{2} \leq 1 \]
Rearranging gives:
\[ -\frac{1}{2}x^2 + x - \frac{7}{2} \leq 0 \]
Multiplying through by -2:
\[ x^2 - 2x + 7 \geq 0 \]
This quadratic also has no real roots, hence it is always positive.

8. **Finding the condition for \(x\):**
We need to find \(x\) such that:
\[ -\frac{1}{2}x^2 + x - \frac{5}{2} = 1 \]
Rearranging gives:
\[ -\frac{1}{2}x^2 + x - \frac{7}{2} = 0 \]
Multiplying by -2:
\[ x^2 - 2x + 7 = 0 \]
Again, this has no real roots.

9. **Setting \(x = 1\):**
We substitute \(x = 1\) into the original equation:
\[ \frac{1^2 + 5}{2} = 1 - 2 \cos(a + b) \]
Simplifying gives:
\[ 3 = 1 - 2 \cos(a + b) \]
Thus:
\[ 2 \cos(a + b) = -2 \quad \Rightarrow \quad \cos(a + b) = -1 \]

10. **Finding \(a + b\):**
The condition \(\cos(a + b) = -1\) occurs when:
\[ a + b = \pi + 2k\pi \quad (k \in \mathbb{Z}) \]
The least value occurs when \(k = 0\):
\[ a + b = \pi \]

11. **Calculating \(\frac{a + b}{\pi}\):**
\[ \frac{a + b}{\pi} = 1 \]

### Final Answer:
The least value of \(\frac{a + b}{\pi}\) is:
\[ \boxed{1} \]