` x^2 + 4 + 3 cos(ax+b) = 2x` has atleast on solution then the value of a+b is :

Options

A. `5pi`
B. `3pi`
C. `2pi`
D. `pi`

Answer

To solve the equation \( x^2 + 4 + 3 \cos(ax + b) = 2x \) for the condition that it has at least one solution, we can follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ x^2 + 4 + 3 \cos(ax + b) = 2x \]
Rearranging gives:
\[ x^2 - 2x + 4 + 3 \cos(ax + b) = 0 \]

### Step 2: Identify the quadratic form
This is a quadratic equation in the form:
\[ x^2 - 2x + (4 + 3 \cos(ax + b)) = 0 \]
For this quadratic to have at least one solution, the discriminant must be non-negative.

### Step 3: Calculate the discriminant
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
In our case:
- \( a = 1 \)
- \( b = -2 \)
- \( c = 4 + 3 \cos(ax + b) \)

Thus, the discriminant is:
\[ D = (-2)^2 - 4(1)(4 + 3 \cos(ax + b)) = 4 - 4(4 + 3 \cos(ax + b)) \]
This simplifies to:
\[ D = 4 - 16 - 12 \cos(ax + b) = -12 - 12 \cos(ax + b) \]

### Step 4: Set the discriminant greater than or equal to zero
For the quadratic to have at least one solution, we require:
\[ -12 - 12 \cos(ax + b) \geq 0 \]
This simplifies to:
\[ -12 \cos(ax + b) \geq 12 \]
Dividing both sides by -12 (and reversing the inequality):
\[ \cos(ax + b) \leq -1 \]

### Step 5: Analyze the cosine function
The cosine function achieves the value of -1 at odd multiples of \( \pi \):
\[ ax + b = (2n + 1)\pi \quad \text{for } n \in \mathbb{Z} \]
Thus, we can set:
\[ a + b = (2n + 1)\pi \]

### Step 6: Find possible values of \( a + b \)
The simplest case is when \( n = 0 \):
\[ a + b = \pi \]
For \( n = 1 \):
\[ a + b = 3\pi \]
For \( n = 2 \):
\[ a + b = 5\pi \]
Thus, the possible values for \( a + b \) are \( \pi, 3\pi, 5\pi \).

### Conclusion
The values of \( a + b \) that satisfy the condition of having at least one solution are \( \pi, 3\pi, \) and \( 5\pi \).