If `a , b in [0,2pi]` and the equation `x^2+4+3sin(a x+b)-2x=0` has at least one solution, then the value of `(a+b)` can be `(7pi)/2` (b) `(5pi)/2` (c) `(9pi)/2` (d) none of these

Options

A. `(7pi)/2`
B. `(5pi)/2`
C. `(9pi)/2`
D. none of these

Explanation

To solve the equation \( x^2 + 4 + 3\sin(ax + b) - 2x = 0 \) and determine the possible values of \( a + b \), we can follow these steps:

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

### Step 2: Complete the square
The quadratic part can be rewritten by completing the square:
\[ x^2 - 2x + 4 = (x - 1)^2 + 3 \]
Thus, our equation becomes:
\[ (x - 1)^2 + 3 + 3\sin(ax + b) = 0 \]

### Step 3: Analyze the equation
The left-hand side, \( (x - 1)^2 + 3 + 3\sin(ax + b) \), is always greater than or equal to \( 3 - 3 = 0 \) since \( (x - 1)^2 \geq 0 \) and \( 3\sin(ax + b) \) can vary between \(-3\) and \(3\).

### Step 4: Determine conditions for solutions
For the equation to have at least one solution, the left-hand side must equal zero:
\[ (x - 1)^2 + 3 + 3\sin(ax + b) = 0 \]
This implies:
\[ (x - 1)^2 + 3 = -3\sin(ax + b) \]
Since the left side is always at least \(3\), we need:
\[ -3\sin(ax + b) \leq 3 \implies \sin(ax + b) \geq -1 \]
This condition is always satisfied since the sine function ranges from \(-1\) to \(1\).

### Step 5: Find specific values of \( a + b \)
For the equation to have a solution, we need:
\[ (x - 1)^2 + 3 + 3\sin(ax + b) = 0 \]
This can only happen if:
\[ 3\sin(ax + b) = -3 \implies \sin(ax + b) = -1 \]
The sine function equals \(-1\) at:
\[ ax + b = \frac{3\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]
This implies:
\[ a + b = \frac{3\pi}{2} + 2n\pi \]

### Step 6: Check the options
We need to find values of \( a + b \) that fit within the given options:
1. \( \frac{7\pi}{2} \)
2. \( \frac{5\pi}{2} \)
3. \( \frac{9\pi}{2} \)

From the equation \( a + b = \frac{3\pi}{2} + 2n\pi \), we can check:
- For \( n = 2 \): \( a + b = \frac{3\pi}{2} + 4\pi = \frac{11\pi}{2} \) (not an option)
- For \( n = 1 \): \( a + b = \frac{3\pi}{2} + 2\pi = \frac{7\pi}{2} \) (this is an option)
- For \( n = 0 \): \( a + b = \frac{3\pi}{2} \) (not an option)
- For \( n = -1 \): \( a + b = \frac{3\pi}{2} - 2\pi = -\frac{\pi}{2} \) (not an option)

Thus, the only valid solution from the options is:
\[ \boxed{\frac{7\pi}{2}} \]