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 (a)`(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 find the value of \(a + b\) given that it has at least one solution, we can follow these steps:

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