For any real number b, let f (b) denotes the maximum of `| sin x+(2)/(3+sin x)+b|AAxx x in R.` Then the minimum value of `f (b)AA b in R`is:

Options

A. (a)`1/2`
B. (b)`3/2`
C. (c)`1/4`
D. (d) `1`

Answer

To solve the problem, we need to find the minimum value of the function \( f(b) = \max_{x \in \mathbb{R}} \left| \sin x + \frac{2}{3 + \sin x} + b \right| \).

### Step 1: Define the function
Let us define:
\[ g(x) = \sin x + \frac{2}{3 + \sin x} \]
Thus, we can express \( f(b) \) as:
\[ f(b) = \max_{x \in \mathbb{R}} \left| g(x) + b \right| \]

### Step 2: Analyze the function \( g(x) \)
To find the maximum value of \( |g(x) + b| \), we first need to analyze \( g(x) \).

### Step 3: Find the derivative of \( g(x) \)
We differentiate \( g(x) \):
\[ g'(x) = \cos x - \frac{2 \cos x}{(3 + \sin x)^2} \]
Setting \( g'(x) = 0 \) to find critical points:
\[ \cos x \left( 1 - \frac{2}{(3 + \sin x)^2} \right) = 0 \]
This gives us two cases:
1. \( \cos x = 0 \) (which gives \( x = \frac{\pi}{2} + n\pi \))
2. \( 1 - \frac{2}{(3 + \sin x)^2} = 0 \)

### Step 4: Solve for critical points
From the second case:
\[ (3 + \sin x)^2 = 2 \implies 3 + \sin x = \pm \sqrt{2} \]
This leads to:
\[ \sin x = \sqrt{2} - 3 \quad \text{or} \quad \sin x = -\sqrt{2} - 3 \]
Since \( \sin x \) is bounded between -1 and 1, we only consider valid values.

### Step 5: Evaluate \( g(x) \) at critical points
Evaluate \( g(x) \) at \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \):
- At \( x = -\frac{\pi}{2} \):
\[ g\left(-\frac{\pi}{2}\right) = -1 + \frac{2}{3 - 1} = -1 + 1 = 0 \]
- At \( x = \frac{\pi}{2} \):
\[ g\left(\frac{\pi}{2}\right) = 1 + \frac{2}{3 + 1} = 1 + \frac{1}{2} = \frac{3}{2} \]

### Step 6: Determine the maximum value of \( g(x) \)
The maximum value of \( g(x) \) in the interval \( [-1, 1] \) is \( \frac{3}{2} \) and the minimum value is \( 0 \).

### Step 7: Find \( f(b) \)
Thus, we have:
\[ f(b) = \max \left( |0 + b|, \left| \frac{3}{2} + b \right| \right) \]

### Step 8: Analyze the cases for \( f(b) \)
1. If \( b \geq 0 \):
\[ f(b) = \max(b, \frac{3}{2} + b) = \frac{3}{2} + b \]
2. If \( b < 0 \):
\[ f(b) = \max(-b, \frac{3}{2} + b) \]

### Step 9: Find the minimum value of \( f(b) \)
To minimize \( f(b) \):
- For \( b = 0 \):
\[ f(0) = \frac{3}{2} \]
- For \( b < 0 \), \( f(b) \) can be minimized at \( b = -\frac{3}{2} \):
\[ f\left(-\frac{3}{2}\right) = \max\left(\frac{3}{2}, 0\right) = \frac{3}{2} \]

### Conclusion
The minimum value of \( f(b) \) is:
\[ \boxed{\frac{3}{2}} \]