Let `f (x) = {{:(2-|x ^(2) +5x +6|, x ne -2),( b ^(2)+1, x =-2):}`
Has relative maximum at `x =-2,` then complete set of values b can take is:

Options

A. `|b| ge 1`
B. `|b| lt 1`
C. ` b gt 1`
D. `b lt 1`

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} 2 - |x^2 + 5x + 6| & \text{if } x \neq -2 \\ b^2 + 1 & \text{if } x = -2 \end{cases} \]

We are tasked with finding the complete set of values for \( b \) such that \( f(x) \) has a relative maximum at \( x = -2 \).

### Step 1: Analyze the function for \( x \neq -2 \)

First, let's simplify the expression inside the modulus:

\[ x^2 + 5x + 6 = (x + 2)(x + 3) \]

The roots of this quadratic are \( x = -2 \) and \( x = -3 \). The quadratic opens upwards (since the coefficient of \( x^2 \) is positive).

### Step 2: Determine the behavior of \( f(x) \)

The function \( f(x) \) can be analyzed in intervals based on the roots:

1. For \( x < -3 \): Both \( x + 2 < 0 \) and \( x + 3 < 0 \) → \( |x^2 + 5x + 6| = -(x^2 + 5x + 6) \)
2. For \( -3 < x < -2 \): \( x + 2 < 0 \) and \( x + 3 > 0 \) → \( |x^2 + 5x + 6| = -(x + 2)(x + 3) \)
3. For \( -2 < x < -1 \): Both \( x + 2 > 0 \) and \( x + 3 > 0 \) → \( |x^2 + 5x + 6| = (x + 2)(x + 3) \)

### Step 3: Find the limit as \( x \) approaches -2

To determine if \( f(x) \) has a relative maximum at \( x = -2 \), we must check the limit as \( x \) approaches -2 from both sides:

\[ \lim_{x \to -2^-} f(x) = 2 - |(-2)^2 + 5(-2) + 6| = 2 - |4 - 10 + 6| = 2 - |0| = 2 \]

\[ \lim_{x \to -2^+} f(x) = b^2 + 1 \]

For \( f(x) \) to have a relative maximum at \( x = -2 \), we require:

\[ b^2 + 1 \leq 2 \]

### Step 4: Solve the inequality

Now we solve the inequality:

\[ b^2 + 1 \leq 2 \implies b^2 \leq 1 \implies -1 \leq b \leq 1 \]

### Conclusion

The complete set of values that \( b \) can take is:

\[ b \in [-1, 1] \]