The solution set satisfying the inequality,`sqrt(21-4a-a^2)/(a+1)<=1`

Options

A. `[-7,-1]`
B. `[2,3]`
C. `[2,oo)`
D. `[-7,0)`

Answer

To solve the inequality \( \frac{\sqrt{21 - 4a - a^2}}{a + 1} \leq 1 \), we will follow these steps:

### Step 1: Rewrite the inequality
We start with the given inequality:
\[ \frac{\sqrt{21 - 4a - a^2}}{a + 1} \leq 1 \]
This can be rewritten as:
\[ \sqrt{21 - 4a - a^2} \leq a + 1 \]

### Step 2: Square both sides
To eliminate the square root, we square both sides of the inequality:
\[ 21 - 4a - a^2 \leq (a + 1)^2 \]
Expanding the right side:
\[ 21 - 4a - a^2 \leq a^2 + 2a + 1 \]

### Step 3: Rearrange the inequality
Now, we will move all terms to one side:
\[ 21 - 4a - a^2 - a^2 - 2a - 1 \leq 0 \]
This simplifies to:
\[ -2a^2 - 6a + 20 \leq 0 \]
Multiplying through by -1 (which reverses the inequality):
\[ 2a^2 + 6a - 20 \geq 0 \]

### Step 4: Factor the quadratic
We can factor the quadratic:
\[ 2(a^2 + 3a - 10) \geq 0 \]
Factoring further:
\[ 2(a + 5)(a - 2) \geq 0 \]

### Step 5: Find the critical points
The critical points are found by setting the factors to zero:
\[ a + 5 = 0 \quad \Rightarrow \quad a = -5 \]
\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \]

### Step 6: Test intervals
We will test the intervals defined by the critical points: \( (-\infty, -5) \), \( (-5, 2) \), and \( (2, \infty) \).

1. **Interval \( (-\infty, -5) \)**: Choose \( a = -6 \):
\[ 2(-6 + 5)(-6 - 2) = 2(-1)(-8) = 16 \geq 0 \quad \text{(True)} \]

2. **Interval \( (-5, 2) \)**: Choose \( a = 0 \):
\[ 2(0 + 5)(0 - 2) = 2(5)(-2) = -20 \geq 0 \quad \text{(False)} \]

3. **Interval \( (2, \infty) \)**: Choose \( a = 3 \):
\[ 2(3 + 5)(3 - 2) = 2(8)(1) = 16 \geq 0 \quad \text{(True)} \]

### Step 7: Combine the results
From the testing, we find that the inequality holds in the intervals \( (-\infty, -5] \) and \( [2, \infty) \).

### Step 8: Check the domain of the original inequality
We need to ensure that the denominator \( a + 1 \neq 0 \), which gives \( a \neq -1 \). Thus, the final solution set is:
\[ (-\infty, -5] \cup [2, \infty) \]

### Final Answer
The solution set satisfying the inequality is:
\[ (-\infty, -5] \cup [2, \infty) \]