`(x+3)/(x ^(2)-2) ge (1)/(x-4)` holds for all x satisfying :

Options

A. ` -2 lt x lt 1 or x gt 4`
B. ` -1 lt x lt 2 or x gt 4`
C. `x lt -1 or 2 lt x lt 4`
D. `x gt -1 or 2 lt x lt 4`

Answer

To solve the inequality \(\frac{x+3}{x^2-2} \geq \frac{1}{x-4}\), we will follow these steps:

### Step 1: Rearrange the inequality
We start by rearranging the inequality to bring all terms to one side:
\[ \frac{x+3}{x^2-2} - \frac{1}{x-4} \geq 0 \]

### Step 2: Find a common denominator
The common denominator of the two fractions is \((x^2 - 2)(x - 4)\). We rewrite the inequality:
\[ \frac{(x+3)(x-4) - (x^2-2)}{(x^2-2)(x-4)} \geq 0 \]

### Step 3: Simplify the numerator
Now, we simplify the numerator:
\[ (x+3)(x-4) = x^2 - 4x + 3x - 12 = x^2 - x - 12 \]
Now, substituting back into the inequality:
\[ \frac{x^2 - x - 12 - (x^2 - 2)}{(x^2-2)(x-4)} \geq 0 \]
This simplifies to:
\[ \frac{-x - 10}{(x^2-2)(x-4)} \geq 0 \]

### Step 4: Factor the numerator
We can factor the numerator:
\[ -x - 10 = -1(x + 10) \]
So, the inequality becomes:
\[ \frac{-1(x + 10)}{(x^2-2)(x-4)} \geq 0 \]
This can be rewritten as:
\[ \frac{(x + 10)}{(x^2-2)(x-4)} \leq 0 \]

### Step 5: Identify critical points
Next, we find the critical points where the expression is zero or undefined:
1. \(x + 10 = 0 \Rightarrow x = -10\)
2. \(x^2 - 2 = 0 \Rightarrow x = \pm \sqrt{2}\)
3. \(x - 4 = 0 \Rightarrow x = 4\)

The critical points are \(x = -10, -\sqrt{2}, \sqrt{2}, 4\).

### Step 6: Test intervals
We will test the intervals determined by these critical points:
- Interval 1: \((- \infty, -10)\)
- Interval 2: \((-10, -\sqrt{2})\)
- Interval 3: \((- \sqrt{2}, \sqrt{2})\)
- Interval 4: \((\sqrt{2}, 4)\)
- Interval 5: \((4, \infty)\)

We will check the sign of the expression in each interval.

### Step 7: Determine the sign of each interval
1. **Interval \((- \infty, -10)\)**: Choose \(x = -11\)
\[ \frac{-1(-11 + 10)}{((-11)^2 - 2)(-11 - 4)} > 0 \quad \text{(positive)} \]

2. **Interval \((-10, -\sqrt{2})\)**: Choose \(x = -9\)
\[ \frac{-1(-9 + 10)}{((-9)^2 - 2)(-9 - 4)} < 0 \quad \text{(negative)} \]

3. **Interval \((- \sqrt{2}, \sqrt{2})\)**: Choose \(x = 0\)
\[ \frac{-1(0 + 10)}{(0^2 - 2)(0 - 4)} < 0 \quad \text{(negative)} \]

4. **Interval \((\sqrt{2}, 4)\)**: Choose \(x = 3\)
\[ \frac{-1(3 + 10)}{(3^2 - 2)(3 - 4)} > 0 \quad \text{(positive)} \]

5. **Interval \((4, \infty)\)**: Choose \(x = 5\)
\[ \frac{-1(5 + 10)}{(5^2 - 2)(5 - 4)} < 0 \quad \text{(negative)} \]

### Step 8: Compile the results
The expression is less than or equal to zero in the intervals:
- \((-10, -\sqrt{2})\)
- \((- \sqrt{2}, \sqrt{2})\)
- \((4, \infty)\)

### Step 9: Exclude points where the expression is undefined
The points where the expression is undefined are \(x = -\sqrt{2}, \sqrt{2}, 4\). Therefore, we exclude these points from our solution.

### Final Solution
The solution to the inequality \(\frac{x+3}{x^2-2} \geq \frac{1}{x-4}\) is:
\[ x \in (-10, -\sqrt{2}) \cup (-\sqrt{2}, \sqrt{2}) \cup (4, \infty) \]