`(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: Move the right-hand side to the left
We start by rewriting the inequality:
\[ \frac{x+3}{x^2-2} - \frac{1}{x-4} \geq 0 \]

### Step 2: Find a common denominator
The common denominator for the 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: Expand the numerator
Now we expand the numerator:
\[ (x+3)(x-4) = x^2 - 4x + 3x - 12 = x^2 - x - 12 \]
So, the numerator becomes:
\[ x^2 - x - 12 - (x^2 - 2) = x^2 - x - 12 - x^2 + 2 = -x - 10 \]
Thus, we have:
\[ \frac{-x - 10}{(x^2-2)(x-4)} \geq 0 \]

### Step 4: Factor the numerator
The numerator can be factored as:
\[ -(x + 10) \]
So the inequality now looks like:
\[ \frac{-(x + 10)}{(x^2-2)(x-4)} \geq 0 \]

### Step 5: Determine the critical points
The critical points occur when the numerator or denominator is zero:
1. \(x + 10 = 0 \Rightarrow x = -10\)
2. \(x^2 - 2 = 0 \Rightarrow x = \sqrt{2}, -\sqrt{2}\)
3. \(x - 4 = 0 \Rightarrow x = 4\)

### Step 6: Analyze the sign of the expression
We will analyze the sign of the expression in the intervals defined by the critical points:
- \((- \infty, -10)\)
- \((-10, -\sqrt{2})\)
- \((- \sqrt{2}, \sqrt{2})\)
- \((\sqrt{2}, 4)\)
- \((4, \infty)\)

### Step 7: Test each interval
1. For \(x < -10\) (e.g., \(x = -11\)):
\[ \frac{-(-11 + 10)}{((-11)^2 - 2)(-11 - 4)} > 0 \]
Positive.

2. For \(-10 < x < -\sqrt{2}\) (e.g., \(x = -5\)):
\[ \frac{-(-5 + 10)}{((-5)^2 - 2)(-5 - 4)} < 0 \]
Negative.

3. For \(-\sqrt{2} < x < \sqrt{2}\) (e.g., \(x = 0\)):
\[ \frac{-(0 + 10)}{(0^2 - 2)(0 - 4)} > 0 \]
Positive.

4. For \(\sqrt{2} < x < 4\) (e.g., \(x = 2\)):
\[ \frac{-(2 + 10)}{(2^2 - 2)(2 - 4)} < 0 \]
Negative.

5. For \(x > 4\) (e.g., \(x = 5\)):
\[ \frac{-(5 + 10)}{(5^2 - 2)(5 - 4)} < 0 \]
Negative.

### Step 8: Combine the results
The solution to the inequality is where the expression is greater than or equal to zero:
- From the intervals, we find that the solution is:
\[ x \in (-\infty, -10] \cup (-\sqrt{2}, \sqrt{2}) \]

### Final Answer
Thus, the inequality \(\frac{x+3}{x^2-2} \geq \frac{1}{x-4}\) holds for all \(x\) satisfying:
\[ x \in (-\infty, -10] \cup (-\sqrt{2}, \sqrt{2}) \]