The solution set of the inequation `(x^2-3x+4)/(x+1) > 1, x in RR`, is

Options

A. `(3, oo)`
B. `(-1, 1) cup (3, oo)`
C. `[-1, 1]cup [3, oo)`
D. none of these

Answer

To solve the inequation

\[ \frac{x^2 - 3x + 4}{x + 1} > 1, \]

we will follow these steps:

### Step 1: Rearrange the Inequation
We start by moving 1 to the left side of the inequality:

\[ \frac{x^2 - 3x + 4}{x + 1} - 1 > 0. \]

### Step 2: Combine the Terms
To combine the terms, we need a common denominator:

\[ \frac{x^2 - 3x + 4 - (x + 1)}{x + 1} > 0. \]

### Step 3: Simplify the Numerator
Now, simplify the numerator:

\[ x^2 - 3x + 4 - x - 1 = x^2 - 4x + 3. \]

So, we rewrite the inequality as:

\[ \frac{x^2 - 4x + 3}{x + 1} > 0. \]

### Step 4: Factor the Numerator
Next, we factor the quadratic expression in the numerator:

\[ x^2 - 4x + 3 = (x - 1)(x - 3). \]

Thus, the inequality becomes:

\[ \frac{(x - 1)(x - 3)}{(x + 1)} > 0. \]

### Step 5: Identify Critical Points
The critical points are found by setting the numerator and denominator to zero:

- From \(x - 1 = 0\), we get \(x = 1\).
- From \(x - 3 = 0\), we get \(x = 3\).
- From \(x + 1 = 0\), we get \(x = -1\).

So, the critical points are \(x = -1\), \(x = 1\), and \(x = 3\).

### Step 6: Test Intervals
We will test the sign of the expression in the intervals defined by these critical points:

1. \( (-\infty, -1) \)
2. \( (-1, 1) \)
3. \( (1, 3) \)
4. \( (3, \infty) \)

- **Interval \( (-\infty, -1) \)**: Choose \(x = -2\):
\[ \frac{(-2 - 1)(-2 - 3)}{-2 + 1} = \frac{(-3)(-5)}{-1} = \frac{15}{-1} < 0. \]

- **Interval \( (-1, 1) \)**: Choose \(x = 0\):
\[ \frac{(0 - 1)(0 - 3)}{0 + 1} = \frac{(-1)(-3)}{1} = 3 > 0. \]

- **Interval \( (1, 3) \)**: Choose \(x = 2\):
\[ \frac{(2 - 1)(2 - 3)}{2 + 1} = \frac{(1)(-1)}{3} = \frac{-1}{3} < 0. \]

- **Interval \( (3, \infty) \)**: Choose \(x = 4\):
\[ \frac{(4 - 1)(4 - 3)}{4 + 1} = \frac{(3)(1)}{5} = \frac{3}{5} > 0. \]

### Step 7: Determine the Solution Set
From the tests, we find that the expression is positive in the intervals:

- \( (-1, 1) \)
- \( (3, \infty) \)

### Step 8: Combine the Intervals
Thus, the solution set for the inequation is:

\[ (-1, 1) \cup (3, \infty). \]

### Final Answer
The solution set of the inequation is:

\[ \boxed{(-1, 1) \cup (3, \infty)}. \]