Solution set of `(5x-1)lt (x+1)^(2)lt (7x -3)` is

Options

A. (1, 4)
B. [2, 4]
C. (2, 4)
D. `(-oo, 1)cup(2, oo)`

Explanation

To solve the inequality \( (5x - 1) < (x + 1)^2 < (7x - 3) \), we will break it down into two parts:

1. Solve \( 5x - 1 < (x + 1)^2 \)
2. Solve \( (x + 1)^2 < 7x - 3 \)

### Step 1: Solve \( 5x - 1 < (x + 1)^2 \)

1. **Expand the right side**:
\[ 5x - 1 < x^2 + 2x + 1 \]

2. **Rearrange the inequality**:
\[ 5x - 1 - x^2 - 2x - 1 < 0 \]
\[ -x^2 + 3x - 2 < 0 \]
Multiplying through by -1 (reversing the inequality):
\[ x^2 - 3x + 2 > 0 \]

3. **Factor the quadratic**:
\[ (x - 1)(x - 2) > 0 \]

4. **Determine the intervals**:
- The roots are \( x = 1 \) and \( x = 2 \).
- Test intervals: \( (-\infty, 1) \), \( (1, 2) \), \( (2, \infty) \).
- For \( x < 1 \): Choose \( x = 0 \) → \( (0 - 1)(0 - 2) > 0 \) → True.
- For \( 1 < x < 2 \): Choose \( x = 1.5 \) → \( (1.5 - 1)(1.5 - 2) < 0 \) → False.
- For \( x > 2 \): Choose \( x = 3 \) → \( (3 - 1)(3 - 2) > 0 \) → True.

5. **Solution for this part**:
\[ x \in (-\infty, 1) \cup (2, \infty) \]

### Step 2: Solve \( (x + 1)^2 < 7x - 3 \)

1. **Expand the left side**:
\[ x^2 + 2x + 1 < 7x - 3 \]

2. **Rearrange the inequality**:
\[ x^2 + 2x + 1 - 7x + 3 < 0 \]
\[ x^2 - 5x + 4 < 0 \]

3. **Factor the quadratic**:
\[ (x - 4)(x - 1) < 0 \]

4. **Determine the intervals**:
- The roots are \( x = 1 \) and \( x = 4 \).
- Test intervals: \( (-\infty, 1) \), \( (1, 4) \), \( (4, \infty) \).
- For \( x < 1 \): Choose \( x = 0 \) → \( (0 - 4)(0 - 1) > 0 \) → False.
- For \( 1 < x < 4 \): Choose \( x = 2 \) → \( (2 - 4)(2 - 1) < 0 \) → True.
- For \( x > 4 \): Choose \( x = 5 \) → \( (5 - 4)(5 - 1) > 0 \) → False.

5. **Solution for this part**:
\[ x \in (1, 4) \]

### Step 3: Combine the solutions

Now we have two conditions:
1. \( x \in (-\infty, 1) \cup (2, \infty) \)
2. \( x \in (1, 4) \)

To find the intersection of these two sets:
- The overlapping region is \( (2, 4) \).

### Final Solution

Thus, the solution set for the inequality \( (5x - 1) < (x + 1)^2 < (7x - 3) \) is:
\[ x \in (2, 4) \]