The relation `p/(P+1) gt 1` is valid when :

Options

A. `p ge -1`
B. `p gt -1`
C. `p lt -1`
D. `-1 lt p lt 0`

Answer

To solve the inequality \( \frac{p}{p+1} > 1 \), we will follow these steps:

### Step 1: Set up the inequality
We start with the given inequality:
\[ \frac{p}{p+1} > 1 \]

### Step 2: Eliminate the fraction
To eliminate the fraction, we can multiply both sides of the inequality by \( p + 1 \). However, we need to consider the sign of \( p + 1 \) because it affects the direction of the inequality. We will analyze two cases: when \( p + 1 > 0 \) and when \( p + 1 < 0 \).

### Case 1: \( p + 1 > 0 \) (i.e., \( p > -1 \))
If \( p + 1 > 0 \), we can multiply both sides by \( p + 1 \) without changing the inequality:
\[ p > 1(p + 1) \]
This simplifies to:
\[ p > p + 1 \]
Subtracting \( p \) from both sides gives:
\[ 0 > 1 \]
This is a contradiction, meaning there are no solutions in this case.

### Case 2: \( p + 1 < 0 \) (i.e., \( p < -1 \))
If \( p + 1 < 0 \), we multiply both sides by \( p + 1 \), which reverses the inequality:
\[ p < 1(p + 1) \]
This simplifies to:
\[ p < p + 1 \]
Subtracting \( p \) from both sides gives:
\[ 0 < 1 \]
This is always true, so all values of \( p \) that satisfy \( p < -1 \) are valid solutions.

### Conclusion
The relation \( \frac{p}{p+1} > 1 \) is valid when:
\[ p < -1 \]

### Final Answer
The condition is \( p < -1 \).

---