For every p, q positive integers at x = 0 or x = 1, the valid relation can be :

Options

A. 1. `p^xq^(1-x) = qx + p^(1-x)`
B. 2. `p^xq^(1-x) = px + q^(1-x)`
C. 3. `p^xq^(1-x) = p^(1-x)qx`
D. 4. either (2) or (3)

Answer

To solve the problem, we need to analyze the given expressions for \( x = 0 \) and \( x = 1 \) and determine which of the relations hold true for positive integers \( p \) and \( q \).

### Step-by-Step Solution:

1. **Examine the First Relation:**
\[ p^x q^{1-x} = q^x + p^{1-x} \]
- **Substituting \( x = 0 \):**
\[ p^0 q^{1-0} = q^0 + p^{1-0} \]
This simplifies to:
\[ 1 \cdot q = 1 + p \implies q = 1 + p \]
This is not valid for all positive integers \( p \) and \( q \).

- **Substituting \( x = 1 \):**
\[ p^1 q^{1-1} = q^1 + p^{1-1} \]
This simplifies to:
\[ p \cdot 1 = q + 1 \implies p = q + 1 \]
This is also not valid for all positive integers \( p \) and \( q \).

**Conclusion:** The first relation is not valid.

2. **Examine the Second Relation:**
\[ p^x q^{1-x} = P^x + Q^{1-x} \]
- **Substituting \( x = 0 \):**
\[ p^0 q^{1-0} = p^0 + q^{1-0} \]
This simplifies to:
\[ 1 \cdot q = 1 + q \implies q = 1 + q \]
This is valid.

- **Substituting \( x = 1 \):**
\[ p^1 q^{1-1} = p^1 + q^{1-1} \]
This simplifies to:
\[ p \cdot 1 = p + 1 \implies p = p + 1 \]
This is also valid.

**Conclusion:** The second relation is valid for both \( x = 0 \) and \( x = 1 \).

3. **Examine the Third Relation:**
\[ p^x q^{1-x} = p^{1-x} q^x \]
- **Substituting \( x = 0 \):**
\[ p^0 q^{1-0} = p^{1-0} q^0 \]
This simplifies to:
\[ 1 \cdot q = p \cdot 1 \implies q = p \]
This is not valid for all positive integers \( p \) and \( q \).

- **Substituting \( x = 1 \):**
\[ p^1 q^{1-1} = p^{1-1} q^1 \]
This simplifies to:
\[ p \cdot 1 = 1 \cdot q \implies p = q \]
This is also not valid for all positive integers \( p \) and \( q \).

**Conclusion:** The third relation is not valid.

### Final Conclusion:
The only valid relation for both \( x = 0 \) and \( x = 1 \) is the second relation. Therefore, the correct option is **B**.

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