`p`, `q` and `r` are distinct integers. If `p` and `q` are both odd and `r` is even, then which of the following expression is/are always false?
(i) `(p+q)p` is odd
(ii) `(p+r)r` is even
(iii) `(p+ r) p^(3)` is odd
(iv) `(p+r)^(3)p` is odd

Options

A. (i)
B. (ii)
C. (iii)
D. (iv)

Answer

To solve the problem, we need to analyze each of the given expressions based on the properties of odd and even integers.

### Step-by-Step Solution:

1. **Understanding the properties of odd and even integers:**
- The sum of two odd integers is even.
- The sum of an odd integer and an even integer is odd.
- The product of any integer with an even integer is even.
- The product of two odd integers is odd.

2. **Analyzing each expression:**

**Expression (i):** \((p + q)p\)
- Since both \(p\) and \(q\) are odd, \(p + q\) is even (odd + odd = even).
- Now, we multiply this even result by \(p\) (which is odd).
- Even × Odd = Even.
- Therefore, \((p + q)p\) is even, not odd.
- **Conclusion:** This expression is always false.

**Expression (ii):** \((p + r)r\)
- Here, \(p\) is odd and \(r\) is even.
- Thus, \(p + r\) is odd (odd + even = odd).
- Now, we multiply this odd result by \(r\) (which is even).
- Odd × Even = Even.
- Therefore, \((p + r)r\) is even.
- **Conclusion:** This expression is true.

**Expression (iii):** \((p + r)p^3\)
- Again, \(p\) is odd and \(r\) is even.
- So, \(p + r\) is odd.
- \(p^3\) (odd raised to any power) is also odd.
- Now, we multiply odd × odd = odd.
- Therefore, \((p + r)p^3\) is odd.
- **Conclusion:** This expression is true.

**Expression (iv):** \((p + r)^3p\)
- As established, \(p + r\) is odd.
- Therefore, \((p + r)^3\) is odd (odd raised to any power).
- Now, we multiply this odd result by \(p\) (which is also odd).
- Odd × Odd = Odd.
- Therefore, \((p + r)^3p\) is odd.
- **Conclusion:** This expression is true.

3. **Final Conclusion:**
- The only expression that is always false is **Expression (i)**: \((p + q)p\).

### Summary:
The expression that is always false is **(i)**: \((p + q)p\).