If `P={x in R : f(x)=0}` and `Q={x in R : g(x)=0 }`, then `PuuQ` is

Options

A. `{x in R : f(x) + g(x)=0}`
B. `{x in R : f(x) g(x)=0}`
C. `{x in R : (f(x))^(2)+(g(x))^(2)=0}`
D. none of these

Answer

To solve the problem, we need to determine the union of the sets \( P \) and \( Q \) defined as follows:

- \( P = \{ x \in \mathbb{R} : f(x) = 0 \} \)
- \( Q = \{ x \in \mathbb{R} : g(x) = 0 \} \)

We want to find \( P \cup Q \).

### Step-by-Step Solution:

1. **Understanding the Sets**:
- The set \( P \) consists of all real numbers \( x \) for which the function \( f(x) \) equals zero.
- The set \( Q \) consists of all real numbers \( x \) for which the function \( g(x) \) equals zero.

2. **Definition of Union**:
- The union of two sets \( P \) and \( Q \), denoted \( P \cup Q \), includes all elements that are in \( P \), in \( Q \), or in both. In terms of functions, this means we are looking for values of \( x \) such that either \( f(x) = 0 \) or \( g(x) = 0 \).

3. **Formulating the Union**:
- Therefore, we can express the union \( P \cup Q \) as:
\[ P \cup Q = \{ x \in \mathbb{R} : f(x) = 0 \text{ or } g(x) = 0 \} \]

4. **Translating to Function Terms**:
- In terms of functions, the condition for \( P \cup Q \) can be stated as:
\[ P \cup Q = \{ x \in \mathbb{R} : f(x) = 0 \lor g(x) = 0 \} \]

5. **Using the Product of Functions**:
- The union can also be represented using the product of the functions:
\[ P \cup Q = \{ x \in \mathbb{R} : f(x) \cdot g(x) = 0 \} \]
- This is because the product \( f(x) \cdot g(x) = 0 \) holds true if either \( f(x) = 0 \) or \( g(x) = 0 \).

6. **Final Representation**:
- Thus, we conclude that:
\[ P \cup Q = \{ x \in \mathbb{R} : f(x) \cdot g(x) = 0 \} \]

### Conclusion:
The final answer for \( P \cup Q \) is:
\[ P \cup Q = \{ x \in \mathbb{R} : f(x) \cdot g(x) = 0 \} \]