If the events A and B are mutually exclusive events such that `P(A) = (1)/(3)(3x+1) and P(B) = (1)/(4) (1-x)` , then the set of possible values of x lies in the interval

Options

A. [0,1]
B. `[(1)/(3),(2)/(9)]`
C. `[-(1)/(3),(5)/(9)]`
D. `[-(7)/(9),(4)/(9)]`

Answer

To solve the problem, we need to find the set of possible values of \( x \) for the mutually exclusive events \( A \) and \( B \) given their probabilities.

### Step-by-Step Solution:

1. **Given Probabilities**:
- \( P(A) = \frac{1}{3}(3x + 1) \)
- \( P(B) = \frac{1}{4}(1 - x) \)

2. **Condition for Probability of A**:
Since \( P(A) \) must be between 0 and 1:
\[ 0 \leq \frac{1}{3}(3x + 1) \leq 1 \]
- Multiply the entire inequality by 3:
\[ 0 \leq 3x + 1 \leq 3 \]
- Subtract 1 from all sides:
\[ -1 \leq 3x \leq 2 \]
- Divide by 3:
\[ -\frac{1}{3} \leq x \leq \frac{2}{3} \]

3. **Condition for Probability of B**:
Similarly, for \( P(B) \):
\[ 0 \leq \frac{1}{4}(1 - x) \leq 1 \]
- Multiply the entire inequality by 4:
\[ 0 \leq 1 - x \leq 4 \]
- Rearranging gives:
\[ -3 \leq -x \leq 1 \]
- Multiplying by -1 (and reversing the inequalities):
\[ 3 \geq x \geq -1 \]
- This can also be written as:
\[ -3 \leq x \leq 1 \]

4. **Condition for the Union of Events**:
Since \( A \) and \( B \) are mutually exclusive, we have:
\[ P(A \cup B) = P(A) + P(B) \leq 1 \]
- Therefore:
\[ \frac{1}{3}(3x + 1) + \frac{1}{4}(1 - x) \leq 1 \]
- Multiply the entire inequality by 12 (the least common multiple of 3 and 4):
\[ 4(3x + 1) + 3(1 - x) \leq 12 \]
- Expanding gives:
\[ 12x + 4 + 3 - 3x \leq 12 \]
- Combine like terms:
\[ 9x + 7 \leq 12 \]
- Subtract 7 from both sides:
\[ 9x \leq 5 \]
- Divide by 9:
\[ x \leq \frac{5}{9} \]

5. **Combining the Ranges**:
We now have three inequalities:
- From \( P(A) \): \( -\frac{1}{3} \leq x \leq \frac{2}{3} \)
- From \( P(B) \): \( -3 \leq x \leq 1 \)
- From \( P(A \cup B) \): \( x \leq \frac{5}{9} \)

The intersection of these ranges is:
- The first range gives \( -\frac{1}{3} \) to \( \frac{2}{3} \).
- The second range gives \( -3 \) to \( 1 \).
- The third range gives \( -\infty \) to \( \frac{5}{9} \).

The intersection of these ranges is:
\[ -\frac{1}{3} \leq x \leq \frac{5}{9} \]

### Final Result:
The set of possible values of \( x \) lies in the interval:
\[ \boxed{\left[-\frac{1}{3}, \frac{5}{9}\right]} \]