If the probability of chossing an interger "k" out of 2m integers `1,2,3,….,2m` is inversely proportional to `k^(4)(1leklem).Ifx_(1)` is the probability that chosen number is odd and `x^(2)` is the probability that chosen number is even, then

Options

A. `x_(1)gt1//2`
B. `x_(1)gt2//3`
C. `x_(2)lt1//2`
D. `x_(2)lt2//3`

Explanation

To solve the problem, we need to find the probabilities of choosing an odd integer \( x_1 \) and an even integer \( x_2 \) from the set of integers \( 1, 2, 3, \ldots, 2m \), given that the probability of choosing an integer \( k \) is inversely proportional to \( k^4 \).

### Step-by-Step Solution:

1. **Define the Probability Function**:
The probability \( P(k) \) of choosing an integer \( k \) is given as:
\[ P(k) = \frac{\lambda}{k^4} \]
where \( \lambda \) is a constant of proportionality.

2. **Normalization Condition**:
The sum of probabilities for all integers from \( 1 \) to \( 2m \) must equal 1:
\[ \sum_{k=1}^{2m} P(k) = 1 \]
This gives us:
\[ \sum_{k=1}^{2m} \frac{\lambda}{k^4} = 1 \]

3. **Calculate the Total Probability**:
We can separate the sum into two parts: odd and even integers.
- Odd integers: \( 1, 3, 5, \ldots, 2m-1 \) (there are \( m \) odd integers)
- Even integers: \( 2, 4, 6, \ldots, 2m \) (there are \( m \) even integers)

Thus, we can express \( x_1 \) and \( x_2 \) as:
\[ x_1 = \sum_{k=1}^{m} P(2k-1) = \sum_{k=1}^{m} \frac{\lambda}{(2k-1)^4} \]
\[ x_2 = \sum_{k=1}^{m} P(2k) = \sum_{k=1}^{m} \frac{\lambda}{(2k)^4} \]

4. **Express \( x_2 \) in terms of \( x_1 \)**:
We can express \( x_2 \) as:
\[ x_2 = \sum_{k=1}^{m} \frac{\lambda}{(2k)^4} = \frac{\lambda}{16} \sum_{k=1}^{m} \frac{1}{k^4} \]

5. **Relate \( x_1 \) and \( x_2 \)**:
From the normalization condition, we know:
\[ x_1 + x_2 = 1 \]
Therefore, we can express \( x_2 \) as:
\[ x_2 = 1 - x_1 \]

6. **Inequalities**:
Since both \( x_1 \) and \( x_2 \) are probabilities, they must satisfy:
\[ 0 < x_1 < 1 \quad \text{and} \quad 0 < x_2 < 1 \]
This leads us to the conclusion that:
\[ x_1 > \frac{1}{2} \quad \text{and} \quad x_2 < \frac{1}{2} \]

### Conclusion:
Thus, we conclude that \( x_1 > \frac{1}{2} \) and \( x_2 < \frac{1}{2} \).