A special die is so constructed that the probabilities of throwing 1, 2, 3, 4, 5 and 6 are `(1-k)/(6), (1+2k)/(6), (1-k)/(6), (1+k)/(6), (1-2k)/(6) and (1+k)/(6)`, respectively. If two such thrown and the probability of getting a sum equal to lies between `(1)/(9) and (2)/(9)`, then the integral value of k is

Answer

To solve the problem step by step, we need to find the integral value of \( k \) such that the probabilities of the sums of two throws of a special die fall between \( \frac{1}{9} \) and \( \frac{2}{9} \).

### Step 1: Define the probabilities

The probabilities for each face of the die are given as follows:

- \( P(1) = \frac{1-k}{6} \)
- \( P(2) = \frac{1+2k}{6} \)
- \( P(3) = \frac{1-k}{6} \)
- \( P(4) = \frac{1+k}{6} \)
- \( P(5) = \frac{1-2k}{6} \)
- \( P(6) = \frac{1+k}{6} \)

### Step 2: Calculate the probability of getting a sum of 9

To get a sum of 9 from two dice, the possible combinations are:

1. (3, 6)
2. (4, 5)
3. (5, 4)
4. (6, 3)

Now, we calculate the probability for each combination:

- For (3, 6): \( P(3) \times P(6) = \left(\frac{1-k}{6}\right) \times \left(\frac{1+k}{6}\right) = \frac{(1-k)(1+k)}{36} = \frac{1-k^2}{36} \)

- For (4, 5): \( P(4) \times P(5) = \left(\frac{1+k}{6}\right) \times \left(\frac{1-2k}{6}\right) = \frac{(1+k)(1-2k)}{36} = \frac{1 - 2k + k - 2k^2}{36} = \frac{1 - k - 2k^2}{36} \)

- For (5, 4): \( P(5) \times P(4) = \left(\frac{1-2k}{6}\right) \times \left(\frac{1+k}{6}\right) = \frac{(1-2k)(1+k)}{36} = \frac{1 - 2k + k - 2k^2}{36} = \frac{1 - k - 2k^2}{36} \)

- For (6, 3): \( P(6) \times P(3) = \left(\frac{1+k}{6}\right) \times \left(\frac{1-k}{6}\right) = \frac{(1+k)(1-k)}{36} = \frac{1-k^2}{36} \)

Now, we sum these probabilities:

\[ P(\text{sum } 9) = 2 \cdot \frac{1-k^2}{36} + 2 \cdot \frac{1-k-2k^2}{36} \]

\[ = \frac{2(1-k^2) + 2(1-k-2k^2)}{36} = \frac{2 + 2 - 2k - 4k^2}{36} = \frac{4 - 2k - 6k^2}{36} = \frac{2 - k - 3k^2}{18} \]

### Step 3: Set up the inequality

We need this probability to lie between \( \frac{1}{9} \) and \( \frac{2}{9} \):

\[ \frac{1}{9} < \frac{2 - k - 3k^2}{18} < \frac{2}{9} \]

### Step 4: Solve the inequalities

1. For the left inequality:

\[ \frac{1}{9} < \frac{2 - k - 3k^2}{18} \]

Multiplying through by 18:

\[ 2 < 2 - k - 3k^2 \]

This simplifies to:

\[ 0 < -k - 3k^2 \implies 3k^2 + k < 0 \]

Factoring gives:

\[ k(3k + 1) < 0 \]

The roots are \( k = 0 \) and \( k = -\frac{1}{3} \). The solution to this inequality is:

\[ -\frac{1}{3} < k < 0 \]

2. For the right inequality:

\[ \frac{2 - k - 3k^2}{18} < \frac{2}{9} \]

Multiplying through by 18:

\[ 2 - k - 3k^2 < 4 \]

This simplifies to:

\[ -k - 3k^2 < 2 \implies 3k^2 + k + 2 > 0 \]

The discriminant of \( 3k^2 + k + 2 \) is:

\[ D = 1^2 - 4 \cdot 3 \cdot 2 = 1 - 24 = -23 \]

Since the discriminant is negative, \( 3k^2 + k + 2 > 0 \) for all \( k \).

### Step 5: Combine the results

From the first inequality, we have:

\[ -\frac{1}{3} < k < 0 \]

Since \( k \) must be an integer, the only integral value for \( k \) in this range is:

\[ k = -1 \]

### Final Answer

Thus, the integral value of \( k \) is:

\[ \boxed{-1} \]