If the solution set of `|x-k| lt 2 ` is a subset of the solution of the inequality `(2x-1)/(x+2) lt 1` , then the number of possible integral value(s) of k is/ are :

Answer

To solve the problem, we need to analyze the two inequalities given in the question.

### Step 1: Solve the first inequality \(|x - k| < 2\)

This inequality can be rewritten as:
\[ -2 < x - k < 2 \]
Adding \(k\) to all parts of the inequality gives:
\[ k - 2 < x < k + 2 \]
This means:
\[ x \in (k - 2, k + 2) \]

### Step 2: Solve the second inequality \(\frac{2x - 1}{x + 2} < 1\)

First, we rearrange the inequality:
\[ \frac{2x - 1}{x + 2} - 1 < 0 \]
This simplifies to:
\[ \frac{2x - 1 - (x + 2)}{x + 2} < 0 \]
\[ \frac{2x - 1 - x - 2}{x + 2} < 0 \]
\[ \frac{x - 3}{x + 2} < 0 \]

### Step 3: Determine the critical points

The critical points occur when the numerator and denominator are zero:
- \(x - 3 = 0 \Rightarrow x = 3\)
- \(x + 2 = 0 \Rightarrow x = -2\)

### Step 4: Test intervals around the critical points

We will test the intervals determined by these points: \((-∞, -2)\), \((-2, 3)\), and \((3, ∞)\).

1. For \(x < -2\) (e.g., \(x = -3\)):
\[ \frac{-3 - 3}{-3 + 2} = \frac{-6}{-1} = 6 > 0 \]
2. For \(-2 < x < 3\) (e.g., \(x = 0\)):
\[ \frac{0 - 3}{0 + 2} = \frac{-3}{2} < 0 \]
3. For \(x > 3\) (e.g., \(x = 4\)):
\[ \frac{4 - 3}{4 + 2} = \frac{1}{6} > 0 \]

Thus, the solution to the inequality \(\frac{x - 3}{x + 2} < 0\) is:
\[ x \in (-2, 3) \]

### Step 5: Set the intervals from both inequalities

From the first inequality, we have:
\[ x \in (k - 2, k + 2) \]
From the second inequality, we have:
\[ x \in (-2, 3) \]

### Step 6: Determine the conditions for \(k\)

Since \((k - 2, k + 2)\) must be a subset of \((-2, 3)\), we have:
1. The lower bound condition:
\[ k - 2 \geq -2 \implies k \geq 0 \]
2. The upper bound condition:
\[ k + 2 \leq 3 \implies k \leq 1 \]

### Step 7: Combine the conditions

Combining the two conditions, we find:
\[ 0 \leq k \leq 1 \]

### Step 8: Identify the integral values of \(k\)

The integral values of \(k\) that satisfy this condition are:
- \(k = 0\)
- \(k = 1\)

### Conclusion

Thus, the number of possible integral values of \(k\) is:
\[ \text{Number of integral values of } k = 2 \]