不等式 $-1 < \frac{1-x}{2} \le k + 1$（其中 $k$ 为常数）的解表示在下方的数轴上。求 $k$ 的值。 答案 $k$ = ...

1. Identify the solution set represented on the number line

From the number line, the solid circle is at $-2$ and the open circle is at $3$. Thus, the solution set is:
$-2 \le x < 3$

2. Solve the given inequality for $x$ in terms of $k$

Given $-1 < \frac{1-x}{2} \le k + 1$
Multiply all parts by 2:
$-2 < 1 - x \le 2k + 2$
Subtract 1 from all parts:
$-3 < -x \le 2k + 1$
Multiply by $-1$ and reverse the inequality signs:
$3 > x \ge -2k - 1$
Which can be written as:
$-2k - 1 \le x < 3$

3. Compare the bounds to find $k$

Comparing $-2k - 1 \le x < 3$ with $-2 \le x < 3$, the upper bounds match ($x < 3$).
Equating the lower bounds:
$-2k - 1 = -2$
$-2k = -1$
$k = 0.5$