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

1. Interpret the solution set from the number line.

From the number line, the solution set is
$-2 \leq x < 3.$

2. Solve the left-hand part of the compound inequality to check the upper bound.

$-1 < \frac{1-x}{2}$
$-2 < 1-x$
$-3 < -x$
$x < 3$
This matches the open circle at $x=3$.

3. Use the lower bound from the number line to find $k$.

$\frac{1-x}{2} \leq k+1$
$1-x \leq 2k+2$
$-x \leq 2k+1$
$x \geq -2k-1$
From the number line, the lower bound is $x \geq -2$, so
$-2k-1=-2$
$-2k=-1$
$k=\frac{1}{2}.$