3. Given that when $-2<x \leqslant 1$, the function value of the linear function $y=$ $(2 k-3) x+|k|$ is always positive. Then the range of $k$ is ( ).
(A) $1<k<2$
(B) $1 \leqslant k<2$
(C) $1 \leqslant k<2$ and $k \neq 1.5$
(D) $1<k \leqslant 2$ and $k \neq 1.5$
3. D.

From the problem, we have
$$
\left\{\begin{array}{l}
2 k-3 \neq 0, \\
(2 k-3) \times(-2)+|k| \geqslant 0, \\
(2 k-3) \times 1+|k|>0 .
\end{array}\right.
$$

From equation (1), we get $k \neq 1.5$.
When $k \geqslant 0$, from equation (2) we solve to get $k \leqslant 2$; from equation (3) we solve to get $k>1$. Therefore, $13$. At this time, there is no value of $k$ that meets the problem's requirements. Hence, the range of values for $k$ is $1<k \leqslant 2$ and $k \neq 1.5$.