4. If the decreasing function $y=f(x)$ defined on $\mathbf{R}$ satisfies the inequality $f\left(x^{2}-2 x\right) \leqslant-f(2 y-$ $y^{2}$ ) for any $x, y \in \mathbf{R}$. Moreover, the graph of the function $y=f(x-1)$ is symmetric about the point $(1,0)$, then when $1 \leqslant x \leqslant 4$, the range of $\frac{y}{x}$ is
A. $\left[-\frac{1}{4}, 1\right)$
B. $\left[-\frac{1}{4}, 1\right]$
C. $\left(-\frac{1}{2}, 1\right]$
D. $\left[-\frac{1}{2}, 1\right]$
4. D Given that the graph of $f(x-1)$ is symmetric about $(1,0)$, we know that $f(x)$ is an odd function. Also, $f(x)$ is a decreasing function on $\mathbf{R}$. Therefore,
$$
f\left(x^{2}-2 x\right) \leqslant -f\left(2 y-y^{2}\right) \Leftrightarrow f\left(x^{2}-2 x\right) \leqslant f\left(y^{2}-2 y\right) \Leftrightarrow x^{2}-
$$
$2 x \geqslant y^{2}-2 y$, hence $x \geqslant y$ or $x+y \leqslant 2$.
When $1 \leqslant x \leqslant 4$,
$$
\frac{-2}{4} \leqslant \frac{y}{x} \leqslant 1
$$

Therefore, the answer is D.