48 Given $2 \leqslant x \leqslant 3$, then the maximum value of the function $y=f(x)=\frac{1-x+\sqrt{2 x^{2}-2 x+1}}{2 x}$ is
$48 \frac{\sqrt{5}-1}{4}$. Since $x \geqslant 2>0$, we have
$$
y=\frac{1}{2}\left(\frac{1}{x}-1+\sqrt{2-\frac{1}{x}+\frac{1}{x^{2}}}\right)=\frac{1}{2}\left(\frac{1}{x}-1+\sqrt{\left(\frac{1}{x}-1\right)^{2}+1}\right) .
$$

Therefore, let $t=\frac{1}{x}-1$, then
$$
y=\frac{1}{2}\left(t+\sqrt{t^{2}+1}\right)=\frac{1}{2} \cdot \frac{1}{\sqrt{t^{2}+1}-t} .
$$

Since $2 \leqslant x \leqslant 3$, we have $-\frac{2}{3} \leqslant t \leqslant-\frac{1}{2}$. On the interval $\left[-\frac{2}{3},-\frac{1}{2}\right]$, $-t>0$ and is decreasing, $\sqrt{t^{2}+1}(>|t|)$ is also decreasing. Therefore, $\frac{1}{\sqrt{t^{2}+1}-t}$ is increasing. When $t=-\frac{1}{2}$, the value of $y$ is maximum $=$ $\frac{\sqrt{5}-1}{4}$.