2. (i) (Grade 11) The range of the function $y=\frac{2 \sin x \cdot \cos x-3}{\sin x+\cos x-2}$ is ( ).
(A) $\left[\frac{1}{2}, 3\right]$
(B) $[2-\sqrt{2}, 2+\sqrt{2}]$
(C) $[\sqrt{2}-1, \sqrt{2}+1]$
(D) $\left[1-\frac{\sqrt{2}}{2}, 1+\frac{\sqrt{2}}{2}\right]$
(ii) (Grade 12) Given that $a, b, c, d$ are distinct positive numbers, and
$$
\begin{array}{l}
a+b=1, x=\sqrt{a c+b d}, \\
y=a \sqrt{c}+b \sqrt{d}, z=\sqrt{a c}+\sqrt{b d} .
\end{array}
$$

Among the following relations, the one that holds is ( ).
(A) $x>y>z$
(B) $z>x>y$
(C) $y>z>x$
(D) $z>y>x$
2. (i)B.

Since $2 \sin x \cdot \cos x - 3 = (\sin x + \cos x)^2 - 4$, therefore, $y = (\sin x + \cos x) + 2$.
Also, $-\sqrt{2} \leq \sin x + \cos x \leq \sqrt{2}$, then
$2 - \sqrt{2} \leq y \leq 2 + \sqrt{2}$.
(ii) B.
$$
\begin{array}{l}
z^{2} - x^{2} = a c + b d + 2 \sqrt{a b c d} - (a c + b d) \\
= 2 \sqrt{a b c d} > 0, \\
x^{2} - y^{2} = (a c + b d) - \left(a^{2} c + b^{2} d + 2 a b \sqrt{c d}\right) \\
= a c (1 - a) + b d (1 - b) - 2 b d \sqrt{c d} \\
= a b c + a b d - 2 a b \sqrt{c d} \\
= a b (c + d - 2 \sqrt{c d}) > 0 .
\end{array}
$$

Therefore, $z > x > y$.