Match the statements given in Column I with the intervals/union of intervals given in Column II.

(A) Given, \(|z|=1 \Rightarrow z \cdot \bar{z}=1\)
\[
\therefore \quad \frac{2 i z}{1-z^2}=\frac{2 i z}{z \cdot \bar{z}-z^2}=\frac{2 i}{\bar{z}-z} \text {, }
\]
Let \(\quad z=x+i y\)
\[
\therefore z-\bar{z}=2 i y=\frac{2 i}{-2 i y}=-\frac{1}{y}
\]
where, \(y=\sqrt{1-x^2}\)
\(\therefore \quad-1 \leq y \leq 1 \Rightarrow-1 \leq y\)
and \(y \leq 1 \Rightarrow-1 \geq \frac{1}{y}\) and \(\frac{1}{y} \geq 1\)
\(\Rightarrow \operatorname{Re}\left(\frac{2 i z}{1-z^2}\right) \in(-\infty,-1] \cup[1, \infty)\)
(B) \(f(x)=\sin ^{-1}\left(\frac{8\left(3^{x-2}\right)}{1-3^{2(x-1)}}\right)\),

\[
\begin{aligned}
& \text { For domain, }-1 \leq \frac{8\left(3^{x-2}\right)}{1-3^{2(x-1)}} \leq 1 \\
& \Rightarrow \quad-1 \leq \frac{9 \cdot\left(3^{x-2}\right)-\left(3^{x-2}\right)}{1-3^{2(x-1)}} \leq 1 \\
& \therefore \quad-1 \leq \frac{3^x-3^{x-2}}{1-3^x \cdot 3^{(x-2)}} \leq 1 \\
& \frac{3^x-3^{x-2}}{1-3^x \cdot 3^{x-2}} \geq-1 \\
& \Rightarrow \frac{\left(3^x-1\right)\left(3^{x-2}-1\right)}{\left(3^{x-1}+1\right)\left(3^{x-1}-1\right)} \geq 0 \\
& \Rightarrow \quad x \in(-\infty, 0] \cup(1, \infty) \\
& \text { and } \frac{3^x-3^{x-2}}{1-3^x \cdot 3^{x-2}} \leq 1 \\
& \Rightarrow \frac{\left(3^{x-2}-1\right)\left(3^x+1\right)}{\left(3^{x-1}+1\right)\left(3^{x-1}-1\right)} \geq 0 \\
&
\end{aligned}
\]



and \(x \in(-\infty, 1) \cup[2, \infty)\)
\(\therefore \quad x \in(-\infty, 0] \cup[2, \infty)\)
(C) \(f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|\)
\[
\begin{aligned}
R_1 & \rightarrow R_1+R_3, \\
f(\theta) & =\left|\begin{array}{ccc}
0 & 0 & 2 \\
-\tan \theta & 1 & \tan \theta \\
-1 & -\tan \theta & 1
\end{array}\right| \\
& =2\left(\tan ^2 \theta+1\right)=2 \sec ^2 \theta \geq 2 \\
f(\theta) & \in[2, \infty)
\end{aligned}
\]
\[
\text { (D) } \begin{aligned}
f(x) & =x^{3 / 2}(3 x-10) ; x \geq 0 \\
f^{\prime}(x) & =x^{3 / 2} \cdot 3+\frac{3}{2} \cdot x^{1 / 2}(3 x-10) \\
& =3 x^{1 / 2}\left\{x+\frac{1}{2}(3 x-10)\right\}
\end{aligned}
\]

\[
\begin{aligned}
& =\frac{3}{2} x^{1 / 2}\{2 x+3 x-10\} \\
& =\frac{15}{2} x^{1 / 2}(x-2) \\
& \\
\therefore \quad & x \geq 2
\end{aligned}
\]