(A) $$z = {{2i(x + iy)} \over {1 - {{(x + iy)}^2}}} = {{2i(x + iy)} \over {1 - ({x^2} - {y^2} + 2ixy)}}$$

Using $$1 - {x^2} = {y^2}$$, we get

$$Z = {{2ix - 2y} \over {2{y^2} - 2ixy}} = - {1 \over y}$$

Since $$ - 1 \le y \le 1 \Rightarrow - {1 \over y} \le - 1$$ or $$ - {1 \over y} \ge 1$$.

(B) For domain :

$$ - 1 \le {{8({3^{x - 2}})} \over {1 - {3^{2(x - 1)}}}} \le 1 \Rightarrow - 1 \le {{{3^x} - {3^{x - 2}}} \over {1 - {3^{2x - 2}}}} \le 1$$

Case 1 : $${{{3^x} - {3^{x - 2}}} \over {1 - {3^{2x - 2}}}} - 1 \le 0$$.

$$ \Rightarrow {{({3^x} - 1)({3^{x - 2}} - 1)} \over {({3^{2x - 2}} - 1)}} \ge 0$$

$$ \Rightarrow x \in ( - \infty ,0] \cup (1,\infty )$$

Case 2 : $${{{3^x} - {3^{x - 2}}} \over {1 - {3^{2x}} - 2}} + 1 \ge 0$$

$$ \Rightarrow {{({x^{x - 2}} - 1)({3^x} + 1)} \over {({3^x}\,.\,{3^{x - 2}} - 1)}} \ge 0$$

$$ \Rightarrow x \in ( - \infty ,1) \cup [2,\infty )$$

So, $$x \in ( - \infty ,0] \cup [2,\infty )$$.

(C) R₁ $$\to$$ R₁ + R₃ :

$$f(\theta ) = \left| {\matrix{ 0 & 0 & 2 \cr { - \tan \theta } & 1 & {\tan \theta } \cr { - 1} & { - \tan \theta } & 1 \cr } } \right| = 2({\tan ^2}\theta + 1) = 2{\sec ^2}\theta $$

(D) $$f'(x) = {3 \over 2}{(x)^{1/2}}(3x - 10) + {(x)^{3/2}} \times 3 = {{15} \over 2}{(x)^{1/2}}(x - 2)$$

Increasing, when $$x \ge 2$$.