Answer:
Option A
Explanation:
(A) Given , $|z|=1 \Rightarrow z. \overline{z}=1$
$\therefore$ $\frac{2iz}{1-z^{2}}=\frac{2iz}{z.\overline{z}-z^{2}}=\frac{2i}{\overline{z}-z},$
Let z=x+iy
$\therefore$ $z-\overline{z}=2iy= \frac{2i}{-2iy}=-\frac{1}{y}$ .......(i)
where, $y=\sqrt{1-x^{2}}$
$\therefore$ $-1\leq y\leq1 \Rightarrow -1 \leq y$
and $y\leq 1 \Rightarrow-1 \geq \frac{1}{y}$ and $\frac{1}{y}\geq 1$
$\Rightarrow$ $Re\left(\frac{2iz}{1-z^{2}}\right)\epsilon (-\infty ,-1)\cup[1, \infty)$
(B) $f(x)= \sin^{-1}\left(\frac{8(3^{x-2})}{1-3^{2(x-1)}}\right)$,
For domain, $-1\leq \frac{8(3^{x-2})}{1-3^{2(x-1)}}\leq1$
$\Rightarrow$ $-1\leq \frac{9.(3^{x-2})-(3^{x-2})}{1-3^{2(x-1)}}\leq1$
$\therefore$ $-1\leq \frac{3^{x}-3^{x-2}}{1-3^{x}. 3^{(x-2)}}\leq1$
$ \frac{3^{x}-3^{x-2}}{1-3^{x}. 3^{(x-2)}}\geq -1$
$\Rightarrow$ $ \frac{(3^{x}-1)(3^{x-2}-1)}{(3^{x+1}+1). 3^{x-1}-1}\geq 0$
$\Rightarrow$ $x\epsilon (-\infty,0] \cup [1, \infty)$
and $\frac{3^{x}-3^{x-2}}{1-3^{x}.3^{x-2}}\leq 1$
$\Rightarrow$ $\frac{(3^{x-2}-1)(3^{x}+1)}{(3^{x-1}+1).(3^{x-1}-1)}\geq 0$
and $x \epsilon (-\infty,1) \cup [2, \infty)$
$\therefore$ $x \epsilon (-\infty,0] \cup [2, \infty)$
(C) $f(\theta)=\begin{bmatrix}1 & \tan \theta&1 \\- \tan \theta & 1& \tan \theta\\ -1& -\tan \theta &1 \end{bmatrix}$
$R_{1} \rightarrow R_{1}+R_{3}$
$f(\theta)=\begin{bmatrix}0 &0 &2 \\- \tan \theta & 1& \tan \theta\\ -1& -\tan \theta &1 \end{bmatrix}$
= $2(\tan^{2} \theta+1)=2 \sec^{2} \theta \geq 2$
$f(\theta) \epsilon [2, \infty)$
(D) $f(x)=x^{3/2}(3x-10);x \geq 0$
$f'(x)=x^{3/2}.3+\frac{3}{2}.x^{1/2}(3x-10)$
$=3x^{1/2}\left\{x+\frac{1}{2}(3x-10)\right\}$
$=\frac{3}{2}x^{1/2}\left\{2x+3x-10\right\}$
$=\frac{15}{2}x^{1/2}(x-2)$
$\therefore$ $x\geq 2$
