Question

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

Column I		Column II
A	The set \left\{\operatorname{Re}\left(\frac{2 i z}{1-z^{2}}\right): z \text { is a complex number, }|z|=1, z \neq \pm 1\right\}	P	$(-\infty ,-1)\cup (1,\infty )$
B	The domain of the function $f(x)={\sin }^{-1}\left(\frac{8(3{)}^{x-2}}{1-3^{2(x-1)}}\right)$ is	Q	$(-\infty ,0)\cup (0,\infty )$
C	If $f(\theta )=\left|\begin{array}{ccc}1& \tan \theta & 1\\ -\tan \theta & 1& \tan \theta \\ -1& -\tan \theta & 1\end{array}\right|$, then the set $\left\{f(\theta ):0\le \theta <\frac{\pi }{2}\right\}$ is	R	$[2,\infty )$
D	If $f(x)=x^{\frac{3}{2}}(3x-10),x\ge 0$, then $f(x)$ is increasing in	S	$(-\infty -1)\cup (1,\infty )$
		T	$(-\infty ,0)\cup (2,\infty )$

• A. $(A)\to (P),(B)\to (Q),(C)\to (R),(D)\to (R)$
• B. $(A)\to (S),(B)\to (T),(C)\to (R),(D)\to (R)$
• C. $(A)\to (Q),(B)\to (T),(C)\to (S),(D)\to (R)$
• D. $(A)\to (R),(B)\to (T),(C)\to (P),(D)\to (S)$

Answer 1

Answer: (B)

(A) $z=\frac{2i(x+iy)}{1-(x+iy{)}^2}=\frac{2i(x+iy)}{1-\left(x^2-y^2+2ixy\right)}$
Using $1-x^2=y^2$, we get
$Z=\frac{2ix-2y}{2y^2-2ixy}=-\frac{1}{y}$
Since $-1\le y\le 1\Rightarrow -\frac{1}{y}$
or $-\frac{1}{y}\ge 1$.
(B) For domain:
$-1\le \frac{8\left(3^{x-2}\right)}{1-3^{2(x-1)}}\le 1$
$\Rightarrow -1\le \frac{3^x-3^{x-2}}{1-3^{2x-2}}\le 1$
Case $1:\frac{3^x-3^{x-2}}{1-3^{2x-2}}-\le 0$
$\Rightarrow \frac{\left(3^x-1\right)\left(3^{x-2}-1\right)}{\left(3^{2x-2}-1\right)}\ge 0$
$\Rightarrow x\in (-\infty ,0]\cup (1,\infty )$
Case 2: $\frac{3^x-3^{x-2}}{1-3^{2x}-2}+1\ge 0$
$\Rightarrow \frac{\left(x^{x-2}-1\right)\left(3^x+1\right)}{\left(3^x\cdot 3^{x-2}-1\right)}\ge 0$
$\Rightarrow x\in (-\infty ,1)\cup [2,\infty )$
So, $x\in (-\infty ,0]\cup [2,\infty )$.
(C) $R{R}_1\to R_1+R_3$ :
$f(\theta )=\left|\begin{array}{ccc}0& 0& 2\\ -\tan \theta & 1& \tan \theta \\ -1& -\tan \theta & 1\end{array}\right|$
$=2\left({\tan }^2\theta +1\right)=2{\sec }^2\theta$
(D) $f^{\prime }(x)=\frac{3}{2}(x{)}^{\frac{1}{2}}(3x-10)+(x{)}^{\frac{3}{2}}\times 3$
$=\frac{15}{2}(x{)}^{\frac{1}{2}}(x-2)$
Increasing, when $x\ge 2$.