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 $f(\theta)=\begin{vmatrix}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{vmatrix}$, then the set $\left\{f(\theta): 0 \leq \theta < \frac{\pi}{2}\right\}$ is	R	$[2, \infty)$
D	If $f(x)=x^{\frac{3}{2}}(3 x-10), x \geq 0$, then $f(x)$ is increasing in	S	$(-\infty -1) \cup(1, \infty)$
		T	$(-\infty, 0) \cup(2, \infty)$

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

Answer 1

Answer: (B)

(A) $z=\frac{2 i(x+i y)}{1-(x+i y)^{2}}=\frac{2 i(x+i y)}{1-\left(x^{2}-y^{2}+2 i x y\right)}$
Using $1-x^{2}=y^{2}$, we get
$Z=\frac{2 i x-2 y}{2 y^{2}-2 i x y}=-\frac{1}{y}$
Since $-1 \leq y \leq 1 \Rightarrow -\frac{1}{y}$
or $-\frac{1}{y} \geq 1$.
(B) For domain:
$-1 \leq \frac{8\left(3^{x-2}\right)}{1-3^{2(x-1)}} \leq 1 $
$\Rightarrow -1 \leq \frac{3^{x}-3^{x-2}}{1-3^{2 x-2}} \leq 1$
Case $1: \frac{3^{x}-3^{x-2}}{1-3^{2 x-2}}-\leq 0$
$\Rightarrow \frac{\left(3^{x}-1\right)\left(3^{x-2}-1\right)}{\left(3^{2 x-2}-1\right)} \geq 0$
$\Rightarrow x \in(-\infty, 0] \cup(1, \infty)$
Case 2: $\frac{3^{x}-3^{x-2}}{1-3^{2 x}-2}+1 \geq 0$
$\Rightarrow \frac{\left(x^{x-2}-1\right)\left(3^{x}+1\right)}{\left(3^{x} \cdot 3^{x-2}-1\right)} \geq 0$
$\Rightarrow x \in(-\infty, 1) \cup[2, \infty)$
So, $x \in(-\infty, 0] \cup[2, \infty)$.
(C) $R R_{1} \rightarrow R_{1}+R_{3}$ :
$f(\theta)=\begin{vmatrix}0 & 0 & 2 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{vmatrix}$
$=2\left(\tan ^{2} \theta+1\right)=2 \sec ^{2} \theta$
(D) $f^{\prime}(x)=\frac{3}{2}(x)^{\frac{1}{2}}(3 x-10)+(x)^{\frac{3}{2}} \times 3$
$=\frac{15}{2}(x)^{\frac{1}{2}}(x-2)$
Increasing, when $x \geq 2$.