Question

Match the statements given List 1 with the intervals/union of intervals given in List 2.

	List 1		List 2
A.	The set $Re\left[\frac{2zi}{1-z^2}\right]$ : z is a complex number, |z|=1,$z\ne \pm 1$} is	1.	$(-\infty ,-1)\cup (1,\infty )$
B.	The domain of the function $f\left(x\right)={\sin }^{-1}\left[\frac{8(3{)}^{x-2}}{1-3^{2(x-1)}}\right]$ is	2	$(-\infty ,0)\cup (0,\infty )$
C.	If $f\left(\theta \right)=\left|\begin{array}{ccc}1& \tan \theta & 1\\ -\tan \theta & 1& \tan \theta \\ -1& -\tan \theta & 1\end{array}\right|$ then the set $\left\{f\right(\theta ):0\le \theta <\frac{\pi }{2}\}$	3.	$[2,\infty )$
D.	lf $f\left(x\right)=x^{\frac{3}{2}}(3x-10)$ , $x\ge 0$, then $f\left(x\right)$ is increasing in	4.	$(-\infty ,-1]\cup [1,\infty )$
		5.	$(-\infty ,0]\cup [2,\infty )$

• A. $A-4,B-3,C-5,D-3$
• B. $A-5,B-4,C-3,D-3$
• C. $A-4,B-5,C-3,D-2$
• D. $A-4,B-5,C-3,D-3$

Answer 1

The correct option is D $A-4,B-5,C-3,D-3$

A) $Re\left(\frac{2iz}{1-z^2}\right)=Re\left(\frac{2i\left(\cos \theta +i\sin \theta \right)}{1-{(\cos \theta +i\sin \theta )}^2}\right)$
$=Re\left(\frac{2i\left(\cos \theta +i\sin \theta \right)}{1-\cos 2\theta -i\sin 2\theta }\right)$
$=-\frac{1}{\sin \theta }=\csc \theta$
Range of $\csc \theta$ is $(-\infty ,-1]\cup [1,\infty ]$
So, $Re\left(\frac{2iz}{1-z^2}\right)\in (-\infty ,-1]\cup [1,\infty ]$
B) Let $y={\sin }^{-1}\left(\frac{8{\left(3\right)}^{x-2}}{1-3^{2(x-1)}}\right)$
$\Rightarrow \sin y=\frac{8{\left(3\right)}^{x-2}}{1-3^{2(x-1)}}$
$-1\le \frac{8{\left(3\right)}^{x-2}}{1-3^{2(x-1)}}\le 1$
$\Rightarrow -1\le \frac{8{\left(3\right)}^x}{9-3^{2x}}\le 1$
Let $3^x=t$
$\Rightarrow -1\le \frac{8t}{9-t^2}\le 1$
$\Rightarrow -1\le \frac{8t}{9-t^2}$ and $\frac{8t}{9-t^2}\le 1$
$\Rightarrow \frac{t^2-8t-9}{t^2-9}\ge 0$ $\Rightarrow \frac{t^2+8t-9}{t^2-9}\ge 0$
$\Rightarrow \frac{(t-9)(t+1)}{(t-3\left)\right(t+3)}\ge 0$ $\Rightarrow \frac{(t+9)(t-1)}{(t-3\left)\right(t+3)}\ge 0$
$\Rightarrow \frac{(t-9)}{(t-3)}\ge 0$ $\Rightarrow \frac{(t-9)}{(t-3)}\ge 0$
$\Rightarrow t<3$ and$t\ge 9$ $\Rightarrow t\le 1$ and $t>3$
$\Rightarrow x\le 1$ and $x>2$ $\Rightarrow x\le 0$ and $x>2$
$\Rightarrow x\in (-\infty ,1)\cup [2,\infty )$ $\Rightarrow x\in (-\infty ,0]\cup (2,\infty )$
So, common solution is $\Rightarrow x\in (-\infty ,0]\cup (2,\infty )$
C) $f\left(\theta \right)=\left|\begin{array}{ccc}1& \tan \theta & 1\\ -\tan \theta & 1& \tan \theta \\ -1& -\tan \theta & 1\end{array}\right|$
$=2{\sec }^2\theta$
Since, ${\sec }^2\theta \ge 1$
$\Rightarrow f\left(\theta \right)\in [2,\infty )$
D) $f\left(x\right)=x^{\frac{3}{2}}(3x-10)=3x^{\frac{5}{2}}-10x^{\frac{3}{2}}$
$\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{15}{2}{x}^{\frac{3}{2}}-15x^{\frac{1}{2}}$
$\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{15}{2}{x}^{\frac{1}{2}}(x-2)$
For increasing , $f^{{}^{\prime }}\left(x\right)\ge 0$
$\Rightarrow x\ge 2$
$f\left(x\right)$ is increasing in $[2,\infty )$