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

• A. A(s),B(t),C(r),D(r)
• B.

Answer 1

The correct option is A A(s),B(t),C(r),D(r)

(A) We have $|z|=1\text{and}z\ne \pm 1$
$z=\cos \theta +i\sin \theta$
$\frac{2z}{1-z^2}=\frac{2(\cos \theta +i\sin \theta )}{1-(\cos \theta +i\sin \theta {)}^2}$

$=\frac{2(\cos \theta +i\sin \theta )}{1-\cos 2\theta -i\sin 2\theta }$

$=\frac{1}{-i\sin \theta }\left(\frac{\cos \theta +i\sin \theta }{\cos \theta +i\sin \theta }\right)$

$=\frac{1}{-i\sin \theta }=\frac{i}{\sin \theta }$

$\Rightarrow R_e\left(\frac{i2z}{1-z^2}\right)=-\frac{1}{\sin \theta }=-\text{cosec}\theta$

$\Rightarrow D_r=(-\infty ,1]\cup [1,\infty )$

(B) For the domain of the given function
$-1\le \frac{{8.3}^{x-2}}{1-3^{2(x-1)}}\le 1$

$\Rightarrow |{8.3}^{x-2}|\le |1-3^{2(x-1)}|$

$\Rightarrow \frac{{8.3}^x}{9}\le 1-\frac{3^{2x}}{9},\frac{3^{2x}}{9}<1$

$\Rightarrow 8a\le 9-a^2,a=3^x$

$\Rightarrow -9\le a\le 1$

$\Rightarrow 3^x\le 3^0$

$\Rightarrow x\le 0$

(C) We have
$f\left(\theta \right)=\left|\begin{array}{ccc}1& \tan \theta & 1\\ \tan \theta & 1& \tan \theta \\ -1& -\tan \theta & 1\end{array}\right|$

$=(1+{\tan }^2\theta )-\tan \theta (-\tan \theta +\tan \theta )+1({\tan }^2\theta +1)$

$=2(1+{\tan }^2\theta )=2{\sec }^2\theta$
$D_r=[2,\infty )$

(D) We have
$f\left(x\right)=x^{\frac{3}{2}}(3x-10)$

$f^{\prime }\left(x\right)=\frac{3}{2}{x}^{\frac{1}{2}}(3x-10)+3.x^{\frac{3}{2}}$

$=\frac{3x^{\frac{1}{2}}}{2}(3x-10+2x)$

$=\frac{15}{2}{x}^{\frac{1}{2}}(x-2)$

Since $f\left(x\right)$ is increasing, hence,
$f^{\prime }\left(x\right)\ge 0$
$\Rightarrow x\ge 2\text{as}x\ge 0$
$D_r\text{is}[2,\infty )$