Match the statements given in Column I with t | vedclass

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(A) Let $z = \cos 	heta + i \sin 	heta$. Then $\frac{2iz}{1-z^2} = \frac{2i(\cos 	heta + i \sin 	heta)}{1-(\cos 2	heta + i \sin 2	heta)} = \frac

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$A-s, B-t, C-r, D-r$

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(A) Let $z = \cos 	heta + i \sin 	heta$. Then $\frac{2iz}{1-z^2} = \frac{2i(\cos 	heta + i \sin 	heta)}{1-(\cos 2	heta + i \sin 2	heta)} = \frac{2i \cos 	heta - 2 \sin 	heta}{2 \sin^2 	heta - i \sin 2	heta} = \frac{2i \cos 	heta - 2 \sin 	heta}{2 \sin 	heta (\sin 	heta - i \cos 	heta)} = \frac{2i(\cos 	heta + i \sin 	heta)}{-2i \sin 	heta (\cos 	heta + i \sin 	heta)} = -\frac{1}{\sin 	heta} = -\csc 	heta$. Since $\sin 	heta \in [-1, 1] \setminus \{0\}$,$-\csc 	heta \in (-\infty, -1] \cup [1, \infty)$. Thus,$(A)-(s)$.$(B)$ For $f(x) = \sin^{-1}(\frac{8 \cdot 3^{x-2}}{1-3^{2x-2}})$,we need $-1 \leq \frac{8 \cdot 3^{x-2}}{1-3^{2x-2}} \leq 1$. Let $u = 3^{x-1}$. Then $\frac{8 \cdot 3^{x-2}}{1-3^{2x-2}} = \frac{8 \cdot 3^{x-1} \cdot 3^{-1}}{1-(3^{x-1})^2} = \frac{8u/3}{1-u^2}$. Solving $-1 \leq \frac{8u/3}{1-u^2} \leq 1$ leads to $u \leq 1/3$,so $3^{x-1} \leq 3^{-1} \Rightarrow x-1 \leq -1 \Rightarrow x \leq 0$. Thus,$(B)-(t)$.$(C)$ $f(	heta) = 1(1 + 	an^2 	heta) - 	an 	heta(-	an 	heta + 	an 	heta) + 1(	an^2 	heta + 1) = 2(1 + 	an^2 	heta) = 2 \sec^2 	heta$. For $0 \leq 	heta $(D)$ $f'(x) = \frac{3}{2}x^{1/2}(3x-10) + x^{3/2}(3) = \frac{3}{2}x^{1/2}(3x-10+2x) = \frac{15}{2}x^{1/2}(x-2)$. For $f(x)$ to be increasing,$f'(x) \geq 0$,which implies $x \geq 2$. Thus,$(D)-(r)$.

Column $I$	Column $II$
$(A)$ The set $\{\operatorname{Re}(\frac{2 i z}{1-z^2}): \|z\|=1, z \neq \pm 1\}$ is	$(p)$ $(-\infty,-1) \cup(1, \infty)$
$(B)$ The domain of $f(x)=\sin ^{-1}(\frac{8(3)^{x-2}}{1-3^{2(x-1)}})$ 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 $\{f(\theta): 0 \leq \theta < \frac{\pi}{2}\}$ is	$(r)$ $[2, \infty)$
$(D)$ If $f(x)=x^{3 / 2}(3 x-10), x \geq 0$,then $f(x)$ is increasing in	$(s)$ $(-\infty,-1] \cup[1, \infty)$
	$(t)$ $(-\infty, 0] \cup[2, \infty)$

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