Mathematics - Functions Question with Solution | TestHub

(A) Given, $|z|=1\Rightarrow z\cdot \bar{z}=1$ $$\therefore \frac{2iz}{1-z^2}=\frac{2iz}{z\cdot \bar{z}-z^2}=\frac{2i}{\bar{z}-z}\text{, }$$ Let $z=x+iy$ $$\therefore z-\bar{z}=2iy=\frac{2i}{-2iy}=-\frac{1}{y}$$ where, $y=\sqrt{1-x^2}$ $\therefore -1\le y\le 1\Rightarrow -1\le y$ and $y\le 1\Rightarrow -1\ge \frac{1}{y}$ and $\frac{1}{y}\ge 1$ $\Rightarrow \mathrm{Re}\left(\frac{2iz}{1-z^2}\right)\in (-\infty ,-1]\cup [1,\infty )$ (B) $f(x)={\sin }^{-1}\left(\frac{8\left(3^{x-2}\right)}{1-3^{2(x-1)}}\right)$, $$\begin{array}{rl}& \text{ For domain, }-1\le \frac{8\left(3^{x-2}\right)}{1-3^{2(x-1)}}\le 1\\ & \Rightarrow -1\le \frac{9\cdot \left(3^{x-2}\right)-\left(3^{x-2}\right)}{1-3^{2(x-1)}}\le 1\\ & \therefore -1\le \frac{3^x-3^{x-2}}{1-3^x\cdot 3^{(x-2)}}\le 1\\ & \frac{3^x-3^{x-2}}{1-3^x\cdot 3^{x-2}}\ge -1\\ & \Rightarrow \frac{\left(3^x-1\right)\left(3^{x-2}-1\right)}{\left(3^{x-1}+1\right)\left(3^{x-1}-1\right)}\ge 0\\ & \Rightarrow x\in (-\infty ,0]\cup (1,\infty )\\ & \text{ and }\frac{3^x-3^{x-2}}{1-3^x\cdot 3^{x-2}}\le 1\\ & \Rightarrow \frac{\left(3^{x-2}-1\right)\left(3^x+1\right)}{\left(3^{x-1}+1\right)\left(3^{x-1}-1\right)}\ge 0\\ & \end{array}$$
and $x\in (-\infty ,1)\cup [2,\infty )$ $\therefore x\in (-\infty ,0]\cup [2,\infty )$ (C) $f(\theta )=\left|\begin{array}{ccc}1& \tan \theta & 1\\ -\tan \theta & 1& \tan \theta \\ -1& -\tan \theta & 1\end{array}\right|$ $$\begin{array}{rl}{R}_1& \to R_1+R_3,\\ f(\theta )& =\left|\begin{array}{ccc}0& 0& 2\\ -\tan \theta & 1& \tan \theta \\ -1& -\tan \theta & 1\end{array}\right|\\ & =2\left({\tan }^2\theta +1\right)=2{\sec }^2\theta \ge 2\\ f(\theta )& \in [2,\infty )\end{array}$$ $$\text{ (D) }\begin{array}{rl}f(x)& =x^{3/2}(3x-10);x\ge 0\\ f^{\prime }(x)& =x^{3/2}\cdot 3+\frac{3}{2}\cdot x^{1/2}(3x-10)\\ & =3x^{1/2}\left\{x+\frac{1}{2}(3x-10)\right\}\end{array}$$ $$\begin{array}{rl}& =\frac{3}{2}{x}^{1/2}\{2x+3x-10\}\\ & =\frac{15}{2}{x}^{1/2}(x-2)\\ & \\ \therefore & x\ge 2\end{array}$$