Match the statements given in column I with the intervals/union of intervals given in column II.

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Professor PC Thomas Classes · Answer

(A)-p; (B)-t; (C)-r; (D)-r\ (A) $\operatorname{Re}\left(\frac{2 i(x+i y)}{1-\left(x^{2}-y^{2}+2 i x yight)}ight)$\ $=\operatorname{Re}\left(\frac{-2 y+2 i x}{2 y(y-i x)}ight)=\operatorname{Re}\left(-\frac{1}{y}ight)=-\frac{1}{y}$ As $-1 \leq y \leq 1$, we have the solution set given by $(-\infty,-1) \cup(1, \infty)$\ (B) To find domain we have\ $-1 \leq \frac{8 \cdot 3^{x-2}}{1-3^{2 x-2}} \leq 1$\ Let $u=3^{x}$ to obtain $-1 \leq \frac{8 u}{9-u^{2}} \leq 1$\ $|8 u| \leq\left(9-u^{2}ight)$\ $\Rightarrow \quad\left(u^{2}+8 u-9ight)\left(u^{2}-8 u-9ight) \geq 0$\ $\Rightarrow \quad(u+9)(u-1)(u-9)(u+1) \geq 0$\ $\Rightarrow \quad(u-1)(u-9) \geq 0$\ As $u$ is always + ve $\Rightarrow u \leq 1$ or $u \geq 9$\ $3^{x} \leq 1$, or $3^{x} \geq 9 \Rightarrow 3^{x} \leq 3^{0}$, or $3^{x} \geq 3^{2}$\ i.e., $x \leq 0$, or $x \geq 2$\ $\Rightarrow x \in(-\infty, 0] \cup[2, \infty)$\ (C) $f(	heta)=\left|\begin{array}{ccc}1 & 	an 	heta & 1 \ -	an 	heta & 1 & 	an 	heta \ -1 & -	an 	heta & 1\end{array}ight|$\ $=\left|\begin{array}{ccc}0 & 0 & 2 \ -	an 	heta & 1 & 	an 	heta \ -1 & -	an 	heta & 1\end{array}ight|=2 \sec ^{2} 	heta$,\ Applying $R_{1} ightarrow R_{1}+R_{3}$.\ (D) $f(x)=x^{3 / 2}(3 x-10), x \geq 0$ $$ \begin{aligned} f^{\prime}(x) & =x^{3 / 2} \cdot 3+\frac{3}{2} x^{1 / 2}(3 x-10) \ & =x^{1 / 2}\left\{3 x+\frac{3}{2}(3 x-10)ight\} \ & =3 x^{1 / 2}\left\{x+\frac{3 x}{2}-5ight\}=3 x^{1 / 2}\left\{\frac{5 x}{2}-5ight\} \ & =\frac{15}{2} x^{1 / 2}(x-2) \end{aligned} $$ $f^{\prime}(x) \geq 0 \Rightarrow x \geq 2$\