\documentclass[]{webquiz} \BreadCrumbs{quizindex | title} \DeclareMathOperator{\cis}{cis} \newcommand{\R}{\mathbb R} \newcommand{\C}{\mathbb C} \usepackage{pst-all} \title{Quiz 1: Numbers and sets} \begin{document} \begin{question} Which of the following are correct ways of writing the set \[ A=\{x \in \mathbb{R} \mid -3 < x \leq -1 \text{ or } x\geq 0\} ?\] \begin{choice}[columns=2] \incorrect $(-3,\infty)$ \feedback The interval $(-3,\infty)$ includes the real numbers between $-1$ and $0$, which do not belong to $A$. \incorrect $[-3,\infty)$ \feedback The interval $[-3,\infty)$ includes $-3$, and the real numbers between $-1$ and $0$, which do not belong to $A$. \incorrect $[-3,-1]\cap[0,\infty)$ \feedback The interval $[-3,-1]\cap[0,\infty)$ is the empty set $\emptyset$. As $A$ is not empty (for example, $A$ includes $-1$), this option cannot be correct. \incorrect $(-3,-1)\cup[0,\infty)$ \feedback The interval $-1$ is not in $(-3,-1)\cup[0,\infty)$, but $-1$ is in $A$. \correct $(-3,-1]\cup[0,\infty)$ \end{choice} \end{question} %%%% 2 \begin{question} What is another way of writing the set \[B= \{x \in {\mathbb R}\ |\ |x-3|<2 \}\ \rm{?}\] \begin{choice}[columns=2] \incorrect $(2,3]$ \feedback For example, $4$ belongs to $B$ but is not in $(2,3]$. \incorrect $[2,4]$ \feedback For example, $1.5$ belongs to $B$ but is not in $[2,4]$. \correct $(1,5)$ \feedback $B$ is the set of all points whose distance from 3 on the number line is less than 2. \\ The solution to $|x-3|<2$ is $1<x<5$. \incorrect $[1,5]$ \feedback Neither $1$ nor $5$ belong to $B$, but both $1$ and $5$ belong to $[1,5]$. \incorrect $[2,3)$ \feedback For example, $4$ belongs to $B$ but is not in $[2,3)$. \end{choice} \end{question} %%%%%%%%%% 3 \begin{question} If $A=\{7,8,9,10\}$ and $B=\{5,6,7,8\}$ then $(A\backslash B)\cup(B\backslash A)$ is \begin{choice}[columns=2] \incorrect $\{5,6,7,8,9,10\}$ \correct $\{5,6,9,10\}$ \feedback $A\backslash B=\{9,10\}$ and $B\backslash A=\{5,6\}$ so $(A\backslash B)\cup(B\backslash A)=\{5,6,9,10\}$. \incorrect $\emptyset$, the empty set. \incorrect $\{7,8\}$ \incorrect None of the above. \end{choice} \end{question} %%%%%%%%%%%%%% 4 \begin{question} The set \(\{0,1,\pm\sqrt{-1},\pi,12\}\) is a subset of \begin{choice}[columns=2] \incorrect $\mathbb N$ \feedback The number \(\pi\) is not a natural number. \incorrect $\mathbb Z$ \feedback The number \(\pi\) is not an integer. \incorrect $\mathbb Q$ \feedback The number \(\pi\) is not a rational number. \incorrect $\mathbb R$ \feedback \(\sqrt{-1}\) is not real. \correct $\mathbb C$ \feedback Since \(\pm\sqrt{-1}\) denotes the two imaginary numbers $i$ and $-i$, the given set cannot be in any of the sets $\mathbb{N,Z,Q}\ \rm{or}\ \mathbb{R}$. \\ Hence the right answer must be ${\mathbb C}$ which contains all imaginary numbers. \end{choice} \end{question} %%%%%%%%%% 5 \begin{question} Which of the following alternatives is the best feedback to `Solve $x^{2}-3x+4=0$ over $\mathbb{C}$'. \begin{choice}[columns=2] \incorrect There are no real solutions. \feedback As the question asks us to solve the equation over $\mathbb C$ (that is, to find all solutions belonging to the set of complex numbers), this is not the best feedback. \incorrect $x=1,4$ \incorrect \(x=\dfrac{3\pm\sqrt{7}}{2}\) \correct \(x=\dfrac{3\pm i\sqrt{7}}{2}\) \feedback Using the quadratic formula, \(x =\dfrac{3\pm\sqrt{9-16}}{2} = \dfrac{3\pm\sqrt{-7}}{2}\). \incorrect None of the above is correct. \end{choice} \end{question} %%%%%%% 6 \begin{question} If $z=9+3i$ and $w=2-i$ then $z+w$ equals \begin{choice}[columns=2] \incorrect $9-i$ \correct $11+2i$ \feedback $z+w=(9+3i)+(2-i)=(9+2)+(3-1)i=11+2i$. \incorrect $6+3i$ \incorrect $8$ \incorrect None of the above \end{choice} \end{question} %%%%%%% 7 \begin{question} If $w=2-i$ then $\overline{w}$ equals \begin{choice}[columns=2] \incorrect $2-i$ \incorrect $2$ \correct $2+i$ \feedback $\overline{w}=\overline{2-i}=2+i$. \incorrect $-2+i$ \incorrect None of the above \end{choice} \end{question} %%%%%%%%% 8 \begin{question} If $p=9+3i$ and $q=2-i$ then $p\overline{q}$ equals \begin{choice}[columns=2] \correct $15+15i$ \feedback $p\overline{q}=(9+3i)\overline{(2-i)}$ \\ $=(9+3i)(2+i)= (18-3)+(6+9)i$\\ $=15+15i$. \incorrect $21+15i$ \incorrect $18+3i$ \incorrect $1-i$ \incorrect None of the above \end{choice} \end{question} %%%%%%%%% 9 \begin{question} If $z=9+3i$ and $w=2-i$ then $\dfrac{z}{w}$ equals \begin{choice}[columns=2] \incorrect $15+15i$ \incorrect $6+3i$ \incorrect $12+15i$ \incorrect $3-3i$ \correct None of the above \feedback \(\dfrac{z}{w}=\dfrac{9+3i}{2-i} =\dfrac{9+3i}{2-i}\times \dfrac{2+i}{2+i} =\dfrac{15+15i}{5}=3+3i\). \end{choice} \end{question} %%%%%%%%%% 10 \begin{question} The shaded region in the graph \begin{center}\begin{pspicture}(-3,-1.5)(3,4) \pscircle[linewidth=2pt,linestyle=dashed,fillcolor=blue,fillstyle=solid](1,1){2} \psaxes[linecolor=red,linewidth=1pt,labels=none]{->}(0,0)(-1.5,-1.5)(3.5,3.5) \rput(3.75,0){$x$} \rput(0,3.85){$iy$} \rput(3,-0.4){3} \rput(-0.4,3){3$i$} \psdots(1,1) \end{pspicture} \end{center} corresponds to which set of complex numbers? \begin{choice}[columns=2] \correct\(\{z \in \C : |z-(i+1)|<2\}\) \incorrect \(\{z \in \C : |z|-|1+i|<2\}\) \feedback This set corresponds to the interior of a circle, centre the origin, radius $2+\sqrt 2$. \incorrect \(\{z \in \C : \text{Re}(z+(i+1))<2 \}\) \feedback This set corresponds to the open half plane containing all complex numbers $z=x+iy$ with $x<1$. \incorrect \(\{z \in \C : |z-2|<|i+1-2|\}\) \feedback This set corresponds to the interior of a circle, centre $2$, radius $\sqrt 2$. \incorrect None of the above. \end{choice} \end{question} \end{document}