\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.6in,right=0.7in,top=0.7in,bottom=0.7in} \pagestyle{fancy} \lhead{\large Exam 2} \cfoot{} \chead{} \rhead{\large February 1985 - page \ \thepage} \textwidth 7.0in \textheight 9.2in \setlength{\headheight}{20pt} \begin{document} \begin{enumerate} \item If \ $y=2x$ \ and \ $z=2y$, \ then \ $x+y+z$ \ equals: A) \ $x$ \ \ \ \ \ \ \ \ B) \ $3x$ \ \ \ \ \ \ \ \ C) \ $5x$ \ \ \ \ \ \ \ \ \ D) \ $7x% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ \ \ \ \ \ \ \ \ E) \ $9x$ Answer: $x+y+z=x+2x+2\left( 2x\right) =\allowbreak 7x$ \item The diagonal of square I is $a+b$. \ The perimeter of square II with twice the area of square I is: A) \ $\left( a+b\right) ^{2}$ \ \ \ \ \ \ \ \ B) \ $\sqrt{2}\left( a+b\right) ^{2}$ \ \ \ \ \ \ \ \ C) \ $2\left( a+b\right) $ \ \ \ \ \ \ \ \ \ D) \ $\sqrt{8}\left( a+b\right) $ \ \ \ \ \ \ \ \ E) \ $4\left( a+b\right) %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ Answer: \ Square I has sides $\dfrac{a+b}{\sqrt{2}}$. \ Square II then has sides $a+b,$ and so its perimeter is $4\left( a+b\right) $. \item What is the $y-$intercept of the line passing through $\left( \sqrt{2}% ,1\right) $ \ and \ $\left( -2,2\right) $? A) \ $1$ \ \ \ \ \ \ \ \ B) \ $1.5$ \ \ \ \ \ \ \ \ C) \ $\sqrt{2}% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ \ \ \ \ \ \ \ \ \ D) \ $\sqrt{3}$ \ \ \ \ \ \ \ \ E) \ $5-1$ Answer: \ Let \ $m$ \ denote the slope and $b$ the $y-$intercept. \ We obtain a system substituting the points given into \ $y=mx+b.$ \begin{equation*} \left\{ \begin{array}{cc} 1=\sqrt{2}m+b & \text{ \ multiply by }\sqrt{2} \\ 2=-2m+b & \end{array}% \right. ~~~~\Longrightarrow ~~~~\left\{ \begin{array}{c} \sqrt{2}=2m+\sqrt{2}b \\ 2=-2m+b% \end{array}% \right. \end{equation*}% We add the two equations and obtain $\sqrt{2}+2=b\left( 1+\sqrt{2}\right) .$ \ Thus \ $b=\dfrac{\sqrt{2}+2}{1+\sqrt{2}}=\dfrac{\sqrt{2}\left( 1+\sqrt{2}% \right) }{1+\sqrt{2}}=\sqrt{2}$ \item For $x>0,$ \ $y>0,$ \ and \ $b>0$ \ where \ $b\not=1$ \ and \ $% y\not=1, $ \ let \ $\log _{b}x=m$ \ and \ $\log _{b}y=n$. \ Three of the following statements are true. \ Which one is, in general, not true? A) \ $\log _{b}xy=nm% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $\ \ \ \ \ \ \ \ \ \ B) \ $\log _{b}x^{p}=pm$ \ \ \ \ \ \ \ \ \ \ C) \ $\log _{y}x=\dfrac{m}{n}$\ \ \ \ \ \ \ \ \ \ \ D) \ $\log _{b}\dfrac{x}{y}=m-n$ \ Answer: $\log _{b}xy=\log _{b}x+\log _{b}y=n+m$ and so A is clearly false. \ \ For example, $b=x=y=5$ \ will result in $n=m=\log _{5}5=\allowbreak 1$ \ and \ $\log _{b}xy=\log _{5}25=2\not=1=nm$. \item How many real solutions has the equation $\left\vert x^{2}-6x\right\vert =9$? A) \ $0$ \ \ \ \ \ \ \ \ \ B) \ $1$ \ \ \ \ \ \ \ \ \ \ C) \ $2$ \ \ \ \ \ \ \ \ \ \ \ \ D) \ $3% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ \ \ \ \ \ \ \ \ \ \ \ \ E) \ $4$ Answer: \ the original equation translates to the pair \ $x^{2}-6x=9$ \ and \ \ $x^{2}-6x=-9$ \ The first equation has two real solutions, $3\pm 3\sqrt{2% }$ \ the second one has one, $3$. \ These are clearly different. \item Find the area of the triangle with vertices $\left( -3,1\right) ,$ \ $% \left( 1,2\right) ,$ \ and \ $\left( 2,-1\right) $. 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3;lineAttributes "Solid";curveColor "[flat::RGB:0x00008000]";curveStyle "Line";valid_file "T";tempfilename 'MEB7TW00.wmf';tempfile-properties "XPR";}} \item Find the positive number $x$ \ for which \ $\sqrt{x}=\sqrt[3]{y}$ \ and \ $\sqrt{y}=8$. A) \ $2$ \ \ \ \ \ \ \ \ \ B) \ $2\sqrt{2}$ \ \ \ \ \ \ \ \ \ \ C) \ $4$ \ \ \ \ \ \ \ \ \ \ \ \ D) \ $16% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ \ \ \ \ \ \ \ \ \ \ \ \ E) \ $64$ Answer: \ $\sqrt{y}=8~~\Longrightarrow ~~y=64~~\Longrightarrow ~~\sqrt[3]{y}% =4~~\Longrightarrow ~~\sqrt{x}=4~~\Longrightarrow ~~x=16$ \item When the three-digit numbers \ $6a3$ \ and \ $2b5$ \ are added together, the answer is a number divisible by $9$. \ The largest value of $% a+b$ \ is: A) \ $2$ \ \ \ \ \ \ \ \ \ B) \ $9$ \ \ \ \ \ \ \ \ \ \ C) \ $11% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ \ \ \ \ \ \ \ \ \ \ \ \ D) \ $17$ \ \ \ \ \ \ \ \ \ \ \ \ E) \ $20$ Answer: \ $6a3+2b5$ is $\ $divisible by $9\ ~~\Longrightarrow ~~6+a+3+2+b+5=\allowbreak a+b+16$ \ is divisible by $9~~\Longrightarrow ~~a+b+7$ \ is divisible by $9$. \ Since $a$ and $b$ are digits, their sum is at most $18$. \ The possible values are $2$ \ and $11.$ \item Find the real part of $\dfrac{i}{1+\dfrac{i}{1+\dfrac{i}{1+i}}}$ A) \ $-\dfrac{1}{4}$ \ \ \ \ \ \ \ \ \ B) \ $\dfrac{1}{3}% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ \ \ \ \ \ \ \ \ \ \ C) \ $\dfrac{4}{5}$ \ \ \ \ \ \ \ \ \ \ \ \ D) \ $% \dfrac{2}{3}$ \ \ \ \ \ \ \ \ \ \ \ \ E) \ none of these Answer:% \begin{eqnarray*} \dfrac{i}{1+\dfrac{i}{1+\dfrac{i}{1+i}}} &=&\dfrac{i}{1+\dfrac{i}{\dfrac{1+2i% }{1+i}}}=\dfrac{i}{1+i\cdot \dfrac{1+i}{1+2i}}=\dfrac{i}{1+\dfrac{i-1}{1+2i}}% =\dfrac{i}{\dfrac{1+2i+i-1}{1+2i}} \\ &=&\dfrac{i}{\dfrac{3i}{1+2i}}=i\cdot \dfrac{1+2i}{3i}=\dfrac{1+2i}{3}=% \dfrac{1}{3}+\dfrac{2}{3}i \end{eqnarray*} \item The solution of the equation $\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-% \sqrt{x-1}}=3$ \ is: A) \ $3$ \ \ \ \ \ \ \ \ \ B) \ $\dfrac{3}{5}$ \ \ \ \ \ \ \ \ \ \ C) \ $% \dfrac{4}{5}$ \ \ \ \ \ \ \ \ \ \ \ \ D) \ $\dfrac{5}{4}$ \ \ \ \ \ \ \ \ \ \ \ \ E) \ $\dfrac{5}{3}% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ Answer: \ \begin{eqnarray*} \dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} &=&3 \\ \sqrt{x+1}+\sqrt{x-1} &=&3\sqrt{x+1}-3\sqrt{x-1} \\ 4\sqrt{x-1} &=&2\sqrt{x+1} \\ 2 &=&\sqrt{\dfrac{x+1}{x-1}} \\ 4 &=&\dfrac{x+1}{x-1} \\ 4x-4 &=&x+1 \\ 3x &=&5 \\ x &=&\dfrac{5}{3} \end{eqnarray*}% \pagebreak \item Which of the following describes the asymptotes for the hyperbola \ $% \dfrac{\left( x-8\right) ^{2}}{16}-\dfrac{\left( y-3\right) ^{2}}{4}=1$ A) \ $y-3=\pm \dfrac{1}{2}\left( x-8\right) %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ \ \ \ \ \ \ \ \ \ B) \ $y-1=\pm \dfrac{3}{8}\left( x-2\right) $ \ \ \ \ \ \ \ \ \ \ C) \ $y+2=\pm \dfrac{3}{8}\left( x-4\right) $ \ \ \ \ \ \ \ \ \ \ \ \ D) \ $y-1=\pm \dfrac{1}{2}\left( x-2\right) $ \ \ \ \ \ \ \ \ \ \ \ \ E) \ $% y+2=\pm \dfrac{1}{4}\left( x-4\right) $ Answer: We solve for $y$ and get \ $y-3=\pm \dfrac{1}{2}\sqrt{x^{2}-16x+48}$ \ which, for large values of $x$ and $y,$ is close to $y-3=\pm \dfrac{1}{2}% \left( x-8\right) $ \item $\sin y+\sin \left( x-y\right) =\sin x$ \ for all \ $y$ \ provided that $x$ is: A) \ $60^{\circ }$ \ \ \ \ \ \ \ \ \ B) \ $90^{\circ }$ \ \ \ \ \ \ \ \ \ \ C) \ $180^{\circ }$ \ \ \ \ \ \ \ \ \ \ \ \ D) \ $270^{\circ }$ \ \ \ \ \ \ \ \ \ \ \ \ E) \ $360^{\circ }% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ Answer: \ \begin{eqnarray*} \sin y+\sin \left( x-y\right) &=&\sin x \\ \sin \left( x-y\right) &=&\sin x-\sin y \end{eqnarray*}% The statement is clearly true for $x=360^{\circ }$. \item In the figure $\overline{AB}=\overline{AC},$ \ angle $BAD=30^{\circ }$% , \ and \ $\overline{AE}=\overline{AD}.$ \ Then $x$ equals:\FRAME{dtbpF}{% 2.1188in}{1.1in}{0pt}{}{}{insert1.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.1188in;height 1.1in;depth 0pt;original-width 2.0799in;original-height 1.0672in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'insert1.bmp';file-properties "XNPEU";}} A) \ $7\dfrac{1}{2}^{\circ }$ \ \ \ \ \ \ \ \ \ B) \ $10^{\circ }$ \ \ \ \ \ \ \ \ \ \ C) \ $12\dfrac{1}{2}^{\circ }$ \ \ \ \ \ \ \ \ \ \ \ \ D) \ $% 15^{\circ }$ \ \ \ \ \ \ \ \ \ \ \ \ E) \ $20^{\circ }$ \item Of the following, which fraction is an integer multiple of each of the fractions $\dfrac{6}{7}$, \ $\dfrac{5}{14}$, $\dfrac{10}{21}$ ? A) \ $\dfrac{7}{30}$ \ \ \ \ \ \ \ \ \ B) \ $\dfrac{7}{15}$ \ \ \ \ \ \ \ \ \ \ C) \ $\dfrac{15}{7}$ \ \ \ \ \ \ \ \ \ \ \ \ D) \ $\dfrac{30}{7}% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ \ \ \ \ \ \ \ \ \ \ \ \ E) \ $\dfrac{80}{21}$ Answer: \ an integer multiple of a fraction can only 'lose' factors from its denominator, never gain. \ Thus we only need to consider C and D. \ Since $6$ is not a factor of $15,$ \ we also rule out C. \item Fifteen billiard balls are lying on a table in such a way that they are just squeezed inside an equilateral triangular frame whose inside perimeter is $876$. \ The radius of a billiard ball is: A) \ $\dfrac{73}{2}$ \ \ \ \ \ \ \ \ \ B) \ $\dfrac{146}{4+\sqrt{3}}$ \ \ \ \ \ \ \ \ \ \ C) \ $\dfrac{146}{2+\sqrt{3}}$ \ \ \ \ \ \ \ \ \ \ \ \ D) \ $% \dfrac{146}{3+\sqrt{3}}% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ \ \ \ \ \ \ \ \ \ \ \ \ E) \ none of these \item The least positive integer which has reminders $1$, $1$, and $5$ when divided by $3$, $5$, and $7$ respectively, is: A) \ $166$ \ \ \ \ \ \ \ \ \ B) \ $151$ \ \ \ \ \ \ \ \ \ \ C) \ $145$ \ \ \ \ \ \ \ \ \ \ \ \ D) \ $131$ \ \ \ \ \ \ \ \ \ \ \ \ E) \ none of these$% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ Answer: \ reminders $1,$ $1$ when divided by $3$ and $5$ translate to reminder $1$ when divided by $15.$ \ Among the first seven such positive numbers, we will find the one with reminder $5$ when divided by $7$. \ We consider \ $k\cdot 15+1,$ \ where $k=0,1,2,3,4,5,6$. \ We obtain $% 1,16,31,46,61,76$. \ The reminders when divided by $7$ are $1,2,3,4,5,6$ and so $61$ is the smallest such number. \item Each valve $A,$ $B,$ and $C,$ when open, releases water into a tank at its own constant rate. \ With all three valves open, the tank fills in one hour, with only valves $A$ and $C$ open it takes $1$ hour and $20$ minutes, and with only valves $B$ and $C$ open it takes $2$ hours. \ The time it takes to fill the tank with only valves $A$ and $B$ open is: A) \ $\dfrac{2}{3}$ hr \ \ \ \ \ \ \ \ \ B) \ $\dfrac{4}{3}$ hr$% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ \ \ \ \ \ \ \ \ \ \ C) \ $\dfrac{3}{2}$ hr \ \ \ \ \ \ \ \ \ \ \ \ D) \ $2$ hr \ \ \ \ \ \ \ \ \ \ \ \ E) \ $\dfrac{9}{4}$ \ hr Answer: Let $a$ denote the time it takes to fill the tank with valve $A$ open only. \ Similarly, let $b$ $\ $and$\ c$ \ denote the time valve $B$ and \ $C$ \ fills the tank. \ Then% \begin{eqnarray*} \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} &=&1 \\ \dfrac{4}{3}\left( \dfrac{1}{a}+\dfrac{1}{c}\right) &=&1 \\ 2\left( \dfrac{1}{b}+\dfrac{1}{c}\right) &=&1 \end{eqnarray*}% \begin{eqnarray*} \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} &=&1 \\ \dfrac{1}{a}+\dfrac{1}{c} &=&\dfrac{3}{4}~~~~\Longrightarrow \dfrac{1}{b}=% \dfrac{1}{4} \\ \dfrac{1}{b}+\dfrac{1}{c} &=&\dfrac{1}{2}~~~~\Longrightarrow \dfrac{1}{a}=% \dfrac{1}{2} \end{eqnarray*} \begin{equation*} \dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{4}\text{ \ \ and so }\dfrac{4}{3}\left( \dfrac{1}{a}+\dfrac{1}{b}\right) =1 \end{equation*} \item In how many different arrangements can a careless office boy place $5$ letters into $5$ mailboxes so that no one gets the right letter? A) \ $32$ \ \ \ \ \ \ \ \ \ B) \ $44% %TCIMACRO{\TeXButton{correct answer}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion $ \ \ \ \ \ \ \ \ \ \ C) \ $60$ \ \ \ \ \ \ \ \ \ \ \ \ D) \ $120$ \ \ \ \ \ \ \ \ \ \ \ \ E) \ $225$ Answer: \ 5-cycles: $4!=\allowbreak 24$ \ 3+2: $\dbinom{5}{3}\left( 2\right) =\allowbreak 20$ \item In the given diagram, points $B$, $C$, and $T$ \ are on the circle, and $AT\ \ $is \ tangent to the circle at $T$ \ \ If \ $AB=3$ \ and \ $BC=4$% , \ find $\dfrac{AB+AT}{AT+AC}$.\FRAME{dtbpF}{1.6561in}{0.9997in}{0pt}{}{}{% insert2.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.6561in;height 0.9997in;depth 0pt;original-width 1.6198in;original-height 0.9669in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'insert2.bmp';file-properties "XNPEU";}} Answer: \ $AT=\sqrt{12}$ and so \ $\dfrac{3+\sqrt{12}}{\sqrt{12}+7}=\dfrac{2% \sqrt{3}+3}{2\sqrt{3}+7}=\dfrac{8\sqrt{3}+9}{37}$ \item Find the limiting value of $\dfrac{1}{9}+\dfrac{3}{27}+\dfrac{5}{81}+% \dfrac{7}{243}+...+\dfrac{2k-1}{3^{k+1}}$ $\dfrac{1}{3}$ $\left( \dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...\right) +\left( {}\right) $ \end{enumerate} \end{document}