From: Simon H. <sim...@us...> - 2010-11-09 16:51:47
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Update of /cvsroot/stack/stack-dev/sample_questions/diagnostictests In directory sfp-cvsdas-3.v30.ch3.sourceforge.com:/tmp/cvs-serv25193/sample_questions/diagnostictests Added Files: Tag: item_state_separation FormulaSheet.tex DiagnosticTests.xml ReadMe.txt Log Message: Brought back up to date with big HEAD merge. --- NEW FILE: DiagnosticTests.xml --- <?xml version="1.0" encoding="UTF-8"?> <mathQuiz version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:dcterms="http://purl.org/dc/terms/" xmlns:lom="http://www.imsglobal.org/xsd/imsmd_v1p2"><assessmentItem version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:dcterms="http://purl.org/dc/terms/" xmlns:lom="http://www.imsglobal.org/xsd/imsmd_v1p2"><questionCasValues><questionStem type="CasText"><castext>Expand \(\left(@apbx@\right)^5\). #ans1# <IEfeedback>ans1</IEfeedback> <PRTfeedback>Result</PRTfeedback> <p>How confident are you (as a %)? #con1# <IEfeedback>con1</IEfeedback></p> </castext><forbidFloats>false</forbidFloats><simplify>true</simplify></questionStem><questionVariables type="RawKeyVal"><rawKeyVals>a = 2*rand(6)-5; b_magnitude = rand(4)+2; dummy1 = block(while a = b_magnitude do b_magnitude: rand(4)+2,return(b_magnitude)); b_sign = 2*rand(2)-1; b = b_magnitude * b_sign; bx = b*x; bx2 = bx^2; bx3 = bx2*bx; bx4 = bx3*bx; bx5 = bx4*bx; apbx = a+bx; a5 = a^5; b5 = b^5; five_a4_b = 5*a^4*b; five_a_b4 = 5*a*b^4; ten_a2_b3 = 10*a^2*b^3; ten_a3_b2 = 10*a^3*b^2; expansion = a^5+5*a^4*b*x+10*a^3*b^2*x^2+10*a^2*b^3*x^3+5*a*b^4*x^4+b^5*x^5</rawKeyVals><forbidFloats>false</forbidFloats><simplify>true</simplify></questionVariables><workedSolution type="CasText"><castext>For \(n\) a positive integer then by the binomial theorem \[ \left(a + b\right)^n = a^n + \frac{n!}{\left(n-1\right)!\cdot 1!}a^{n-1}\cdot b + \ldots+\frac{n!}{\left(n-r\right)!\cdot r!}a^{n-r}\cdot b^r + \ldots + b^n\mbox{.} \] In this case \(a=@a@\), \(b=@b@\) and \(n=5\); therefore \[\begin{array}{rcl}\displaystyle \left(@apbx@\right)^5 = @a@^5 + 5\cdot @a@^4 \cdot @b@ \cdot x + [...1545 lines suppressed...] \[ x=\frac{@dg1@ - @bg2@}{@ad@ - @bc@} = \frac{@xnum@}{@xden@} = @x_value@ \] and then substituting \(x=@x_value@\) in the former equation we see \[y = \frac{@dg1@ - @ad@ \times @x@}{@bd@} = \frac{@ynum@}{@bd@} = @y_value@\mbox{.}\]</castext><forbidFloats>false</forbidFloats><simplify>true</simplify></workedSolution><questionNote type="CasText"><castext>\[\begin{array}{rcl}\displaystyle @lhs1@&\displaystyle =&\displaystyle @g1@\mbox{,}\\ @lhs2@&\displaystyle =&\displaystyle @g2@\mbox{,}\\ \displaystyle x & \displaystyle = & \displaystyle @x_value@\mbox{,}\\ \displaystyle y & \displaystyle = & \displaystyle @y_value@\mbox{.} \end{array}\]</castext><forbidFloats>false</forbidFloats><simplify>true</simplify></questionNote></questionCasValues><questionparts><questionpart><name>ans1</name><inputType type="Meta"><selection>Algebraic 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= a*ans1+b*ans2; ansg2 = c*ans1+d*ans2; xy_answer = [ans1,ans2]</feedbackVariables><description type="Meta"><selection></selection><default></default><values/></description><PotentialResponses><PR id="0"><answerTest>AlgEquiv</answerTest><teachersAns>xy_values</teachersAns><studentAns>xy_answer</studentAns><testoptions></testoptions><quietAnsTest></quietAnsTest><true><rawModMark>=</rawModMark><rawMark>1</rawMark><feedback><p>You can check your answer by substituting for \(x\) and \(y\) and confirming that the equations balance.</p></feedback><ansnote>EQN-SIM-TRUE</ansnote><nextPR>-1</nextPR></true><false><rawModMark>=</rawModMark><rawMark>0</rawMark><feedback>Substituting \(x=@ans1@\) and \(y=@ans2@\) \begin{eqnarray*} @lhs1@&=&@ansg1@\mbox{,}\\ @lhs2@&=&@ansg2@\mbox{,} \end{eqnarray*} but substituting in the unique correct values for \(x\) and \(y\) the above equations should evaluate to @g1@ and @g2@ 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--- NEW FILE: FormulaSheet.tex --- \documentclass{article} \usepackage{amsmath,amssymb,latexsym,a4wide,times} \usepackage{epsfig} \usepackage{graphicx} \parindent=0pt \parskip=2mm \columnsep=1cm \newcommand{\R}{{\mathbb R}} \newcommand{\N}{{\mathbb N}} \newcommand{\Z}{{\mathbb Z}} \newcommand{\Q}{{\mathbb Q}} \newcommand{\C}{{\mathbb C}} \renewcommand{\d}{\mathrm{d}} \pagestyle{empty} \begin{document} \begin{center} {\Large\bf Facts and formulae} \end{center} Indices: \[ a^ma^n = a^{m+n},\quad \frac{a^m}{a^n} = a^{m-n},\quad (a^m)^n = a^{mn}\] \[ a^0 = 1,\quad a^{-m} = \frac{1}{a^m},\quad a^{\frac{1}{n}} = \sqrt[n]{a},\quad a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m.\] Logarithms: for any $b>0$, $b \neq 1$: $\log_b(a) = c$, means $a = b^c$. \[\log_b(a) + \log_b(b) = \log_b(ab),\quad \log_b(a) - \log_b(b) = \log_b\left(\frac{a}{b}\right),\quad n\log_b(a) = \log_b\left(a^n\right)\] \[\log_b(1) = 0,\quad \log_b(b) = 1,\quad \log_a(x) = \frac{\log_b(x)}{\log_b(a)}\] Natural logarithms use base $e\approx 2.718$, and are denoted $\log_e$ or alternatively $\ln$. Standard Trigonometric Values \[ \sin(45^\circ)={1\over \sqrt{2}}, \qquad \cos(45^\circ) = {1\over \sqrt{2}},\qquad \tan( 45^\circ)=1 \] \[ \sin (30^\circ)={1\over 2}, \qquad \cos (30^\circ)={\sqrt{3}\over 2},\qquad \tan (30^\circ)={1\over \sqrt{3}}\] \[ \sin (60^\circ)={\sqrt{3}\over 2}, \qquad \cos (60^\circ)={1\over 2},\qquad \tan (60^\circ)={ \sqrt{3}} \] Standard Trigonometric Identities \[\sin(a\pm b)\ = \ \sin(a)\cos(b)\ \pm\ \cos(a)\sin(b)\] \[\cos(a\ \pm\ b)\ = \ \cos(a)\cos(b)\ \mp \sin(a)\sin(b)\] \[\tan (a\ \pm\ b)\ = \ {\tan (a)\ \pm\ \tan (b)\over1\ \mp\ \tan (a)\tan (b)}\] \[2\sin(a)\cos(b)\ = \ \sin(a+b)\ +\ \sin(a-b)\] \[2\cos(a)\cos(b)\ = \ \cos(a-b)\ +\ \cos(a+b)\] \[2\sin(a)\sin(b) \ = \ \cos(a-b)\ -\ \cos(a+b)\] \[\sin^2(a)+\cos^2(a)\ = \ 1\] \[1+{\rm cot}^2(a)\ = \ {\rm cosec}^2(a),\quad \tan^2(a) +1 \ = \ \sec^2(a)\] \[\cos(2a)\ = \ \cos^2(a)-\sin^2(a)\ = \ 2\cos^2(a)-1\ = \ 1-2\sin^2(a)\] \[\sin(2a)\ = \ 2\sin(a)\cos(a)\] \[\sin^2(a) \ = \ {1-\cos (2a)\over 2}, \qquad \cos^2(a)\ = \ {1+\cos(2a)\over 2}\] Hyperbolic Functions \[\cosh(x) = \frac{e^x+e^{-x}}{2}, \qquad \sinh(x)=\frac{e^x-e^{-x}}{2}\] \[\tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{{e^x-e^{-x}}}{e^x+e^{-x}}\] \[{\rm sech}(x) ={1\over \cosh(x)}={2\over {\rm e}^x+{\rm e}^{-x}}, \qquad {\rm cosech}(x)= {1\over \sinh(x)}={2\over {\rm e}^x-{\rm e}^{-x}}\] \[{\rm coth}(x) ={\cosh(x)\over \sinh(x)} = {1\over {\rm tanh}(x)} ={{\rm e}^x+{\rm e}^{-x}\over {\rm e}^x-{\rm e}^{-x}}\] Hyperbolic Identities \[{\rm e}^x=\cosh(x)+\sinh(x), \quad {\rm e}^{-x}=\cosh(x)-\sinh(x)\] \[\cosh^2(x) -\sinh^2(x) = 1$$ $$1-{\rm tanh}^2(x)={\rm sech}^2(x)\] \[{\rm coth}^2(x)-1={\rm cosech}^2(x)$$ $$\sinh(x\pm y)=\sinh(x)\ \cosh(y)\ \pm\ \cosh(x)\ \sinh(y)\] \[\cosh(x\pm y)=\cosh(x)\ \cosh(y)\ \pm\ \sinh(x)\ \sinh(y)\] \[\sinh(2x)=2\,\sinh(x)\cosh(x)\] \[\cosh(2x)=\cosh^2(x)+\sinh^2(x)\] \[\cosh^2(x)={\cosh(2x)+1\over 2}\] \[\sinh^2(x)={\cosh(2x)-1\over 2}\] Inverse Hyperbolic Functions \[\cosh^{-1}(x)=\ln\left(x+\sqrt{x^2-1}\right) \quad \mbox{ for } x\geq 1\] \[ \sinh^{-1}(x)=\ln\left(x+\sqrt{x^2+1}\right)\] \[\tanh^{-1}(x) = \frac{1}{2}\ln\left({1+x\over 1-x}\right) \quad \mbox{ for } -1< x < 1\] Calculus rules The Product Rule: \[\frac{\mathrm{d}}{\mathrm{d}{x}} \big(f(x)g(x)\big) = f(x) \cdot \frac{\mathrm{d} g(x)}{\mathrm{d}{x}} + g(x)\cdot \frac{\mathrm{d} f(x)}{\mathrm{d}{x}}.\] The Quotient Rule: \[\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{g(x)\cdot\frac{df(x)}{dx}\ \ - \ \ f(x)\cdot \frac{dg(x)}{dx}}{g(x)^2}. \] The Chain Rule \[\frac{df(g(x))}{dx} = \frac{dg(x)}{dx}\cdot\frac{df(u)}{du}.\] Integration by Substitution: \[\int f(u){{\rm d}u\over {\rm d}x}{\rm d}x=\int f(u){\rm d}u \quad\hbox{and}\quad \int_a^bf(u){{\rm d}u\over {\rm d}x}\,{\rm d}x = \int_{u(a)}^{u(b)}f(u){\rm d}u.\] Integration by Parts: \[\int_a^b u{{\rm d}v\over {\rm d}x}{\rm d}x=\left[uv\right]_a^b- \int_a^b{{\rm d}u\over {\rm d}x}v\,{\rm d}x.\] \begin{center} \begin{tabular}{ll} $f(x)$ & $f'(x)$\\ $k$, constant & $0$ \\ $x^n$, any constant $n$ & $nx^{n-1}$\\ $e^x$ & $e^x$\\ $\ln(x)=\log_{\rm e}(x)$ & $\frac{1}{x}$ \\ $\sin(x)$ & $\cos(x)$ \\ $\cos(x)$ & $-\sin(x)$ \\ $\tan(x) = \frac{\sin(x)}{\cos(x)}$ & $\sec^2(x)$ \\ $\mbox{cosec}(x)=\frac{1}{\sin(x)}$ & $-\mbox{cosec}(x)\cot(x)$ \\ $\sec(x)=\frac{1}{\cos(x)}$ & $\sec(x)\tan(x)$ \\ $\cot(x)=\frac{\cos(x)}{\sin(x)}$ & $-\mbox{cosec}^2(x)$ \\ $\cosh(x)$ & $\sinh(x)$ \\ $\sinh(x)$ & $\cosh(x)$ \\ $\tanh(x)$ & $\mbox{sech}^2(x)$ \\ $\mbox{sech}(x)$ & $-\mbox{sech}sech(x)\tanh(x)$ \\ $\mbox{cosech}(x)$ & $-\mbox{cosech}(x)\coth(x)$ \\ $coth(x)$ & $-\mbox{cosech}^2(x)$ \\ \end{tabular} \end{center} \[\frac{d}{dx}\left(\sin^{-1}(x)\right) = \frac{1}{\sqrt{1-x^2}},\quad \frac{d}{dx}\left(\cos^{-1}(x)\right) = \frac{-1}{\sqrt{1-x^2}},\quad \frac{d}{dx}\left(\tan^{-1}(x)\right) = \frac{1}{1+x^2} \] \[\frac{d}{dx}\left(\cosh^{-1}(x)\right) = \frac{1}{\sqrt{x^2-1}},\quad \frac{d}{dx}\left(\sinh^{-1}(x)\right) = \frac{1}{\sqrt{x^2+1}},\quad \frac{d}{dx}\left(\tanh^{-1}(x)\right) = \frac{1}{1-x^2}\] \begin{center} \begin{tabular}{lll} $f(x)$ & $\int f(x)\ dx$\\ $e^x$ & $e^x+c$ & \\ $\cos(x)$ & $\sin(x)+c$ & \\ $\sin(x)$ & $-\cos(x)+c$ & \\ $\tan(x)$ & $\ln(\sec(x))+c$ & $-\frac{\pi}{2} < x < \frac{\pi}{2}$\\ $\sec x$ & $\ln (\sec(x)+\tan(x))+c$ & $-{\pi\over 2}< x < {\pi\over 2}$\\ cosec$\, x$ & $\ln ($cosec$(x)-\cot(x))+c$ & $0 < x < \pi$\\ cot$\,x$ & $\ln(\sin(x))+c$ & $0< x< \pi$ \\ $\cosh(x)$ & $\sinh(x)+c$ & \\ $\sinh(x)$ & $\cosh(x) + c$ & \\ $\tanh(x)$ & $\ln(\cosh(x))+c$ & \\ coth$(x)$ & $\ln(\sinh(x))+c $ & $x>0$\\ ${1\over x^2+a^2}$ & ${1\over a}\tan^{-1}{x\over a}+c$ & $a>0$\\ [2pt] ${1\over x^2-a^2}$ & ${1\over 2a}\ln{x-a\over x+a}+c$ & $|x|>a>0$\\ [2pt] ${1\over a^2-x^2}$ & ${1\over 2a}\ln{a+x\over a-x}+c$ & $|x|<a$\\ [3pt] ${1\over \sqrt{x^2+a^2}}$ & $\sinh^{-1}\left(\frac{x}{a}\right) + c$ & $a>0$ \\ ${1\over \sqrt{x^2-a^2}}$ & $\cosh^{-1}\left(\frac{x}{a}\right) + c$ & $x\geq a > 0$ \\ ${1\over \sqrt{x^2+k}}$ & $\ln (x+\sqrt{x^2+k})+c$ & \\ ${1\over \sqrt{a^2-x^2}}$ & $\sin^{-1}\left(\frac{x}{a}\right)+c$ & $-a\leq x\leq a$ \end{tabular} \end{center} \vfill {\scriptsize C J Sangwin, \verb$C.J...@bh...$, \today. This formula sheet is released under Creative Commons Attribution-Share Alike.\\ \includegraphics[width=1.5cm]{88x31.png}} \end{document} --- NEW FILE: ReadMe.txt --- Stimulating Techniques in Entry-level Mathematics with the STACK CAA system In April 2010 the National HE STEM Programme (http://www.stemprogramme.com) funded a mini-project stimulating Techniques in Entry-level Mathematics (STEM) with the STACK computer aided assessment (CAA) system}. The aims of this project are to take existing diagnostic tests in core mathematics and develop similar automatic tests for the STACK computer aided assessment system. The outcome of these tests will be a user profile which links outcomes to existing online learning materials. See http://web.mat.bham.ac.uk/C.J.Sangwin/projects/2010STEM/ The questions in this folder are the results of this project. These questions files are released under Creative Commons Attribution-Share Alike. http://creativecommons.org/licenses/by-sa/2.0/uk/ |