// Numbas version: finer_feedback_settings {"name": "2hrs 2024-25 WM175 Assessment 1", "metadata": {"description": "
Topics:
Algebra, Calculus, Trigonometeric Equations, Complex numbers & Partial Derivatives
Students must complete the exam within 120 mins (standard time).
Questions have variables to produce randomised questions.
Some quadratics are to be solved by factorising
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\nHence solve $\\simplify{{a[0]*c[0]}x^2+{a[0]*d[0]+b[0]*c[0]}x+ {b[0]*d[0]}}=0$.
\n[[1]]
\nUse fractions or integers for your answers.
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\nHence solve $\\simplify{{a[1]*c[1]}x^2+{a[1]*d[1]+b[1]*c[1]}x+ {b[1]*d[1]}}=0$.
\n[[1]]
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", "templateType": "anything", "can_override": false}, "k": {"name": "k", "group": "part a", "definition": "random(1..6)", "description": "Constant part of the LHS of the first equation in part a
", "templateType": "anything", "can_override": false}, "m": {"name": "m", "group": "part a", "definition": "h*x1 + k*y1", "description": "RHS of the first equation in part a
", "templateType": "anything", "can_override": false}, "d": {"name": "d", "group": "Part b", "definition": "random(a+1..9 except map(j*a,j,0..10/a))", "description": "$x$ coefficient in the second equation of part b. Never an integer multiple of the $x$ coefficient in the first equation.
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", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Part b", "definition": "a*x2+b*y2", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Part b", "definition": "random(a+1..7)", "description": "Coefficient of $y$ in the first equation of part b.
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\\simplify{{h}x+{k}y} &= \\var{m} \\text{,} \\\\
\\simplify{{j}x+{l}y} &= \\var{n} \\text{.}
\\end{align}
$x =$ [[0]]
\n$y =$ [[1]]
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Find the derivative of $y$ using the function of a function rule (also known as the chain rule).
\nNote, you should check that the preview of your answer shown is correct before submitting.
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\n\n$\\displaystyle{u}=$ [[0]]
\n$\\displaystyle \\frac{\\mathrm{d}u}{\\mathrm{d}x}=$ [[1]]
\n$\\displaystyle \\frac{\\mathrm{d}y}{\\mathrm{d}u}=$ [[2]]
\nHence,
\n$\\displaystyle \\frac{\\mathrm{d}y}{\\mathrm{d}x}=$ [[3]]
\n\nFor help, you may wish to click on Show steps - but be aware this will cost you 2 marks.
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Make sure your calculator is in radians for this question.
\n", "advice": "(a) and (b):
\n\\[\\int_\\var{d-2}^\\var{d}\\simplify[all]{{a}*x-sin({b}*x)}\\,dx=\\left[\\simplify{{a}*x^2/2+cos({b}*x)/{b}}\\right]_\\var{d-2}^\\var{d}=\\var{a1}\\]
\n(c) and (d):
\n\\[\\int_\\var{d}^\\var{d+3}\\simplify[all]{{c}/x^{a+1}+{b}*sqrt(x)}\\,dx=\\left[\\simplify{{-c/a}/x^{a}+{2*b/3}*x^(3/2)}\\right]_\\var{d}^\\var{d+3}=\\var{a2}\\]
\n(e) and (f):
\n\\[\\int_{-2}^\\var{a-2}\\simplify[all]{{c+1}*exp(x/{b})-{a}}\\,dx=\\left[\\simplify{{b*(c+1)}*exp(x/{b})-{a}*x}\\right]_{-2}^\\var{a-2}=\\var{a3}\\]
\n(g) and (h):
\n\\[\\int_\\var{b}^\\var{b+a}\\simplify[all]{{c}/({b}*x)+{b}/{a+1}*cos({c}*x)}\\,dx=\\left[\\simplify{{c}/{b}*ln(x)+{b}/{c*(a+1)}*sin({c}*x)}\\right]_\\var{b}^\\var{b+a}=\\var{a4}\\]
\n(i) and (j):
\n\\[\\int_\\var{a}^\\var{a+1}\\simplify[all]{x^{b-1}/{b+2}-{d}+{c}*exp({-a}*x)}\\,dx=\\left[\\simplify{x^{b}/{b*(b+2)}-{d}*x-{c/a}*exp({-a}*x)}\\right]_\\var{a}^\\var{a+1}=\\var{a5}\\]
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the integral of $x^n$ is $x^{n+1} / {n+1}$ , and
the integral of sin$(ax)$ is $-$cos$(ax) / a$
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\n\\[\\int_\\var{d-2}^\\var{d}\\simplify[all]{{a}*x-sin({b}*x)}\\,dx\\]
\n\nRemember to use radians and give your answer correct to 3 significant figures.
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e.g.
$ \\int_1^4\\ {{x}^4}\\,dx = 4^5 / 5 - 1^4/5=204.6=205$ (3 significant figures)
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\n(Note: you should leave your answer as a fraction in terms of $\\pi,$ which can be entered by typing \"pi\" without the quotation marks)
\nAn angle of $\\frac{\\pi}{\\var{radians}}$ radians corresponds to [[1]] degrees.
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\nGive your answers correct to 2 decimal places in ascending order.
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\n$\\sin(\\theta)=$ [[0]]
\n$\\theta=$ [[1]] or [[2]]
\nDiscard any solutions out of range.
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\nGive your final answers correct to the nearest degree in ascending order.
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\n(Discard any solutions not in range.)
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\n\n
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Do not include decimals in your answers, only fractions or integers. Also do not include brackets in your answers.
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$z=\\;\\;$[[0]] + [[1]] i
\\[z=(\\var{z3})^{\\var{-d1}}\\]
$z=\\;\\;$[[0]] + [[1]] i
Make sure that you input the real and imaginary parts as fractions and not as decimals
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\n\nwhen \\[z=i^0\\] $z=\\;\\;$ [[0]]
\nwhen \\[z=i^1\\] $z=\\;\\;$ [[1]]
\nwhen \\[z=i^2\\] $z=\\;\\;$ [[2]]
\nwhen \\[z=i^3\\]$z=\\;\\;$ [[3]]
\nHence, or otherwise calculate
\n\\[z=i^{\\var{n}}\\]
$z=\\;\\;$[[4]]
Finding the modulus and argument (in radians) of four complex numbers; the arguments between $-\\pi$ and $\\pi$ and careful with quadrants!
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\nInput final answers to 3 decimal places.
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\nInput both answers to 3 decimal places.
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Remember that
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"variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Given $\\displaystyle f(z) = \\simplify[std]{z ^ 3 + {( -2) * a1 -c1}*z ^ 2 + {2 * a1 * c1 + a1 ^ 2 + b1 ^ 2} * z -{c1 * (a1 ^ 2 + b1 ^ 2)}}$, one of the following complex numbers is a root $z_1$ of the equation $f(z)=0$.
\nChoose the correct value for $z_1$:[[0]]
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\n$z_2=\\;\\;$[[0]] (enter the complex root here)
\n$z_3=\\;\\;$[[1]] (enter the real root here)
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Consider both methods shown to get the correct answer to the question below.
After looking through both methods, choose which one you prefer and say why, giving a full explanation of your choice.
Remember to compare efficiency, clarity of setting out, the reader's ability to follow the logic used easily in your answer.
Question
Given $\\displaystyle f(z) = \\simplify[std]{z ^ 4+ 4*z ^ 3+ 28 * z^2 +240z+800}$, one of the following complex numbers is a root $z_1$ of the equation $f(z)=0$.
\nChoose the correct value for $z_1$: (i) $1-4i$ or (ii) $-4+2i$
\n
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Please give your full reasoning for which of the methods shown above you prefer.
You should aim to write about 50 words in your answer.
\nThen send it to Shaheen Charlwood via Teams direct chat before the submission time given (usually the end time of the exam).
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "In the following question find the $(x,y)$ coordinates of the single stationary point $(a,b)$ of the function
\n\\[f(x,y)=\\simplify[std]{{a}*x^2+{b}*x*y+{c1}*y^2+{d}*x+{f}*y}\\]
", "advice": "The $(x,y)$ coordinates of the stationary point of a function of 2 variables $f(x,y)$ are given by solving
the following 2 equations for $x$ and $y$:
\\[\\begin{eqnarray*} \\partial f \\over \\partial x &=&0\\\\ \\\\ \\partial f \\over \\partial y &=&0 \\end{eqnarray*} \\]
\nIn this case you get two linear equations to solve for $x$ and $y$:
\n\\[\\begin{eqnarray*} \\simplify[std]{{2*a}x+{b}y+{d}}&=&0\\\\ \\\\ \\simplify[std]{{b}x+{2*c1}y+{f}}&=&0 \\end{eqnarray*} \\]
On solving these we get \\[ x = \\simplify[std]{{2*c1*d-b*f}/{b^2-4*a*c1}},\\;\\;\\;y=\\simplify[std]{{2*a*f-b*d}/{b^2-4*a*c1}}\\]
On substituting these values into $f(x,y)$ we get:
\\[f\\left(\\simplify[std]{{2*c1*d-b*f}/{b^2-4*a*c1}},\\simplify[std]{{2*a*f-b*d}/{b^2-4*a*c1}}\\right) = \\var{rawstatval}\\approx\\var{statval}\\]
to 2 decimal places.
Input both cooordinates as fractions or integers and not decimals.
\n$x$–coordinate, $a=$ [[0]].
\n$y$–coordinate, $b=$ [[1]].
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "\nAnswer the following questions about the function:
\n\\[f(x,y)=\\simplify[std]{ ({a} / 3) * x ^ 3 + ({b} / 2) * x ^ 2 * y + {c} * y ^ 2 * x + {d} * y}\\]
\n ", "advice": "\\[\\begin{eqnarray*} {\\partial f \\over \\partial x} &=&\\simplify[std]{(({a} * (x ^ 2)) + ({b} * x * y) + ({c} * (y ^ 2)))}\\\\ \\\\ \\partial f \\over \\partial y &=&\\simplify[std]{((({b} / 2) * (x ^ 2)) + ({(2 * c)} * x * y) + {d})} \\end{eqnarray*}\\]
\n$(a,b)$ is a stationary point for the function $f(x,y)$ if $f_x=0,\\;\\;f_y=0$,where the partial derivatives are evaluated at $x=a,\\;\\;y=b$.
So you have to make sure that both these partial derivatives are $0$ at the stationary point.
For this example we have from the above equations that:
\\[\\begin{eqnarray*} \\simplify[std]{(({a} * (x ^ 2)) + ({b} * x * y) + ({c} * (y ^ 2)))}&=&0,\\qquad &\\mathbf{(1)}&\\\\ \\\\ \\simplify[std]{((({b} / 2) * (x ^ 2)) + ({(2 * c)} * x * y) + {d})}&=&0, \\qquad &\\mathbf{(2)}& \\end{eqnarray*}\\]
The left hand side of equation (1) can be factorised as:
\n\\[\\simplify[std]{({a1}x+{b1}y)*({c1}x+{d1}y)=0}\\]
\nand so we have:
\\[y=\\simplify[std]{{-a1}/{b1}*x},\\mbox{ or } y= \\simplify[std]{{-c1}/{d1}*x}\\]
Substituting this into equation (2) gives:
\\[\\simplify[std]{{b}/2*x^2-{2c*a1}/{b1}*x^2+{d}}=0 \\Rightarrow \\simplify[std]{{-b*b1+4*c*a1}/{2*b1}*x^2={d}}\\]
Hence $x=\\var{m}\\mbox{ or } x = \\var{-m}$ and the stationary points which are on the list and which you had to choose are:
\\[\\left(\\var{m},\\simplify[std]{-{a1*m}/{b1}}\\right)\\mbox{ and }\\left(\\var{-m},\\simplify[std]{{a1*m}/{b1}}\\right)\\]
{check}
\nSubstituting this into equation (2) gives:
\\[\\simplify[std]{{b}/2*x^2-{2c*c1}/{d1}*x^2+{d}}=0 \\Rightarrow \\simplify[std]{{-b*d1+4*c*c1}/{2*d1}*x^2={d}}\\]
{other}
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$\\displaystyle { \\partial f \\over \\partial x}=$ [[0]]
\n$\\displaystyle {\\partial f \\over \\partial y}=$ [[1]]
", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 2, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "(({a} * (x ^ 2)) + ({b} * x * y) + ({c} * (y ^ 2)))", "answerSimplification": "std", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": true, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}, {"name": "y", "value": ""}]}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 2, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "((({b} / 2) * (x ^ 2)) + ({(2 * c)} * x * y) + {d})", "answerSimplification": "std", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": true, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}, {"name": "y", "value": ""}]}], "sortAnswers": false}, {"type": "m_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "\nTick the two choices which give stationary points for $f(x,y)$.
\nNote that the easiest way to do this question is to substitute the values for $x$ and for $y$ into the expressions for $\\displaystyle {\\partial f \\over \\partial x}$ and $\\displaystyle{\\partial f \\over \\partial y}$ and see if you get $0$ for both.
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the following 2 equations for $x$ and $y$
\\[\\begin{eqnarray*} \\partial f \\over \\partial x &=&0\\\\ \\\\ \\partial f \\over \\partial y &=&0 \\end{eqnarray*} \\]
\nIn this case you get two equations to solve for $x$ and $y$
\n\\[\\begin{eqnarray*} \\simplify[std]{{-2*b}*(x-{c})*e^(-(x-{c})^2-(y-{d})^2)}&=&0\\\\ \\\\ \\simplify[std]{{-2*b}*(y-{d})*e^(-(x-{c})^2-(y-{d})^2)}&=&0 \\end{eqnarray*} \\]
We can cancel off the term $\\simplify[std]{e^(-(x-{c})^2-(y-{d})^2)}$ in both equations as $\\simplify[std]{e^(-(x-{c})^2-(y-{d})^2)} \\neq 0,\\;\\forall x,\\;y$.
On solving these we get \\[ x = \\var{c},\\;\\;\\;y=\\var{d}\\]
\nSo the stationary point is $(\\var{c},\\var{d}) \\in D$.
\nOn substituting these values into $f(x,y)$ we get:
\n\\[f(\\var{c},\\var{d})=\\simplify[std]{{a}+{b}*e^0={a+b}}\\]
\n ", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"stepsPenalty": 0, "prompt": "$x$–coordinate, $a=$ [[0]]
\n$y$–coordinate, $b=$ [[1]]
\nInput the value of $f(x,y)$ at $(a,b)$:
\n$f(a,b)=$ [[2]]
\nIf you want some help, click on Show steps. You will not lose any marks if you do so.
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the following 2 equations for $x$ and $y$
\\[\\begin{eqnarray*}\n \n \\partial f \\over \\partial x &=&0\\\\\n \n \\\\\n \n \\partial f \\over \\partial y &=&0\n \n \\end{eqnarray*}\n \n \\]
\n \n \n \nIn this case you get two equations to solve for $x$ and $y$
\n \n \n ", "marks": 0}], "type": "gapfill"}], "statement": "In the following question find the $(x,y)$ coordinates of the single stationary point $(a,b) \\in D$ of the continuous function $f: D \\rightarrow \\mathbb{R}$:
\n\\[f(x,y) = \\simplify[std]{{a} + {b}*e^(-(x-{c})^2-(y-{d})^2)}\\]
\nwhere \\[D = \\{(x,y): \\simplify[std]{(x-{c})^2+(y-{d})^2}\\} \\le \\var{r}\\]
\nThat is, $D$ is a disk of radius $\\simplify[std]{sqrt({r})}$ and centre $(\\var{c},\\var{d})$.
\nInput both cooordinates as fractions or integers and not decimals.
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\n \t\tAdded tags.
Question appears to be working correctly.
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Find the coordinates of the stationary point for $f: D \\rightarrow \\mathbb{R}$: $f(x,y) = a + be^{-(x-c)^2-(y-d)^2}$, $D$ is a disk centre $(c,d)$.
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You have 5 minutes left.
\nPlease ensure you have submitted all answers and sent your explanation for Question 15 to Shaheen via Teams direct chat now!
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