// Numbas version: exam_results_page_options {"question_groups": [{"pickQuestions": 1, "name": "Group", "pickingStrategy": "all-ordered", "questions": [{"name": "Differentiation: coordinates of stationary points from a graph", "extensions": ["geogebra", "jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Lovkush Agarwal", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1358/"}], "functions": {"plotgraph": {"language": "javascript", "parameters": [["a", "number"], ["b", "number"], ["c", "number"], ["d", "number"]], "definition": "// This functions plots a cubic with coefficients a,b,c,d\n// It creates the board, sets it up, then returns an\n// HTML div tag containing the board.\n\n\n// Max and min x and y values for the axis.\nvar x_min = -6;\nvar x_max = 6;\nvar y_min = -10;\nvar y_max = 10;\n\n\n// First, make the JSXGraph board.\nvar div = Numbas.extensions.jsxgraph.makeBoard(\n '500px',\n '600px',\n {\n boundingBox: [x_min,y_max,x_max,y_min],\n axis: false,\n showNavigation: true,\n grid: true\n }\n);\n\n\n\n\n// div.board is the object created by JSXGraph, which you use to \n// manipulate elements\nvar board = div.board; \n\n// create the x-axis.\nvar xaxis = board.create('line',[[0,0],[1,0]], { strokeColor: 'black', fixed: true});\nvar xticks = board.create('ticks',[xaxis,1],{\n drawLabels: true,\n label: {offset: [-4, -10]},\n minorTicks: 0\n});\n\n// create the y-axis\nvar yaxis = board.create('line',[[0,0],[0,1]], { strokeColor: 'black', fixed: true });\nvar yticks = board.create('ticks',[yaxis,1],{\ndrawLabels: true,\nlabel: {offset: [-20, 0]},\nminorTicks: 0\n});\n\n\n\n\n// Plot the function.\n board.create('functiongraph',\n [function(x){ return a*x*x*x+b*x*x+c*x + d;},x_min,x_max]);\n\n\n\n\nreturn div;", "type": "html"}}, "variablesTest": {"condition": "max(abs(ymin),abs(ymax))<10", "maxRuns": 100}, "metadata": {"description": "

A cubic with a maximum and minimum point is given. Question is to determine coordinates of the minimum and maximum point. Non-calculator. Advice is given.

{plotgraph(a,b,c,d)}

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Above is the graph of some function $f$.

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What are the coordinates of its maximum point? ([[0]],[[1]])

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What are the coordinates of its minimum point? ([[2]],[[3]])

", "showFeedbackIcon": true, "scripts": {}, "showCorrectAnswer": true, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "variableReplacementStrategy": "originalfirst"}], "tags": [], "advice": "

(i) A maximum point is a point where regardless if you move right or left, the height will decrease.  A visual analogy would be a hill: if you're at the top of a hill, no matter which direction you go your height will decrease.  So you're looking for a part of the graph which is 'like a hill', and in this graph the point is at $(\\var{xmax}, \\var{ymax})$.

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(ii) A minimum point is the opposite of a maximum point (or an upside-down version of a maximum point, if you like).  The analogy in this case would be a valley: no matter which direction you go your height will increase.  In this graph, the minimum point is at $(\\var{xmin}, \\var{ymin})$.

", "statement": "

Finding stationary points on a graph.

", "type": "question"}, {"name": "Differentiation: stationary points of a cubic", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Lovkush Agarwal", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1358/"}], "advice": "

A stationary point is where the gradient is zero, i.e. when $f'$ is equal to zero.  So first we determine $f'$ by differentiating.

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$f'(x) = \\simplify{3{a}x^2 + 2{b}x + {c}}$.

\n

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When we are at a stationary point, $f'(x)$ has to equal 0, so we get the quadratic equation:

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$0 = f'(x) = \\simplify{3{a}x^2 + 2{b}x + {c}}$.

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To solve this, we can factorise the quadratic to get:

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$0 = (\\simplify{3x-{r1}}) (\\simplify{{a}x-{r2}})$.

\n

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So the $x$-coordinates are $\\var{x1}$ and $\\var{x2}$.

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To find the $y$-coordinates, we plug these $x$-coordinates into the original function $f$ to get $\\var{y1}$ and $\\var{y2}$.

\n

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After rounding appropriately, we get the coordinates.

\n

\n

\n

(Note, this answer may not be the most efficient.  In some versions of the randomised question, you can divide the whole equation by a number before you factorise the quadratic.  For example, if the quadratic equation is $0 = 2x^2 + 6x + 4$, so you would divide everything by 2 to get: $0 = x^2 + 3x + 2$,  so then $0=(x+2)(x+1)$.)

Let $f(x)=\\simplify[all,!collectNumbers,!noleadingminus]{{a}x^3+{b}x^2+{c}x+{d}}$. Find the coordinates of the stationary points and determine whether they are a minimum or a maximum point.

\n

Give all answers to three significant figures.

\n

\n

Stationary Point 1:

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The coordinates are ([[0]],[[1]]) and it is a [[2]] point.

\n

\n

Stationary Point 2:

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The coordinates are ([[3]],[[4]]) and it is a [[5]] point.

\n

\n

(So the computer marks your answer correctly, make sure that Stationary Point 1 has a smaller $x$ coordinate than Stationary Point 2).

"}], "metadata": {"licence": "None specified", "description": "

A cubic is given which has two stationary points.  Students are asked to determine the coordinates and the type of each stationary point. Calculator question. Detailed advice is included.

"}, "statement": "

Determine the coordinates and types of stationary points of a function.

", "type": "question"}, {"name": "Differentiation: chain rule, inside is at+b, advice included", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Lovkush Agarwal", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1358/"}], "functions": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "metadata": {"description": "

Differentiating eight functions using the chain rule, where the 'inside' is of the form at+b. Non-calculator. Advice included

Differentiate the following with respect to $t$.

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(i) $\\sin(\\simplify{{a1}*t+{b1}})$ [[0]]

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(ii) $\\var{c2} \\sin(\\simplify{{a2}*t+{b2}})$ [[1]]

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(iii) $\\cos(\\simplify{{a3}*t+{b3}})$ [[2]]

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(iv) $\\var{c4}\\cos(\\simplify{{a4}*t+{b4}})$ [[3]]

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(v) $e^{\\simplify{{a5}*t+{b5}}}$ [[4]]

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(vi) $\\var{c6} e^{\\simplify{{a6}*t+{b6}}}$ [[5]]

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(vii) $\\ln(\\simplify{{a7}*t+{b7}})$ [[6]]

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(viii) $\\var{c8} \\ln(\\simplify{{a8}*t+{b8}})$ [[7]]

", "showFeedbackIcon": true, "scripts": {}, "showCorrectAnswer": true, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "variableReplacementStrategy": "originalfirst"}], "tags": [], "advice": "

(Note, these are not 'model' answers, but explanations of how I obtain an answer.)

\n

\n

\n

\n

(i) The inside' is $\\simplify{{a1}*t+{b1}}$ and differentiating this we get $\\simplify{{a1}}$.

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Differentiating the outside' while keeping the inside the same gives $\\simplify{cos({a1}*t+{b1})}$.

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Multiplying these two together gives the answer: $\\simplify{{a1}*cos({a1}*t+{b1})}$.

\n

\n

\n

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(ii) The inside' is $\\simplify{{a2}*t+{b2}}$ and differentiating this we get $\\simplify{{a2}}$.

\n

Differentiating the outside' while keeping the inside the same gives $\\simplify{{c2}*cos({a2}*t+{b2})}$.

\n

Multiplying these two together gives the answer: $\\simplify{{c2}*{a2}*cos({a2}*t+{b2})}$.

\n

\n

\n

\n

(iii) The inside' is $\\simplify{{a3}*t+{b3}}$ and differentiating this we get $\\simplify{{a3}}$.

\n

Differentiating the outside' while keeping the inside the same gives $\\simplify{-sin({a3}*t+{b3})}$.

\n

Multiplying these two together gives the answer: $\\simplify{-{a3}*sin({a3}*t+{b3})}$.

\n

\n

\n

\n

(iv) The inside' is $\\simplify{{a4}*t+{b4}}$ and differentiating this we get $\\simplify{{a4}}$.

\n

Differentiating the outside' while keeping the inside the same gives $\\simplify{-{c4}sin({a4}*t+{b4})}$.

\n

Multiplying these two together gives the answer: $\\simplify{-{c4}{a4}*sin({a4}*t+{b4})}$.

\n

\n

\n

(v) The inside' is $\\simplify{{a5}*t+{b5}}$ and differentiating this we get $\\simplify{{a5}}$.

\n

Differentiating the outside' while keeping the inside the same gives $\\simplify{e^({a5}*t+{b5})}$.

\n

Multiplying these two together gives the answer: $\\simplify{{a5}*e^({a5}*t+{b5})}$.

\n

\n

\n

\n

(vi) The inside' is $\\simplify{{a6}*t+{b6}}$ and differentiating this we get $\\simplify{{a6}}$.

\n

Differentiating the outside' while keeping the inside the same gives $\\simplify{{c6}*e^({a6}*t+{b6})}$.

\n

Multiplying these two together gives the answer: $\\simplify{{c6}*{a6}*e^({a6}*t+{b6})}$.

\n

\n

\n

\n

(vii) The inside' is $\\simplify{{a7}*t+{b7}}$ and differentiating this we get $\\simplify{{a7}}$.

\n

Differentiating the outside' while keeping the inside the same gives $\\simplify{1/({a7}*t+{b7})}$.

\n

Multiplying these two together gives the answer: $\\simplify{{a7}/({a7}*t+{b7})}$.

\n

\n

\n

(vii) The inside' is $\\simplify{{a8}*t+{b8}}$ and differentiating this we get $\\simplify{{a8}}$.

\n

Differentiating the outside' while keeping the inside the same gives $\\simplify{{c8}/({a8}*t+{b8})}$.

\n

Multiplying these two together gives the answer: $\\simplify{{a8}*{c8}/({a8}*t+{b8})}$.

\n

\n

", "statement": "

Differentiation using the Chain Rule

", "type": "question"}, {"name": "Differentiation: product and chain rule, (a+bx)^m e^(nx), factorise answer", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Lovkush Agarwal", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1358/"}], "rulesets": {}, "metadata": {"description": "

Differentiate the function $f(x)=(a + b x)^m e ^ {n x}$ using the product and chain rule. Find $g(x)$ such that $f^{\\prime}(x)= (a + b x)^{m-1} e ^ {n x}g(x)$. Non-calculator. Advice is given.

", "licence": "Creative Commons Attribution 4.0 International"}, "parts": [{"type": "gapfill", "prompt": "

$\\simplify{f(x) = ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$

\n

You are told that $\\simplify{Diff(f,x,1) = ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) * g(x)}$, for a polynomial $g(x)$.

\n

\n

You have to find $g(x)$.

\n

$g(x)=\\;$[[0]]

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\n

$f(x)$ is the product of the two functions $\\simplify{({a} + {b}*x)^{m}}$ and $\\simplify{e ^ ({n} * x)}$, so we need to use the product rule.

\n

\n

Differentiating the first part, keeping the second half the same, gives the term: $\\simplify{{m} *{ b} * ({a} + {b} * x) ^ {m -1}} \\times \\simplify{e ^ ({n} * x)}$.

\n

Note that that we needed the chain rule to do this differentiation.

\n

\n

\n

Differentiating the second part, keeping the first half the same, gives the term: $\\simplify{{n} * e ^ ({n} * x)} \\times \\simplify{({a} + {b}x)^{m}}$.

\n

Again, we needed the chain rule to do this differentiation.

\n

\n

Hence, $\\simplify{Diff(f,x,1) = {m * b} * ({a} + {b} * x) ^ {m -1} * e ^ ({n} * x) + {n} * ({a} + {b} * x) ^ {m} * e ^ ({n} * x)}$.

\n

$= \\simplify{({a} + {b} * x) ^ {m -1} * ({m * b + n * a} + {n * b} * x) * e ^ ({n} * x)}$, (by doing some factorising)

\n

\n

Hence, $\\simplify{g(x) = {m * b + n * a} + {n * b} * x}$.

", "statement": "

Differentiate the following function $f(x)$.

Let $f(x)=\\simplify{(e^x)({a}x^2+{b}x)}$. Find the coordinates of the stationary points.

\n

Give all answers to three significant figures.

\n

\n

Stationary Point 1:

\n

The coordinates are ([[0]],[[1]]).

\n

\n

Stationary Point 2:

\n

The coordinates are ([[2]],[[3]]).

\n

\n

(So the computer marks your answer correctly, make sure that Stationary Point 1 has a smaller $x$ coordinate than Stationary Point 2).

", "unitTests": [], "showCorrectAnswer": true}], "variablesTest": {"maxRuns": 100, "condition": "x1 <> x2"}, "advice": "

A stationary point is where the gradient is zero, i.e. when $f'$ is equal to zero.  So first we determine $f'$, using the product rule.

\n

$f'(x) = \\simplify{e^(x)({a}x^2 + {b}x)+(e^(x))*(2*{a}*x +{b})}$.

\n

\n

\n

When we are at a stationary point, $f'(x)$ has to equal 0, so:

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$0 = f'(x) = \\simplify{e^(x)({a}x^2 + {b}x)+(e^(x))*(2*{a}*x +{b})}$.

\n

\n

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To solve this equation, we can first divide both sides by $e^{x}$ to get:

\n

$0 = \\simplify{{a}x^2 + {b}x + 2*{a}*x +{b} }$.

\n

\n

After expanding brackets and simplifying this becomes the quadratic equation:

\n

$0 = \\simplify{{a}x^2 + {2a+b}x + {b}}$.

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Using the quadratic formula, we solve this equation to find the $x$-coordinates, which are: $\\var{x1}$ and $\\var{x2}$.

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To find the $y$-coordinates, we plug these $x$-coordinates into the original function $f$ to get $\\var{y1}$ and $\\var{y2}$.

\n

\n

\n

After rounding appropriately, we get the answer.

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Determine the coordinates of the stationary points of a function.

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A function of the form $e^{x}(ax^2+bx)$ is given which has two stationary points.  Students are asked to determine the coordinates of each stationary point. Calculator. Advice is provided.

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Questions on differentiation and stationary points.

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