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Try these questions as a little refresher on what you did in first year. These are the type of thing you should know going into second year. If you find any questions tricky then Maths Cafe is a great place to go and get a little support.

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Multiplication of $2 \\times 2$ matrices.

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Do the following matrix problems :
Let
\$A=\\begin{pmatrix} \\var{a11}&\\var{a12}&\\var{a13}\\\\ \\var{a21}&\\var{a22}&\\var{a23}\\\\ \\end{pmatrix},\\;\\; B=\\begin{pmatrix} \\var{b11}&\\var{b12}\\\\ \\var{b21}&\\var{b22}\\\\ \\var{b31}&\\var{b32}\\end{pmatrix},\\;\\; \$
Calculate the following products of these matrices:

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"", "definition": "random(1..3)", "group": "Ungrouped variables", "name": "a22"}, "a23": {"templateType": "anything", "description": "", "definition": "random(-2..3)", "group": "Ungrouped variables", "name": "a23"}}, "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers"]}, "parts": [{"marks": 0, "unitTests": [], "sortAnswers": false, "scripts": {}, "useCustomName": false, "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "prompt": "

$AB = \\begin{pmatrix} \\var{a11}&\\var{a12}&\\var{a13}\\\\ \\var{a21}&\\var{a22}&\\var{a23}\\\\ \\end{pmatrix}\\begin{pmatrix} \\var{b11}&\\var{b12}\\\\ \\var{b21}&\\var{b22}\\\\ \\var{b31}&\\var{b32}\\end{pmatrix} =$ [[0]]

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$BA = \\begin{pmatrix} \\var{b11}&\\var{b12}\\\\ \\var{b21}&\\var{b22}\\\\ \\var{b31}&\\var{b32}\\end{pmatrix}\\begin{pmatrix} \\var{a11}&\\var{a12}&\\var{a13}\\\\ \\var{a21}&\\var{a22}&\\var{a23}\\\\ \\end{pmatrix}=$ [[0]]

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Consider the following system of equations:

\n

\\begin{align}
\\simplify{{a}x+{b}y}&=\\var{c}\\text{,}\\\\
\\simplify{{a1}x+{b1}y}&=\\var{c1}\\text{.}
\\end{align}

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Solve to find the values of $x$ and $y$.

\n

$x=$ [[0]]

\n

$y=$ [[1]]

\n

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Multiply the first equation by $\\var{b1}$ and the second equation by $\\var{b}$ so they both have the same $y$ coefficient:

\n

\\begin{align}
\\simplify{{a*b1}x+{b*b1}y} &= \\var{c*b1} \\\\
\\simplify{{a1*b}x+{b1*b}y} &= \\var{c1*b}
\\end{align}

\n

Next, subtract the second equation from the first to get

\n

\$\\simplify[std]{{a*b1-a1*b}x} = \\var{c*b1-c1*b} \$

\n

So $x = \\displaystyle \\simplify[std]{{(c*b1-c1*b)/(a*b1-a1*b)}}$.

\n

Substitute this value of $x$ into the first equation and rearrange to obtain $y$:

\n

\\begin{align}
\\simplify[std]{{a}*{(c*b1-c1*b)/(a*b1-a1*b)} + {b}y} &= \\var{c} \\\\
\\simplify[std]{{b}y} &= \\simplify[std]{{c}-{a*(c*b1-c1*b)/(a*b1-a1*b)}} \\\\
y &= \\simplify[std]{{(c-a*(c*b1-c1*b)/(a*b1-a1*b))/b}}
\\end{align}

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Solve a system of three simultaneous linear equations

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Solve the following system of three simultaneous linear equations:

\n

\$$\\var{a1}x+2y+4z=\\var{r1}\$$

\n

and

\n

\$$2x+\\var{b1}y+3z=\\var{r2}\$$

\n

and

\n

\$$5x+6y+\\var{c1}z=\\var{r3}\$$

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(i)    \$$\\var{a1}x+2y+4z=\\var{r1}\$$

\n

(ii)   \$$2x+\\var{b1}y+3z=\\var{r2}\$$

\n

(iii)  \$$5x+6y+\\var{c1}z=\\var{r3}\$$

\n

First reduce the three equations in three unknowns to a two equations in two unknowns problem by eliminating one of the variables.

\n

We can eliminate \$$x\$$ using equations (i) and (ii)

\n

2*(i)     \$$\\simplify{2*{a1}}x+4y+8z=\\simplify{2*{r1}}\$$

\n

\$$\\var{a1}\$$*(ii)    \$$\\simplify{2*{a1}}x+\\simplify{{a1}*{b1}}y+\\simplify{3*{a1}}z=\\simplify{{a1}*{r2}}\$$

\n

Subtracting gives us a new equation

\n

(iv)    \$$\\simplify{(4-{a1}{b1})y+(8-3*{a1})z}=\\simplify{2*{r1}-{a1}*{r2}}\$$

\n

We can also eliminate \$$x\$$ using equations (ii) and (iii)

\n

5*(ii)    \$$10x +\\simplify{5*{b1}}y+15z=\\simplify{5*{r2}}\$$

\n

2*(iii)   \$$10x+12y+\\simplify{2*{c1}}z=\\simplify{2*{r3}}\$$

\n

Subtracting gives us another new equation

\n

(v)     \$$\\simplify{(5*{b1}-12)y+(15-2*{c1})z}=\\simplify{5*{r2}-2*{r3}}\$$

\n

We could then eliminate the \$$y\$$ from these two new equations

\n

\$$\\simplify{5*{b1}-12}\$$*(iv)    \$$\\simplify{(5*{b1}-12)*(4-{a1}{b1})y+(5*{b1}-12)*(8-3*{a1})z}=\\simplify{(5*{b1}-12)*(2*{r1}-{a1}*{r2})}\$$

\n

\$$\\simplify{4-{a1}{b1}}\$$*(v)    \$$\\simplify{(4-{a1}{b1})*(5*{b1}-12)y+(4-{a1}{b1})*(15-2*{c1})z}=\\simplify{(4-{a1}{b1})*(5*{r2}-2*{r3})}\$$

\n

Subtracting gives us

\n

\$$\\simplify{(5*{b1}-12)*(8-3*{a1})-(4-{a1}{b1})*(15-2*{c1})}z=\\simplify{(5*{b1}-12)*(2*{r1}-{a1}*{r2})-(4-{a1}{b1})*(5*{r2}-2*{r3})}\$$

\n

Thus

\n

\$$z=\\frac{\\simplify{(5*{b1}-12)*(2*{r1}-{a1}*{r2})-(4-{a1}{b1})*(5*{r2}-2*{r3})}}{\\simplify{(5*{b1}-12)*(8-3*{a1})-(4-{a1}{b1})*(15-2*{c1})}}=\\simplify{decimal{((5*{b1}-12)*(2*{r1}-{a1}*{r2})-(4-{a1}*{b1})*(5*{r2}-2*{r3}))/( (5*{b1}-12)*(8-3*{a1})-(4-{a1}*{b1})*(15-2*{c1}))}}\$$

\n

We can now back substitute this value for \$$z\$$ into equation (iv) to find the correct value for \$$y\$$ and then back substitute both these values into equation (i) to calculate \$$x\$$.

\n

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Input the value of \$$x\$$ that satisfies the three equations.

\n

\$$x = \$$ [[0]]

\n

Input the value of \$$y\$$ that satisfies the three equations.

\n

\$$y = \$$ [[1]]

\n

Input the value of \$$z\$$ that satisfies the three equations.

\n

\$$z = \$$ [[2]]

Given the matrix

\n

\$$A =\\begin{pmatrix} \\var{a11}&\\var{a12}\\\\ \\var{a21}&\\var{a22}\\\\ \\end{pmatrix}\$$

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The eigenvalues of a matrix are the values of \$$\\lambda\$$ that satisfy the relation

\n

\$$|A-\\lambda I| = 0\$$

\n

\$$\\begin{vmatrix} \\var{a11}-\\lambda&\\var{a12}\\\\ \\var{a21}&\\var{a22}-\\lambda\\\\ \\end{vmatrix}=0\$$

\n

This gives:

\n

\$$(\\var{a11}-\\lambda)*(\\var{a22}-\\lambda)-(\\var{a12})*(\\var{a21})=0\$$

\n

\$$\\lambda^2-\\simplify{{a11}+{a22}}\\lambda+\\simplify{{a11}*{a22}-{a21}*{a12}}=0\$$

\n

This can be solved using factorisation or by the quadratic formula to give:

\n

\$$\\lambda_1 =\\var{lambda1}\$$ and \$$\\lambda_2 =\\var{lambda2}\$$

\n

An eigenvector \$${\\bf v}=\\begin{pmatrix} x\\\\ y\\\\ \\end{pmatrix}\$$ corresponding to an eigenvalue \$$\\lambda\$$ must satisfy the relation:  \$$(A-\\lambda I){\\bf v} = {\\bf 0}\$$

\n

so for \$$\\lambda_1=\\var{lambda1}\$$

\n

\$$\\begin{pmatrix} \\simplify{{a11}-{lambda1}}&\\var{a12}\\\\ \\var{a21}&\\simplify{{a22}-{lambda1}}\\\\ \\end{pmatrix}\\begin{pmatrix} x\\\\ \\var{a21}\\\\ \\end{pmatrix}={\\bf 0}\$$

\n

thus

\n

\$$\\var{a21}x+\\simplify{{a22}-{lambda1}}*\\var{a21}=0\$$

\n

\$$\\Rightarrow ~ \\var{a21}x=-\\simplify{({a22}-{lambda1})*{a21}}\$$

\n

\$$\\Rightarrow ~ x=-\\simplify{({a22}-{lambda1})}\$$

\n

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This question concerns the evaluation of the eigenvalues and corresponding eigenvectors of a 2x2 matrix.

Calculate the eigenvalues of the matrix A

\n

\$$\\lambda_1\$$ is the lesser of the two eigenvalues and \$$\\lambda_2\$$ is the greater of the two eigenvalues;

\n

\$$\\lambda_1\$$ = [[0]]

\n

\$$\\lambda_2\$$ = [[1]]

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For the lesser eigenvalue \$$\\lambda_1\$$ the corresponding eigenvector is \$$v_1=\\begin{pmatrix}x\\\\ \\var{a21}\\\\ \\end{pmatrix}\$$

\n

Enter the value for \$$x=\$$ [[0]]

\n

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The answer is either \"Yes\" or \"No\".

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For help go to the maths cafe

", "metadata": {"description": "", "licence": "None specified"}, "functions": {}, "variable_groups": [], "statement": "

Which of these statements are true in general:

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\$(x+y)^2 = x^2+y^2\$

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\$\\sqrt{x^2+y^2} = x+y\$

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\$\\frac{a^2+2a}{4a+b} = \\frac{a+2}{4+b}\$

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\$s(s^2+1)-3(s^2+2) +3 = (s-3)(s^2+1)\$

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The student is asked to factorise a quadratic $x^2 + ax + b$. A custom marking script uses pattern matching to ensure that the student's answer is of the form $(x+a)(x+b)$, $(x+a)^2$, or $x(x+a)$.

\n

To find the script, look in the Scripts tab of part a.

"}, "rulesets": {}, "statement": "", "tags": [], "functions": {}, "variables": {"b": {"description": "", "templateType": "anything", "definition": "random(-5..5 except 0)", "group": "Ungrouped variables", "name": "b"}, "a": {"description": "", "templateType": "anything", "definition": "random(-5..5)", "group": "Ungrouped variables", "name": "a"}}, "variable_groups": [], "parts": [{"extendBaseMarkingAlgorithm": true, "checkingAccuracy": 0.001, "useCustomName": false, "scripts": {"mark": {"order": "after", "script": "// Parse the student's answer as a syntax tree\nvar studentTree = Numbas.jme.compile(this.studentAnswer,Numbas.jme.builtinScope);\n\n// Create the pattern to match against \n// we just want two sets of brackets, each containing two terms\n// or one of the brackets might not have a constant term\n// or for repeated roots, you might write (x+a)^2\nvar rule = Numbas.jme.compile('m_any( (x+??)(x+??), (x+??)^2, x*(x+??) )');\n\n// Check the student's answer matches the pattern. \nvar m = Numbas.jme.display.matchTree(rule,studentTree,true);\n// If not, take away marks\nif(!m) {\n this.multCredit(0.5,'Your answer is not in the form $(\\\\dots)(\\\\dots)$.');\n}\n"}}, "customName": "", "marks": "1", "showCorrectAnswer": true, "unitTests": [], "type": "jme", "answer": "(x+{a})(x+{b})", "customMarkingAlgorithm": "", "failureRate": 1, "variableReplacements": [], "checkingType": "absdiff", "adaptiveMarkingPenalty": 0, "vsetRange": [0, 1], "prompt": "

Factorise $\\simplify{x^2+{a+b}x+{a*b}}$

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When quadratic equations can't be factorised, or if equations are difficult to factorise (perhaps if the coefficients are large), we need to use the quadratic formula to solve the equations.

\n

Use the quadratic formula to calculate values for $x$ in these equations. Input the possible values as $x_1$ and $x_2$, where $x_1<x_2$.

", "preamble": {"js": "", "css": ""}, "parts": [{"sortAnswers": false, "type": "gapfill", "marks": 0, "scripts": {}, "showFeedbackIcon": true, "variableReplacements": [], "stepsPenalty": 0, "extendBaseMarkingAlgorithm": true, "steps": [{"extendBaseMarkingAlgorithm": true, "type": "information", "showCorrectAnswer": true, "prompt": "

An equation of the form

\n

\$ax^2+bx+c=0\\text{,}\$

\n

\n

can be solved using the quadratic formula

\n

\$x={\\frac {-b\\pm\\sqrt{b^2-4\\times a\\times c}}{2a}}\\text{.}\$

\n

", "marks": 0, "showFeedbackIcon": true, "unitTests": [], "customMarkingAlgorithm": "", "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst"}], "showCorrectAnswer": true, "prompt": "

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

\n

$x_1=$ [[0]]

\n

$x_2=$ [[1]]

\n

$\\simplify{{a1}x^2+{a2}x+{a3}={a4}}$

\n

$x_1=$ [[0]]

\n

$x_2=$ [[1]]

\n

$\\simplify{{b1}x^2+{b2}x+{b3}={b4}x}$

\n

$x_1=$ [[0]]

\n

$x_2=$ [[1]]

Apply the quadratic formula to find the roots of a given equation. The quadratic formula is given in the steps if the student requires it.

"}, "ungrouped_variables": ["a1", "a2", "a3", "a4", "b1", "b2", "b3", "b4", "x1", "p1", "p2", "x2", "a", "m"], "advice": "

\n

\$x={\\frac {-b\\pm\\sqrt{b^2-4\\times a\\times c}}{2a}}\\text{.}\$

\n

#### (a)

\n

From the equation, we can read off values for $a$, $b$ and $c$:

\n

\\\begin{align} a&=1\\text{,}\\\\ b&=\\var{a+m}\\text{,}\\\\ c&=\\var{a*m} \\text{.} \\end{align}\

\n

Substituting these values into the quadratic formula,

\n

\$x = \\frac {-\\var{a+m}\\pm\\sqrt{\\var{a+m}^2-4\\times \\var{a*m}}}{2}\\text{.}\$

\n

Note the $\\pm$ symbol in the formula. This means there are two solutions: one using $+$, the other using $-$.

\n

The two solutions are

\n

\\\begin{align} x_1&=\\var{m}\\text{,}\\\\ x_2&=\\var{a}\\text{.} \\end{align}\

\n

\n

#### (b)

\n

Note that the right-hand side of the given equation is not zero. We need to rewrite it in the form $ax^2+bx+c=0$:

\n

\\\begin{align} \\simplify{{a1}x^2+{a2}x+{a3}}&=\\var{a4}\\\\ \\simplify{{a1}x^2+{a2}x+{a3-a4}}&=0\\text{.} \\end{align}\

\n

Then we can read off values for $a$, $b$ and $c$:

\n

\\\begin{align} a&=\\var{a1}\\\\ b&=\\var{a2}\\\\ c&=\\var{a3-a4} \\text{.} \\end{align}\

\n

We can now substitute these values into the quadratic formula:

\n

\$x = {\\frac {-\\var{a2}\\pm\\sqrt{\\var{a2}^2-4\\times \\var{a1}\\times \\var{a3-a4}}}{2\\times\\var{a1}}}\\text{.}\$

\n

So the two solutions are

\n

\\\begin{align} x_1&=\\var{dpformat(x1,2)}\\\\ x_2&=\\var{dpformat(x2,2)}\\text{.} \\end{align}\

\n

\n

#### (c)

\n

We first rearrange our equation into the form $ax^2+bx+c=0$:

\n

\\\begin{align} \\simplify{{b1}x^2+{b2}x+{b3}}&=\\var{b4}x\\\\ \\simplify{{b1}x^2+{b2-b4}x+{b3}}&=0\\text{.} \\end{align}\

\n

We can then read off the values for $a, b$ and $c$, which are

\n

\\\begin{align} a&=\\var{b1}\\text{,}\\\\ b&=\\var{b2-b4}\\text{,}\\\\ c&=\\var{b3}\\text{.} \\end{align}\

\n

Substituting these values into the quadratic formula,

\n

\$x = {\\frac {-\\var{b2-b4}\\pm\\sqrt{\\var{b2-b4}^2-4\\times \\var{b1}\\times \\var{b3}}}{2\\times\\var{b1}}},\$

\n

we obtain solutions

\n

\\\begin{align} x_1&=\\var{dpformat(p1,2)}\\text{,}\\\\ x_2&=\\var{dpformat(p2,2)}\\text{.} \\end{align}\

", "variable_groups": [{"name": "part 2", "variables": ["b", "c"]}]}, {"name": "Differentiation: simple", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Sanka Balasuriya", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/738/"}], "functions": {}, "ungrouped_variables": ["a", "b", "c", "d", "p", "q"], "tags": [], "preamble": {"css": "", "js": ""}, "advice": "", "rulesets": {}, "parts": [{"prompt": "

$f'(x)=\\;$ [[0]]

", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"vsetrangepoints": 5, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "scripts": {}, "answer": "{a}*{p}*x^({p}-1)+{b}*{q}*x^({q}-1)+{c}", "marks": 1, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}], "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "gapfill"}], "statement": "

Find $f'(x),$ the derivate of $f(x),$ if $\\simplify{f(x)={a}x^{p}+{b}x^{q}+{c}x+{d}}.$

", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"a": {"definition": "random(-10..10 except 0)", "templateType": "anything", "group": "Ungrouped variables", "name": "a", "description": ""}, "c": {"definition": "random(-100..100 except 0)", "templateType": "anything", "group": "Ungrouped variables", "name": "c", "description": ""}, "b": {"definition": "random(-10..10 except 0)", "templateType": "anything", "group": "Ungrouped variables", "name": "b", "description": ""}, "d": {"definition": "random(-100..100 except 0)", "templateType": "anything", "group": "Ungrouped variables", "name": "d", "description": ""}, "q": {"definition": "random(-10..10 except 0 except 1 except p)", "templateType": "anything", "group": "Ungrouped variables", "name": "q", "description": ""}, "p": {"definition": "random(-10..10 except 0 except 1)", "templateType": "anything", "group": "Ungrouped variables", "name": "p", "description": ""}}, "metadata": {"notes": "", "description": "

Finding the derivative  of a simple polynomial function.

", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Differentiation: Product rule", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}], "variable_groups": [], "variables": {"s1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s1"}, "b": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s1*random(1..9)", "description": "", "name": "b"}, "a": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..9)", "description": "", "name": "a"}, "m": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..9)", "description": "", "name": "m"}}, "ungrouped_variables": ["a", "s1", "b", "m"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "functions": {}, "showQuestionGroupNames": false, "parts": [{"stepsPenalty": 0, "scripts": {}, "gaps": [{"answer": "{m}x ^ {m-1} * cos({a} * x+{b})-{a}x^{m} * sin({a} * x+{b})", "showCorrectAnswer": true, "vsetrange": [0, 1], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "answersimplification": "std", "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n\t\t\t

$\\simplify[std]{f(x) = x ^ {m} * cos({a} * x+{b})}$

\n\t\t\t

$\\displaystyle \\frac{df}{dx}=\\;$[[0]]

\n\t\t\t

Clicking on Show steps gives you more information, you will not lose any marks by doing so.

\n\t\t\t", "steps": [{"type": "information", "prompt": "

The product rule says that if $u$ and $v$ are functions of $x$ then
\$\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\$

", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "

Differentiate the following function $f(x)$ using the product rule.

", "tags": ["calculus", "Calculus", "checked2015", "derivative of a product", "differentiating a product", "differentiating trigonometric functions", "differentiation", "MAS1601", "mas1601", "product rule", "Steps", "steps", "trigonometric functions"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"], "surdf": [{"result": "(sqrt(b)*a)/b", "pattern": "a/sqrt(b)"}]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n\t\t

31/07/2012:

\n\t\t

\n\t\t

\n\t\t

Steps problem to be addressed. Now resolved.

\n\t\t

Checked calculation.OK.

\n\t\t

Improved prompt display.

\n\t\t

Clicking on Show steps does not lose any marks.

\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "

Differentiate $x^m\\cos(ax+b)$

"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n\t \n\t \n\t

The product rule says that if $u$ and $v$ are functions of $x$ then
\$\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\$

\n\t \n\t \n\t \n\t

For this example:

\n\t \n\t \n\t \n\t

\$\\simplify[std]{u = x ^ {m}}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {m}x ^ {m -1}}\$

\n\t \n\t \n\t \n\t

\$\\simplify[std]{v = cos({a} * x+{b})} \\Rightarrow \\simplify[std]{Diff(v,x,1) = -{a} * sin({a} * x+{b})}\$

\n\t \n\t \n\t \n\t

Hence on substituting into the product rule above we get:

\n\t \n\t \n\t \n\t

\$\\simplify[std]{Diff(f,x,1) = {m}x ^ {m-1} * cos({a} * x+{b})-{a}x^{m} * sin({a} * x+{b})}\$

\n\t \n\t \n\t"}, {"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}, "contributors": [{"name": "Lovkush Agarwal", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1358/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}], "variable_groups": [], "preamble": {"js": "", "css": ""}, "ungrouped_variables": ["a", "s1", "b", "m", "n"], "functions": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "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"}, "statement": "

Differentiate the following function $f(x)$.

", "variables": {"b": {"templateType": "anything", "description": "", "definition": "s1*random(1..5)", "group": "Ungrouped variables", "name": "b"}, "m": {"templateType": "anything", "description": "", "definition": "random(2..8)", "group": "Ungrouped variables", "name": "m"}, "a": {"templateType": "anything", "description": "", "definition": "random(1..4)", "group": "Ungrouped variables", "name": "a"}, "s1": {"templateType": "anything", "description": "", "definition": "random(1,-1)", "group": "Ungrouped variables", "name": "s1"}, "n": {"templateType": "anything", "description": "", "definition": "random(2..6)", "group": "Ungrouped variables", "name": "n"}}, "tags": [], "parts": [{"customName": "", "customMarkingAlgorithm": "", "sortAnswers": false, "scripts": {}, "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]]

", "extendBaseMarkingAlgorithm": true, "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "type": "gapfill", "showFeedbackIcon": true, "marks": 0, "unitTests": [], "useCustomName": false, "gaps": [{"customName": "", "customMarkingAlgorithm": "", "failureRate": 1, "checkingType": "absdiff", "showCorrectAnswer": true, "checkVariableNames": false, "vsetRange": [0, 1], "showFeedbackIcon": true, "vsetRangePoints": 5, "checkingAccuracy": 0.001, "scripts": {}, "answer": "({((m * b) + (n * a))} + ({(n * b)} * x))", "valuegenerators": [{"value": "", "name": "x"}], "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "type": "jme", "answerSimplification": "all", "marks": "4", "unitTests": [], "showPreview": true, "useCustomName": false, "variableReplacements": []}], "variableReplacements": []}], "rulesets": {}, "advice": "

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

"}, {"name": "Basic Integration", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Nick Walker", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2416/"}, {"name": "Thomas Waters", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3649/"}], "metadata": {"description": "

Integration techniques for monomials and simple polynomials.

", "licence": "Creative Commons Attribution 4.0 International"}, "rulesets": {}, "variable_groups": [], "statement": "

Find the following definite integrals.

", "preamble": {"css": "", "js": ""}, "ungrouped_variables": ["a", "b", "n"], "variablesTest": {"condition": "", "maxRuns": 100}, "functions": {}, "advice": "

$\\int_a^bx^n=\\frac{b^{n+1}-a^{n+1}}{n+1}$

", "tags": [], "variables": {"b": {"templateType": "randrange", "description": "

upper bound of integration

", "definition": "random(4..8#1)", "group": "Ungrouped variables", "name": "b"}, "n": {"templateType": "randrange", "description": "

the power on x

", "definition": "random(2..10#1)", "group": "Ungrouped variables", "name": "n"}, "a": {"templateType": "randrange", "description": "

lower bound of integration

", "definition": "random(0..3#1)", "group": "Ungrouped variables", "name": "a"}}, "type": "question", "parts": [{"marks": 1, "correctAnswerFraction": false, "scripts": {}, "minValue": "1/(n+1)", "notationStyles": ["plain", "en", "si-en"], "mustBeReduced": false, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "prompt": "

$\\int_{0}^{1}{x^{\\var{n}}}$

", "allowFractions": true, "mustBeReducedPC": 0, "variableReplacements": [], "correctAnswerStyle": "plain", "showCorrectAnswer": true, "type": "numberentry", "maxValue": "1/(n+1)"}, {"marks": 1, "correctAnswerFraction": false, "scripts": {}, "minValue": "(b^(n+1)-a^(n+1))/(n+1)", "notationStyles": ["plain", "en", "si-en"], "mustBeReduced": false, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "prompt": "

$\\int_{\\var{a}}^{\\var{b}}{x^{\\var{n}}}$

", "allowFractions": true, "mustBeReducedPC": 0, "variableReplacements": [], "correctAnswerStyle": "plain", "showCorrectAnswer": true, "type": "numberentry", "maxValue": "(b^(n+1)-a^(n+1))/(n+1)"}]}, {"name": "Indefinite Integration", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}, {"name": "Violeta CIT", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1030/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}, {"name": "Jo-Ann Lyons", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2630/"}], "tags": [], "parts": [{"checkVariableNames": false, "variableReplacementStrategy": "originalfirst", "failureRate": 1, "marks": 1, "unitTests": [], "prompt": "

$\\int{(\\frac{1}{4}\\sqrt{x}-3\\sqrt{x^5})}\\mathrm{dx}$

", "answer": "1/6x^(3/2)-6/7x^(7/2)+c", "showCorrectAnswer": true, "checkingType": "absdiff", "expectedVariableNames": [], "vsetRangePoints": 5, "checkingAccuracy": 0.001, "vsetRange": [0, 1], "showFeedbackIcon": true, "customMarkingAlgorithm": "malrules:\n [\n [\"1/6x^(3/2)-6/7x^(7/2)\",\"Don't forget the constant of integration!\",0.9],\n [\"1/6x^(3/2)-2(x^5)^(3/2)+C\", \"Check the second term again. Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"1/6x^(3/2)-2(x^5)^(3/2)\", \"Check the second term again. Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"1/4x^(1/2)-2(x^5)^(3/2)\", \"You need to look at both terms again. First term: $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated. Second term: Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$ before integrating. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"1/4x^(1/2)-2(x^5)^(3/2)+C\", \"You need to look at both terms again. First term: $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated. Second term: Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$ before integrating. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"1/4x^(1/2)-9/2(x^5)^(3/2)\", \"You need to look at both terms again. First term: $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated. Second term: Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$ before integrating. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"1/4x^(1/2)-9/2(x^5)^(3/2)+C\", \"You need to look at both terms again. First term: $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated. Second term: Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$ before integrating. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"1/4x^(1/2)-6/7x^(7/2)\", \"You need to look at the first term again. $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated.\",0],\n [\"1/4x^(1/2)-6/7x^(7/2)+C\", \"You need to look at the first term again. $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated.\",0],\n [\"1/4x^(1/2)-21/2x^(7/2)\", \"You need to look at both terms again. First term: $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated. Second term: It looks like you have multiplied by the new power of $\\\\frac{7}{2}$...\",0],\n [\"1/4x^(1/2)-21/2x^(7/2)+C\", \"You need to look at both terms again. First term: $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated. Second term: It looks like you have multiplied by the new power of $\\\\frac{7}{2}$...\",0],\n [\"1/2x^(1/2)-2(x^5)^(3/2)\", \"You need to look at both terms again. First term: $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated. Second term: Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$ before integrating. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"1/2x^(1/2)-2(x^5)^(3/2)+C\", \"You need to look at both terms again. First term: $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated. Second term: Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$ before integrating. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"1/2x^(1/2)-9/2(x^5)^(3/2)\", \"You need to look at both terms again. First term: $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated. Second term: Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$ before integrating. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"1/2x^(1/2)-9/2(x^5)^(3/2)+C\", \"You need to look at both terms again. First term: $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated. Second term: Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$ before integrating. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"1/2x^(1/2)-6/7x^(7/2)\", \"You need to look at the first term again. $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated.\",0],\n [\"1/2x^(1/2)-6/7x^(7/2)+C\", \"You need to look at the first term again. $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated.\",0],\n [\"1/2x^(1/2)-21/2x^(7/2)\", \"You need to look at both terms again. First term: $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated. Second term: It looks like you have multiplied by the new power of $\\\\frac{7}{2}$...\",0],\n [\"1/2x^(1/2)-21/2x^(7/2)+C\", \"You need to look at both terms again. First term: $\\\\sqrt{x}=x^{\\\\frac{1}{2}}$ but this has not actually been integrated. Second term: It looks like you have multiplied by the new power of $\\\\frac{7}{2}$...\",0],\n [\"3/8x^(3/2)-2(x^5)^(3/2)\", \"You need to look at both terms again. First term: It looks like you have multiplied by the new power of $\\\\frac{3}{2}$. Second term: Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$ before integrating. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"3/8x^(3/2)-2(x^5)^(3/2)+C\", \"You need to look at both terms again. First term: It looks like you have multiplied by the new power of $\\\\frac{3}{2}$. Second term: Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$ before integrating. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"3/8x^(3/2)-9/2(x^5)^(3/2)\", \"You need to look at both terms again. First term: It looks like you have multiplied by the new power of $\\\\frac{3}{2}$. Second term: Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$ before integrating. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"3/8x^(3/2)-9/2(x^5)^(3/2)+C\", \"You need to look at both terms again. First term: It looks like you have multiplied by the new power of $\\\\frac{3}{2}$. Second term: Try to write the power on the $x$ as a single power rather than $(x^5)^{\\\\frac{1}{2}}$ before integrating. Remember, if the powers are side by side, multiply them: $\\\\sqrt{x^5}=(x^5)^{\\\\frac{1}{2}}=x^{5 \\\\times \\\\frac{1}{2}}=x^{\\\\frac{5}{2}}$.\",0],\n [\"3/8x^(3/2)-6/7x^(7/2)\", \"You need to look at the first term again. It looks like you have multiplied by the new power of $\\\\frac{3}{2}$.\",0],\n [\"3/8x^(3/2)-6/7x^(7/2)+C\", \"You need to look at the first term again. It looks like you have multiplied by the new power of $\\\\frac{3}{2}$.\",0],\n [\"3/8x^(3/2)-21/2x^(7/2)\", \"You need to look at both terms again. First term: It looks like you have multiplied by the new power of $\\\\frac{3}{2}$. Second term: It looks like you have multiplied by the new power of $\\\\frac{7}{2}$...\",0],\n [\"3/8x^(3/2)-21/2x^(7/2)+C\", \"You need to look at both terms again. First term: It looks like you have multiplied by the new power of $\\\\frac{3}{2}$. Second term: It looks like you have multiplied by the new power of $\\\\frac{7}{2}$...\",0],\n [\"1/6x^(3/2)-21/2x^(7/2)\", \"You need to look at the second term again. It looks like you have multiplied by the new power of $\\\\frac{7}{2}$...\",0],\n [\"1/6x^(3/2)-21/2x^(7/2)+C\", \"You need to look at the second term again. It looks like you have multiplied by the new power of $\\\\frac{7}{2}$...\",0]\n ]\n\n\nparsed_malrules: \n map(\n [\"expr\":parse(x[0]),\"feedback\":x[1],\"credit\":x[2]],\n x,\n malrules\n )\n\nagree_malrules (Do the student's answer and the expected answer agree on each of the sets of variable values?):\n map(\n len(filter(not x ,x,map(\n try(\n resultsEqual(unset(question_definitions,eval(studentexpr,vars)),unset(question_definitions,eval(malrule[\"expr\"],vars)),settings[\"checkingType\"],settings[\"checkingAccuracy\"]),\n message,\n false\n ),\n vars,\n vset\n )))Simple Indefinite Integrals

\n

"}, "variable_groups": [], "ungrouped_variables": ["a", "c", "b", "d", "f"], "statement": "

Solve the following indefinite integrals, using $C$ to represent an unknown constant.

", "functions": {}, "preamble": {"css": "", "js": ""}, "variables": {"f": {"group": "Ungrouped variables", "description": "", "definition": "random(1..8 except d)", "name": "f", "templateType": "anything"}, "a": {"group": "Ungrouped variables", "description": "", "definition": "random(2..9)", "name": "a", "templateType": "anything"}, "b": {"group": "Ungrouped variables", "description": "", "definition": "random(2..9 except a)", "name": "b", "templateType": "anything"}, "c": {"group": "Ungrouped variables", "description": "", "definition": "random(1..9 except a except b)", "name": "c", "templateType": "anything"}, "d": {"group": "Ungrouped variables", "description": "", "definition": "random(1..8)", "name": "d", "templateType": "anything"}}, "type": "question", "rulesets": {}, "advice": "

Indefinite Integrals

", "variablesTest": {"condition": "", "maxRuns": 100}}, {"name": "Integration by parts", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}], "variable_groups": [], "variables": {"s1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s1"}, "b": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s1*random(1..9)", "description": "", "name": "b"}, "c": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s3*random(2..5)", "description": "", "name": "c"}, "s3": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s3"}, "a": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..5)", "description": "", "name": "a"}, "s2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s2"}, "a1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s1*random(1..9)", "description": "", "name": "a1"}, "a2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s2*random(1..9)", "description": "", "name": "a2"}}, "ungrouped_variables": ["a", "c", "b", "s3", "s2", "s1", "a1", "a2"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "functions": {}, "showQuestionGroupNames": false, "parts": [{"stepsPenalty": 1, "scripts": {}, "gaps": [{"answer": "({a}/{c})*x+{c*b-a}/{c^2}", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "showCorrectAnswer": true, "expectedvariablenames": [], "notallowed": {"message": "

Do not input numbers as decimals, only as integers without the decimal point, or fractions

", "showStrings": false, "partialCredit": 0, "strings": ["."]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "answersimplification": "all", "marks": 2, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n\t\t\t

$I=\\displaystyle \\int \\simplify[std]{({a}x+{b})*e^({c}x)} dx$
You are given that the answer is of the form \$I=g(x)e^{\\var{c}x}+C\$ for a polynomial $g(x)$. You have to find $g(x)$.

\n\t\t\t

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

\n\t\t\t

Input all numbers as fractions or integers and not decimals.

\n\t\t\t

You can get help by clicking on Show steps. You will lose 1 mark if you do.

\n\t\t\t", "steps": [{"type": "information", "prompt": "\n\t\t\t\t\t \n\t\t\t\t\t \n\t\t\t\t\t

The formula for integrating by parts is
\$\\int u\\frac{dv}{dx} dx = uv - \\int v \\frac{du}{dx} dx. \$

\n\t\t\t\t\t \n\t\t\t\t\t \n\t\t\t\t\t", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}, {"scripts": {}, "gaps": [{"answer": "{a^2}/{c}*x^2+{2*a*b*c-2*a^2}/{c^2}*x+{b^2*c^2-2*a*b*c+2*a^2}/{c^3}", "vsetrange": [0, 1], "checkingaccuracy": 0.001, "showCorrectAnswer": true, "expectedvariablenames": [], "notallowed": {"message": "

Do not input numbers as decimals, only as integers without the decimal point, or fractions

", "showStrings": false, "partialCredit": 0, "strings": ["."]}, "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "checkvariablenames": false, "type": "jme", "answersimplification": "all", "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n\t\t\t

Use the result from the first part to find:

\n\t\t\t

$\\displaystyle I=\\int \\simplify[std]{({a}x+{b})^2*e^({c}x)} dx$

\n\t\t\t

You are given that the answer is of the form \$I=h(x)e^{\\var{c}x}+C\$ for a polynomial $h(x)$. You have to find $h(x)$.

\n\t\t\t

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

\n\t\t\t

Input all numbers as fractions or integers and not decimals.

\n\t\t\t", "showCorrectAnswer": true, "marks": 0}], "statement": "\n\t

Find the following indefinite integrals.

\n\t

Input all numbers as fractions or integers and not decimals.

\n\t", "tags": ["Calculus", "MAS1601", "Steps", "algebraic manipulation", "checked2015", "exponential function", "integration", "integration by parts", "integration of exponential function"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n\t\t

3/08/2012:

\n\t\t

\n\t\t

\n\t\t

Checked calculation. OK.

\n\t\t

Got rid of redundant instructions about inputting constant of integration.

\n\t\t

Penalised use of steps in first part, 1 mark. Added message to that effect in first part.

\n\t\t

\n\t\t

Changed marks reflecting the use of steps and degree of difficulty in second part.

\n\t\t

\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "

Given $\\displaystyle \\int (ax+b)e^{cx}\\;dx =g(x)e^{cx}+C$, find $g(x)$. Find $h(x)$, $\\displaystyle \\int (ax+b)^2e^{cx}\\;dx =h(x)e^{cx}+C$.

"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n\t

a)

\n\t

The formula for integrating by parts is
\$\\int u\\frac{dv}{dx} dx = uv - \\int v \\frac{du}{dx} dx. \$

\n\t

We choose $u = \\simplify[std]{{a}x+{b}}$ and $\\displaystyle\\frac{dv}{dx} = \\simplify[std]{e^({c}x)}$.

\n\t

So $\\displaystyle \\frac{du}{dx} = \\var{a}$ and $\\displaystyle v = \\simplify[std]{(1/{c})*e^({c}*x)}$.

\n\t

Hence,
\$\\begin{eqnarray} \\int \\simplify[std]{({a}*x+{b})*e^({c}*x)} dx &=& uv - \\int v \\frac{du}{dx} dx \\\\ &=& \\simplify[std]{({1}/{c})*({a}x+{b})*e^({c}x) - (1/{c})*Int(({a})*e^({c}x),x)} \\\\ &=& \\simplify[std]{(1/{c})*({a}x+{b})*e^({c}x) -({a}/{c^2})*e^({c}x) + C}\\\\ &=& \\simplify[std]{(({a}x+{b})/{c}-{a}/{c^2})*e^({c}*x) + C}\\\\ &=& \\simplify[std]{(({a}/{c})x+{b*c-a}/{c^2})*e^({c}*x) + C} \\end{eqnarray} \$

\n\t

Hence $\\displaystyle \\simplify[std]{g(x)=({a}/{c})*x+{c*b-a}/{c^2}}$

\n\t

b)

\n\t

For this part we choose $u = \\simplify[std]{({a}x+{b})^2}$ and $\\displaystyle \\frac{dv}{dx} = \\simplify[std]{e^({c}x)}$.

\n\t

So $\\displaystyle \\frac{du}{dx}$ = $\\simplify[std]{{2*a}*({a}*(x)+{b})}$ and $\\displaystyle v = \\simplify[std]{(1/{c})*e^({c}*x)}$.

\n\t

Hence,
\$\\begin{eqnarray*}I= \\int \\simplify[std]{({a}*x+{b})^2*e^({c}*x)} dx &=& uv - \\int v \\frac{du}{dx} dx \\\\ &=& \\simplify[std]{({1}/{c})*({a}x+{b})^2*e^({c}x) - (1/{c})*Int({2*a}*({a}x+{b})*e^({c}x),x)} \\\\ &=& \\simplify[std]{(1/{c})*({a}x+{b})^2*e^({c}x) -({2*a}/{c})*Int(({a}x+{b})*e^({c}x),x)}\\dots (*) \\end{eqnarray*}\$

\n\t

But in Part a we have aready worked out $\\displaystyle \\simplify[std]{Int(({a}x+{b})*e^({c}*x),x)=(({a}/{c})*x+({c*b-a}/{c^2}))*e^({c}*x)+C}$

\n\t

So on substituting this in equation (*) we find:
\$\\begin{eqnarray*}I&=& \\simplify[std]{(1/{c})*({a}x+{b})^2*e^({c}x) -({2*a}/{c})*(({a}/{c})*x+({c*b-a}/{c^2}))*e^({c}*x)+C}\\\\ &=& \\simplify[std]{({a^2}/{c}*x^2+{2*a*b*c-2*a^2}/{c^2}*x+{b^2*c^2-2*a*b*c+2*a^2}/{c^3})*e^({c}x) +C} \\end{eqnarray*}\$

\n\t

Hence $\\displaystyle \\simplify[std]{h(x)={a^2}/{c}*x^2+{2*a*b*c-2*a^2}/{c^2}*x+{b^2*c^2-2*a*b*c+2*a^2}/{c^3}}$

\n\t"}, {"name": "Indefinite integration using standard integrals", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}, {"name": "Thomas Waters", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3649/"}], "statement": "\n

Integrate the following function $f(x)$.

\n

You must input the constant of integration as $C$.

\n ", "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["a", "b", "s3", "s2", "s1", "s5", "s4", "a1", "a2", "b1", "c3"], "metadata": {"description": "

Integrate $f(x) = ae ^ {bx} + c\\sin(dx) + px^q$. Must input $C$ as the constant of integration.

", "licence": "Creative Commons Attribution 4.0 International"}, "variable_groups": [], "preamble": {"js": "", "css": ""}, "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "tags": [], "advice": "\n

Splitting the integral into three parts and using the information in Steps we have:

\n

\$\\begin{eqnarray*}\\simplify[std]{Int({b} * e ^ ({a}*x) + {b1} * Sin({a1}*x) + {a2} * x ^ {c3},x)}&=&\\simplify[std]{Int({b} * e ^ ({a}*x),x)+Int({b1} * Sin({a1}*x),x)+Int({a2} * x ^ {c3},x) }\\\\ &=&\\simplify[std]{({b}/{a}) * (e ^({a}*x)) + (({(-b1)}/{a1}) * Cos({a1}*x)) + ({a2}/{c3+1}) * (x ^ {(c3 + 1)})+C} \\end{eqnarray*}\$

\n ", "parts": [{"stepsPenalty": 0, "adaptiveMarkingPenalty": 0, "scripts": {}, "variableReplacements": [], "marks": 0, "useCustomName": false, "extendBaseMarkingAlgorithm": true, "customMarkingAlgorithm": "", "unitTests": [], "prompt": "\n

$\\simplify[std]{f(x) = {b} * e ^ ({a}*x) + {b1} * Sin({a1}*x) + {a2} * x ^ {c3}}$

\n

$\\displaystyle \\int\\;f(x)\\,dx=\\;$[[0]]

\n

Enter all numbers as integers or fractions and not as decimals.

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Enter all numbers as integers or fractions and not as decimals.

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Note that \$\\begin{eqnarray*} &\\int& \\;x^n\\;dx&=&\\frac{x^{n+1}}{n+1}+C,\\;\\;n \\neq -1\\\\ &\\int& \\;\\sin(ax)\\;dx &=& -\\frac{1}{a}\\cos(ax)+C\\\\ &\\int& \\;e^{ax}\\;dx &=& \\frac{1}{a}e^{ax}+C\\\\ \\end{eqnarray*}\$

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(i) Definition: A 'root' of a function $f(x)$ is a value of $x$ which makes $f(x)=0$.  Visually a root can be found be seeing when the $y$-coordinate of the graph is $0$, i.e., when the graph crosses the $x$-axis. Therefore, to count the roots, you need to count how many times the graph crosses the $x$-axis.  In this question, the graph crosses the $x$-axis $\\var{num_roots}$ time(s), so there are $\\var{num_roots}$ roots.

\n

(ii) Definition: A 'stationary point' of a function is a point on the graph where $f'(x)=0$.  Remember that $f'$ tells us the gradient of $f$, so visually a stationary point is where the gradient of the curve is 0.  In this question, there is/are $\\var{num_stat}$ place(s) where the gradient of the graph is $0$, so the answer is $\\var{num_stat}$.

A graph (of a cubic) is given. The question is to determine the number of roots and number of stationary points the graph has. Non-calculator. Advice is given.

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Finding the number of roots and turning points based on a graph.

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{plotgraph(num_stat,num_roots, a, hshift, vshift)}

\n

Above is the graph of some function $f$.

\n

How many roots does $f$ have? [[0]]

\n

How many stationary points does $f$ have? [[1]]

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The number of roots.

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Random amount of vertifical shift for sake of variability.

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Number of stationary points

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Random amount of horizontal shift to create variability.

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Coefficient of x^3

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