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DIAGNOSYS is a knowledge-based test of mathematics background knowledge for first-year university students, created by John Appleby at Newcastle University.

The questions have been translated directly into Numbas, with as few changes as possible.

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Calculate: $(\\var{a}) \\times (\\var{b})$

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Calculate: $\\var{a} \\times (\\var{b})$

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Calculate: $(\\var{a}) + (\\var{b})$

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Which is the largest of the following?

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Which of the following ratios is not equal to the others?

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Which of the following are true?

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Add: $\\simplify[]{ {a}/{b} + {c}/{d}}$ leaving your answer in the simplest possible form.

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Select all of the following that are true.

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If a car takes {hours1} hours for a journey travelling at {speed1} miles per hour (mph), how many hours would it take if it travelled at {speed2} mph?

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Cancel all possible common factors to leave the fraction in its simplest form.

$\\frac{\\var{a}}{\\var{b}} =$ [[0]]

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Calculate $\\simplify[]{ {a}/{b} - {c}/{d} }$.

Enter your answer in the simplest possible form.

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The following inequality can be solved to give $x < a$.

\\[ \\simplify[]{ {a}-{b} < {c} - {d}x } \\]

Enther the number $a$.

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Enter the number $\\var{n}$ rounded to {dp} decimal places.

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Enter the value of $\\var{a}^\\var{b}$.

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Enter the number $\\var{n}$ rounded to {sf} significant figures.

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Enter (as a fraction) the number given by $\\var{a}^{\\var{-b}}$.

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If $\\simplify[]{ {x}^{a} * {x}^{b} = ({x}^{c})^n}$, what is the number $n$?

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We can write the number $\\var{n}$ in the form $\\var{significand} \\times 10^n$. Enter the number $n$.

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Express the following in its simplest form (without powers/indices):

\\[ \\var{expr} \\]

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If $\\var{expr}$, what is the value of $n$?

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If $\\var[fractionnumbers]{expr}$ give the value of $a$.

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Give the value of $\\simplify[fractionnumbers,flatfractions]{ {a}^{b} }$.

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The number $\\frac{\\var{a} \\times 10^{\\var{b}}}{\\var{c} \\times 10^{\\var{d}}}$ can be simplified to give $\\var{significand} \\times 10^n$.

Enter the value of $n$.

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If $\\simplify{ {a}ln({b}) - {c}ln({d})} = \\ln(a)$, give the number $a$.

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Solving $a\\log(x)+\\log(b)=\\log(c)$ for $x$, where $a$, $b$ and $c$ are positive integers.

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Solve for $x$:

\\[ \\var{a}\\log(x)+\\log(\\var{b})=\\log(\\var{c}). \\]

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To solve $\\var{a}\\log(x)+\\log(\\var{b})=\\log(\\var{c})$ for $x$, we want to use the following logarithm rules:

$\\log(a^n) = n\\log(a)$;
$\\log(a)+\\log(b) = \\log(ab)$.

Hence,

\\[ \\begin{split} \\var{a}\\log(x)+\\log(\\var{b}) &\\,=\\log(\\var{c}) \\\\ \\log(x^\\var{a})+\\log(\\var{b}) &\\,= \\log(\\var{c}) \\\\ \\log(\\var{b}x^\\var{a}) &\\,= \\log(\\var{c}). \\end{split} \\]

If $\\log(a)=\\log(b)$ then this implies $a=b$. Therefore,

\\[ \\begin{split} \\var{b}x^\\var{a} &\\,=\\var{c} \\\\ x^\\var{a} &\\,= \\simplify[fractionNumbers]{{c/b}} \\\\ x &\\,= \\simplify[fractionNumbers]{({c/b})^(1/{a})} \\\\ x &\\,= \\var{sol} \\text{ (2 d.p.)}\\end{split} \\]

Use this link to find rsources to help you revise how the rules of logarithms to help you solve logarithmic equations.

$x=$ [[0]] (Give you answer to 2 decimal places where necessary)

Finding $x$ from a logarithmic equation of the form $\\log_a\\left(\\frac{1}{x}\\right) = b$, where $a$ is a positive integer and $b$ is a negative integer.

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Solve for $x$:

\\[ \\log_\\var{a}\\left(\\frac{1}{x}\\right) = \\var{b}. \\]

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Recall that $\\log_a(n)=b$ is equivalent to $n=a^b$.

Therefore,

\\[ \\log_\\var{a}\\left(\\frac{1}{x}\\right) = \\var{b} \\quad \\implies \\quad \\frac{1}{x} = \\var{a}^\\var{b}.\\]

Hence,

\\[ \\frac{1}{x} = \\frac{1}{\\var{a}^\\var{-b}},\\]

So, \\[ x=\\var{sol}.\\]

$x=$ [[0]]

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In the following, the factor $w^\\var{a}$ has been taken out of the left-hand side to give the right-hand side.

\\[ \\var{expr} \\]

Enter the value of the number $n$.

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Collect terms in the following expression

\\[ \\var{expr} \\]

leaving your answer in the simplest possible form.

Simplify the following by collecting like terms.

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$\\simplify[!collectnumbers]{{a[1]}x+{b[1]}y+{c[1]}z+{b[2]}y+{a[2]}x+{c[2]}z+{a[0]}x+{c[0]}z+{b[0]}y}$ = [[0]]

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Like terms are terms where the variable part is the same. For example, $4x$ and $-x$ have the same variable part $x$. However, $3x$ and $-2y$ have different variable parts and are therefore unlike terms (or not like terms).

We can only collect like terms! Just like we can't say 2 m + 3 cm equals 5 m or 5 cm, we can't say $2x+3y$ equals $5x$ or $5y$! We can, however, say $2a+3a=5a$.

In our question we look at all the terms with a variable part of $x$ and add up all the corresponding coefficients, we do the same for the $y$ terms and the $z$ terms:

\\[\\simplify[!collectnumbers]{{a[1]}x+{b[1]}y+{c[1]}z+{b[2]}y+{a[2]}x+{c[2]}z+{a[0]}x+{c[0]}z+{b[0]}y}=\\simplify[basic]{({a[1]}+{a[2]}+{a[0]})x+({b[1]}+{b[2]}+{b[0]})y+({c[1]}+{c[2]}+{c[0]})z}\\]

We present this as the sum of three unlike terms:

\\[\\simplify{{sum(a)}x+{sum(b)}y+{sum(c)}z}\\]

$\\simplify[!collectnumbers]{{d[1]}x^2+{f[1]}x+{g[1]}+{d[0]}x^2+{f[0]}x+{g[0]}}$ = [[0]]

Like terms are terms where the variable part is the same. For example, $4x$ and $-x$ have the same variable part $x$. However, $3x^2$ and $-2x$ have different variable parts and are therefore unlike terms (or not like terms).

We can only collect like terms! Just like we can't say 2 m + 3 cm equals 5 m or 5 cm, we can't say $2x^2+3x$ equals $5x^2$ or $5x$! We can, however, say $2x^2+3x^2=5x^2$.

In our question we look at all the terms with a variable part of $x^2$ and add up all the corresponding coefficients (the numbers in front of the variables), we do the same for the $x$ terms and the constant terms (the terms with no variable part):

\\[\\simplify[!collectnumbers]{{d[1]}x^2+{f[1]}x+{g[1]}+{d[0]}x^2+{f[0]}x+{g[0]}}=\\simplify[basic]{({d[1]}+{d[0]})x^2+({f[1]}+{f[0]})x+({g[1]}+{g[0]})}\\]

We present this as the sum of three unlike terms:

\\[\\simplify{{sum(d)}x^2+{sum(f)}x+{sum(g)}}\\]

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Solve the equation, and give the value of $x$:

\\[ \\var{expr} \\]

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Simplify: $\\simplify[]{ {a} - {b}*{c} }$.

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Evaluate $\\var{a} + \\var{b}\\var{x}$ if $\\var{x} = \\var{c}$.

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What is $\\var{a} \\times \\var{b}$?

Enter the other factor in the equation:

\\[ \\var{expr}( ? ) \\]

Factorise the following, taking out the highest factor possible:

\\[ \\var{expr} \\]

Expand the bracket in:

\\[ \\var{expr} \\]

Things like \"expand 4(5a-3)\"

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The number in front of the bracket is multiplying the bracketed term, that is, each term in the brackets. Also, recall that a negative multiplied by a negative is a positive.

$\\begin{align*}\\simplify{{nmult}({nxcoeff}a+{nconstant})}&=\\simplify[expandBrackets]{{nmult}({nxcoeff}a+{nconstant})}\\\\&=\\simplify[!noLeadingMinus]{{nmult*nxcoeff}a+{nmult*nconstant}}\\end{align*}$

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The expression $\\simplify{{nmult}({nxcoeff}a+{nconstant})}$ is factorised (written as a product). We can expand the expression (so it is written as a sum) to get

[[0]]$a$ + [[1]]

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Collect the terms in the following expression:

\\[ \\var{expr} \\]

Solve the equation:

\\[ \\var{expr} \\]

Enter the value of $\\var{x}$.

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If $\\var{expr}$ then what is $\\var{x}$?

Enter an expression for $\\var{x}$.

If $Q = \\var{expr}$ and $\\var{values_string}$, then what value has $Q$?

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To calculate $\\var{expr}$ you press a sequence of keys on your calculator.

Which one of the following would give you the WRONG answer?

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If $\\var{expr1}$ is divided by $\\var{expr2}$ then the result (when simplified) is:

Two fractions are put over a common denominator as shown:

\\[ \\var{expr} = \\frac{?}{\\var{denom}} \\]

Enter the numerator (shown as $?$) in simplified form:

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Which one of the following statements about this quadratic equation is true?

\\[ \\var{expr} \\]

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The quadratic equation $x^2+ax+b=0$ has roots $x = \\var{x1}$ and $x=\\var{x2}$.

Enter the values of $a$ and $b$.

$a = $ [[0]]

$b = $ [[1]]

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Factorise the following expression into two brackets:

\\[ \\var{expr} \\]

that is, into the form $(\\var{x}\\ldots)(\\var{x}\\ldots)$.

Expand the brackets and collect terms:

\\[ \\var{expr} \\]

Factorise the following expression

\\[ \\var{expr} \\]

Solve the simultaneous equations

\\begin{align}
\\simplify{ {x1}x + {y1}y} &=\\var{c1} \\\\
\\simplify{ {x2}x + {y2}y} &= \\var{c2}
\\end{align}

Enter the values for $x$ and $y$:

$x = $ [[0]]

$y = $ [[1]]

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Solve $\\var{expr}$

Enter the value of $y$.

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If we solve the quadratic equation

\\[ \\simplify{ x^2 + {a}x + {b} = 0 } \\]

we obtain two solutions in the form $x=\\var{c} \\pm \\sqrt{n}$.

Enter the value of $n$.

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Simplify $(\\var{a}) - (\\var{b})$.

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When we wish to add these fractions, we put them over a lowest common denominator.

\\[ \\var{expr} \\]

Enter the lowest common denominator (in factorised form):

Which of the following are correct?

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If we solve the equation $\\simplify{{x}^2+{c1}{x}+{c2}=0}$ by completing the square, we get an answer in the form $(\\var{x}-a) = \\pm b$.

Enter the number $b$.

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Put the expression $\\simplify{ x^2 + {b}x + {c}}$ into the form $(\\ldots)^2 + \\text{number}$ by completing the square.

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Calculate the product of complex numbers: $\\simplify[]{ {a}*{b} }$

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Which of the following have NO solution (can not be satisfied for any value of $x$?)

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If $f(z) = \\var{f}$, what is $f(\\var{val})$?

How many real solutions are there to the equation $\\simplify{x^2+{b}x+{c}}=0$?

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Solve the following equation for $T$ in terms of the constant $a$:

(i.e. make $T$ the subject of the formula):

\\[ a = \\var{expr} \\]

Enter your expression for $T$.

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Differentiate $\\var{x}^\\var{n}$ with respect to $\\var{x}$.

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What value of $x$ gives the minimum of the function $f(x) = \\var{expr}$?

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A Geometric Progression has first term $\\var{a}$ and ratio $\\var{r}$.

What is the 4^th term?

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Differentiate $\\var{expr}$ with respect to $x$.

If $\\frac{\\mathrm{d}y}{\\mathrm{d}t} = \\simplify{ {a}t^{b} }$ find $y$, given $y=0$ when $t=0$.

Enter $y$.

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Enter the $x$ and $y$ coordinates of the point shown:

{graph(x,y)}

( [[0]] , [[1]] )

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What is the gradient of the straight line joing the points $\\var[rowvector]{a}$ and $\\var[rowvector]{b}$?

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The equation (in $x$ and $y$) of the straight line joining the points $\\var[rowvector]{a}$ and $\\var[rowvector]{b}$ is $y = ?$

Enter the right-hand side of the equation.

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Choose the function whose graph looks like this.

{graph()}

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Choose the function whose graph looks like this.

{graph()}

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Enter the number $n$ (under the square root):

{max_width(20,graph)}

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Enter the {fn} of the angle $\\theta$

{max_width(20,graph)}

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What is {a}% of {b}?

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A circle with radius $\\var{r}$ with centre at $x=\\var{x}$, $y=\\var{y}$, has equation

\\[ ? = \\var{r^2} \\]

Enter the left-hand side of the equation.

Which of the following statements are true for all values of $x$?

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Express the angle $\\simplify{ {a}pi/{b} }$ radians in degrees.

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The value of a {vehicle} is initially {currency(price,\"£\",\".\")}. If the value {direction1} by {change1}%, then {direction2} by {change2}%, what is the final value?

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What is the radius of a circle with equation $\\simplify{ x^2 + {xc}x + y^2 + {yc}y + {c}}=0$?

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What is the period of the function $\\var{expr}$?

($\\pi$ should appear in your answer)

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What is the area of the triangle shown?

{max_width(30,drawing)}

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What is the area of the trapezium shown?

{max_width(30,drawing)}

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What is the {measure1} of a circle with {measure2} $\\var{v}$?

{max_width(20,drawing)}

", "answer": "{answer}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": true, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "valuegenerators": []}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "exam"}]}, {"name": "263, Similar triangles", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": [""], "variable_overrides": [[]], "questions": [{"name": "Similar triangles", "extensions": ["eukleides", "jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "http://localhost:8000/accounts/profile/1/"}, {"name": "Christian Lawson-Perfect", "profile_url": "http://localhost:8000/accounts/profile/1/"}, {"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "tags": ["category: area+volume", "leads to: 362, Area/Length relationship", "skill: 263, Similar triangles"], "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"drawingA": {"name": "drawingA", "group": "Ungrouped variables", "definition": "eukleides(\"Two similar triangles, with sides 4,7,a and 12,b,27\",\nlet(\np1,point(2,2),\np2,point(5,2),\np3,point(7,7),\np4,point(13,2),\np5,point(22,2),\np6,point(28,17),\ns1,label(4),\ns2,label(7),\ns3,label(\"a\") italic,\ns4,label(12),\ns5,label(\"b\") italic,\ns6,label(27),\nstroke,size(4),\n[\n p1..p2..p3 stroke, p4..p5..p6 stroke,\np2..p1 s1 stroke,\np3..p2 s2 stroke,\np1..p3 s3 stroke,\np5..p4 s4 stroke,\np6..p5 s5 stroke,\np4..p6 s6 stroke\n]\n))", "description": "", "templateType": "anything", "can_override": false}, "groups": {"name": "groups", "group": "Ungrouped variables", "definition": "[\n [drawingA,9,21],\n [drawingB,4,24],\n [drawingC,11,24],\n [drawingD,5,21],\n [drawingE,7,25]\n]", "description": "", "templateType": "anything", "can_override": false}, "drawingB": {"name": "drawingB", "group": "Ungrouped variables", "definition": "eukleides(\"Two similar triangles, with sides a,5,6 and 16,20,b\",\nlet(\np1,point(2,2),\np2,point(5,2),\np3,point(7,7),\np4,point(13,2),\np5,point(22,2),\np6,point(28,17),\ns1,label(\"a\") italic,\ns2,label(5),\ns3,label(6),\ns4,label(16),\ns5,label(20),\ns6,label(\"b\") italic,\nstroke,size(4),\n[\n p1..p2..p3 stroke, p4..p5..p6 stroke,\np2..p1 s1 stroke,\np3..p2 s2 stroke,\np1..p3 s3 stroke,\np5..p4 s4 stroke,\np6..p5 s5 stroke,\np4..p6 s6 stroke\n]\n))", "description": "", "templateType": "anything", "can_override": false}, "drawingC": {"name": "drawingC", "group": "Ungrouped variables", "definition": "eukleides(\"Two similar triangles, with sides 5,8,a and 15,b,33\",\nlet(\np1,point(2,2),\np2,point(5,2),\np3,point(7,7),\np4,point(13,2),\np5,point(22,2),\np6,point(28,17),\ns1,label(5),\ns2,label(8),\ns3,label(\"a\") italic,\ns4,label(15),\ns5,label(\"b\") italic,\ns6,label(33),\nstroke,size(4),\n[\n p1..p2..p3 stroke, p4..p5..p6 stroke,\np2..p1 s1 stroke,\np3..p2 s2 stroke,\np1..p3 s3 stroke,\np5..p4 s4 stroke,\np6..p5 s5 stroke,\np4..p6 s6 stroke\n]\n))", "description": "", "templateType": "anything", "can_override": false}, "drawingD": {"name": "drawingD", "group": "Ungrouped variables", "definition": "eukleides(\"Two similar triangles, with sides a,6,7 and 15,18,b\",\nlet(\np1,point(2,2),\np2,point(5,2),\np3,point(7,7),\np4,point(13,2),\np5,point(22,2),\np6,point(28,17),\ns1,label(\"a\") italic,\ns2,label(6),\ns3,label(7),\ns4,label(15),\ns5,label(18),\ns6,label(\"b\") italic,\nstroke,size(4),\n[\n p1..p2..p3 stroke, p4..p5..p6 stroke,\np2..p1 s1 stroke,\np3..p2 s2 stroke,\np1..p3 s3 stroke,\np5..p4 s4 stroke,\np6..p5 s5 stroke,\np4..p6 s6 stroke\n]\n))", "description": "", "templateType": "anything", "can_override": false}, "drawingE": {"name": "drawingE", "group": "Ungrouped variables", "definition": "eukleides(\"Two similar triangles, with sides 3,5,a and 15,b,35\",\nlet(\np1,point(2,2),\np2,point(5,2),\np3,point(7,7),\np4,point(13,2),\np5,point(22,2),\np6,point(28,17),\ns1,label(3),\ns2,label(5),\ns3,label(\"a\") italic,\ns4,label(15),\ns5,label(\"b\") italic,\ns6,label(35),\nstroke,size(4),\n[\n p1..p2..p3 stroke, p4..p5..p6 stroke,\np2..p1 s1 stroke,\np3..p2 s2 stroke,\np1..p3 s3 stroke,\np5..p4 s4 stroke,\np6..p5 s5 stroke,\np4..p6 s6 stroke\n]\n))", "description": "", "templateType": "anything", "can_override": false}, "group": {"name": "group", "group": "Ungrouped variables", "definition": "random(groups)", "description": "", "templateType": "anything", "can_override": false}, "drawing": {"name": "drawing", "group": "Ungrouped variables", "definition": "group[0]", "description": "", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "Ungrouped variables", "definition": "group[1]", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "group[2]", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["drawingA", "drawingB", "drawingC", "drawingD", "drawingE", "groups", "group", "drawing", "a", "b"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "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": "

We have two similar triangles as shown:

{max_width(30,drawing)}

Enter the lengths of sides $a$ and $b$:

$a = $ [[0]]

$b = $ [[1]]

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What is the volume of a cylinder with radius {r} and height {h}?

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We have two similar triangles as shown:

{max_width(30,drawing)}

The ratio of the lengths of their sides is $\\var{r}:1$.

The ratio of their areas is $a:1$. Enter the number $a$.

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What is the area of the coloured part in terms of $x$?

{max_width(30,drawing)}

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What is the surface area of a cylinder with radius {r} and height {h} (including both ends)?

", "answer": "{2r^2 + 2r*h}*pi", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}]}, {"name": "462, Volume Area Length relationships", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": [""], "variable_overrides": [[]], "questions": [{"name": "Volume Area Length relations", "extensions": ["eukleides", "jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "http://localhost:8000/accounts/profile/1/"}, {"name": "Christian Lawson-Perfect", "profile_url": "http://localhost:8000/accounts/profile/1/"}, {"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "tags": ["category: area+volume", "skill: 462, Volume Area Length relationships"], "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"groups": {"name": "groups", "group": "Ungrouped variables", "definition": "[\n [\n \"doubled\",\n [\n \"the surface area is doubled\",\n \"the volume is 8 times as great\",\n \"the surface area is multiplied by 4\",\n \"the volume is multiplied by 16\",\n \"the volume is doubled\"\n ],\n [0,1,1,0,0]\n ],\n [\n \"multiplied by three\",\n [\n \"the volume is 27 times as great\",\n \"the surface area is tripled\",\n \"the volume is multiplied by 81\",\n \"the volume is tripled\",\n \"the surface area is multiplied by 9\"\n ],\n [1,0,0,0,1]\n ]\n]", "description": "", "templateType": "anything", "can_override": false}, "group": {"name": "group", "group": "Ungrouped variables", "definition": "random(groups)", "description": "", "templateType": "anything", "can_override": false}, "action": {"name": "action", "group": "Ungrouped variables", "definition": "group[0]", "description": "", "templateType": "anything", "can_override": false}, "choices": {"name": "choices", "group": "Ungrouped variables", "definition": "group[1]", "description": "", "templateType": "anything", "can_override": false}, "choice_marks": {"name": "choice_marks", "group": "Ungrouped variables", "definition": "group[2]", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["groups", "group", "action", "choices", "choice_marks"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"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": "

If the dimensions of a cube are {action} which of these are true?

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What is the mean (average) of the numbers given:

{table([data])}

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What is the range of the discrete data given?

{table([data])}

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A fair coin is tossed twice with equal probability of 'head' or 'tail'.

What is the probability of obtaining {statement}?

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The percentage of students on a course with one, both or neither of A-Level Mathematics and A-Level Physics is shown by the diagram below.

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What is the probability that a randomly chosen student has only one of these A-Levels?

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A family has two children.

{statement}

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The percentage of students on a course with one, both or neither of A-level Mathematics and A-level Physics is shown by the diagram below.

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If a randomly chosen student has A-level Physics, what is the probability he/she also has A-level Mathematics?

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Match the graphs to the functions. No randomisation. Multiple choice.

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This is about knowledge of graphs. Generally with trigonometric graphs it is best to start with making sure you know and understand the graphs of the functionts $\\sin(x)$, $\\cos(x)$ and $\\tan(x)$. From there you can use knowledge of where they are zero to work out the position of the asymptotes in the graphs of $\\sec(x)$, $\\text{cosec}(x)$ and $\\cot(x)$. However, you still need really to be able to recall the shape of each graph for some purposes and be confident about where the zeros and turning points are.

Use this link to find some resources to help you familiarise yourself with these graphs.

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Match the graph to its function.

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Rewrite the expression $\\frac{mx^2+nx+k}{(x+a)(x^2+bx+c)}$ as partial fractions in the form $\\frac{A}{x+a}+\\frac{Bx+C}{x^2+bx+c}$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rewrite the following expression as partial fractions:

\\[ \\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))}. \\]

", "advice": "

To express \\[ \\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} \\] as partial fractions, we want to set this equal to the sum of two fractions with denominators $\\simplify{x+{a}}$ and $\\simplify{x^2+{b}x+{c}}$. Since we have a linear factor and a quadratic factor, this tells us that the form of the partial fractions will be

\\[ \\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\simplify{A/(x+{a}) + (B*x+C)/(x^2+{b}x+{c})},\\]

where $A$, $B$, and $C$ are constants.

To find the values of $A$, $B$, and $C$, we want to first multiply this equation by the denominator of the left-hand side. This gives

\\[ \\simplify{{m}x^2+{n}x+{k}=A(x^2+{b}x+{c})+B*x(x+{a}) + C(x+{a})}.\\]

(Note: To find $A$, $B$, and $C$, we will use a combination of choosing suitable values of $x$ to eliminate terms, and equating coefficients. It can be solved by only equating coefficients, but this is a more efficient process.)

To find $A$, we can eliminate $B$ and $C$ by setting $x=\\var{-a}$:

\\[ \\simplify{{m*a^2-n*a+k}=A{(a^2-b*a+c)}} \\implies A=\\simplify[fractionNumbers]{{Asol}}.\\]

To find $C$, we can eliminate $B$ by setting $x=0$ and substituting in the result of $A$:

\\[ \\simplify{{k}={c}A+{a}C} \\implies C=\\simplify[all,fractionNumbers]{({k}-{c}A)/{a}}.\\]

Hence,

\\[ C = \\simplify[fractionNumbers]{{Csol}}.\\]

Finally, by equating coefficients of the $x^2$-terms we can find $B$:

\\[ (x^2): \\quad \\var{m} = \\simplify{A+B} \\implies B=\\var{m}-A. \\]

Therefore, \\[ B=\\simplify[fractionNumbers]{{Bsol}}, \\]

and

{check}

Use this link to find some resources which will help you revise this topic.

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\\\\[ \\\\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\\\simplify{{Asol}/(x+{a})+({Bsol}x+{Csol})/(x^2+{b}x+{c})}.\\\\]

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\\\\[ \\\\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\\\simplify[all,fractionNumbers]{{m*a^2-n*a+k}/({a^2-a*b+c}(x+{a}))+({m*c-m*b*a+n*a-k}x+{k*(a-b)-m*a*c+n*c})/({a^2-a*b+c}(x^2+{b}x+{c}))}.\\\\]

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\\\\[ \\\\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\\\simplify[all,fractionNumbers]{{m*a^2-n*a+k}/({a^2-a*b+c}(x+{a}))+({(m*c-m*b*a+n*a-k)/simp2}x+{(k*(a-b)-m*a*c+n*c)/simp2})/({(a^2-a*b+c)/simp2}(x^2+{b}x+{c}))}.\\\\]

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[[0]]

Solving an equation of the form $a^x=b$ using logarithms to find $x$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve for $x$:

\\[ \\var{a}^x = \\var{b} \\,. \\]

", "advice": "

To solve $\\var{a}^x = \\var{b}$ for $x$, since $x$ is the exponent we want to make use of the following logarithm rule:

$\\log(a^n) = n \\log(a)$.

By taking the logarithm of each side and applying the above rule:

\\[ \\begin{split}\\var{a}^x &\\,= \\var{b} \\\\ \\log_{10}(\\var{a}^x) & \\,= \\log_{10}(\\var{b})\\\\ x \\log_{10}(\\var{a}) &\\,= \\log_{10}(\\var{b}) \\\\\\\\ x&\\,=\\simplify{log({b})/log({a})} \\\\\\\\ x &\\,= \\var{sol} \\text{ (2 d.p.)}. \\end{split} \\]

Use this link to find resources to help you revise how logarithms.

$x=$ [[0]] (Give you answer to 2 decimal places where necessary)

Finding $x$ from a logarithmic equation of the form $\\log_ax = b$, where $a$ and $b$ are positive integers.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the value of $x$:

\\[ \\log_\\var{a}x = \\var{n} \\]

", "advice": "

To find the value of $x$, recall that $\\log_a(x)=b$ is equivalent to $x=a^b$.

Therefore, \\[\\log_\\var{a}(x) = \\var{n} \\implies \\simplify[!collectNumbers]{x={a}^{n}}.\\]

Hence, \\[x=\\var{a^n}\\,.\\]

Use this link to find resources to help you revise logarithms.

$x=$ [[0]]

Using the Table of Integrals/Antiderivatives, calculate the integral of $y=\\simplify[unitFactor, unitPower,fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}}$.

", "advice": "

From the Table of Integrals we see that a function of the form \\[ f(x)=x^n \\] has the integral \\[ \\int x^n dx = \\frac{x^{n+1}}{n+1}+ c,\\]

and \\[\\int kf(x) dx = k \\int f(x) dx.\\]

Additionally, the integral of the sum or difference of two or more functions is equal to the sum or difference of the integrals of each function: \\[ \\int(f(x)\\pm g(x))\\, dx = \\int f(x)\\, dx \\pm \\int g(x) \\, dx.\\]

So, for the function

\\[y=\\simplify[all,fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}},\\]

the integral is

\\[ \\begin{split}\\simplify[all,fractionNumbers]{int({a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3},x)} = \\simplify[all,fractionNumbers]{{a_1}int(x^{b_1},x)+{a_2}int(x^{b_2},x)+{a_3}int(x^{b_3},x)} &\\,= \\simplify[all,fractionNumbers,!collectNumbers,!simplifyFractions,!noLeadingMinus]{{a_1}(x^({b_1+1})/{b_1+1}) +{a_2}(x^({b_2+1})/{b_2+1}) +{a_3}(x^({b_3+1})/{b_3+1})} + c,\\\\ \\\\&\\,= \\simplify[all,fractionNumbers]{{a_1/(b_1+1)}x^{b_1+1}+{a_2/(b_2+1)}x^{b_2+1}+{a_3/(b_3+1)}x^{b_3+1}}+c.\\end{split} \\]

Note: You only need to put one $c$ term here, you do not need to put a separate constant term for each calculation.

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[[0]]

Using the Table of Integrals/Antiderivatives, calculate the integral of $f(x)=\\var{a}e^{\\simplify[fractionNumbers]{{b}x}}$

", "advice": "

From the Table of Integrals we see that a function of the form \\[ f(x)= e^{mx} \\] has the integral \\[ \\int e^{mx} dx = \\frac{1}{m}e^{mx}+c,\\]

and \\[ \\int kf(x) \\,dx = k \\int f(x) \\, dx.\\]

So, for the function

\\[f(x)=\\simplify[fractionNumbers]{{a}e^({b}x)},\\]

the integral is

\\[ \\begin{split} \\int\\simplify[fractionNumbers]{{a}e^({b}x)} dx \\,= \\simplify[unitFactor,fractionNumbers]{{a}int(e^({b}x),x)} &\\,=\\var{a}\\times\\simplify[unitFactor,fractionNumbers,unitDenominator]{(e^({b}x)/{b})} +c, \\\\ &\\,=\\simplify[unitFactor,fractionNumbers]{{a/b} e^({b}x)+c}. \\end{split} \\]

[[0]]

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Calculating the definite integral $\\int_{n_1}^{n_2}a_1x^{b_1}+a_2x^{b_2}+a_3x^{b_3} dx$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate \\[ \\int_{\\var{n_1}}^{\\var{n_2}}\\simplify[unitFactor, unitPower, fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}} \\,dx.\\]

", "advice": "

Integrating a function of the form \\[ f(x)=x^n \\] has the integral \\[ \\int_a^b x^n dx = \\left[\\frac{x^{n+1}}{n+1}\\right]_a^b,\\]

and \\[\\int_a^b kf(x) dx = k \\int_a^b f(x) dx.\\]

Therefore,

\\[ \\begin{split}\\simplify[unitFactor,unitPower]{defint({a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3},x,{n_1},{n_2})} &\\,= \\simplify{{a_1}defint(x^{b_1},x,{n_1},{n_2})+{a_2}defint(x^{b_2},x,{n_1},{n_2})+{a_3}defint(x^{b_3},x,{n_1},{n_2})} \\\\ &\\,= \\left[\\simplify[all,fractionNumbers]{{a_1}x^{b_1+1}/{b_1+1}+{a_2}x^{b_2+1}/{b_2+1}+{a_3}x^{b_3+1}/{b_3+1}}\\right]_\\var{n_1}^\\var{n_2} \\\\ &\\,= \\left[\\simplify[all,fractionNumbers,!collectNumbers]{{a_1*n_2^(b_1+1)}/{b_1+1}+{a_2*n_2^(b_2+1)}/{b_2+1}+{a_3*n_2^(b_3+1)}/{b_3+1}}\\right] -\\left[\\simplify[all,fractionNumbers,!collectNumbers]{{a_1*n_1^(b_1+1)}/{b_1+1}+{a_2*n_1^(b_2+1)}/{b_2+1}+{a_3*n_1^(b_3+1)}/{b_3+1}}\\right] \\\\ &\\,= \\simplify[!collectNumbers]{{eval2a}-{eval1a}} \\\\ &\\,=\\var{sol1} \\end{split} \\]

Use this link to find some resources on areas under curves which will help you revise this topic.

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[[0]] (Give answers to 2 decimal places where necessary)

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Calculating the integral of a function of the form $\\frac{nx^{n-1}}{x^n+a}$ using integration by substitution.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate \\[ \\simplify[all]{int(({n}x^{n-1})/(x^{n}+{a}),x)}\\]

by using the substitution \\[ \\simplify[all]{u=x^{n}+{a}}.\\]

", "advice": "

Since this integral is of the form \\[ \\int g'(x)f(g(x))\\,dx,\\] we can use the method of substitution to calculate the solution.

Firstly, we must make a change of variables from $x$ to $u$, where $u$ is equal to the 'inner' function $g(x)$.

So, for \\[\\simplify[fractionNumbers]{int(({n}x^{n-1})/((x^{n}+{a})),x)}\\]

let $\\color{red}{u=\\simplify[fractionNumbers]{x^{n}+{a}}}.$

Now, we need to calculate the differential, $du$, where \\[ du = \\left(\\frac{du}{dx}\\right)dx. \\]

Differentiating $u$ with respect to $x$:

\\[ \\frac{du}{dx}= \\simplify[fractionNumbers]{{n}x^{n-1}}.\\]

Therefore, \\[ \\color{blue}{du = \\simplify[fractionNumbers]{{n}x^{n-1}}\\, dx}.\\]

We can now rewrite the original integral in terms of $u$:

\\[ \\int \\frac{\\color{blue}{\\simplify{{n}x^{n-1}}}}{\\color{red}{\\simplify{x^{n}+{a}}}}\\color{blue}{\\text{d}x} = \\int \\frac{1}{\\color{red}{u}}\\color{blue}{\\text{d}u}.\\]

(Note: It is important to see that both the function we are integrating, and the variable we are integrating with respect to, has changed.)

\\[ \\simplify[fractionNumbers]{int(1/u,u) = ln(abs(u)) + c}.\\]

Finally, we must rewrite our solution back in terms of the original variable $x$:

\\[ \\simplify[fractionNumbers]{ln(abs(u)) + c = ln(abs(x^{n}+{a})) + c}.\\]

Use this link to find some resources which will help you revise this topic.

[[0]]

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Correct answer", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": true, "customName": "Alternative using brackets", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

Technically we should use the absolute value symbols for the logs. This can be done in NUMBAS by using \"abs(*function*)\".

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Technically we should use the absolute value symbols for the logs. This can be done in NUMBAS by using \"abs(*function*)\".

", "useAlternativeFeedback": false, "answer": "ln(x^{n}+{a})+k", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "k", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Forgotten constant", "marks": "0.5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

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Calculating the integral of a function of the form $ax^2 \\cos(bx)$ using integration by parts.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the integral \\[ \\simplify{int({a}x^2 cos({b}x),x)}\\]

", "advice": "

If we have a function of $x$ which is the product of two functions of $x$, to integrate such a function it is often necessary to use Integration by Parts. The formula for Integration by Parts is:

\\[ \\int u(x) \\frac{dv}{dx} dx = u(x)v(x) - \\int v(x) \\frac{du}{dx} dx.\\]

Using this method can be broken down into steps:

Identify $u(x)$ and $\\tfrac{dv}{dx}$ (The function you pick for each is important, in general you want $u(x)$ to become simpler when differentiating it, and you must be able to integrate $\\tfrac{dv}{dx}$ to find $v(x)$);
Calculate $\\tfrac{du}{dx}$ and $v(x)$;
Put the functions $u(x)$, $v(x)$, and their derivatives into the Integration by Parts formula;
Calculate the integral $\\int v(x) \\tfrac{du}{dx} dx$ (This may require you to use Integration by Parts again, this is OK!);
Simplify your answer where possible and don't forget to add the constant of integration.

For the integral

\\[ \\simplify{int({a}x^2 cos({b}x),x)},\\]

we must first identify $u(x)$ and $\\tfrac{dv}{dx}$. In this case, let \\[ u(x)=\\simplify{{a}x^2},\\quad \\frac{dv}{dx}= \\simplify{cos({b}x)}. \\]

Next, we need to calculate $\\tfrac{du}{dx}$ and $v(x)$:

\\[ \\begin{split} u(x) = \\var{a}x^2 \\quad &\\implies \\frac{du}{dx} = \\simplify{{2a}x}; \\\\ \\frac{dv}{dx} = \\cos(\\var{b}x) &\\implies v(x) = \\simplify[fractionNumbers]{1/{b} sin({b}x)}. \\end{split} \\]

Plugging these 4 terms into the integration by parts formula:

\\[ \\begin{split} \\simplify{int({a}x^2 cos({b}x),x)} &\\,= \\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) - int({2a/b}x sin({b}x),x)}, \\\\ \\\\ &\\,= \\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) -{2a/b}int(x sin({b}x),x)}.\\end{split} \\]

Since the integral on the right-hand side is still the product of two functions of $x$, we need to use integration by parts again.

So, for

\\[ \\simplify{int(x sin({b}x),x)}, \\]

Let $u=x$ and $\\tfrac{dv}{dx} = \\sin(\\var{b}x)$. Therefore, $\\tfrac{du}{dx}=1$ and $v(x)=\\simplify{-1/{b} cos({b}x)}$.

Hence,

\\[ \\begin{split} \\simplify{int(x sin({b}x),x)} &\\,= \\simplify{-1/{b}x cos({b}x)- int(-1/{b} cos({b}x),x)} \\\\ \\\\ &\\,= \\simplify{-1/{b}x cos({b}x)+1/{b^2}sin({b}x)}. \\end{split}\\]

Plugging this back into the original calculation:

\\[ \\begin{split} \\simplify{int({a}x^2 cos({b}x),x)} &\\,= \\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) -{2a/b}int(x cos({b}x),x)} \\\\ \\\\ &\\,= \\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) -{2a/b}[-1/{b}x cos({b}x)+1/{b^2}sin({b}x)]} \\\\ \\\\ &\\,=\\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) +{2a/b^2}x cos({b}x)-{2a/b^3}sin({b}x)} + c.\\end{split} \\]

Use this link to find some resources which will help you revise this topic.

[[0]]

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It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

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Calculating the integral of a function of the form $\\frac{c}{(x+a)(x+b)}$ using partial fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the integral

\\[ \\simplify{int({c}/((x^2+{aPlusb}x+{ab})),x)} .\\]

", "advice": "

In order to integrate the function \\[ \\simplify{int({c}/((x^2+{aPlusb}x+{ab})),x)}, \\] we want to rewrite it in terms of its partial fractions.

First we need to factorise the denominator so we have

\\[ \\simplify{{c}/((x+{a})(x+{b}))}. \\]

Now to write this as a partial fraction, we want to set the function equal to the sum of 2 fractions with denominators $\\simplify{x+{a}}$ and $\\simplify{x+{b}}$. Since these are both distinct linear factors, this tells us that the numerators will be constants, which we will call $A$ and $B$:

\\[ \\simplify{{c}/((x+{a})(x+{b}))} = \\simplify{A/(x+{a}) + B/(x+{b})}.\\]

To find the values of $A$ and $B$, we want to multiply this equation by the denominator of the left-hand side. This gives

\\[ \\simplify{{c}=A(x+{b})+B(x+{a})}.\\]

To find $A$, we can eliminate $B$ by setting $\\simplify{x={-a}}$:

\\[ \\simplify{{c}=A{b-a}} \\implies \\simplify[fractionNumbers]{A={c/(b-a)}}.\\]

Similarly, to find B, we can eliminate $A$ by setting $\\simplify{x={-b}}$:

\\[ \\simplify{{c}=B{a-b}} \\implies \\simplify[fractionNumbers]{B={c/(a-b)}}.\\]

Therefore,

{check1}

and

{check2}

Use this link to find some resources which will help you revise this topic.

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\\\\[ \\\\simplify{{c}/((x+{a})(x+{b}))} = \\\\simplify[all,fractionNumbers]{{Asol}/(x+{a})+{Bsol}/(x+{b})},\\\\]

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\\\\[ \\\\simplify{{c}/((x+{a})(x+{b}))} = \\\\simplify[all,fractionNumbers]{{c}/(({b-a})(x+{a}))+{c}/(({a-b})(x+{b}))},\\\\]

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\\\\[ \\\\begin{split} \\\\simplify{int({c}/((x+{a})(x+{b})),x)} &\\\\,= \\\\simplify[all,fractionNumbers]{int({Asol}/(x+{a})+{Bsol}/(x+{b}),x)}\\\\\\\\\\\\\\\\ &\\\\,=\\\\simplify[all,fractionNumbers]{{Asol} int(1/(x+{a}),x)+{Bsol} int(1/(x+{b}),x)} \\\\\\\\\\\\\\\\ &\\\\,=\\\\simplify[all,fractionNumbers]{{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b})) + C}. \\\\end{split}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "int2": {"name": "int2", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\begin{split} \\\\simplify{int({c}/((x+{a})(x+{b})),x)} &\\\\,= \\\\simplify[all,fractionNumbers]{int({c}/(({b-a})(x+{a}))+{c}/(({a-b})(x+{b})),x)} \\\\\\\\\\\\\\\\ &\\\\,=\\\\simplify[basic,fractionNumbers,zeroFactor,noLeadingMinus]{{Asol} int(1/(x+{a}),x)+{Bsol} int(1/(x+{b}),x)} \\\\\\\\ \\\\\\\\ &\\\\,=\\\\simplify[basic,fractionNumbers,zeroFactor,noLeadingMinus]{{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b})) + C}. \\\\end{split}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "check2": {"name": "check2", "group": "Ungrouped variables", "definition": "if(Asol=round(Asol),'{int1}','{int2}')", "description": "", "templateType": "anything", "can_override": false}, "ab": {"name": "ab", "group": "Ungrouped variables", "definition": "a*b", "description": "", "templateType": "anything", "can_override": false}, "aPlusb": {"name": "aPlusb", "group": "Ungrouped variables", "definition": "a+b", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "100"}, "ungrouped_variables": ["b", "a", "c", "Bsol", "Asol", "check1", "Sol1", "Sol2", "check2", "int1", "int2", "ab", "aPlusb"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "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": "\n

[[0]]

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Correct answer", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": true, "customName": "brackets", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

Technically we should use the absolute value symbols for the logs. This can be done in NUMBAS by using \"abs(*function*)\".

", "useAlternativeFeedback": true, "answer": "{Asol} ln (x+{a})+{Bsol} ln (x+{b}) + c", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Alt constant \"+k\"", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": true, "answer": "{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b})) + k", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "k", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Alt constant \"+k\" brackets", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

Technically we should use the absolute value symbols for the logs. This can be done in NUMBAS by using \"abs(*function*)\".

", "useAlternativeFeedback": true, "answer": "{Asol} ln (x+{a})+{Bsol} ln (x+{b}) + k", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "k", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Forgotten constant", "marks": "0.5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

", "useAlternativeFeedback": false, "answer": "{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b}))", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b})) + c", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}]}, {"name": "341b, Differentiation of polynomials", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": [""], "variable_overrides": [[]], "questions": [{"name": "Differentiating polynomials 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Krishna Kedia", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/25454/"}], "tags": [], "metadata": {"description": "

Differentiate a polynomial expression involving coefficients and, negative and fractional indices.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the derivative of $y=\\simplify[unitFactor, fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}}$.

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=kx^n \\] has a derivative \\[ \\frac{df}{dx} = knx^{n-1}. \\]

Additionally, the derivative of the sum or difference of two or more functions is equal to the sum or difference of the derivatives of each function: \\[ \\frac{d}{dx}(f(x)\\pm g(x)) = \\frac{df}{dx} \\pm \\frac{dg}{dx}.\\]

{advice}

Use this link to find some resources which will help you revise this topic.

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} +(\\\\var[fractionNumbers]{a_2}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} +(\\\\var[fractionNumbers]{a_3}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutionb": {"name": "solutionb", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} -(\\\\var[fractionNumbers]{abs(a_2)}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} +(\\\\var[fractionNumbers]{a_3}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutionc": {"name": "solutionc", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} +(\\\\var[fractionNumbers]{a_2}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} -(\\\\var[fractionNumbers]{abs(a_3)}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutiond": {"name": "solutiond", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} -(\\\\var[fractionNumbers]{abs(a_2)}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} -(\\\\var[fractionNumbers]{abs(a_3)}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

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$\\frac{dy}{dx}=$[[0]]

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Calculating the derivative of an exponential function of the form $ae^{bx}$, using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the derivative of $y=\\simplify[all]{{a}*e^({b}x)}.$

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=a e^{kx} \\] has a derivative \\[ak e^{kx}.\\]

Therefore, the function \\[y=\\simplify[unitFactor]{{a}*e^({b}x)}\\] has a derivative\\[ \\begin{split} \\frac{dy}{dx} &=(\\var{a}\\times \\var{b})e^{\\simplify[unitFactor]{{b}x}}\\\\ &= \\simplify[unitFactor]{{a*b}e^({b}x)}.\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Find the derivative of a function of the form $y=a \\sin(bx+c)$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Using the Table of Derivatives, calculate the derivative of $y=\\simplify[unitFactor]{{a}sin({b}x+{c})}.$

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=a \\sin(kx+c) \\] has a derivative \\[ak \\cos (kx+c).\\]

Therefore, the function \\[y=\\simplify[unitFactor]{{a}*sin({b}x+{c})}\\] has a derivative\\[ \\begin{split} \\frac{dy}{dx} &=(\\var{a}\\times \\var{b})\\cos(\\simplify[unitFactor]{{b}x+{c}})\\\\ &= \\simplify[unitFactor]{{a*b}cos({b}x+{c})}.\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Find the derivative of a function of the form $y=a \\tan(bx+c)$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Using the Table of Derivatives, calculate the derivative of $y=\\simplify[unitFactor]{{a}tan({b}x+{c})}.$

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=a \\tan(kx+c) \\] has a derivative \\[ak \\sec^2(kx+c).\\]

Therefore, the function \\[y=\\simplify[unitFactor]{{a}*tan({b}x+{c})}\\] has a derivative\\[ \\begin{split} \\frac{dy}{dx} &=(\\var{a}\\times \\var{b})\\sec^2(\\simplify[unitFactor]{{b}x+{c}})\\\\ &= \\simplify[unitFactor]{{a*b}}\\sec^2(\\simplify[unitFactor]{{b}x+{c}}).\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

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Calculating the derivative of a function of the form $a \\ln(bx)$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the derivative of $y=\\simplify[unitFactor]{{a}*ln({a_1}*x^2+{a_2}*x+{a_3})}.$

", "advice": "

From the Table of Derivatives and the chain rule we see that a function of the form \\[ f(x)=a \\ln(g(x)) \\] has a derivative \\[\\frac{df}{dx}=\\frac{g'(x)}{g(x)}.\\]

In this case $g(x)=\\var{a_1}x^2+\\var{a_2}x+\\var{a_3}$ so

\\[g'(x)=\\var{2*a_1}x+\\var{a_2}\\]

Therefore, the function \\[ \\simplify[unitFactor]{y={a}ln({a_1}*x^2+{a_2}*x+{a_3})}\\] has a derivative \\[(\\var{a*a_1*2}x+\\var{a*a_2})/(\\var{a_1}x^2+\\var{a_2}x+\\var{a_3})\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

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Calculating the derivative of a function of the form $\\sin(ax^m+bx^n)$ using the chain rule.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the derivative of $y=\\simplify[all]{sin({a}*x^{n}+{b}*x^{m})}$.

", "advice": "

If we have a function of the form $y=f(g(x))$, sometimes described as a function of a function, to calculate its derivative we need to use the chain rule:

\\[ \\frac{dy}{dx} = \\frac{du}{dx} \\times \\frac{dy}{du}.\\]

This can be split up into steps:

Let $u=g(x)$;
Rewrite $y$ in terms of $u$, such that $y=f(u)$;
Calculate $\\frac{du}{dx}$ and $\\frac{dy}{du}$;
Write $\\frac{dy}{dx}$ as a product of $\\frac{du}{dx}$ and $\\frac{dy}{du}$;
Make sure $\\frac{dy}{dx}$ is only in terms of $x$. Ensure any $u$ terms have been replaced using the initial substitution.

Following this process, we must first identify $g(x)$. Since the function is of the form $y=f(g(x))$, we are looking for the 'inner' function.

So, for $y=\\simplify[all,fractionNumbers]{sin({a}*x^{n}+{b}*x^{m})}$, \\[g(x)=\\simplify[all, fractionNumbers, unitFactor]{{a}*x^{n}+{b}*x^{m}}.\\]

If we now set $u=g(x)$, we can rewrite $y$ in terms of $u$ such that $y=f(u)$:

\\[y=\\simplify[all, fractionNumbers,unitFactor]{sin(u)}.\\]

Next, we calculate the two derivatives $\\frac{du}{dx}$ and $\\frac{dy}{du}$:

\\[\\frac{du}{dx}=\\simplify[all,fractionNumbers]{{a*n}x^{n-1}+{b*m}x^{m-1}}, \\quad \\frac{dy}{du}=\\simplify[all, fractionNumbers, unitFactor]{cos(u)}.\\]

Plugging these into the chain rule:

\\[ \\begin{split} \\frac{dy}{dx} &= \\frac{du}{dx} \\times \\frac{dy}{du}, \\\\&=(\\simplify[all,fractionNumbers]{{a*n}x^{n-1}+{b*m}x^{m-1}}) \\times\\simplify[all, fractionNumbers, unitFactor]{cos(u)}. \\end{split} \\]

Finally, we need to express $\\frac{dy}{dx}$ only in terms of $x$, so we must replace the $u$ term using the initial substitution $u=\\simplify[all, fractionNumbers, unitFactor]{{a}*x^{n}+{b}*x^{m}}$:

\\[ \\frac{dy}{dx} =(\\simplify[all,fractionNumbers]{{a*n}x^{n-1}+{b*m}x^{m-1}})\\simplify[all, fractionNumbers, unitFactor]{cos({a}*x^{n}+{b}*x^{m})}.\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

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Calculating the derivative of a function of the form $\\frac{ax^n}{bx+c}$ using the quotient rule.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the derivative of \\[ \\simplify{y={a}x^{n}/({b}x+{c})}. \\]

", "advice": "

If we have a function of the form $y=\\tfrac{u(x)}{v(x)}$, to calculate its derivative we need to use the quotient rule:

\\[ \\dfrac{dy}{dx} = \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2}\\,.\\]

This can be split up into steps:

Identify the functions $u(x)$ and $v(x)$;
Calculate their derivatives $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$;
Substitute these into the formula for the quotient rule to obtain an expression for $\\tfrac{dy}{dx}$;
Simplify $\\tfrac{dy}{dx}$ where possible.

Following this process, we must first identify $u(x)$ and $v(x)$.

As \\[ \\simplify{y={a}x^{n}/({b}x+{c})}, \\]

let \\[ u(x) = \\simplify{{a}x^{n}} \\quad \\text{and} \\quad v(x)=\\simplify{{b}x+{c}}.\\]

Next, we need to find the derivatives, $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$:

\\[ \\dfrac{du}{dx} = \\simplify{{a*n}x^{n-1}}\\quad \\text{and} \\quad\\dfrac{dv}{dx}=\\simplify{{b}}.\\]

Substituting these results into the quotient rule formula we can obtain an expression for $\\tfrac{dy}{dx}$:

\\[ \\begin{split} \\dfrac{dy}{dx} &\\,= \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2} \\\\ \\\\&\\,=\\dfrac{(\\simplify{{b}x+{c}}) \\times\\simplify{{a*n}x^{n-1}} - \\simplify{{a}x^{n}} \\times \\simplify{{b}}}{\\simplify{({b}x+{c})^2}}. \\end{split}\\]

Simplifying,

\\[ \\begin{split} \\dfrac{dy}{dx} &\\,=\\dfrac{(\\simplify{{b}x+{c}})\\simplify{{a*n}x^{n-1}} - \\simplify{{b*a}x^{n}}}{\\simplify{({b}x+{c})^2}} \\\\ \\\\&\\,=\\dfrac{\\simplify[all,!cancelTerms]{{b*a*n}x^{n}+{c*a*n}x^{n-1} - {b*a}x^{n}}}{\\simplify{({b}x+{c})^2}}\\\\ \\\\ &\\,=\\dfrac{\\simplify{{b*a*n}x^{n}+{c*a*n}x^{n-1} - {b*a}x^{n}}}{\\simplify{({b}x+{c})^2}} \\\\ \\\\ &\\,=\\dfrac{\\simplify{{simp}x^{n-1}({(b*a*n-b*a)/simp}x+{c*a*n/simp})}}{\\simplify{({b}x+{c})^2}} \\end{split} \\]

Use this link to find some resources which will help you revise this topic.

$\\dfrac{dy}{dx}=$[[0]]

The number of patients arriving at a dentist’s surgery each afternoon follows
a Poisson distribution, with a mean of four patients per hour.
Calculate the probability that in a particular one-hour period

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

The number of patients arriving at a dentist’s surgery each afternoon follows a Poisson distribution, with a mean of $\\var{l}$ patients per hour.

The Poisson distribution formula: $P(r)=\\frac{\\lambda^re^{-\\lambda}}{r!}$ or $P(r)=e^{-\\lambda}\\left[\\frac{\\lambda^r}{r!}\\right]$

Please give your answer to at least 3 decimal places.

", "advice": "

Part (a)

Remember that for a Poisson random variable:
\\begin{align}
\\operatorname{P}(X=x)&=\\dfrac{\\lambda^x\\times e^{-\\lambda}}{x!}\\\\
\\end{align}

1.\\[ \\begin{eqnarray*}\\operatorname{P}(X = \\var{x}) &=& \\frac{\\var{l} ^ {\\var{x}}e ^ { -\\var{l}}} {\\var{x}!}\\\\& =& \\var{answer1} \\end{eqnarray*} \\] to 3 decimal places.

Part (b)

The probability that in a particular one hour period, the number of patients entering the waiting room will be between $\\var{x}$ and $\\var{y}$ inclusive is given by:

$P(\\var{x} \\leq X\\leq\\var{y}) = P(X=\\var{x}) + P(X=\\var{x+1}) +P(X=\\var{y})$

where

$P(X=\\var{x}) =\\frac{\\var{l}^{\\var{x}}e^{-\\var{l}}}{\\var{x}!}=\\var{prx}$

$P(X=\\var{x+1}) =\\frac{\\var{l}^{\\var{x+1}}e^{-\\var{l}}}{\\var{x+1}!}=\\var{prx1}$

$P(X=\\var{y}) =\\frac{\\var{l}^{\\var{y}}e^{-\\var{l}}}{\\var{y}!}=\\var{pry}$

Hence

$P(\\var{x} \\leq X \\leq \\var{y})=$ $\\var{prx}+\\var{prx1}+\\var{pry}=\\var{answer2}$

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time interval

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upper value of X

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time hour

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number of customers entering the shop

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average, lambda

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Write down the value of $\\lambda$ for one hour.

$\\lambda = $ [[0]]

Calculate the probability that in a particular one-hour period exactly $\\var{x}$ patients will arrive. Please give your answer to at least 3 decimal places.

$P(r = \\var{x}) =$ [[0]]

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Find the probability that in a particular one hour period, the number of patients entering the waiting room will be between $\\var{x}$ and $\\var{y}$ inclusive. Please give your answer to at least 3 decimal places.

$P(\\var{x}\\leq r \\leq\\var{y}) =$ [[0]]

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It is estimated that 30% of all CIT students cycle to college. If a random sample of eight CIT students is chosen, calculate the probability that...

rebelmaths

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Please give your answer to at least 3 decimal places.

It is estimated that $\\var{p_perc}$% of all CIT students cycle to college. A random sample of $\\var{n}$ CIT students is chosen.

", "advice": "

Part (a)

If a random variable $X$ follows a binomial distribution with parameters $n$ and $p$. The probability of $r$ successes out of $n$ trials is given by:

$P(X=r)=P(r,n)=C^n_{r}p^{r}q^{n-r}$

where $p$ is the probability of success for each trial and $q$ is the probability of failure for each trial.

The probability that a student cycles to college is $\\var{p}$, therefore $p=\\var{p}$ and $q=1-\\var{p}=\\var{q}$.

We are interested in claculating the probability that none of the sample of $\\var{n}$ students cycle to college so $r=0$ and $n=\\var{n}$

$P(\\var{r0}, \\var{n})= C^\\var{n}_{\\var{r0}}$ $\\var{p}^\\var{r0}$ $\\var{q}^{\\var{n}-\\var{r0}}$

$P(\\var{r0}, \\var{n})= \\var{pr0}$

Part (b)

We are interested in claculating the probability that at least $\\var{r}$ of the $\\var{n}$ students cycle to college. Let $X$ represent the number of students that cycle to college. We need to calculate:

$P(X \\geq \\var{r}) = P(X= \\var{r}) + P(X= \\var{r+1})+...+ P(X=\\var{n})$

Since $P(X=\\var{r0})+P(X=\\var{r0+1})+...+P(X=\\var{n})=\\var{r0+1}$

We may write

$P(X \\geq \\var{r}) = 1-P(X= \\var{r0}) - P(X=\\var{r0+1})-...- P(X=\\var{r-1})$

where

$P(X= \\var{r0})=P(\\var{r0}, \\var{n})= C^\\var{n}_{\\var{r0}}$ $\\var{p}^\\var{r0}$ $\\var{q}^{\\var{n}-\\var{r0}}=\\var{pr0}$

$P(X=1) =P(1, \\var{n})= C^\\var{n}_{1}$ $\\var{p}^{1}$ $\\var{q}^{\\var{n}-1}$ $=\\var{pr1}$

$P(X=2) = P(2, \\var{n})=$ $C^\\var{n}_{2}$ $\\var{p}^{2}$ $\\var{q}^{\\var{n}-2}$ $=\\var{pr2}$

Then

$P(X \\geq \\var{r}) = 1-\\var{qn}-\\var{pr1}-\\var{pr2}=\\var{answer2}$

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probability that r = 1

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percentage of students that cycle to college

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probability that r = 2

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probability tha an individual does not cycle to college

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probability that r = 3

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more than r of the students cycle to college

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the probability that an individual student cycles to college

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sample size

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probability that r = 0

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Calculate the probability that none of the $\\var{n}$ students in the sample cycle to college.

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Calculate the probability that at least $\\var{r}$ of the $\\var{n}$ students cycle to college.

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Calculate probability using P(A) = 1-P(not A)

", "licence": "None specified"}, "statement": "

Two $\\var{n}$-sided dice, each with sides labelled $1,2...\\var{n}$ are rolled and their scores are added.

", "advice": "

part a)

We are looking for pairs of numbers than total to make $\\var{min[choice]}$ or less, which is quite a small total, so it doesn't take too long to find that the pairs of numbers are:

\\[ \\var{pairstruncate} \\]

so there {isorare} $\\var{pairschoice[choice]+1}$ {event} we are interested in, and therefore the probability is

\\[ P(\\text{total is} \\leq \\var{min[choice]}) = \\frac{\\var{pairschoice[choice]+1}}{\\var{total}} \\]

part b)

There are a very large number of ways to totals of $\\var{min[choice]+1}$ or greater, so in this case we use the complement rule;

\\[P(A') = 1-P(A)\\]

which in this case means

\\[P(\\text{total is}\\ge \\var{min[choice]+1}) = 1-P(\\text{total is} \\leq \\var{min[choice]}) = 1-\\frac{\\var{pairschoice[choice]+1}}{\\var{total}} = \\var[fractionnumbers]{1-options[choice]/{total}}\\]

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What is the probability that the score is less than or equal to $\\var{min[choice]}$?

[[0]] (Give your answer as a fraction)

What is the probability that the score is $\\var{min[choice]+1}$ or greater?

Hint: It will probably save you time if you use the complement rule.

[[0]] (Give your answer as a fraction)

Find the $x$ and $y$ components of the resultant force on an object, when multiple forces are applied at different angles.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

$F = \\var{force1} \\, \\mathrm{N}$ at an angle of $\\theta = \\var{theta1}^{\\circ}$ clockwise from the left horizonatal.
$P = \\var{force2} \\, \\mathrm{N}$ vertically upwards.
$Q = \\var{force3} \\, \\mathrm{N}$ at an able of $\\theta^{\\ast} = \\var{theta2}^{\\circ}$ clockwise from the right horizontal.

Give your answers to the following questions in Newtons to 3 decimal places.

", "advice": "

a) - c)

Resolve each force in the HORIZONTAL:

$P$ doesn't contribute.

The force $F$ is at $\\var{theta1}^{\\circ}$ to the horizontal, therefore has a contribution of $F \\times \\cos \\var{theta1}^{\\circ} = \\var{force1} \\times \\cos \\var{theta1}^{\\circ} = \\var{precround(force1*cos(radians(theta1)),3)}$. This force is acting to the left.

The force $Q$ is at $\\var{theta2}^{\\circ}$ to the horizontal, therefore has a contribution of $Q \\times cos \\var{theta2}^{\\circ} = \\var{force3} \\times \\cos\\var{theta2}^{\\circ} = \\var{precround(force3*cos(radians(theta2)),3)}$. This is acting to the right.

Therefore the sum of components in the $x$-direction is

forces acting to the right - forces acting to the left

\\[\\var{precround(force3*cos(radians(theta2)),3)} - \\var{precround(-force1*cos(radians(180-theta1)),3)}\\]
\\[= \\var{precround(force1*cos(radians(180-theta1)) + force3*cos(radians(theta2)),3)}\\]

(if this value is positive then it is acting to the right, if it is negative it is acting to the left)

d) - g)

Resolve each force from the VERTICAL:

For the force $P$ this is acting completely in the positive direction, at no angle. Therefore it's contribution is $\\var{force2}$.

The force $F$ is at $\\var{theta1}^{\\circ}$ to the horizontal, therefore has a contribution of $F \\times \\sin \\var{theta1}^{\\circ} = \\var{force1} \\times \\sin \\var{theta1}^{\\circ} = \\var{precround(force1*sin(radians(theta1)),3)}$. This force is acting upwards.

Therefore the sum of components in the $y$-direction is

forces acting up - forces acting down

\\[\\var{precround(force3*sin(radians(theta2)),3)} - \\var{precround(-force1*sin(radians(180-theta1)),3)}\\]
\\[= \\var{precround(force1*sin(radians(180-theta1)) + force3*sin(radians(theta2)),3)}\\]

(if this value is positive then it is acting to the right, if it is negative it is acting to the left)

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Find the component of $F$ in the $x$-direction

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You have not given your answer to the correct precision.

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Find the component of $Q$ in the $x$-direction.

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Find the resultant force in the $x$-direction.

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Find the component of $P$ in the $y$-direction.

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Find the component of $F$ in the $y$-direction.

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Find the component of $Q$ in the $y$-direction.

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Find the resultant force in the $y$-direction.

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Find the $x$ and $y$ components of a force which is applied at an angle to a particle. Resolve using $F \\cos \\theta$. The force acts in the positive $x$ and positive $y$ direction.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

In the above diagram the force $F=\\var{force} \\ \\mathrm{N}$ and the angle $\\theta = \\var{angle}^{\\circ}$.

", "advice": "

a)

The component of force in the $x$-direction can be found using $\\cos\\theta \\times F$. Remember to set your calculator to use degrees and not radians.

\\begin{align} \\text{component in the }x \\text{-direction } & = F \\cos \\theta \\\\
& = \\var{force} \\times \\cos \\var{angle} \\\\
& = \\var{precround(force*cos(radians(angle)),3)}\\end{align}

b)

This time we are using sin

\\begin{align} \\text{component in the y-direction } & = F \\sin \\theta \\\\
& = \\var{force} \\times \\sin \\var{angle} \\\\
& = \\var{precround(force*sin(radians(angle)),3)}\\end{align}

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Find the component of the force in the $x$-direction, in Newtons to 3 decimal places.

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Find the component of the force in the $y$-direction, in Newtons to 3 decimal places.

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Calculate the magnitude of a 3-dimensional vector $\\mathbf v$, where $\\mathbf v$ is written in the form $v_1 \\mathbf i+v_2 \\mathbf j + v_3 \\mathbf k$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the magnitude of the vector $\\mathbf v =\\simplify[all,!noLeadingMinus]{{vi}*mathbf:i+{vj}*mathbf:j+{vk}*mathbf:k}$ .

", "advice": "

For a vector of the form $\\mathbf a = \\simplify{a_1*mathbf:i+a_2*mathbf:j+a_3*mathbf:k}$, the magnitude is found by calculating the square root of the sum of the components squared:

\\[ |\\mathbf a| = \\sqrt{a_1^2+a_2^2+a_3^2}.\\]

Therefore, for the vector $\\mathbf v =\\simplify[!noLeadingMinus]{{vi}*mathbf:i+{vj}*mathbf:j+{vk}*mathbf:k}$,

{advice}

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\\\\[ \\\\begin{split} |\\\\mathbf v| &\\\\,= \\\\simplify[!collectNumbers]{sqrt({vi}^2+{vj}^2+{vk}^2)} \\\\\\\\ &\\\\,= \\\\sqrt{\\\\var{vi^2+vj^2+vk^2}} \\\\\\\\ &\\\\, = \\\\var{magv2} .\\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "sol2": {"name": "sol2", "group": "Ungrouped variables", "definition": "\"

\"", "description": "", "templateType": "long string", "can_override": false}, "vk": {"name": "vk", "group": "Ungrouped variables", "definition": "random(-10..10)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["vi", "vj", "vk", "magv", "magv2", "advice", "sol1", "sol2"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "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": "

$|\\mathbf v |= $[[0]]

(Give your answer to 2 decimal places where necessary)

Given 3 vectors $\\mathbf v$, $\\mathbf a$ and $\\mathbf b$, find the constants $c_1$ and $c_2$ such that $\\mathbf v = c_1 \\mathbf a + c_2 \\mathbf b$ .

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the values of the constants $c_1$ and $c_2$ such that the vector $\\mathbf v = \\var{v}$ is a linear combination of the vectors $\\mathbf a=\\var{a}$ and $\\mathbf b=\\var{b}$: \\[ \\mathbf v = c_1 \\mathbf a + c_2 \\mathbf b. \\]

", "advice": "

We are trying to find the constants $c_1$ and $c_2$ such that we can express $\\mathbf v$ as a linear combination of the vectors $\\mathbf a$ and $\\mathbf b$:

\\[ \\var{v} = c_1 \\var{a} + c_2 \\var{b} ,\\]

We can calculate $c_1$ and $c_2$ by solving the simultaneous equations

\\[ \\begin{split} \\var{v[0]} &\\,= \\simplify[all,!noLeadingMinus]{{a[0]}c_1 + {b[0]} c_2} \\\\ \\var{v[1]} &\\,= \\simplify[all,!noLeadingMinus]{{a[1]}c_1 + {b[1]} c_2} . \\end{split} \\]

We find $c_1 = \\var{c1}$ and $c_2 = \\var{c2}$, and therefore,

\\[\\var{v} = \\simplify[all,!noLeadingMinus]{{c1}{a}+{c2}{b}}. \\]

$c_1= $[[0]]

$c_2= $[[1]]

Calculating several linear combinations of three 2-dimensional vectors.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Given three vectors,

\\[ \\mathbf a = \\var{a},\\quad \\mathbf b = \\var{b}, \\quad \\mathbf c = \\var{c}, \\]

calculate the following:

", "advice": "

When multiplying a vector by a scalar, and when adding or subtracting vectors, we must use the following rules:

For the vectors $\\mathbf v = \\begin{pmatrix} v_1 \\\\ v_2 \\end{pmatrix}$ and $\\mathbf w = \\begin{pmatrix} w_1 \\\\ w_2 \\end{pmatrix}$,

$\\alpha \\,\\mathbf v = \\begin{pmatrix} \\alpha \\,v_1 \\\\ \\alpha \\, v_2 \\end{pmatrix},\\, \\text{ where $\\alpha$ is a constant,}$

\n
$\\mathbf v \\pm \\mathbf w = \\begin{pmatrix} v_1 \\pm w_1 \\\\ v_2 \\pm w_2 \\end{pmatrix}.$
\n

Combining these rules:

\\[ \\alpha \\,\\mathbf v \\pm \\beta \\, \\mathbf w= \\begin{pmatrix} \\alpha \\,v_1 \\\\ \\alpha \\, v_2 \\end{pmatrix} \\pm \\begin{pmatrix} \\beta \\, w_1 \\\\ \\beta \\, w_2 \\end{pmatrix} = \\begin{pmatrix} \\alpha \\,v_1 \\pm \\beta \\, w_1 \\\\ \\alpha \\, v_2 \\pm \\beta \\, w_2 \\end{pmatrix}. \\]

Part a)

{check1}

Part b)

{check2}

Part c)

{check3}

Part d)

{check4}

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\\\\[ \\\\begin{split} \\\\simplify{{m1}*mathbf:a +{n1}*mathbf:b} \\\\, &\\\\,= \\\\simplify{{m1*a}+{n1*b}} \\\\\\\\ &\\\\,= \\\\var{sola} \\\\end{split} \\\\]

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$\\simplify{{k}*(mathbf:a+mathbf:b)+{p3}*mathbf:c}=$ [[0]]

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Given the coordinates of three 2-dimensional points $A$, $B$ and $C$, find the vectors $\\vec{AB}$, $\\vec{AC}$ and $\\vec{CB}$.

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The points A, B and C have coordinates $(\\var{a[0]},\\, \\var{a[1]})$, $(\\var{b[0]},\\, \\var{b[1]})$, and $(\\var{c[0]},\\, \\var{c[1]})$ respectively.

Find:

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To find the vector between two points, it can help to mark the points and their corresponding position vectors on a graph. In this case, let $\\bf a$, $\\bf b$ and $\\bf c$ be the position vectors of $A$, $B$, and $C$, such that

\\[ \\mathbf a = \\var{a}, \\quad \\mathbf b = \\var{b}, \\quad \\mathbf c = \\var{c}.\\]

{geogebra_applet('https://www.geogebra.org/m/jrrcctdj',defs)}

To find the vector $\\vec{AB}$, consider how you can get between $A$ and $B$ only 'travelling' along the position vectors. Firstly, we need to go from $A$ to the origin $O$, which is the vector $\\mathbf{-a}$, and then from the $O$ to $B$, which is the vector $\\mathbf b$. Therefore, $\\vec{AB} = \\mathbf{-a}+\\mathbf{b}$, or more simply, \\[ \\vec{AB} = \\mathbf {b} - \\mathbf{a}.\\] The same method applies for finding the vectors $\\vec{AC}$ and $\\vec{CB}$.

Therefore,

\\[ \\vec{AB} = \\mathbf{b}-\\mathbf{a} = \\var{sola}, \\]

\\[ \\vec{AC} = \\mathbf{c}-\\mathbf{a} = \\var{solb}, \\]

\\[ \\vec{CB} = \\mathbf{b}-\\mathbf{c} = \\var{solc}, \\]

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$\\vec{AB}$

[[0]]

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$\\vec{AC}$

[[0]]

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$\\vec{CB}$

[[0]]

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The purpose of this test is to establish your background knowledge of mathematics before beginning a university degree course.

The system chooses which questions to ask you based on your answers to previous questions.

If the test feels too long, use the PAUSE button and save the page (don't close the tab!); you can get back to it at a later time.

When the test is over, you will be shown a summary of the system's estimates of your understanding of each of the topics covered in the test.

The test is anonymous and won't appear in any record, so don't be afraid to test your skills!

", "end_message": "

Depending on your results, you may wish to access the Math helpdesk here, or book a one to one appointment with a tutor. They can help you assess the best way to improve your skills and fill in the gaps. Save the page with your results to have a reference on what are your weaker and stronger topics, and bring it to the appointment.

Use the \"print this result summary\" button and save the file, it will contain all your attempted questions with your answers; it will be useful to go through the questions with the tutor.

Click here to book your appointment.

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Can the student find the factors of a given integer?

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