// Numbas version: finer_feedback_settings {"name": "Skills Audit for Maths and Stats - Mathematics (MAS106)", "metadata": {"description": "

Skills Audit for Maths and Stats for MAS106 students.

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perform a calculation involving negative indices.

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

Evaluate and simplify the following expression:

\\[\\frac{\\var{x}^\\var{n}}{\\var{y}^\\var{m}}\\]

", "advice": "

To simplify this expression we use the rule $a^{-n}=\\frac1{a^n}$.

\\[\\frac{\\var{x}^\\var{n}}{\\var{y}^\\var{m}}=\\frac{\\var{y}^\\var{-m}}{\\var{x}^\\var{-n}}=\\frac{\\var{y^-m}}{\\var{x^-n}}=\\simplify{{y^-m}/{x^-n}}\\]

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

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Solving a pair of simultaneous equations of the form $a_1x+y=c_1$ and $a_2x^2+b_2xy=c_2$ by forming a quadratic equation.

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

Solve the following simultaneous equations:

\\[ \\begin{split} \\simplify{{a1}x+y} &\\,= \\var{c1} \\\\ \\simplify{{a2}x^2+{b2}x*y} &\\,= \\var{c2} \\end{split} \\]

Give your answers to 2 decimal places where necessary.

", "advice": "

To solve a pair of simultaneous equations of this type we want to rearrange the linear equation such that $y$ is the subject, which we can then substitute into the equation with the quadratic $x$-term. This will result in a quadratic equation in terms of $x$ only.

For the equations

\\[ \\begin{split} \\simplify{{a1}x+y} &\\,= \\var{c1} \\qquad \\qquad &(1) \\\\\\simplify{{a2}x^2+{b2}x*y} &\\,= \\var{c2} \\qquad \\qquad &(2) \\end{split} \\]

we can rearrange equation (1) to make $y$ the subject:

\\[ y = \\simplify{{c1}-{a1}x}. \\qquad\\qquad (3)\\]

Substituting this into equation (2):

\\[ \\begin{split}\\simplify{{a2}x^2+{b2}x({c1}-{a1}x)} &\\,=\\var{c2} \\\\ \\simplify[!cancelTerms,unitFactor]{{a2}x^2+{b2*c1}x-{b2*a1}x^2} &\\,=\\var{c2}. \\end{split} \\]

Collecting similar terms:

\\[ \\simplify{({a2}-{b2*a1})x^2+{b2*c1}x-{c2}} =0. \\qquad\\qquad (4) \\]

Using the quadratic formula, we find two solutions for $x$:

{check}

Therefore, the 2 pairs of solutions for these simultaneous equations are

\\[ (x_1,y_1) = (\\var{x1dp},\\var{y1dp}) \\] and \\[ (x_2,y_2) = (\\var{x2dp},\\var{y2dp}). \\]

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

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\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp} \\\\, \\\\text{ (2 d.p.)} \\\\]

\\n

To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:

\\n

\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\, \\\\text{(2 d.p.)} \\\\]

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\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\text{ (2 d.p.)}\\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp} \\\\]

\\n

To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:

\\n

\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\text{(2 d.p.)} \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "text3": {"name": "text3", "group": "Ungrouped variables", "definition": "\"

\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\text{ (2 d.p.)}\\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp} \\\\, \\\\text{(2 d.p.)} \\\\]

\\n

To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:

\\n

\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\text{(2 d.p.)} \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\, \\\\text{(2 d.p.)} \\\\]

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

\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp}\\\\]

\\n

To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:

\\n

\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\]

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$(x_1,y_1)=$[[0]]

$(x_2,y_2)=$[[1]]

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Rearrange a specific formula. No randomisation.

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

Rearrange the following equation, to make $y$ the subject:

\\[{cy -b = 3x}\\]

", "advice": "

In order to rearrange the equation so that it is in terms of $y$, we must first add $b$ to both sides, and then divide both sides of the equation by $c$:

\\begin{split} cy-b &= 3x \\\\ cy &= 3x + b \\\\ y &=\\frac{3x+b}{c} \\end{split}

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

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$y=$ [[0]]

Rearrange expressions in the form $ax^2+bx+c$ to $a(x+b)^2+c$.

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

We can rewrite quadratic equations given in the form $ax^2+bx+c$ as a square plus another term - this is called \"completing the square\".

This can be useful when it isn't obvious how to fully factorise a quadratic equation.

Rewrite the following expressions in the form \\[(x+b)^2-c\\]

", "advice": "

Completing the square works by noticing that

\\[ (x+a)^2 = x^2 + 2ax + a^2 \\]

So when we see an expression of the form $x^2 + 2ax$, we can rewrite it as $(x+a)^2-a^2$.

Replace $x^2+\\var{evens2}x$ with $(x+\\var{evens2/2})^2 - \\var{evens2/2}^2$. Remember to keep the $\\var{evens2-evens1}$ term on the end!

\\begin{align}
\\simplify[basic]{ x^2 + {evens2}x + {evens2-evens1}} &= \\simplify[basic]{ (x+{evens2/2})^2 - {evens2/2}^2 + {evens2-evens1} } \\\\
&= \\simplify[basic]{ (x+{evens2/2})^2 + {evens2-evens1 - evens2^2/4} }
\\end{align}

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

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$\\simplify {x^2+ {evens2}x +{evens2-evens1}} =$ [[0]]

It doesn't look like you've completed the square.

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Factorise a quadratic equation where the coefficient of the $x^2$ term is greater than 1 and then write down the roots of the equation

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

As this question involves a number greater than $1$ before the $x^2$ value it has a factorised form $(ax+b)(cx+d)$.

To find $a$ and $c$, we need to consider the factors of $\\var{a*c}$.

You may have to test a a few different options before you find one that works. In this case $a$ and $c$ are $\\var{a}$ and $\\var{c}$.

This means our factorised equation must take the form

\\[(\\var{a}x+b)(\\var{c}x+d)=0\\text{.}\\]

This expands to

\\[ \\simplify{ {a*c}x^2 + ({a}*d+{c}*b)x + a*b} \\]

So we must find two numbers which add together to make $\\var{a*d+b*c}$, and multiply together to make $\\var{b*d}$.

Therefore $b$ and $d$ must satisfy

\\begin{align}
b \\times d &=\\var{b*d}\\\\
\\simplify{{a}d+{c}b} &= \\var{a*d+b*c}\\text{.}
\\end{align}

$b = \\var{b}$ and $d = \\var{d}$ satisfy these equations:

\\begin{align}
\\var{b} \\times \\var{d} &=\\var{b*d}\\\\
\\simplify[]{ {a}*{d} + {b}*{c} } &= \\var{a*d+b*c}
\\end{align}

So the factorised form of the equation is

\\[ \\simplify{({a}x+{b})({c}x+{d}) = 0} \\text{.}\\]

$\\simplify{({a}x+{b})({c}x+{d}) = 0}$ when either $\\var{a}x+\\var{b} = 0$ or $\\var{c}x+ \\var{d} = 0$.

So the roots of the equation are $\\var[fractionnumbers]{-b/a}$ and $\\var[fractionnumbers]{-d/c}$.

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

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$b$ in $(ax+b)(cx+d)$

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$c$ in $(ax+b)(cx+d)$

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$a$ in $(ax+b)(cx+d)$

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The roots of the equation

", "templateType": "anything", "can_override": false}, "d": {"name": "d", "group": "last q", "definition": "random(-8..8 except 0)", "description": "

$d$ in $(ax+b)(cx+d)$

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Solve the following equation by factorisation to find $x$.

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

Input your answers in ascending order.

$x=$ [[0]]

$x=$ [[1]]

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Factorising a quadratic expression of the form $a^2x^2-b^2$ to $(ax+b)(ax-b)$, using the difference of two squares formula.

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

Factorise the following quadratic expression:

\\[ \\simplify[unitFactor]{{a^2}x^2-{c^2}} \\]

", "advice": "

For a quadratic expression of this form we can make use of the Difference of Squares formula, which states that \\[a^2-b^2 = (a+b)(a-b).\\]

Therefore,

\\[ \\simplify[unitFactor]{{a^2}x^2-{c^2} = ({a}x+{c})({a}x-{c})}. \\]

Use this link to find resources to help you revise how to factorise a quadratic equation using the difference of two squares formula.

Simplify the sum of two algebraic fractions where spotting factorising of both numerators and denominators can reduce the work massively.

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

Write the following as a single fraction $\\frac{\\var{num1}}{\\var{den1}}+\\frac{\\var{num2}}{\\var{den2}}$ simplifying as much as possible. Your answer should be in the form $\\frac{\\alpha\\var{v}+\\beta}{\\delta\\var{v}^2-\\gamma}.$

", "advice": "

To write the following as a single fraction $\\frac{\\var{num1}}{\\var{den1}}+\\frac{\\var{num2}}{\\var{den2}}$ first factorise as much as possible and look for any cancellations:

\\[\\begin{split}
&\\frac{\\var{a}\\times\\var{b}}{\\var{den1fact}} + \\frac{\\var{num2}}{\\var{den2fact}}\\\\
& = \\frac{\\var{b}}{\\var{den1simp}} + \\frac{1}{\\var{f1c}}.
\\end{split}\\]

Then get a common denominator for the two fractions and combine into a single fraction:

\\[\\begin{split}
&\\frac{\\var{b}}{\\var{den1simp}} + \\frac{\\var{f1}}{\\var{den1simp}}\\\\
& = \\frac{\\var{b}+\\var{f1}}{\\var{den1simp}}\\\\
& = \\var{ans}.
\\end{split}\\]

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

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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.

\\\\[ \\\\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})}.\\\\]

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

\\\\[ \\\\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]]

Basic calculation from a sum given in Sigma notation.

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

Calculate:

\\[\\displaystyle{\\Sigma_{n=1}^3} \\var{b}n.\\]

", "advice": "

The sigma notation $\\displaystyle\\sum_{n=1}^{3}\\var{b}n$ is asking us to find the sum of the first three terms of the sequence $\\var{b}n$.

\\[\\begin{split}\\Sigma_{n=1}^3 \\var{b}n &\\, = (\\var{b}\\times 1) + (\\var{b}\\times 2) + (\\var{b}\\times 3) \\\\ &\\, = \\var{b1} + \\var{b2} + \\var{b3} \\\\ &\\, = \\var{sum}.\\end{split}\\]

Use this link to find resources to help you revise sigma notation.

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

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

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.

", "minMarks": 0, "maxMarks": 0, "minAnswers": 0, "maxAnswers": 0, "shuffleChoices": true, "shuffleAnswers": true, "displayType": "radiogroup", "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["$\\sec(x)$", "$\\text{cosec}(x)$", "$\\cot(x)$"], "matrix": [["1", "0", 0], [0, "1", 0], ["0", 0, "1"]], "layout": {"type": "all", "expression": ""}, "answers": ["{geogebra_applet('https://www.geogebra.org/m/h9d8hzna')}", "{geogebra_applet('https://www.geogebra.org/m/kqnrbjzy')}", "{geogebra_applet('https://www.geogebra.org/m/xm44vcwe')}"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "GB5 Trigonometric Identities 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": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Oliver Spenceley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23557/"}], "tags": [], "metadata": {"description": "

Rewriting a trigonometric expression of the form $A\\cos(\\theta)\\pm B\\sin(\\theta)$ to either $R\\sin(\\theta+\\alpha)$ or $R\\cos(\\theta+\\alpha)$ by calculating $R$ and $\\alpha$.

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

{question}

find the values for $R$ and $\\alpha$, given $R>0$ and $0<\\alpha<\\frac{\\pi}{2}$.

", "advice": "

{answer}

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

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\\\\[ \\\\begin{split} R^2\\\\cos^2(\\\\alpha) + R^2 \\\\sin^2(\\\\alpha) &\\\\,= \\\\var{A}^2+\\\\var{B}^2 \\\\\\\\ R^2 (\\\\cos^2(\\\\alpha) +\\\\sin^2(\\\\alpha)) &\\\\,= \\\\var{A^2+B^2} \\\\\\\\ R &\\\\,= \\\\var{R}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "Rsol2": {"name": "Rsol2", "group": "Ungrouped variables", "definition": "\"

\"", "description": "", "templateType": "long string", "can_override": false}, "alpharound": {"name": "alpharound", "group": "Ungrouped variables", "definition": "precround(alpha,2)", "description": "", "templateType": "anything", "can_override": false}, "question": {"name": "question", "group": "Ungrouped variables", "definition": "if(Q=1,'{q1}','{q2}')", "description": "", "templateType": "anything", "can_override": false}, "Q": {"name": "Q", "group": "Ungrouped variables", "definition": "random(1,2)", "description": "", "templateType": "anything", "can_override": false}, "sign": {"name": "sign", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "templateType": "anything", "can_override": false}, "q1": {"name": "q1", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\simplify[unitFactor]{{A}sin(theta)+{sign*B}cos(theta)} = \\\\simplify[unitFactor]{R sin (theta+{sign}*alpha)},\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "q2": {"name": "q2", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\simplify[unitFactor]{{A}cos(theta)-{sign*B}sin(theta)} = \\\\simplify[unitFactor]{R cos (theta+{sign}*alpha)},\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "a1": {"name": "a1", "group": "Ungrouped variables", "definition": "\"

To find $R$ and $\\\\alpha$ we want to first rewrite our equation using the double-angle formula, $\\\\simplify[unitFactor]{sin(a+{sign}*b)=sin(a)cos(b)+{sign}*sin(b)cos(a)}$:

\\n

\\\\[ \\\\begin{split}\\\\simplify[unitFactor]{{A}sin(theta)+{sign*B}cos(theta)} &\\\\,= \\\\simplify{R sin(theta+{sign}*alpha)} \\\\\\\\ &\\\\,= \\\\simplify{R(sin(theta)cos(alpha) + {sign}*sin(alpha)cos(theta))} \\\\\\\\ &\\\\,= \\\\simplify{Rsin(theta)cos(alpha) + {sign}*R sin(alpha)cos(theta)}. \\\\end{split} \\\\]

\\n

By comparing the coefficients of $\\\\sin(\\\\theta)$ and $\\\\cos(\\\\theta)$, we find that

\\n

\\\\[ R\\\\cos(\\\\alpha) = \\\\var{A},\\\\quad \\\\text{and} \\\\quad R\\\\sin(\\\\alpha) = \\\\var{B}. \\\\]

\\n

To calculate $R$, we want to square these results and add them together, allowing us to make use of $\\\\sin^2(\\\\alpha)+\\\\cos^2(\\\\alpha) = 1$:

\\n

{Rsol}

\\n

Similarly, to find $\\\\alpha$ we can divide $R\\\\sin(\\\\alpha) = \\\\var{B}$ by $R\\\\cos(\\\\alpha) = \\\\var{A}$, and use the identity $\\\\tan(\\\\alpha) = \\\\frac{\\\\sin(\\\\alpha)}{\\\\cos(\\\\alpha)}$:

\\n

\\\\[ \\\\frac{R\\\\sin(\\\\alpha)}{R\\\\cos(\\\\alpha)} = \\\\frac{\\\\var{B}}{\\\\var{A}} \\\\implies \\\\tan(\\\\alpha) = \\\\simplify[fractionNumbers]{{B/A}}.\\\\]

\\n

Therefore, \\\\[ \\\\begin{split} \\\\alpha &\\\\,= \\\\tan^{-1}\\\\left(\\\\simplify[fractionNumbers]{{B/A}}\\\\right) \\\\\\\\ &\\\\,= \\\\var{alpharound} \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

\\n

\"", "description": "", "templateType": "long string", "can_override": false}, "a2": {"name": "a2", "group": "Ungrouped variables", "definition": "\"

To find $R$ and $\\\\alpha$ we want to first rewrite our equation using the double-angle formula, $\\\\simplify{cos(a+{sign}*b)=cos(a)cos(b)-{sign}*sin(a)sin(b)}$:

\\n

\\\\[ \\\\begin{split}\\\\simplify[unitFactor]{{A}cos(theta)-{sign*B}sin(theta)} &\\\\,= \\\\simplify[unitFactor]{R cos (theta + {sign}*alpha)} \\\\\\\\ &\\\\,= \\\\simplify{R(cos(theta)cos(alpha) - {sign}*sin(theta)sin(alpha))} \\\\\\\\ &\\\\,= \\\\simplify{Rcos(theta)cos(alpha) - {sign}*R sin(theta)sin(alpha)}. \\\\end{split} \\\\]

\\n

By comparing the coefficients of $\\\\cos(\\\\theta)$ and $\\\\sin(\\\\theta)$, we find that

\\n

\\\\[ R\\\\cos(\\\\alpha) = \\\\var{A},\\\\quad \\\\text{and} \\\\quad R\\\\sin(\\\\alpha) = \\\\var{B}. \\\\]

\\n

To calculate $R$, we want to square these results and add them together, allowing us to make use of $\\\\sin^2(\\\\alpha)+\\\\cos^2(\\\\alpha) = 1$:

\\n

{Rsol}

\\n

\\\\[ \\\\frac{R\\\\sin(\\\\alpha)}{R\\\\cos(\\\\alpha)} = \\\\frac{\\\\var{B}}{\\\\var{A}} \\\\implies \\\\tan(\\\\alpha) = \\\\simplify[fractionNumbers]{{B/A}}.\\\\]

\\n

Therefore, \\\\[ \\\\begin{split} \\\\alpha &\\\\,= \\\\tan^{-1}\\\\left(\\\\simplify[fractionNumbers]{{B/A}}\\\\right) \\\\\\\\ &\\\\,= \\\\var{alpharound} \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

\\n

\"", "description": "", "templateType": "long string", "can_override": false}, "answer": {"name": "answer", "group": "Ungrouped variables", "definition": "if(Q=1,'{a1}','{a2}')", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["Q", "A", "B", "sign", "R", "Rround", "alpha", "alpharound", "Rsol", "Rsol1", "Rsol2", "question", "q1", "q2", "answer", "a1", "a2"], "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": "

$R=$[[0]]

$\\alpha=$[[1]]

(Give your answers to 2 decimal places where necessary.)

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Gap 0", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{Rround}", "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": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{alpharound}", "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": []}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "CA3 - Graphs of trig functions (sin, cos, tan)", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Match the relevant graph (sin, cos, tan) with its equation.

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

This is about core knowledge of graphs. You should know the shapes of the fundamental trig graphs, if you don't familiarize yourself with them from the resources linked below. In this setting the $x$-axis is given with a scale in radians but you might also find some where it is given in degrees. You should also be aware of the difference between those two different units of angles.

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.

", "minMarks": 0, "maxMarks": 0, "minAnswers": 0, "maxAnswers": 0, "shuffleChoices": true, "shuffleAnswers": true, "displayType": "radiogroup", "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["$\\sin(x)$", "$\\cos(x)$", "$\\tan(x)$"], "matrix": [["1", 0, 0], [0, "1", 0], [0, 0, "1"]], "layout": {"type": "all", "expression": ""}, "answers": ["{geogebra_applet('https://www.geogebra.org/m/ntqvuwqr')}", "{geogebra_applet('https://www.geogebra.org/m/fsqmnhsc')}", "{geogebra_applet('https://www.geogebra.org/m/yg6f9eqz')}"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "CA4 Function notation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Evaluating a linear function for a given value of $x$.

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

Given $f(x)=\\simplify{{m}x+{c}}$, find $f(\\var{n})$.

", "advice": "

If $f(x)=\\simplify{{m}x+{c}}$, to find $f(\\var{n})$ we need to evaluate $f(x)$ when $x=\\var{n}$:

\\[ \\begin{split} f(\\var{n}) &\\,= \\simplify[alwaysTimes]{{m}({n})+{c}} \\\\ &\\,= \\simplify[!collectNumbers]{{m*n}+{c}} \\\\ &\\,= \\simplify{{m*n+c}}. \\end{split} \\]

Use this link to find resources to help you revise function notation.

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$f(\\var{n})=$[[0]]

Determining the range of a function of the form $f = m|x| + a$.

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

The range is the set of values that can be taken by $f(x)$, i.e. the values on the $y$-axis.

{geogebra_applet('https://www.geogebra.org/m/aqcgkurg',[a: a, m: m])}

Therefore, for $f(x)=\\simplify{{m}x^2+{a}}$, the range is $[\\var{a}, \\infty)$.

Use this link to find some resources to help you revise how to find the domain and range of a function.

Given $f(x)=\\simplify{{m}x^2+{a}}$

What is the range of $f(x)$?

", "minMarks": 0, "maxMarks": 0, "shuffleChoices": false, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["$\\mathbb{R}$", "$[\\var{a},\\infty)$", "$\\left[\\simplify{{a}/{m}},\\infty\\right)$", "$(-\\infty,\\var{a}]$", "$(\\var{a},\\infty)$"], "matrix": [0, "1", 0, 0, 0], "distractors": ["", "", "", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "CA6 Inverse functions", "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": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Finding the inverse of a function of the form $f(x)=\\frac{mx+c}{x+a},\\,x\\neq-a$.

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

If $f(x)=\\simplify{({m}x+{c})/(x+{a})},\\,x\\neq \\simplify{{-a}}$, find the inverse function, $f^{-1}(x)$.

", "advice": "

To find $f^{-1}x$, it can help to first set $f(x)$ to a different variable, which we will call $y$:

\\[ y = f(x) = \\simplify{({m}x+{c})/(x+{a})}\\]

Since the function $f(x)$ takes us from $x$ to $y$, the inverse function $f^{-1}$ will take us from $y$ to $x$. So to obtain $f^{-1}$, we want to rearrange $y=\\simplify{({m}x+{c})/(x+{a})}$ so that it is $x$ as a function of $y$:

\\[ \\begin{split} y &\\,= \\simplify{({m}x+{c})/(x+{a})} \\\\\\\\ \\simplify{(x+{a})y} &\\,= \\simplify{{m}x+{c}} \\\\\\\\ \\simplify{x*y+{a}y} &\\,= \\simplify{{m}x+{c}} \\\\\\\\ \\simplify{x*y - {m}x} &\\,= \\simplify{{c}- {a}y} \\\\ \\\\ \\simplify{x(y-{m})} &\\,= \\simplify{{c}-{a}y} \\\\\\\\ x&\\,= \\simplify{({c}-{a}y)/(y-{m})}. \\end{split} \\]

Hence, $f^{-1}(y) =\\simplify{({c}-{a}y)/(y-{m})}$, and therefore \\[ f^{-1}(x) =\\simplify{({c}-{a}x)/(x-{m})}.\\]

(Note: The inverse is valid for all values of $x$ except $x=\\var{m}$, since this would make the denominator equal to 0.)

Use this link to find resources to help you revise how to find the inverse of functions.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"m": {"name": "m", "group": "Ungrouped variables", "definition": "random(2..9)", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "random(-9..9 except 0)", "description": "", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "Ungrouped variables", "definition": "random(-8..8)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "abs(m)-abs(c)>0 or abs(m)-abs(c)<0", "maxRuns": 100}, "ungrouped_variables": ["m", "c", "a"], "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": "

$f^{-1}(x)=$[[0]]

Finding composite functions of 2 linear functions.

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

If $f(x)=\\simplify{{m}x+{c}}$ and $g(x)=\\simplify{{n}x+{d}}$, find expressions for $f\\circ g(x)$ and $g \\circ f(x)$.

Recall: $f \\circ g(x) \\equiv f(g(x))$ and $g \\circ f(x) \\equiv g(f(x))$.

", "advice": "

To find the composition $f \\circ g(x)$ we are substituting the expression for $g(x)$ into the function $f(x)$, replacing the $x$-terms with the function $g(x)$. Similarly, to find the composition $g \\circ f(x)$ we are substituting the expression for $f(x)$ into the function $g(x)$, replacing the $x$-terms with the function $f(x)$.

So, for $f(x)=\\simplify{{m}x+{c}}$ and $g(x)=\\simplify{{n}x+{d}}$,

\\[ \\begin{split} f \\circ g(x) \\equiv f(g(x)) &\\,= \\simplify{{m}({n}x+{d})+{c}} \\\\ &\\,=\\simplify[!collectNumbers,unitFactor]{{m*n}x+{m*d}+{c}} \\\\ &\\,=\\simplify{{m*n}x+{m*d+c}}, \\end{split} \\]

and

\\[ \\begin{split} g \\circ f(x) \\equiv g(f(x)) &\\,= \\simplify{{n}({m}x+{c})+{d}} \\\\ &\\,=\\simplify[!collectNumbers,unitFactor]{{m*n}x+{n*c}+{d}} \\\\ &\\,=\\simplify{{m*n}x+{n*c+d}}. \\end{split} \\]

Use this link to find resources to help you revise how to find composite functions.

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$f \\circ g(x)=$[[0]]

$g \\circ f(x)=$[[1]]

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Gap 0", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{m*n}x+{m*d+c}", "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": ""}]}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{n*m}x+{n*c+d}", "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": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "CA8 Logs - definition", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

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]]

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

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

Solve for $x$:

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

", "advice": "

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)

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)

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}

\"", "description": "", "templateType": "long string", "can_override": false}, "advice2": {"name": "advice2", "group": "Ungrouped variables", "definition": "if(a_2<0 and a_3>0,'{solutionb}',{advice3})", "description": "", "templateType": "anything", "can_override": false}, "advice3": {"name": "advice3", "group": "Ungrouped variables", "definition": "if(a_2>0 and a_3<0,'{solutionc}','{solutiond}')", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "random(2..10)", "description": "", "templateType": "anything", "can_override": false}, "b_3": {"name": "b_3", "group": "Ungrouped variables", "definition": "b/c", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "gcd(b,c)=1", "maxRuns": "100"}, "ungrouped_variables": ["a_1", "a_2", "a_3", "b_1", "b_2", "b_3", "b", "c", "advice", "advice2", "advice3", "solutiona", "solutionb", "solutionc", "solutiond"], "variable_groups": [{"name": "Unnamed group", "variables": []}], "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": "

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

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.\\]

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.\\]

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)

Finding the stationary points of a cubic equation and determining their nature.

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

Given the function \\[ \\simplify{y={a}x^3+{b}x^2+{c}x+{d}} ,\\] find its stationary points and determine their nature.

", "advice": "

To find the stationary points of the function, we must solve $\\tfrac{dy}{dx}=0$ for $x$. For the function $\\simplify{y={a}x^3+{b}x^2+{c}x+{d}}$,

\\[ \\frac{dy}{dx} = \\simplify{{3a}x^2+{2b}x+{c}}. \\]

Setting $\\frac{dy}{dx}=0$ and solving for $x$:

\\[ \\simplify{{3a}x^2+{2b}x+{c}} =0 \\\\ \\\\ \\implies x=\\var{solx1dp} \\var{x1} \\text{ and } x=\\var{solx2dp} \\var{x2}. \\]

Hence, the function has two stationary points at $x=\\var{solx1dp}$ and $x=\\var{solx2dp}$. To find the corresponding $y$-coordinates, we want to plug these values back into the initial equation.

When $x=\\var{solx1dp}$,

\\[ \\begin{split} y &\\,= \\simplify[unitFactor,!cancelTerms]{{a}*({solx1dp})^3+{b}*({solx1dp})^2+{c}*({solx1dp})+{d}} \\\\ &\\,=\\simplify{{soly1dp}} \\var{y1}. \\end{split} \\]

When $x=\\var{solx2dp}$,

\\[ \\begin{split} y &\\,= \\simplify[unitFactor,!cancelTerms]{{a}*({solx2dp})^3+{b}*({solx2dp})^2+{c}*({solx2dp})+{d}} \\\\ &\\,=\\simplify{{soly2dp}} \\var{y2}. \\end{split} \\]

Therefore, the stationary points of $y=\\simplify{{a}x^3+{b}x^2+{c}x+{d}}$ are

\\[ (\\simplify{{solx1dp}},\\, \\simplify{{soly1dp}}) \\, , \\,(\\simplify{{solx2dp}},\\, \\simplify{{soly2dp}}). \\]

Finally, we need to determine the nature of the stationary points. To do this we want to calculate the second derivative of the initial function and then evaluate it for each $x$-value of the stationary points.

Recall:

If $\\tfrac{d^2y}{dx^2}<0$ the stationary point is a maximum;
If $\\tfrac{d^2y}{dx^2}>0$ the stationary point is a minimum;
If $\\tfrac{d^2y}{dx^2}=0$ the stationary point is a point of inflection.

To calculate $\\tfrac{d^2y}{dx^2}$, we want to differentiate $\\tfrac{dy}{dx}$ again with respect to $x$:

\\[ \\begin{split} &\\frac{dy}{dx} = \\simplify{{3a}x^2+{2b}x+{c}}, \\\\ \\\\\\implies &\\frac{d^2y}{dx^2} = \\simplify{{6a}x+{2b}}. \\end{split}\\]

For $(\\simplify{{solx1dp}},\\, \\simplify{{soly1dp}})$, $\\frac{d^2y}{dx^2} = \\simplify{{check}}$, so it is a minimum.

For $(\\simplify{{solx2dp}},\\, \\simplify{{soly2dp}})$, $\\frac{d^2y}{dx^2} = \\simplify{{check2}}$, so it is a maximum.

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

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There is a minimum point at ([[0]], [[1]]) and a maximum point at ([[2]] , [[3]]).

(Give the coordinates of the stationary points to 2 decimal places where necessary.)

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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]]

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Gap 0", "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": false, "customName": "", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "answer": "{a*b}sec^2({b}x+{c})", "answerSimplification": "fractionNumbers, basic, unitFactor", "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": "sec", "value": ""}, {"name": "x", "value": ""}]}], "answer": "{a*b}sec({b}x+{c})^2", "answerSimplification": "fractionNumbers, basic, unitFactor", "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": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "CB9 Differentiating with Exponentials", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

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]]

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]]

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]]

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

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

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

", "advice": "

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

\\[ \\dfrac{dy}{dx} = u(x) \\times \\dfrac{dv}{dx} + v(x) \\times\\dfrac{du}{dx}.\\]

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 product 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} sin({b}x)}, \\]

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

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}cos({b}x)}.\\]

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

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

Simplifying,

\\[\\dfrac{dy}{dx} = \\simplify{{n*a}x^{n-1}sin({b}x) + {a*b}x^{n}cos({b}x)}. \\]

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

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]]

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": "CD2 - Integration - trig identities", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Using the various versions of $\\cos{2x}$ identity to integrate $\\sin^2{x}$ and $\\cos^2{x}$.

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

Integrate $f(x)=\\var{Func}$.

", "advice": "

We can't integrate $\\var{Coeff}\\sin^2(x)$ directly so first we have to use the double angle formula $\\cos(2x)=1-2\\sin^2(x)$. We re-arrange using the double angle formula to give us,

\\[\\var{Coeff}\\sin^2(x)=\\frac{\\var{Coeff}}2-\\frac{\\var{Coeff}}2\\cos(2x).\\]

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

So, for the function

\\[f(x)=\\simplify[unitFactor,fractionNumbers]{{-Coeff/2}cos(2x)},\\]

the integral is

\\[ \\begin{split} \\int\\simplify[unitFactor,fractionNumbers]{{-Coeff/2}cos(2x)} dx \\,= \\simplify[unitFactor,fractionNumbers]{{-Coeff/2}int(cos(2x),x)} &\\,=\\simplify[unitFactor,fractionNumbers]{{-Coeff/2}(1/2 sin({2}x))} +c, \\\\ &\\,=\\simplify[unitFactor,fractionNumbers]{{-Coeff/4} sin(2x)+c}. \\end{split} \\]

The integral of $\\frac{\\var{Coeff}}2$ is

\\[\\int\\frac{\\var{Coeff}}2dx=\\frac{\\var{Coeff}}2x+c,\\]

so combining these our final answer is

\\[\\int\\frac{\\var{Coeff}}2-\\frac{\\var{Coeff}}2\\cos(2x)dx=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}x-{Coeff/4} sin(2x)+c}\\]

We can't integrate $\\var{Coeff}\\cos^2(x)$ directly so first we have to use the double angle formula $\\cos(2x)=2\\cos^2(x)-1$. We re-arrange using the double angle formula to give us,

\\[\\var{Coeff}\\cos^2(x)=\\frac{\\var{Coeff}}2+\\frac{\\var{Coeff}}2\\cos(2x).\\]

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

So, for the function

\\[f(x)=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}cos(2x)},\\]

the integral is

\\[ \\begin{split} \\int\\simplify[unitFactor,fractionNumbers]{{Coeff/2}cos(2x)} dx \\,= \\simplify[unitFactor,fractionNumbers]{{Coeff/2}int(cos(2x),x)} &\\,=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}(1/2 sin({2}x))} +c, \\\\ &\\,=\\simplify[unitFactor,fractionNumbers]{{Coeff/4} sin(2x)+c}. \\end{split} \\]

The integral of $\\frac{\\var{Coeff}}2$ is

\\[\\int\\frac{\\var{Coeff}}2dx=\\frac{\\var{Coeff}}2x+c,\\]

so combining these our final answer is

\\[\\int\\frac{\\var{Coeff}}2+\\frac{\\var{Coeff}}2\\cos(2x)dx=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}x+{Coeff/4} sin(2x)+c}\\]

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

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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 $\\frac{nx^{n-1}}{x^n+a}$ 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]]

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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*)\".

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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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This is a tool for you! It is here to help you diagnose whether there are any maths or statistics pre-requisites for your course that you may want to brush up on. If at any point you are struggling with any question you should find a link at the end of the \"reveal answer\" section that will take you to some recommended online resources on that subject area. You can also always contact the Maths and Stats Help team (MaSH) to arrange a one to one appointment or check out our workshop timetable to see if you can access the support you need that way. Find all this information via our website here!

", "end_message": "

Thanks for completing the Skills Audit. You can attempt this as many times as you need. Remember the score is not what matters - this is in no way assessed work - this is simply a tool for working out whether you may need to brush up on anything to ensure that you can access all the material on your course and get off to the best possible start.

Don't forget to look up what support is available to you through our web pages here!

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