A Level Math RevisionA Level Maths · AQA, Edexcel and OCR · Year 12 and 13
A Level Maths revision notes for every topic.
52 short notes covering Pure, Statistics and Mechanics: the key formulae, worked examples with every answer checked, and a box for your board saying where the topic is examined and what is different.
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Download PDF AQA AS (Year 12) notes · 19 pages · 1.2 MB · free with a free account
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Download PDF Edexcel AS (Year 12) notes · 19 pages · 1.2 MB · free with a free account
Download PDF Edexcel A2 (Year 13) notes · 20 pages · 1.5 MB · free with a free account
Download PDF OCR A AS (Year 12) notes · 19 pages · 1.2 MB · free with a free account
Download PDF OCR A A2 (Year 13) notes · 20 pages · 1.5 MB · free with a free account
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Year 12 (AS) notes
The AS content, examined on the AS papers and again in the full A Level.
Pure (14)
Proof: deduction, exhaustion and counter-example
AS proof uses three methods. Proof by deduction builds a chain of correct steps from facts you know. Proof by exhaustion splits the claim into a finite number of cases and checks every one. Disproof by counter-example finds one case that meets the conditions but breaks the conclusion. Always finish with a sentence that says what has been proved.
Key formulae
- An even integer is $2n$ and an odd integer is $2n+1$, where $n$ is an integer
- Consecutive integers: $n,\ n+1,\ n+2$. Consecutive even integers: $2n,\ 2n+2$
- A multiple of $k$ can be written as $k \times (\text{an integer})$
- Difference of two squares: $a^2-b^2=(a-b)(a+b)$
- Completing the square shows a quadratic is always positive: $x^2-6x+10=(x-3)^2+1>0$
Worked examples
- Prove that the sum of any three consecutive integers is a multiple of $3$. Let them be $n$, $n+1$ and $n+2$. The sum is $3n+3=3(n+1)$, and $n+1$ is an integer, so the sum is a multiple of $3$.
- Prove by exhaustion that $n^2+3n+1$ is odd for every integer $n$. If $n$ is even, $n^2$ and $3n$ are both even, so $n^2+3n+1$ is even $+$ even $+\,1$, which is odd. If $n$ is odd, $n^2$ and $3n$ are both odd, so their sum is even and adding $1$ gives an odd number. Every integer is even or odd, so the statement is true.
- Disprove: 'if $p$ is prime, then $2^p-1$ is prime'. Try $p=11$: $2^{11}-1=2047=23\times 89$, which is not prime. So $p=11$ is a counter-example.
AQA 7357 · A1
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- Proof (section A) can be asked on any of the three papers, because Papers 2 and 3 can include any Paper 1 content.
Edexcel 9MA0 · Pure 1.1
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- Proof is Pure topic 1. Proof by contradiction is the only A Level addition.
OCR A H240 · 1.01a–c
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- OCR expects you to read and use the logic symbols $\Rightarrow$, $\Leftarrow$ and $\Leftrightarrow$ and the words 'if … then' and 'if and only if'.
- OCR defines a counter-example exactly as a value for which the condition is true but the conclusion is false.
Indices and surds
The laws of indices work for any rational power, so fractional and negative powers can be rewritten as roots and reciprocals. Surds are roots left in exact form. Simplify them by taking out square factors, and rationalise a denominator by multiplying top and bottom by a suitable surd or conjugate.
Key formulae
- $a^m\times a^n=a^{m+n},\quad a^m\div a^n=a^{m-n},\quad (a^m)^n=a^{mn}$
- $a^0=1,\quad a^{-n}=\dfrac{1}{a^n},\quad a^{\frac{m}{n}}=\left(\sqrt[n]{a}\right)^m$
- $\sqrt{ab}=\sqrt a\,\sqrt b,\quad \sqrt{\dfrac ab}=\dfrac{\sqrt a}{\sqrt b}$
- Rationalising: $\dfrac{1}{\sqrt a}=\dfrac{\sqrt a}{a}$ and $\dfrac{1}{a+\sqrt b}=\dfrac{a-\sqrt b}{a^2-b}$
Worked examples
- Evaluate $27^{-\frac23}$. First $27^{\frac13}=3$, so $27^{\frac23}=9$ and $27^{-\frac23}=\dfrac19$.
- Simplify $\sqrt{75}-2\sqrt{12}$. $\sqrt{75}=5\sqrt3$ and $2\sqrt{12}=4\sqrt3$, so the answer is $\sqrt3$.
- Rationalise $\dfrac{6}{3-\sqrt5}$. Multiply by $\dfrac{3+\sqrt5}{3+\sqrt5}$: the denominator is $9-5=4$, so the answer is $\dfrac{6(3+\sqrt5)}{4}=\dfrac{9+3\sqrt5}{2}$.
- Solve $4^{x}=8^{x-1}$. Write both as powers of $2$: $2^{2x}=2^{3x-3}$, so $2x=3x-3$ and $x=3$.
AQA 7357 · B1–B2
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- No board-specific differences. Exact surd answers are expected wherever a question says 'exact' or 'show that'.
Edexcel 9MA0 · Pure 2.1–2.2
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- Edexcel questions that say 'solutions relying entirely on calculator technology are not acceptable' need every step of the surd or index work shown.
OCR A H240 · 1.02a, b
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- When an OCR question says 'In this question you must show detailed reasoning', a calculator answer alone scores nothing: write the surd or index steps out.
Quadratics and the discriminant
A quadratic can be solved by factorising, completing the square or the formula. Completing the square also gives the turning point. The discriminant $b^2-4ac$ tells you how many real roots there are without solving, which is how 'find the range of values of $k$' questions are answered.
Key formulae
- $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
- Discriminant: $b^2-4ac>0$ two distinct real roots, $=0$ one repeated root, $<0$ no real roots
- Completing the square: $x^2+bx+c=\left(x+\tfrac b2\right)^2+c-\tfrac{b^2}4$; turning point $\left(-\tfrac b2,\ c-\tfrac{b^2}4\right)$
- Quadratics in disguise: $x^4-5x^2+4=0$ is a quadratic in $u=x^2$
Worked examples
- Write $2x^2-12x+23$ in the form $a(x+p)^2+q$. $2(x^2-6x)+23=2(x-3)^2-18+23=2(x-3)^2+5$. The minimum point is $(3,\ 5)$.
- Find the values of $k$ for which $x^2+kx+9=0$ has two distinct real roots. Need $k^2-36>0$, so $k<-6$ or $k>6$.
- Solve $x^4-5x^2+4=0$. Let $u=x^2$: $u^2-5u+4=0$ gives $u=1$ or $u=4$, so $x=\pm1$ or $x=\pm2$.
AQA 7357 · B3
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- None of the three boards prints the quadratic formula: learn it.
Edexcel 9MA0 · Pure 2.3
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- Edexcel's booklet does not print the quadratic formula either. 'Range of values of $k$' answers are expected in set or inequality form.
OCR A H240 · 1.02d–f
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- OCR excludes quadratic equations with complex roots: 'no real roots' is the answer when the discriminant is negative.
Simultaneous equations and inequalities
Solve one linear and one quadratic equation by substituting the linear one into the quadratic. The number of solutions tells you how a line and curve meet. For a quadratic inequality, find the critical values, sketch the curve and read off where it is above or below the axis. Give the answer with correct inequality or set notation.
Key formulae
- Substitute the linear equation into the quadratic; the discriminant of the result shows whether the line cuts ($>0$), touches ($=0$) or misses ($<0$) the curve
- For $(x-a)(x-b)<0$ with $a<b$: $a<x<b$. For $(x-a)(x-b)>0$: $x<a$ or $x>b$
- Set notation: $\{x: x<a\}\cup\{x: x>b\}$; 'and' is $\cap$, 'or' is $\cup$
- Multiplying or dividing an inequality by a negative number reverses the sign
Worked examples
- Solve $y=x+1$ and $x^2+y^2=13$. Substituting, $x^2+(x+1)^2=13$, so $2x^2+2x-12=0$, giving $x^2+x-6=0$ and $x=2$ or $x=-3$. The solutions are $(2,\ 3)$ and $(-3,\ -2)$.
- Solve $x^2-2x-15\ge0$. Critical values: $(x-5)(x+3)=0$, so $x=5$ or $x=-3$. The curve is above the axis outside the roots: $\{x: x\le-3\}\cup\{x: x\ge5\}$.
- Find where the line $y=2x+c$ is a tangent to $y=x^2+4$. $x^2-2x+4-c=0$ has a repeated root when $4-4(4-c)=0$, so $c=3$.
AQA 7357 · B4–B5
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- AQA's specification asks you to interpret inequalities graphically, including regions defined by inequalities (shading).
Edexcel 9MA0 · Pure 2.4–2.5
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- Edexcel mark schemes accept inequality or set notation, but a question that says 'using set notation' needs $\{x: \ldots\}$ with $\cup$ or $\cap$.
OCR A H240 · 1.02c, g–i
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- OCR 1.02h names set notation for solution sets explicitly, so expect 'give your answer in set notation'.
Polynomials and the factor theorem
A polynomial can be expanded, simplified and divided by a linear factor. The factor theorem says $(x-a)$ is a factor of $f(x)$ exactly when $f(a)=0$, so test small values to find one factor of a cubic, then divide or compare coefficients to find the rest.
Key formulae
- Factor theorem: $f(a)=0 \iff (x-a)$ is a factor of $f(x)$
- More generally $(ax-b)$ is a factor when $f\!\left(\tfrac ba\right)=0$
- Algebraic division: $f(x)=(x-a)\,q(x)$ when $(x-a)$ is a factor
- Simplify algebraic fractions by factorising and cancelling common factors
Worked examples
- Show that $(x-2)$ is a factor of $f(x)=x^3-3x^2-4x+12$ and factorise fully. $f(2)=8-12-8+12=0$, so $(x-2)$ is a factor. Dividing gives $x^2-x-6$, so $f(x)=(x-2)(x-3)(x+2)$.
- $(2x+1)$ is a factor of $g(x)=2x^3+x^2+kx+3$. Find $k$. $g\!\left(-\tfrac12\right)=-\tfrac14+\tfrac14-\tfrac k2+3=0$, so $k=6$.
- Simplify $\dfrac{x^2-9}{x^2+x-6}$. This is $\dfrac{(x-3)(x+3)}{(x+3)(x-2)}=\dfrac{x-3}{x-2}$.
AQA 7357 · B6
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- AQA (B6) names expanding, factorising, simple algebraic division and the factor theorem; there is no separate remainder theorem.
Edexcel 9MA0 · Pure 2.6
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- Edexcel expects division by a linear factor; 'show that … is a factor' must show $f(a)=0$ and a conclusion.
OCR A H240 · 1.02j, k
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- OCR expects the factor theorem with $(x-a)$ and $(ax-b)$, normally for cubics.
- At A Level OCR adds dividing by $(ax-b)$ to find the quotient and the remainder (1.02k), which AQA and Edexcel do not name.
Graphs, proportion and transformations
Sketch polynomials from their roots and the sign of the leading term, and reciprocal graphs from their asymptotes. Use intersections of graphs to count or solve equations. A transformation of $y=f(x)$ moves the graph in a predictable way: changes outside $f$ act vertically, changes inside act horizontally and the 'wrong way'.
Key formulae
- $y=f(x)+a$: up $a$. $y=f(x+a)$: left $a$. $y=af(x)$: vertical stretch, factor $a$. $y=f(ax)$: horizontal stretch, factor $\tfrac1a$
- $y=-f(x)$: reflect in the $x$-axis. $y=f(-x)$: reflect in the $y$-axis
- Repeated root: the curve touches the axis there. $y=\dfrac kx$ and $y=\dfrac k{x^2}$ have asymptotes $x=0$ and $y=0$
- Direct proportion $y=kx$; inverse proportion $y=\dfrac kx$
Worked examples
- Sketch $y=(x+1)^2(3-x)$. Roots: it touches the axis at $x=-1$ and crosses at $x=3$. The $y$-intercept is $(1)^2(3)=3$. The $x^3$ coefficient is $-1$, so the curve goes from top left to bottom right.
- The curve $y=f(x)$ has a maximum at $(2,\ 5)$. For $y=f(2x)$ the maximum is at $(1,\ 5)$; for $y=3f(x)$ it is at $(2,\ 15)$; for $y=f(x-4)$ it is at $(6,\ 5)$.
- How many solutions does $x^2-4=\dfrac1x$ have? Sketch both graphs: the parabola meets the left branch of $y=\tfrac1x$ twice and the right branch once, so there are $3$ solutions.
AQA 7357 · B7, B9
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- AQA describes translations with words or a vector; give the vector $\binom{a}{b}$ when asked to 'describe fully'.
Edexcel 9MA0 · Pure 2.7, 2.9
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- No board-specific differences. When asked for a transformed curve, label the new turning point and intercepts with coordinates.
OCR A H240 · 1.02m–r, w
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- At AS, OCR asks for single transformations only (1.02w); combinations are A Level (1.02x).
Straight lines
A straight line is fixed by a point and a gradient. Parallel lines have equal gradients and perpendicular gradients multiply to $-1$. Straight-line models link two quantities that change at a constant rate; interpret the gradient and intercept in context.
Key formulae
- $y-y_1=m(x-x_1)$, $\quad y=mx+c$, $\quad ax+by+c=0$
- Gradient $m=\dfrac{y_2-y_1}{x_2-x_1}$; perpendicular lines: $m_1m_2=-1$
- Midpoint $\left(\tfrac{x_1+x_2}2,\ \tfrac{y_1+y_2}2\right)$; distance $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Worked examples
- Find the perpendicular bisector of $A(1,\ 2)$ and $B(7,\ 6)$. Midpoint $(4,\ 4)$; gradient of $AB$ is $\tfrac46=\tfrac23$, so the bisector has gradient $-\tfrac32$: $y-4=-\tfrac32(x-4)$, or $3x+2y-20=0$.
- A plumber charges £$45$ for 1 hour and £$105$ for 3 hours. Modelled as linear, $C=30t+15$: the gradient $30$ is the charge per hour and the intercept $15$ is a call-out fee.
AQA 7357 · C1
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- No board-specific differences.
Edexcel 9MA0 · Pure 3.1
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- No board-specific differences; the straight-line formulae are not in the booklet.
OCR A H240 · 1.03a–c
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- OCR 1.03c names straight-line models in context: be ready to criticise a linear model, e.g. outside the data range.
Circles
The circle with centre $(a,b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$. Complete the square to find the centre and radius from expanded form. Three facts solve most problems: a tangent is perpendicular to the radius, the angle in a semicircle is $90^\circ$, and the perpendicular from the centre bisects a chord.
Key formulae
- $(x-a)^2+(y-b)^2=r^2$
- $x^2+y^2+2fx+2gy+c=0$ has centre $(-f,\ -g)$ and radius $\sqrt{f^2+g^2-c}$
- Tangent at $P$ is perpendicular to the radius to $P$
- Line and circle: substitute and use the discriminant (meets twice, touches, misses)
Worked examples
- Find the centre and radius of $x^2+y^2-6x+4y-12=0$. $(x-3)^2-9+(y+2)^2-4-12=0$, so $(x-3)^2+(y+2)^2=25$: centre $(3,\ -2)$, radius $5$.
- Find the tangent to $(x-3)^2+(y+2)^2=25$ at $P(6,\ 2)$. Radius gradient $\tfrac{2-(-2)}{6-3}=\tfrac43$, so the tangent gradient is $-\tfrac34$: $y-2=-\tfrac34(x-6)$, or $3x+4y-26=0$.
AQA 7357 · C2
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- No board-specific differences; the circle equation is not printed in AQA's booklet.
Edexcel 9MA0 · Pure 3.2
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- No board-specific differences; the circle equation is not in Pearson's booklet.
OCR A H240 · 1.03d–f
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- OCR lists the circle facts it expects: angle in a semicircle, perpendicular from centre to chord, tangent perpendicular to radius (1.03f).
Binomial expansion (positive integer powers)
$(a+b)^n$ for a positive integer $n$ expands into $n+1$ terms whose coefficients are $\binom nr$, the numbers in Pascal's triangle. Most questions ask for the first few terms or for one coefficient, sometimes followed by an approximation.
Key formulae
- $(a+b)^n=a^n+\binom n1a^{n-1}b+\binom n2a^{n-2}b^2+\cdots+b^n$
- $\binom nr={}^nC_r=\dfrac{n!}{r!(n-r)!}$; the term in $b^r$ is $\binom nr a^{n-r}b^r$
- Link to probability: $\binom nr$ is the number of ways of choosing $r$ successes in $n$ trials
Worked examples
- Find the first three terms of $(2-3x)^5$ in ascending powers of $x$. $2^5+5(2^4)(-3x)+10(2^3)(-3x)^2=32-240x+720x^2$.
- Find the coefficient of $x^3$ in $\left(1+\tfrac x2\right)^8$. $\binom83\left(\tfrac12\right)^3=56\times\tfrac18=7$.
- Use the first three terms of $(1+2x)^6$ with $x=0.01$ to estimate $1.02^6$. $1+12x+60x^2=1+0.12+0.006=1.126$ (the true value is $1.1262$ to 4 d.p.).
AQA 7357 · D1
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- AQA's formulae booklet prints the binomial expansion for positive integer $n$ and $\binom nr$.
Edexcel 9MA0 · Pure 4.1
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- Pearson's booklet prints the expansion and $\binom nr$; Edexcel calls it Pure topic 4.1.
OCR A H240 · 1.04a, b
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- OCR prints the expansion in each paper. OCR may ask for a particular term, but not for a formula for the general term (1.04a).
Trigonometry: rules, graphs, identities and equations
In Year 12 trigonometry is in degrees. Use the sine and cosine rules for non-right-angled triangles and $\tfrac12ab\sin C$ for area. The graphs of sine, cosine and tangent are periodic, so a trig equation has several solutions in an interval: find the calculator value, then use symmetry. Two identities let you change between functions.
Key formulae
- Sine rule $\dfrac a{\sin A}=\dfrac b{\sin B}$; cosine rule $a^2=b^2+c^2-2bc\cos A$; area $\tfrac12ab\sin C$
- $\tan\theta\equiv\dfrac{\sin\theta}{\cos\theta}$, $\quad\sin^2\theta+\cos^2\theta\equiv1$
- $\sin\theta=\sin(180^\circ-\theta)$, $\cos\theta=\cos(360^\circ-\theta)$, $\tan\theta=\tan(180^\circ+\theta)$
- Exact values: $\sin30^\circ=\tfrac12$, $\cos30^\circ=\tfrac{\sqrt3}2$, $\tan45^\circ=1$, $\tan60^\circ=\sqrt3$
Worked examples
- In triangle $ABC$, $AB=8$ cm, $AC=5$ cm and angle $A=40^\circ$. $BC^2=64+25-80\cos40^\circ=27.72$, so $BC=5.26$ cm. Area $=\tfrac12(8)(5)\sin40^\circ=12.9$ cm².
- Solve $2\sin^2\theta=3\cos\theta$ for $0\le\theta<360^\circ$. Using $\sin^2\theta=1-\cos^2\theta$: $2\cos^2\theta+3\cos\theta-2=0$, so $(2\cos\theta-1)(\cos\theta+2)=0$. $\cos\theta=-2$ is impossible, so $\cos\theta=\tfrac12$ and $\theta=60^\circ$ or $300^\circ$.
- Solve $\tan2x=1$ for $0\le x<180^\circ$. $2x$ runs over $0\le 2x<360^\circ$: $2x=45^\circ, 225^\circ$, so $x=22.5^\circ$ or $112.5^\circ$.
AQA 7357 · E1, E3, E5, E7
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- The sine rule, cosine rule and the two identities are not in AQA's booklet; learn them.
Edexcel 9MA0 · Pure 5.1, 5.3, 5.5, 5.7
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- Pearson's booklet does not give the sine and cosine rules or $\sin^2\theta+\cos^2\theta\equiv1$.
OCR A H240 · 1.05a–c, f, j, o
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- OCR's printed formulae do not include the sine and cosine rules either. OCR 1.05b names the ambiguous case of the sine rule.
Exponentials and logarithms
$y=a^x$ and $y=\log_a x$ are inverse functions; $e^x$ and $\ln x$ are the natural pair, and $e^{kx}$ has gradient $ke^{kx}$. Use the log laws to solve equations with the unknown in a power. Exponential models describe growth or decay at a rate proportional to size, and log graphs turn $y=ax^n$ and $y=kb^x$ into straight lines.
Key formulae
- $a^x=b\iff x=\log_ab$; $\quad e^{\ln x}=x$, $\ln e^x=x$
- $\log_a(xy)=\log_ax+\log_ay$, $\log_a\!\left(\tfrac xy\right)=\log_ax-\log_ay$, $\log_ax^k=k\log_ax$
- $\dfrac{d}{dx}e^{kx}=ke^{kx}$
- $y=ax^n\Rightarrow\log y=\log a+n\log x$ (gradient $n$); $y=kb^x\Rightarrow\log y=\log k+x\log b$ (gradient $\log b$)
Worked examples
- Solve $3^{2x-1}=20$. $2x-1=\dfrac{\ln20}{\ln3}=2.727$, so $x=1.86$ (3 s.f.).
- A car's value is modelled by $V=18000e^{-0.15t}$ after $t$ years. When is it first below £$9000$? $e^{-0.15t}<\tfrac12$ gives $t>\dfrac{\ln2}{0.15}=4.62$, so during the 5th year.
- A graph of $\log_{10}y$ against $x$ is a straight line through $(0,\ 0.7)$ with gradient $0.3$. Then $y=kb^x$ with $k=10^{0.7}=5.01$ and $b=10^{0.3}=2.00$.
AQA 7357 · F1–F7
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- The log laws are not in AQA's booklet; learn them.
Edexcel 9MA0 · Pure 6.1–6.7
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- Pearson's booklet prints the change-of-base rule and $a^x=e^{x\ln a}$; the log laws must still be learnt.
OCR A H240 · 1.06a–i
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- OCR excludes change of base. OCR papers often set log questions in a modelling context with 'comment on the model' parts.
Differentiation
The derivative is the gradient of the tangent, defined as a limit (first principles). Differentiate $x^n$ by bringing the power down and reducing it by one, after writing roots and fractions as powers. Use the derivative for tangents and normals, stationary points, increasing and decreasing functions, and simple optimisation.
Key formulae
- $f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$
- $\dfrac{d}{dx}x^n=nx^{n-1}$ for any rational $n$
- Tangent $y-y_1=f'(x_1)(x-x_1)$; normal gradient $-\dfrac1{f'(x_1)}$
- Stationary where $f'(x)=0$: maximum if $f''(x)<0$, minimum if $f''(x)>0$. Increasing where $f'(x)>0$
Worked examples
- Differentiate $x^2$ from first principles. $\dfrac{(x+h)^2-x^2}{h}=\dfrac{2xh+h^2}{h}=2x+h\to2x$ as $h\to0$.
- $y=4\sqrt x-\dfrac2x$. Find the tangent at $x=4$. $y=4x^{\frac12}-2x^{-1}$, so $\dfrac{dy}{dx}=2x^{-\frac12}+2x^{-2}=1+\tfrac18=\tfrac98$ at $x=4$. There $y=8-\tfrac12=7.5$, so the tangent is $y-7.5=\tfrac98(x-4)$.
- An open box has a square base of side $x$ cm and volume $500$ cm³. Its surface area is $A=x^2+\dfrac{2000}x$. $A'=2x-\dfrac{2000}{x^2}=0$ gives $x^3=1000$, $x=10$; $A''=2+\dfrac{4000}{x^3}>0$, so the minimum area is $300$ cm².
AQA 7357 · G1–G3
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- AQA's booklet does not print the first-principles definition; learn it.
Edexcel 9MA0 · Pure 7.1–7.3
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- Pearson's booklet prints the first-principles definition, but you still need to show the limit.
OCR A H240 · 1.07a–e, g, i, m–o
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- OCR prints the first-principles definition in each paper. At AS, OCR asks for first principles only for small positive integer powers of $x$ (1.07g).
Integration
Integration reverses differentiation: raise the power by one and divide by the new power, and add $+c$ for an indefinite integral. A definite integral gives the signed area between a curve and the $x$-axis. Split the region where the curve crosses the axis, because area below the axis counts as negative.
Key formulae
- $\displaystyle\int x^n\,dx=\frac{x^{n+1}}{n+1}+c,\ n\ne-1$
- $\displaystyle\int_a^bf'(x)\,dx=f(b)-f(a)$ (fundamental theorem of calculus)
- Area $=\displaystyle\int_a^by\,dx$ when $y\ge0$ on $[a,\,b]$
Worked examples
- A curve has $\dfrac{dy}{dx}=6x^2-4x$ and passes through $(1,\ 5)$. Then $y=2x^3-2x^2+c$ and $5=2-2+c$, so $y=2x^3-2x^2+5$.
- Find the area enclosed by $y=x(4-x)$ and the $x$-axis. $\displaystyle\int_0^4(4x-x^2)\,dx=\left[2x^2-\tfrac{x^3}3\right]_0^4=32-\tfrac{64}3=\tfrac{32}3$.
- Find the total area between $y=x^2-1$ and the $x$-axis for $0\le x\le2$. $\displaystyle\int_0^1=-\tfrac23$ and $\displaystyle\int_1^2=\tfrac43$, so the area is $\tfrac23+\tfrac43=2$ (the single integral would give $\tfrac23$).
AQA 7357 · H1–H3
Examined in: Papers 1, 2 and 3; AS 7356 Papers 1 and 2
- No board-specific differences at AS.
Edexcel 9MA0 · Pure 8.1–8.3
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- No board-specific differences; Edexcel often combines this with a straight line to find a shaded region.
OCR A H240 · 1.08a, b, d, e
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- OCR 1.08f includes areas below the $x$-axis; 'detailed reasoning' questions need the integration shown, not a calculator value.
Vectors in two dimensions
A vector has size and direction. Write it in $\mathbf i,\mathbf j$ form or as a column vector, find its magnitude with Pythagoras and its direction with trigonometry. Position vectors give points, so $\overrightarrow{AB}=\mathbf b-\mathbf a$. Vector methods prove geometric facts such as two lines being parallel.
Key formulae
- $\mathbf a=x\mathbf i+y\mathbf j=\begin{pmatrix}x\\y\end{pmatrix}$, $\ |\mathbf a|=\sqrt{x^2+y^2}$, direction $\arctan\dfrac yx$ from $\mathbf i$
- $\overrightarrow{AB}=\mathbf b-\mathbf a$; distance $AB=|\mathbf b-\mathbf a|$
- Unit vector $\hat{\mathbf a}=\dfrac{\mathbf a}{|\mathbf a|}$; parallel vectors: $\mathbf b=\lambda\mathbf a$
Worked examples
- Find the magnitude and direction of $\mathbf v=-5\mathbf i+12\mathbf j$. $|\mathbf v|=13$. It points up and left, at $180^\circ-\arctan\tfrac{12}5=112.6^\circ$ anticlockwise from $\mathbf i$.
- $A(1,\ 2)$, $B(4,\ 6)$, $C(10,\ 14)$. $\overrightarrow{AB}=3\mathbf i+4\mathbf j$ and $\overrightarrow{BC}=6\mathbf i+8\mathbf j=2\overrightarrow{AB}$, so the vectors are parallel and share $B$: $A$, $B$ and $C$ are collinear, with $AB=5$.
AQA 7357 · J1–J5
Examined in: Paper 2 (pure and mechanics); AS 7356 Paper 1
- AQA examines vectors in A Level Paper 2 (with mechanics) and AS Paper 1, not in Paper 3.
Edexcel 9MA0 · Pure 10.1–10.5
Examined in: Papers 1 and 2 (pure); AS 8MA0 Paper 1
- Edexcel examines vectors in the pure papers; vectors in forces and kinematics appear in Paper 3 mechanics.
OCR A H240 · 1.10a, c–g
Examined in: Paper 1, and the pure parts of Papers 2 and 3; AS H230 Papers 1 and 2
- OCR accepts $\overrightarrow{AB}$ or a bold letter, and asks you to distinguish vectors by hand (underline). At AS, 3D vectors are not included.
Statistics (6)
Populations and sampling
A population is everything you want to know about; a census measures all of it, a sample only part. A good sample is representative and the method should fit the situation. Know how to carry out and criticise each sampling method, and remember that different samples can give different conclusions.
Key formulae
- Simple random: every sample of size $n$ is equally likely (number the population, use random numbers)
- Systematic: every $k$th item from a random start, $k=\dfrac{\text{population}}{\text{sample size}}$
- Stratified: sample each group in proportion, $\dfrac{\text{group size}}{\text{population}}\times n$
- Quota and opportunity: non-random; quick and cheap, but may be biased
Worked examples
- A college has $360$ Year 12 and $240$ Year 13 students. A stratified sample of $50$: Year 12 $\tfrac{360}{600}\times50=30$, Year 13 $20$.
- A systematic sample of $20$ from a list of $400$: $k=20$, so pick a random start from $1$ to $20$, then every $20$th name.
AQA 7357 · K1
Examined in: Paper 3 (pure and statistics); AS 7356 Paper 2
- AQA's specification names simple random and opportunity sampling; questions on AQA's large data set also use stratified, systematic and quota samples.
Edexcel 9MA0 · Statistics 1.1
Examined in: Paper 3, Section A (statistics); AS 8MA0 Paper 2, Section A
- Edexcel expects you to know simple random, stratified, systematic, quota and opportunity sampling.
OCR A H240 · 2.01a–d
Examined in: Paper 2 (pure and statistics); AS H230 Paper 1
- OCR lists systematic, stratified, cluster and quota sampling: know and criticise them, but you won't be asked to carry them out (2.01d). Assume the population is large enough to sample without replacement.
Data presentation and interpretation
Describe data with a measure of location (mean, median, mode) and spread (range, interquartile range, standard deviation). In a histogram the area of each bar is proportional to the frequency. Spot outliers with a rule, clean the data, and interpret scatter diagrams and correlation in context, remembering that correlation does not imply causation.
Key formulae
- $\bar x=\dfrac{\sum x}n$, $\quad\sigma^2=\dfrac{\sum x^2}n-\bar x^2$ (or $\dfrac{S_{xx}}n$ with $S_{xx}=\sum x^2-\tfrac{(\sum x)^2}n$)
- Frequency density $=\dfrac{\text{frequency}}{\text{class width}}$
- A common outlier rule: below $Q_1-1.5\times\text{IQR}$ or above $Q_3+1.5\times\text{IQR}$
Worked examples
- $n=10$, $\sum x=172$, $\sum x^2=3020$. Mean $=17.2$; variance $=302-17.2^2=6.16$; standard deviation $=2.48$.
- $Q_1=12$, $Q_3=20$. Outliers are below $12-12=0$ or above $20+12=32$, so a value of $35$ is an outlier.
- Classes $10\le x<15$ (frequency $30$) and $15\le x<25$ (frequency $40$) have frequency densities $6$ and $4$. Estimate how many values lie between $12$ and $20$: $\tfrac35\times30+\tfrac5{10}\times40=38$.
AQA 7357 · L1–L4
Examined in: Paper 3 (pure and statistics); AS 7356 Paper 2
- AQA's booklet prints the standard deviation formula. Calculations involving regression lines are excluded: interpret them only.
Edexcel 9MA0 · Statistics 2.1–2.4
Examined in: Paper 3, Section A (statistics); AS 8MA0 Paper 2, Section A
- Edexcel uses $S_{xx}$ notation and expects $\sqrt{S_{xx}/n}$ (the $n-1$ version is accepted). It adds coding, interpolation for percentiles and log changes of variable to straighten data.
OCR A H240 · 2.02a–j
Examined in: Paper 2 (pure and statistics); AS H230 Paper 1
- OCR uses divisor $n$ and excludes formal estimation of a population variance; know which symbol your calculator uses. Calculating regression lines is excluded.
The large data set
Each board publishes a large data set of real data that you study during the course. You won't have it in the exam, but some statistics questions are set in its context and reward knowing it: what one row represents, the variables and units, how missing values are shown, and the main patterns. The data set is different for every board, so learn yours.
Key formulae
- Clean the data first: remove or adjust missing and impossible values, and reduce $n$ to match
- Compare groups of different sizes with proportions or percentages, not raw counts
- Use the context: say what a statistic means for the places, cars or areas in the data
Worked examples
- A sample of $8$ daily readings has one value recorded as missing. The other seven total $91$, so the mean is $\tfrac{91}7=13$ (not $\tfrac{91}8$). These numbers are made up for practice.
- Area A: $612$ of $9000$ workers cycle; area B: $455$ of $5200$. Percentages: A $6.8\%$, B $8.75\%$, so B has the higher proportion even though A has more cyclists. These numbers are made up for practice.
AQA 7357 · 3.21 (large data set)
Examined in: Paper 3 (pure and statistics); AS 7356 Paper 2
- AQA: cars data, an extract of the Department for Transport's Stock Vehicle Database, used since the 2020 exams. Questions are in Paper 3.
- Several fields (propulsion type, body type) are numeric codes, and some emission values are blank or zero: decide whether each is missing or genuine (an electric car has no exhaust emissions).
Edexcel 9MA0 · Statistics 1, 2 (large data set)
Examined in: Paper 3, Section A (statistics); AS 8MA0 Paper 2, Section A
- Edexcel: Met Office weather data for May to October in 1987 and 2015, at five UK stations and three overseas (Beijing, Jacksonville, Perth). Questions are in Paper 3, Section A.
- 'tr' means a trace of rain (under 0.05 mm) and 'n/a' means not available. Perth is in the southern hemisphere, so its May to October is autumn and winter.
OCR A H240 · 2.01, 2.02 (pre-release data set)
Examined in: Paper 2 (pure and statistics); AS H230 Paper 1
- OCR A: 2001 and 2011 census data for England and Wales by local authority: method of travel to work and age structure. Questions are in Paper 2. (London Underground data belongs to OCR's other specification, MEI, not OCR A.)
- Age groups have unequal widths, so use frequency density for histograms; ages are 'age last birthday', so '20–24' means $20\le$ age $<25$.
Probability: Venn diagrams, trees and independence
Mutually exclusive events can't happen together; independent events don't affect each other. Draw a Venn diagram, tree diagram or two-way table to organise the information, then add along branches for 'or' and multiply for 'and'.
Key formulae
- $P(A\cup B)=P(A)+P(B)-P(A\cap B)$; mutually exclusive: $P(A\cap B)=0$
- Independent: $P(A\cap B)=P(A)\times P(B)$
- $P(A')=1-P(A)$
Worked examples
- $P(A)=0.4$, $P(B)=0.5$, $P(A\cup B)=0.7$. Then $P(A\cap B)=0.4+0.5-0.7=0.2=0.4\times0.5$, so $A$ and $B$ are independent.
- A bag has $5$ red and $3$ blue counters. Two are taken without replacement. $P(\text{same colour})=\tfrac58\cdot\tfrac47+\tfrac38\cdot\tfrac27=\tfrac{13}{28}=0.464$.
- Of $30$ students, $18$ play football, $12$ play tennis and $5$ play both. Neither: $30-(18+12-5)=5$.
AQA 7357 · M1
Examined in: Paper 3 (pure and statistics); AS 7356 Paper 2
- AQA's booklet prints $P(A\cup B)$ and $P(A\cap B)=P(A)P(B\mid A)$.
Edexcel 9MA0 · Statistics 3.1
Examined in: Paper 3, Section A (statistics); AS 8MA0 Paper 2, Section A
- Pearson's booklet prints these and the independence condition.
OCR A H240 · 2.03a–c
Examined in: Paper 2 (pure and statistics); AS H230 Paper 1
- OCR prints the addition and multiplication rules in each paper.
Discrete distributions and the binomial distribution
A discrete random variable has a list of values with probabilities that add to $1$. The binomial distribution $B(n,p)$ models the number of successes in $n$ independent trials, each with the same probability $p$ of success. Use your calculator for $P(X=x)$ and cumulative probabilities $P(X\le x)$.
Key formulae
- $X\sim B(n,p)$: $P(X=x)=\binom nx p^x(1-p)^{n-x}$
- $P(X\ge k)=1-P(X\le k-1)$; $\ P(X<k)=P(X\le k-1)$
- Conditions: fixed $n$, two outcomes, independent trials, constant $p$
Worked examples
- $X\sim B(12,\ 0.3)$. $P(X=4)=0.231$, $P(X\le2)=0.253$, $P(X\ge5)=1-P(X\le4)=0.276$.
- A spinner gives a $6$ with probability $\tfrac16$. In $10$ spins, $P(\text{at least two }6\text{s})=1-P(X\le1)=0.515$.
- $P(X=x)=kx$ for $x=1,2,3,4$. The probabilities sum to $1$: $10k=1$, so $k=0.1$.
AQA 7357 · N1
Examined in: Paper 3 (pure and statistics); AS 7356 Paper 2
- AQA's booklet prints the binomial probability formula, mean and variance. Calculating the mean and variance of a general discrete random variable is excluded.
Edexcel 9MA0 · Statistics 4.1
Examined in: Paper 3, Section A (statistics); AS 8MA0 Paper 2, Section A
- Pearson's booklet prints the binomial formula, mean and variance, plus cumulative binomial tables.
OCR A H240 · 2.04a–d
Examined in: Paper 2 (pure and statistics); AS H230 Paper 1
- OCR prints the binomial formula in each paper and excludes the mean and variance of a general discrete random variable (2.04a).
Hypothesis testing with the binomial distribution
A hypothesis test uses a sample to judge a claim about a population. State $H_0$ and $H_1$ in terms of the parameter $p$, assume $H_0$ is true, and find how likely a result at least as extreme as the one observed would be. Compare this $p$-value with the significance level, or find the critical region, and conclude in context.
Key formulae
- $H_0: p=p_0$; $\ H_1: p<p_0$ or $p>p_0$ (one-tailed) or $p\ne p_0$ (two-tailed)
- Reject $H_0$ if $P(X\le x)$ (or $P(X\ge x)$) is less than the significance level; for two-tailed tests use half the level in each tail
- Critical region: the values that lead to rejecting $H_0$; actual significance level $=P(X\text{ in critical region})$
Worked examples
- A supplier claims $80\%$ of seeds germinate. In a sample of $20$, only $12$ do. Test at $5\%$: $H_0: p=0.8$, $H_1: p<0.8$. $P(X\le12)=0.0321<0.05$, so reject $H_0$: there is evidence that the germination rate is lower than claimed.
- For the same test, $P(X\le13)=0.0867>0.05$, so the critical region is $X\le12$ with actual significance level $3.21\%$.
- Test whether a coin is biased, $n=20$, $5\%$ two-tailed. $P(X\le5)=0.0207<0.025$ and $P(X\le6)=0.0577$, so the critical region is $X\le5$ or $X\ge15$.
AQA 7357 · O1–O2
Examined in: Paper 3 (pure and statistics); AS 7356 Paper 2
- AQA (O2) expects you to appreciate that the significance level is the probability of wrongly rejecting $H_0$.
Edexcel 9MA0 · Statistics 5.1–5.2
Examined in: Paper 3, Section A (statistics); AS 8MA0 Paper 2, Section A
- Edexcel asks for critical regions and the actual significance level; its booklet includes cumulative binomial tables.
OCR A H240 · 2.05a–c
Examined in: Paper 2 (pure and statistics); AS H230 Paper 1
- OCR excludes the normal approximation in binomial tests and notes that the probability of the test statistic being in the rejection region is at most the significance level.
Mechanics (5)
Units, modelling and kinematics graphs
Mechanics uses SI units: metres, seconds and kilograms, and units built from them such as m s$^{-1}$ and newtons. Models simplify reality, for example treating an object as a particle. On a displacement–time graph the gradient is velocity; on a velocity–time graph the gradient is acceleration and the area is displacement.
Key formulae
- Base units: m, s, kg. Derived: velocity m s$^{-1}$, acceleration m s$^{-2}$, force N $=$ kg m s$^{-2}$
- Velocity–time graph: gradient $=$ acceleration, area $=$ displacement
- Average speed $=\dfrac{\text{total distance}}{\text{total time}}$
- Modelling words: particle (no size), light (no mass), inextensible (fixed length), smooth (no friction)
Worked examples
- Convert $54$ km h$^{-1}$ to m s$^{-1}$: $54\times\dfrac{1000}{3600}=15$ m s$^{-1}$.
- A cyclist accelerates uniformly from rest to $15$ m s$^{-1}$ in $5$ s, rides at that speed for $20$ s, then slows uniformly to rest in $10$ s. Distance $=\tfrac12(5)(15)+20(15)+\tfrac12(10)(15)=412.5$ m. Accelerations: $3$ m s$^{-2}$ then $-1.5$ m s$^{-2}$. Average speed $=\tfrac{412.5}{35}=11.8$ m s$^{-1}$.
AQA 7357 · P1, Q1–Q2
Examined in: Paper 2 (pure and mechanics); AS 7356 Paper 1
- AQA examines mechanics in A Level Paper 2 alongside vectors, and in AS Paper 1.
Edexcel 9MA0 · Mechanics 6.1, 7.1–7.2
Examined in: Paper 3, Section B (mechanics); AS 8MA0 Paper 2, Section B
- Edexcel puts all mechanics in Paper 3, Section B (AS: Paper 2, Section B).
OCR A H240 · 3.01a–c, 3.02a–c
Examined in: Paper 3 (pure and mechanics); AS H230 Paper 2
- OCR puts mechanics in Paper 3 with pure (AS: H230 Paper 2). OCR 3.01 says length, time and mass are independent base quantities.
Constant acceleration (suvat)
When acceleration is constant, five equations link displacement $s$, initial velocity $u$, final velocity $v$, acceleration $a$ and time $t$. List the three you know and the one you want, then choose the equation without the fifth. For vertical motion under gravity, $a=g$ downwards.
Key formulae
- $v=u+at$, $\quad s=ut+\tfrac12at^2$, $\quad s=vt-\tfrac12at^2$, $\quad s=\tfrac12(u+v)t$, $\quad v^2=u^2+2as$
- Under gravity: $a=-9.8$ m s$^{-2}$ if up is positive
Worked examples
- A car at $20$ m s$^{-1}$ brakes uniformly and stops in $50$ m. $0=400+2a(50)$ gives $a=-4$ m s$^{-2}$, and $0=20-4t$ gives $t=5$ s.
- A ball is thrown up at $14.7$ m s$^{-1}$ from $2$ m above the ground. Greatest height above the ground $=\dfrac{14.7^2}{2(9.8)}+2=13.0$ m. It lands when $-2=14.7t-4.9t^2$, so $t=3.13$ s.
AQA 7357 · Q3
Examined in: Paper 2 (pure and mechanics); AS 7356 Paper 1
- AQA's booklet prints the constant-acceleration formulas.
Edexcel 9MA0 · Mechanics 7.3
Examined in: Paper 3, Section B (mechanics); AS 8MA0 Paper 2, Section B
- Pearson's booklet prints them too; the default is $g=9.8$, but a question may give another value such as $g=10$.
OCR A H240 · 3.02d, e
Examined in: Paper 3 (pure and mechanics); AS H230 Paper 2
- OCR prints the formulas in each paper and may ask you to derive them, by integration, from a velocity–time graph, or by substitution (3.02d).
Variable acceleration in one dimension
When acceleration changes with time, use calculus. Differentiate displacement to get velocity and velocity to get acceleration; integrate to go back, using initial conditions for the constants. To find distance travelled, check whether the particle changes direction (where $v=0$).
Key formulae
- $v=\dfrac{ds}{dt}$, $\quad a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}$
- $s=\displaystyle\int v\,dt$, $\quad v=\int a\,dt$ (plus a constant each time)
Worked examples
- $v=3t^2-12t+9$ for $t\ge0$. The particle is at rest when $t=1$ and $t=3$. $a=6t-12$, so $a=0$ at $t=2$.
- For the same particle, starting at $s=0$: $s=t^3-6t^2+9t$. $s(1)=4$ and $s(3)=0$, so in the first $3$ s it travels $4+4=8$ m but its displacement is $0$.
- $a=4-2t$ and $v=3$ when $t=0$. Then $v=3+4t-t^2$, and the maximum velocity is $7$ m s$^{-1}$, at $t=2$.
AQA 7357 · Q4
Examined in: Paper 2 (pure and mechanics); AS 7356 Paper 1
- The calculus kinematics results are not in AQA's booklet; AQA writes $r$ for displacement.
Edexcel 9MA0 · Mechanics 7.4
Examined in: Paper 3, Section B (mechanics); AS 8MA0 Paper 2, Section B
- Edexcel's guidance says the calculus will match the level of the pure content.
OCR A H240 · 3.02f
Examined in: Paper 3 (pure and mechanics); AS H230 Paper 2
- OCR 3.02f writes these with $s$; 2D vector versions are Year 13 (3.02g).
Forces and Newton's laws
Draw a force diagram showing every force on the object: weight $mg$, normal reaction, tension or thrust, resistance and any applied force. Newton's first law says no resultant force means constant velocity; the second says $F=ma$ for the resultant force; the third says that when two bodies interact, the forces they exert on each other are equal and opposite.
Key formulae
- $F=ma$ (resultant force in the direction of the acceleration)
- $W=mg$; on a horizontal surface with only vertical forces, $R=mg$
- Forces as vectors: $\mathbf F=m\mathbf a$
Worked examples
- A lift of mass $600$ kg accelerates upwards at $0.5$ m s$^{-2}$. $T-600g=600(0.5)$, so $T=6180$ N.
- A $5$ kg box is pulled across a floor by a horizontal $20$ N force against a resistance of $8$ N. $a=\tfrac{20-8}5=2.4$ m s$^{-2}$.
- Forces $(3\mathbf i+4\mathbf j)$ N and $(\mathbf i-\mathbf j)$ N act on a $2$ kg particle. $\mathbf a=\tfrac12(4\mathbf i+3\mathbf j)$, magnitude $2.5$ m s$^{-2}$.
AQA 7357 · R1–R3
Examined in: Paper 2 (pure and mechanics); AS 7356 Paper 1
- At AS, AQA restricts forces to two perpendicular directions or simple 2D vectors. Friction is not in AQA's AS at all; it is A Level (R6).
Edexcel 9MA0 · Mechanics 8.1–8.3
Examined in: Paper 3, Section B (mechanics); AS 8MA0 Paper 2, Section B
- Edexcel's AS problems have forces parallel or perpendicular to the motion, or given in $\mathbf i,\mathbf j$ form. Default $g=9.8$ unless stated.
OCR A H240 · 3.03a–l, r
Examined in: Paper 3 (pure and mechanics); AS H230 Paper 2
- OCR's AS includes friction as a force you are given (3.03r), as in the second example; the coefficient of friction is A Level.
Connected particles and pulleys
Particles joined by a light inextensible string (or a tow bar) move with the same acceleration. Apply $F=ma$ to each particle separately, or to the whole system for the acceleration, then to one particle for the tension. Over a smooth pulley the tension is the same on both sides.
Key formulae
- Whole system: (driving force $-$ resistances) $=$ (total mass) $\times a$
- Each particle: resultant force $=ma$; a tow bar can be in tension or thrust
- Smooth light pulley: same tension in both parts of the string
Worked examples
- Particles of mass $3$ kg and $2$ kg hang over a smooth pulley. $a=\dfrac{(3-2)g}5=1.96$ m s$^{-2}$ and $T=2(9.8+1.96)=23.5$ N.
- A $1200$ kg car tows a $400$ kg trailer. The driving force is $2400$ N; resistances are $400$ N and $200$ N. $a=\dfrac{2400-600}{1600}=1.125$ m s$^{-2}$, and for the trailer $T-200=400(1.125)$, so $T=650$ N.
AQA 7357 · R4
Examined in: Paper 2 (pure and mechanics); AS 7356 Paper 1
- AQA's AS version is 'smooth pulleys and connected particles' with forces in two perpendicular directions.
Edexcel 9MA0 · Mechanics 8.4
Examined in: Paper 3, Section B (mechanics); AS 8MA0 Paper 2, Section B
- Edexcel includes lift problems; particles on inclined planes connected over pulleys are A Level.
OCR A H240 · 3.03m–o
Examined in: Paper 3 (pure and mechanics); AS H230 Paper 2
- At A Level OCR adds resolving forces in connected-particle problems (3.03o).
Year 13 (A2) notes
The content that is in the full A Level only.
Pure (18)
Proof by contradiction
To prove a statement by contradiction, assume it is false and show that this leads to something impossible. The assumption must then be wrong, so the statement is true. The two classic results are that $\sqrt2$ is irrational and that there are infinitely many primes; learn the structure of both.
Key formulae
- Structure: 'Assume the opposite … this gives a contradiction … so the original statement is true'
- A rational number can be written $\dfrac pq$ with integers $p,\ q$, $q\ne0$, in lowest terms
- If $p^2$ is a multiple of a prime $k$, then $p$ is a multiple of $k$
- Infinitely many primes: assume a finite list $p_1,\dots,p_n$; then $p_1p_2\cdots p_n+1$ leaves remainder $1$ on division by each
Worked examples
- Prove that $\sqrt3$ is irrational. Assume $\sqrt3=\dfrac pq$ in lowest terms. Then $p^2=3q^2$, so $p^2$ and hence $p$ is a multiple of $3$: $p=3m$. Then $9m^2=3q^2$, so $q^2=3m^2$ and $q$ is a multiple of $3$ too. So $p$ and $q$ share the factor $3$, which contradicts 'lowest terms'. Hence $\sqrt3$ is irrational.
- Prove that there are no integers $a$ and $b$ with $6a+9b=2$. Assume there are. The left side is $3(2a+3b)$, a multiple of $3$, but $2$ is not a multiple of $3$. This is a contradiction, so no such integers exist.
- Prove that if $n^2$ is odd then $n$ is odd. Assume $n^2$ is odd but $n$ is even, $n=2k$. Then $n^2=4k^2=2(2k^2)$ is even, which contradicts $n^2$ being odd.
AQA 7357 · A1
Examined in: Papers 1, 2 and 3
- AQA follows the DfE content: the proofs that $\sqrt2$ is irrational and that there are infinitely many primes are named.
Edexcel 9MA0 · Pure 1.1
Examined in: Papers 1 and 2 (pure)
- Edexcel follows the DfE content, which names the proofs that $\sqrt2$ is irrational and that there are infinitely many primes.
OCR A H240 · 1.01d
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR says contradiction questions are set on content you already know from GCSE, AS or A Level, and names the $\sqrt2$ and primes proofs (1.01d).
Modulus, composite and inverse functions
A function maps each input in its domain to exactly one output; the set of outputs is the range. A composite $fg(x)$ means 'do $g$, then $f$'. An inverse $f^{-1}$ undoes $f$, exists only for one-to-one functions, swaps domain and range, and its graph is the reflection of $y=f(x)$ in $y=x$. The modulus $|x|$ is the size of $x$.
Key formulae
- $fg(x)=f(g(x))$; generally $fg\ne gf$
- $f^{-1}$: swap $x$ and $y$ and rearrange; $ff^{-1}(x)=x$; domain of $f^{-1}$ = range of $f$
- $|x|<a\iff -a<x<a$; $\quad |a|=|b|\iff a^2=b^2$
- $y=|ax+b|$: reflect the part below the $x$-axis in the $x$-axis
Worked examples
- $f(x)=2x+3$ and $g(x)=x^2$. Then $fg(2)=f(4)=11$ and $gf(2)=g(7)=49$.
- Find $h^{-1}$ for $h(x)=\dfrac{x+1}{x-2}$, $x\ne2$. From $y(x-2)=x+1$: $x(y-1)=2y+1$, so $h^{-1}(x)=\dfrac{2x+1}{x-1}$, $x\ne1$.
- Solve $|2x-1|=x+4$. Either $2x-1=x+4$, giving $x=5$, or $2x-1=-(x+4)$, giving $x=-1$. Both satisfy the original equation.
AQA 7357 · B7–B8
Examined in: Papers 1, 2 and 3
- AQA (B7) asks you to sketch the modulus of a linear function; harder modulus graphs come from transformations (B9).
Edexcel 9MA0 · Pure 2.7–2.8
Examined in: Papers 1 and 2 (pure)
- Edexcel's guidance asks you to sketch $y=|ax+b|$ and use it to solve equations and inequalities such as $|2x-1|>x$.
OCR A H240 · 1.02l, s, u, v
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR limits modulus graphs to $y=|ax+b|$ with a single modulus sign; graphs of the modulus of other functions are excluded (1.02s).
Combined transformations and modelling with functions
Apply combined transformations one at a time, in an order that matches the algebra: changes inside the bracket first, then outside. Functions are also used as models, so expect to fit parameters, use the model, and comment on its limitations.
Key formulae
- $y=af(bx+c)+d$: translate left $c$, stretch horizontally by $\tfrac1b$, stretch vertically by $a$, translate up $d$
- $y=|f(x)|$: reflect the parts below the $x$-axis; $y=f(|x|)$: reflect the right half in the $y$-axis
- A model's domain should match the real situation; check predictions against the context
Worked examples
- $y=f(x)$ has a minimum at $(3,\ -2)$. Find the minimum of $y=4f(x+1)-5$. Translate left $1$: $(2,\ -2)$. Stretch vertically by $4$: $(2,\ -8)$. Translate down $5$: $(2,\ -13)$.
- The height of a ball is modelled by $h=-4.9t^2+14t+1.5$ metres. Its greatest height is at $t=\tfrac{14}{9.8}=1.43$ s, when $h=11.5$ m; it lands when $h=0$, at $t=2.96$ s.
AQA 7357 · B9, B11
Examined in: Papers 1, 2 and 3
- AQA (B11) asks for the limitations and refinements of function models.
Edexcel 9MA0 · Pure 2.9, 2.11
Examined in: Papers 1 and 2 (pure)
- Edexcel's guidance lists combinations such as $y=2f(3x)$ and $y=f(-x)+1$ for any function in the specification, including trig and exponential graphs.
OCR A H240 · 1.02x, z
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR 1.02x allows combined transformations at A Level (only single ones at AS); translations may be given as a vector.
Algebraic and partial fractions
Partial fractions split a fraction with a factorised denominator into simpler fractions, which is the key step for some integrals and binomial expansions. Use one term per linear factor, and two terms (powers 1 and 2) for a repeated factor. Find the constants by substituting convenient values of $x$ or comparing coefficients.
Key formulae
- $\dfrac{px+q}{(x-a)(x-b)}=\dfrac A{x-a}+\dfrac B{x-b}$
- $\dfrac{\cdots}{(x-a)^2(x-b)}=\dfrac A{x-a}+\dfrac B{(x-a)^2}+\dfrac C{x-b}$
- If the degree of the numerator is not less than the degree of the denominator, divide first
Worked examples
- Express $\dfrac{5x+1}{(x-1)(x+2)}$ in partial fractions. $5x+1=A(x+2)+B(x-1)$. $x=1$: $6=3A$, $A=2$. $x=-2$: $-9=-3B$, $B=3$. So it is $\dfrac2{x-1}+\dfrac3{x+2}$.
- Express $\dfrac{x+5}{(x+1)^2(x-3)}$ in partial fractions. $x+5=A(x+1)(x-3)+B(x-3)+C(x+1)^2$. $x=3$: $8=16C$, $C=\tfrac12$. $x=-1$: $4=-4B$, $B=-1$. Coefficient of $x^2$: $0=A+C$, $A=-\tfrac12$.
AQA 7357 · B10
Examined in: Papers 1, 2 and 3
- AQA limits denominators to at most three linear terms, including a squared linear term (B10).
Edexcel 9MA0 · Pure 2.10
Examined in: Papers 1 and 2 (pure)
- Edexcel's guidance gives denominators such as $(ax+b)(cx+d)(ex+f)$ and $(ax+b)(cx+d)^2$.
OCR A H240 · 1.02y
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR 1.02y lists the same denominators: products of two or three linear factors, or one squared linear factor.
Parametric equations
A parametric curve gives $x$ and $y$ separately in terms of a parameter, often $t$ or $\theta$. Convert to a Cartesian equation by eliminating the parameter, by substitution or with a trig identity. Parametric equations model motion, where the parameter is time.
Key formulae
- Eliminate $t$: make $t$ the subject of one equation and substitute into the other
- With trig parameters use $\sin^2\theta+\cos^2\theta=1$ or $\sec^2\theta=1+\tan^2\theta$
- The domain and range of the Cartesian form come from the allowed values of the parameter
Worked examples
- $x=2t-1$, $y=t^2+3$. Then $t=\dfrac{x+1}2$, so $y=\dfrac{(x+1)^2}4+3$.
- $x=3\cos\theta$, $y=2\sin\theta$. Then $\dfrac{x^2}9+\dfrac{y^2}4=\cos^2\theta+\sin^2\theta=1$, an ellipse.
- Where does $x=t^2$, $y=2t$ meet the line $y=x-3$? $2t=t^2-3$ gives $t=3$ or $t=-1$: the points $(9,\ 6)$ and $(1,\ -2)$.
AQA 7357 · C3–C4
Examined in: Papers 1, 2 and 3
- No board-specific differences.
Edexcel 9MA0 · Pure 3.3–3.4
Examined in: Papers 1 and 2 (pure)
- No board-specific differences.
OCR A H240 · 1.03g, h
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR 1.03h names modelling with parametric equations, for example the path of a moving object.
Binomial expansion for any rational power
For a negative or fractional power, $(1+x)^n$ becomes an infinite series that is valid only for $|x|<1$. To expand $(a+bx)^n$, take out $a^n$ first. The expansion gives approximations, and it can be combined with partial fractions.
Key formulae
- $(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\cdots$, valid for $|x|<1$
- $(a+bx)^n=a^n\left(1+\tfrac bax\right)^n$, valid for $|x|<\left|\tfrac ab\right|$
Worked examples
- Expand $(1-2x)^{-3}$ up to $x^2$. $1+(-3)(-2x)+\dfrac{(-3)(-4)}2(-2x)^2=1+6x+24x^2$, valid for $|x|<\tfrac12$.
- Expand $\sqrt{4+x}$ up to $x^2$. $2\left(1+\tfrac x4\right)^{\frac12}=2\left(1+\tfrac x8-\tfrac{x^2}{128}\right)=2+\tfrac x4-\tfrac{x^2}{64}$, valid for $|x|<4$.
- Use $x=0.04$ in $\sqrt{4+x}\approx2+\tfrac x4-\tfrac{x^2}{64}$ to estimate $\sqrt{4.04}$: $2+0.01-0.000025=2.009975$ (true value $2.0099751$).
AQA 7357 · D1
Examined in: Papers 1, 2 and 3
- AQA's booklet prints the expansion for rational $n$, with its validity condition.
Edexcel 9MA0 · Pure 4.1
Examined in: Papers 1 and 2 (pure)
- Pearson's booklet prints the general expansion; Edexcel often follows it with partial fractions or an approximation.
OCR A H240 · 1.04c, d
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR prints the expansion in each paper and 1.04d asks you to know it is valid for $|x|<1$; the proof is not required.
Sequences and series
An arithmetic sequence adds a fixed difference; a geometric sequence multiplies by a fixed ratio. Learn the $n$th-term and sum formulas, the sum to infinity for $|r|<1$, sigma notation and recurrence relations. Many questions model savings, populations or repeated doses.
Key formulae
- Arithmetic: $u_n=a+(n-1)d$, $S_n=\tfrac n2\big(2a+(n-1)d\big)=\tfrac n2(a+l)$
- Geometric: $u_n=ar^{n-1}$, $S_n=\dfrac{a(1-r^n)}{1-r}$, $S_\infty=\dfrac a{1-r}$ for $|r|<1$
- $\displaystyle\sum_{r=1}^n r=\tfrac12n(n+1)$; recurrence $u_{n+1}=f(u_n)$; sequences can be increasing, decreasing or periodic
Worked examples
- An arithmetic series has first term $7$ and common difference $4$. Find $\displaystyle\sum$ of the first $20$ terms: $S_{20}=10(14+76)=900$.
- A geometric series has $a=48$ and $S_\infty=64$. Then $1-r=\tfrac{48}{64}$, so $r=0.25$, and $u_4=48(0.25)^3=0.75$.
- Savings of £$500$ in year 1 grow by $3\%$ a year. In which year does the yearly saving first exceed £$800$? $500(1.03)^{n-1}>800$ gives $n-1>\dfrac{\ln1.6}{\ln1.03}=15.9$, so year $17$.
AQA 7357 · D2–D6
Examined in: Papers 1, 2 and 3
- AQA's booklet prints the arithmetic and geometric sums and the sum to infinity, but not the $n$th-term formulas.
Edexcel 9MA0 · Pure 4.2–4.6
Examined in: Papers 1 and 2 (pure)
- Edexcel's guidance says the proof of the geometric sum formula should be known, and that logs may be needed to find $n$; the sums are in the booklet.
OCR A H240 · 1.04e–k
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR prints the arithmetic and geometric sums in each paper; 1.04f includes increasing, decreasing and periodic sequences.
Radians, arcs, sectors and small angles
In Year 13 angles are usually in radians: $\pi$ radians is $180^\circ$. Arc length and sector area have simple formulas in radians, and so do the small-angle approximations, which replace $\sin\theta$, $\cos\theta$ and $\tan\theta$ by polynomials when $\theta$ is small.
Key formulae
- $\pi\text{ rad}=180^\circ$; arc $s=r\theta$; sector $A=\tfrac12r^2\theta$
- Segment area $=\tfrac12r^2(\theta-\sin\theta)$
- For small $\theta$ in radians: $\sin\theta\approx\theta$, $\cos\theta\approx1-\tfrac{\theta^2}2$, $\tan\theta\approx\theta$
Worked examples
- A sector has radius $6$ cm and angle $1.2$ rad. Arc $=7.2$ cm, area $=\tfrac12(36)(1.2)=21.6$ cm², perimeter $=19.2$ cm.
- Find the area of the segment cut off by a chord subtending $\tfrac\pi3$ at the centre of a circle of radius $10$: $50\left(\tfrac\pi3-\tfrac{\sqrt3}2\right)=9.06$.
- For small $\theta$, approximate $\dfrac{1-\cos2\theta}{\theta\tan\theta}$. Using $\cos2\theta\approx1-2\theta^2$ and $\tan\theta\approx\theta$: $\dfrac{2\theta^2}{\theta^2}=2$.
AQA 7357 · E1–E2
Examined in: Papers 1, 2 and 3
- AQA's booklet prints the small-angle approximations but not the arc and sector formulas.
Edexcel 9MA0 · Pure 5.1–5.2
Examined in: Papers 1 and 2 (pure)
- Pearson's booklet prints the small-angle approximations; the arc and sector formulas must be learnt.
OCR A H240 · 1.05d, e, g
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR prints the small-angle approximations and expects you to know $s=r\theta$ and $A=\tfrac12r^2\theta$ (1.05d), plus exact values in radians (1.05g).
Reciprocal and inverse trigonometric functions
$\sec\theta$, $\csc\theta$ and $\cot\theta$ are the reciprocals of $\cos\theta$, $\sin\theta$ and $\tan\theta$. Two new identities follow from $\sin^2\theta+\cos^2\theta=1$. The inverse functions $\arcsin$, $\arccos$ and $\arctan$ need restricted domains so that they are functions.
Key formulae
- $\sec\theta=\dfrac1{\cos\theta}$, $\csc\theta=\dfrac1{\sin\theta}$, $\cot\theta=\dfrac1{\tan\theta}=\dfrac{\cos\theta}{\sin\theta}$
- $\sec^2\theta\equiv1+\tan^2\theta$, $\quad\csc^2\theta\equiv1+\cot^2\theta$
- Ranges: $\arcsin x\in[-\tfrac\pi2,\tfrac\pi2]$, $\arccos x\in[0,\pi]$, $\arctan x\in(-\tfrac\pi2,\tfrac\pi2)$
Worked examples
- Solve $\tan^2\theta+2\sec\theta=-1$ for $0\le\theta<2\pi$. Using $\tan^2\theta=\sec^2\theta-1$: $\sec^2\theta+2\sec\theta=0$, so $\sec\theta=0$ (impossible) or $\sec\theta=-2$, $\cos\theta=-\tfrac12$: $\theta=\tfrac{2\pi}3$ or $\tfrac{4\pi}3$.
- Find the exact value of $\cos\left(\arcsin\tfrac35\right)$. If $\sin\alpha=\tfrac35$ with $\alpha$ in $[-\tfrac\pi2,\tfrac\pi2]$, then $\cos\alpha=\tfrac45$.
AQA 7357 · E4–E5
Examined in: Papers 1, 2 and 3
- The identities $\sec^2\theta=1+\tan^2\theta$ and $\csc^2\theta=1+\cot^2\theta$ are not in AQA's booklet.
Edexcel 9MA0 · Pure 5.4–5.5
Examined in: Papers 1 and 2 (pure)
- Pearson's booklet does not print these identities either; Edexcel writes cosec, as in the booklet.
OCR A H240 · 1.05h, i, k
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR expects the graphs, ranges and domains of $\arcsin$, $\arccos$ and $\arctan$ (1.05i) and writes cosec.
Compound and double angles, and the R form
The compound-angle formulas expand $\sin(A\pm B)$, $\cos(A\pm B)$ and $\tan(A\pm B)$; setting $A=B$ gives the double-angle formulas. Any $a\sin\theta+b\cos\theta$ can be written as a single wave $R\sin(\theta+\alpha)$, which gives maximum and minimum values and solves equations.
Key formulae
- $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$, $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$
- $\sin2A=2\sin A\cos A$, $\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A$, $\tan2A=\dfrac{2\tan A}{1-\tan^2A}$
- $a\sin\theta+b\cos\theta=R\sin(\theta+\alpha)$ with $R=\sqrt{a^2+b^2}$, $\tan\alpha=\tfrac ba$
Worked examples
- Find the exact value of $\sin75^\circ$. $\sin(45^\circ+30^\circ)=\tfrac{\sqrt2}2\cdot\tfrac{\sqrt3}2+\tfrac{\sqrt2}2\cdot\tfrac12=\dfrac{\sqrt6+\sqrt2}4$.
- Write $3\sin\theta+4\cos\theta$ as $R\sin(\theta+\alpha)$. $R=5$ and $\tan\alpha=\tfrac43$, $\alpha=53.13^\circ$. The maximum value is $5$, when $\theta=36.87^\circ$.
- Solve $\cos2x=\sin x$ for $0\le x<360^\circ$. $1-2\sin^2x=\sin x$, so $(2\sin x-1)(\sin x+1)=0$: $x=30^\circ, 150^\circ, 270^\circ$.
AQA 7357 · E6
Examined in: Papers 1, 2 and 3
- AQA's booklet prints the compound-angle formulas but not the double-angle formulas; learn those.
Edexcel 9MA0 · Pure 5.6
Examined in: Papers 1 and 2 (pure)
- Pearson's booklet prints the compound-angle formulas only; the double-angle formulas and the $R$ form must be learnt.
OCR A H240 · 1.05l–n
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR prints the compound-angle formulas and 1.05m asks you to understand their geometric proofs.
Trigonometric proofs, equations and modelling
A Level trig questions combine the identities: prove an identity by working on one side until it matches the other, then use it to solve an equation. Equations in radians and with transformed arguments need the interval adjusted. Trig functions also model periodic situations such as tides and daylight hours.
Key formulae
- Prove identities with $\equiv$: start from the more complicated side and use known identities
- For $\sin(ax+b)=k$ on $p\le x<q$, solve for $ax+b$ on $ap+b\le ax+b<aq+b$
- Model $y=A\sin(\omega t+c)+D$: amplitude $A$, period $\tfrac{2\pi}\omega$, mean level $D$
Worked examples
- Prove $\dfrac{\sin2\theta}{1+\cos2\theta}\equiv\tan\theta$. The left side is $\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta$.
- Solve $\sin\left(2x+\tfrac\pi6\right)=\tfrac12$ for $0\le x<\pi$. $2x+\tfrac\pi6$ lies in $[\tfrac\pi6,\ \tfrac{13\pi}6)$, so $2x+\tfrac\pi6=\tfrac\pi6$ or $\tfrac{5\pi}6$ (the next solution, $\tfrac{13\pi}6$, is just outside). So $x=0$ or $x=\tfrac\pi3$.
- The depth of water in a harbour is $d=3\sin(0.5t)+5$ metres, $t$ hours after midnight. Maximum depth $8$ m, minimum $2$ m, period $4\pi=12.6$ hours.
AQA 7357 · E7–E9
Examined in: Papers 1, 2 and 3
- AQA (E9) includes trig in vector and kinematics contexts, so trig equations can appear in Paper 2 mechanics.
Edexcel 9MA0 · Pure 5.7–5.9
Examined in: Papers 1 and 2 (pure)
- No board-specific differences. Identity proofs must show every step: a proof that only checks values scores nothing.
OCR A H240 · 1.05o–q
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR's guidance lists equations such as $6\sin^2\theta+\cos\theta-4=0$ and $\tan3\theta=-1$; detailed reasoning is often required.
Further differentiation: standard functions and the chain, product and quotient rules
Year 13 adds the derivatives of $e^{kx}$, $a^{kx}$, $\ln x$ and the trig functions, and three rules for combining them. Use the chain rule for a function of a function, the product rule for a product and the quotient rule for a fraction. The derivatives of $\sin x$ and $\cos x$ can be found from first principles using small-angle results.
Key formulae
- $\dfrac d{dx}e^{kx}=ke^{kx}$, $\dfrac d{dx}a^{kx}=ka^{kx}\ln a$, $\dfrac d{dx}\ln x=\dfrac1x$
- $\dfrac d{dx}\sin kx=k\cos kx$, $\dfrac d{dx}\cos kx=-k\sin kx$, $\dfrac d{dx}\tan kx=k\sec^2kx$
- Chain $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}$; product $(uv)'=uv'+vu'$; quotient $\left(\dfrac uv\right)'=\dfrac{vu'-uv'}{v^2}$
Worked examples
- Differentiate $y=x^2e^{3x}$. Product rule: $2xe^{3x}+3x^2e^{3x}=xe^{3x}(2+3x)$.
- Differentiate $y=\ln(\cos x)$. Chain rule: $\dfrac{-\sin x}{\cos x}=-\tan x$.
- Find the stationary point of $y=\dfrac{\ln x}{x}$. Quotient rule: $\dfrac{1-\ln x}{x^2}=0$, so $x=e$ and $y=\dfrac1e$.
AQA 7357 · G2, G4
Examined in: Papers 1, 2 and 3
- AQA's booklet prints the quotient rule and the derivatives of $\tan$, $\sec$, $\cot$ and $\csc$, but not the product or chain rule.
Edexcel 9MA0 · Pure 7.2, 7.4
Examined in: Papers 1 and 2 (pure)
- Pearson's booklet prints the quotient rule and the less common trig derivatives; the product and chain rules must be learnt.
OCR A H240 · 1.07h, j–l, q, r
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR prints the quotient rule in each paper. OCR 1.07h asks for first-principles derivatives of $\sin x$ and $\cos x$.
Implicit and parametric differentiation, convexity and rates of change
Differentiate implicitly when $x$ and $y$ are mixed: each $y$ term picks up a factor $\dfrac{dy}{dx}$. For parametric curves divide $\dfrac{dy}{dt}$ by $\dfrac{dx}{dt}$. The second derivative shows where a curve is convex or concave, with points of inflection where it changes sign. Connected rates of change link derivatives with the chain rule.
Key formulae
- $\dfrac d{dx}f(y)=f'(y)\dfrac{dy}{dx}$, $\dfrac d{dx}(xy)=x\dfrac{dy}{dx}+y$
- $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$
- Convex where $f''(x)>0$, concave where $f''(x)<0$; point of inflection where $f''$ changes sign
- Connected rates: $\dfrac{dV}{dt}=\dfrac{dV}{dr}\cdot\dfrac{dr}{dt}$
Worked examples
- Find $\dfrac{dy}{dx}$ for $x^2+xy+y^2=7$ at $(1,\ 2)$. $2x+x\dfrac{dy}{dx}+y+2y\dfrac{dy}{dx}=0$, so $\dfrac{dy}{dx}=-\dfrac{2x+y}{x+2y}=-\dfrac45$.
- $x=t^2+1$, $y=t^3-3t$. $\dfrac{dy}{dx}=\dfrac{3t^2-3}{2t}$, so the tangents are horizontal at $t=\pm1$: the points $(2,\ -2)$ and $(2,\ 2)$.
- A spherical balloon's volume increases at $50$ cm³/s. When $r=10$ cm, $\dfrac{dr}{dt}=\dfrac{50}{4\pi(10)^2}=0.0398$ cm/s.
- Find where $y=x^4-6x^2$ is concave. $y''=12x^2-12<0$ for $-1<x<1$; the points of inflection are at $x=\pm1$.
AQA 7357 · G3, G5–G6
Examined in: Papers 1, 2 and 3
- None of the implicit, parametric or connected-rates results is in AQA's booklet. Parametric and implicit differentiation is for first derivatives only (G5).
Edexcel 9MA0 · Pure 7.3, 7.5–7.6
Examined in: Papers 1 and 2 (pure)
- Edexcel 7.6 asks you to form simple differential equations from a description, such as a rate proportional to size.
OCR A H240 · 1.07f, p, s, t
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR (1.07f, p) uses the words 'convex' and 'concave' and asks for points of inflection.
Integration techniques and areas
Year 13 integration adds $e^{kx}$, $\dfrac1x$ and the trig functions, integration by substitution and by parts, partial fractions and trig identities. Recognise which technique fits, then find areas between curves. Integration is also the limit of a sum of thin strips.
Key formulae
- $\displaystyle\int e^{kx}dx=\tfrac1ke^{kx}+c$, $\displaystyle\int\frac1x dx=\ln|x|+c$, $\displaystyle\int\cos kx\,dx=\tfrac1k\sin kx+c$
- $\displaystyle\int\frac{f'(x)}{f(x)}dx=\ln|f(x)|+c$
- By parts: $\displaystyle\int u\frac{dv}{dx}dx=uv-\int v\frac{du}{dx}dx$
- Area between curves $=\displaystyle\int_a^b(y_{\text{top}}-y_{\text{bottom}})\,dx$
Worked examples
- $\displaystyle\int_0^1xe^{2x}dx$ by parts with $u=x$: $\left[\tfrac x2e^{2x}\right]_0^1-\int_0^1\tfrac12e^{2x}dx=\tfrac{e^2}2-\tfrac{e^2-1}4=\dfrac{e^2+1}4$.
- $\displaystyle\int\frac{6x}{x^2+4}dx=3\ln(x^2+4)+c$ (numerator is $3\times$ the derivative of the denominator).
- Use $u=1+x^2$ to find $\displaystyle\int_0^2x\sqrt{1+x^2}\,dx=\int_1^5\tfrac12u^{\frac12}du=\tfrac13\left(5\sqrt5-1\right)=3.39$.
- Find the area between $y=x^2$ and $y=x+2$. They meet at $x=-1$ and $x=2$: $\displaystyle\int_{-1}^2(x+2-x^2)dx=4.5$.
AQA 7357 · H2–H6
Examined in: Papers 1, 2 and 3
- AQA's booklet prints integration by parts; $\int\frac{f'}{f}\,dx$ and $\int\tan x\,dx$ are not printed.
Edexcel 9MA0 · Pure 8.2–8.6
Examined in: Papers 1 and 2 (pure)
- Pearson's booklet prints integration by parts and extra trig integrals such as $\int\tan x\,dx$; Edexcel uses the trapezium rule and area questions together.
OCR A H240 · 1.08c, f–j
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR prints integration by parts and the reverse-chain results $\int\frac{f'}{f}$ and $\int f'f^n$. OCR excludes reduction formulae.
Differential equations
A first-order differential equation links $\dfrac{dy}{dx}$ to $x$ and $y$. If it separates, collect the $y$ terms with $dy$ and the $x$ terms with $dx$, integrate both sides and use a given condition to find the constant. Interpret the solution in context, including what happens in the long term.
Key formulae
- $\dfrac{dy}{dx}=f(x)g(y)\Rightarrow\displaystyle\int\frac{1}{g(y)}dy=\int f(x)\,dx$
- 'Rate proportional to amount': $\dfrac{dN}{dt}=kN\Rightarrow N=Ae^{kt}$
- Add one constant of integration, then use the initial condition
Worked examples
- Solve $\dfrac{dy}{dx}=xy^2$ with $y=1$ at $x=0$. $\displaystyle\int y^{-2}dy=\int x\,dx$: $-\dfrac1y=\dfrac{x^2}2+c$, and $c=-1$, so $y=\dfrac2{2-x^2}$.
- A liquid cools so that $\dfrac{d\theta}{dt}=-0.1(\theta-20)$ with $\theta=90$ at $t=0$. Then $\theta=20+70e^{-0.1t}$. It reaches $50^\circ$C when $t=10\ln\tfrac73=8.47$ minutes; in the long term $\theta\to20$.
AQA 7357 · H7–H8
Examined in: Papers 1, 2 and 3
- AQA's differential equations are first order and separable; forming the equation is in G6.
Edexcel 9MA0 · Pure 8.7–8.8
Examined in: Papers 1 and 2 (pure)
- Edexcel 8.8 asks you to interpret the solution in context, including its limitations; links to kinematics are included.
OCR A H240 · 1.08k, l
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR 1.08l asks you to interpret the solution and its limitations in context.
Roots: change of sign, iteration and Newton–Raphson
When an equation can't be solved exactly, approximate its roots. A change of sign of a continuous function shows a root lies in an interval. Fixed-point iteration $x_{n+1}=g(x_n)$ may converge (staircase or cobweb) or diverge. Newton–Raphson uses tangents and usually converges fast, but fails near a stationary point.
Key formulae
- If $f$ is continuous and $f(a)$, $f(b)$ have opposite signs, a root lies between $a$ and $b$
- Iteration: rearrange $f(x)=0$ as $x=g(x)$ and use $x_{n+1}=g(x_n)$
- Newton–Raphson: $x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$
Worked examples
- Show that $f(x)=x^3-2x-5$ has a root between $2$ and $3$. $f(2)=-1<0$ and $f(3)=16>0$, $f$ is continuous, so there is a root.
- Use $x_{n+1}=\sqrt[3]{2x_n+5}$ with $x_0=2$: $x_1=2.0801$, $x_2=2.0924$, $x_3=2.0942$.
- Newton–Raphson for $f(x)=x^3-2x-5$ from $x_0=2$: $x_1=2-\dfrac{-1}{10}=2.1$, $x_2=2.0946$. The root is $2.0946$ (4 d.p.).
AQA 7357 · I1–I2
Examined in: Papers 1, 2 and 3
- AQA's booklet prints the Newton–Raphson formula; AQA's calculator rules require an iterative function.
Edexcel 9MA0 · Pure 9.1–9.3
Examined in: Papers 1 and 2 (pure)
- Pearson's booklet prints Newton–Raphson. Edexcel numbers iteration and Newton–Raphson as separate topics (9.2 and 9.3).
OCR A H240 · 1.09a–e
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR prints Newton–Raphson in each paper and 1.09e asks you to show how such methods can fail, with diagrams.
Numerical integration and numerical methods in context
The trapezium rule estimates an integral by joining points on the curve with straight lines. More strips give a better estimate. If the curve is convex the rule overestimates; if concave it underestimates. Numerical methods are used when an integral or equation from a real problem has no exact solution.
Key formulae
- $\displaystyle\int_a^by\,dx\approx\tfrac h2\big[y_0+y_n+2(y_1+\cdots+y_{n-1})\big]$, $h=\dfrac{b-a}n$
- Convex curve ($f''>0$): overestimate. Concave ($f''<0$): underestimate
Worked examples
- Estimate $\displaystyle\int_0^2\sqrt{1+x^3}\,dx$ with 4 strips. $h=0.5$; $y$ values $1,\ 1.0607,\ 1.4142,\ 2.0917,\ 3$. Estimate $=0.25\big[1+3+2(4.5665)\big]=3.283$ (3 d.p.).
- Explain whether $\displaystyle\int_1^3\ln x\,dx$ is over- or under-estimated by the trapezium rule. $\ln x$ is concave ($f''=-\tfrac1{x^2}<0$), so the rule underestimates.
AQA 7357 · I3–I4
Examined in: Papers 1, 2 and 3
- AQA's booklet prints the trapezium rule.
Edexcel 9MA0 · Pure 9.4–9.5
Examined in: Papers 1 and 2 (pure)
- Pearson's booklet prints the trapezium rule; Edexcel may ask how to improve the estimate.
OCR A H240 · 1.09f, g
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR prints the trapezium rule; Simpson's rule is excluded. OCR 1.09f includes the limits the area must lie between.
Vectors in three dimensions
Year 13 extends vectors to three dimensions with $\mathbf i$, $\mathbf j$ and $\mathbf k$. Magnitude, position vectors, distances and parallel vectors work exactly as in 2D. Use them to solve geometric problems such as finding the fourth vertex of a parallelogram or proving points are collinear.
Key formulae
- $|x\mathbf i+y\mathbf j+z\mathbf k|=\sqrt{x^2+y^2+z^2}$
- $\overrightarrow{AB}=\mathbf b-\mathbf a$; midpoint $\tfrac12(\mathbf a+\mathbf b)$
- $ABCD$ is a parallelogram when $\overrightarrow{AB}=\overrightarrow{DC}$
Worked examples
- $A(1,\ 0,\ 2)$ and $B(3,\ -2,\ 3)$. $\overrightarrow{AB}=2\mathbf i-2\mathbf j+\mathbf k$ and $AB=\sqrt{4+4+1}=3$.
- $A(1,\ 0,\ 2)$, $B(3,\ -2,\ 3)$ and $C(4,\ 1,\ 0)$ are three vertices of parallelogram $ABCD$. $\mathbf d=\mathbf a+\overrightarrow{BC}=(1,0,2)+(1,3,-3)=(2,\ 3,\ -1)$.
AQA 7357 · J1–J5
Examined in: Paper 2 (pure and mechanics)
- AQA examines 3D vectors in Paper 2, alongside mechanics.
Edexcel 9MA0 · Pure 10.1–10.5
Examined in: Papers 1 and 2 (pure)
- Edexcel examines 3D vectors in the pure papers; the scalar product and vector lines are not in 9MA0.
OCR A H240 · 1.10b, h
Examined in: Paper 1, and the pure parts of Papers 2 and 3
- OCR excludes the direction of a vector in three dimensions (1.10b).
Statistics (4)
Conditional probability and modelling
$P(A\mid B)$ is the probability of $A$ given that $B$ has happened: restrict attention to $B$. Find it from a two-way table, a tree diagram or a Venn diagram, or with the formula. Probability models rest on assumptions; questions may ask you to criticise them.
Key formulae
- $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$
- $P(A\cap B)=P(A)P(B\mid A)$; independent if $P(A\mid B)=P(A)$
Worked examples
- Of $200$ people, $120$ were vaccinated; $18$ vaccinated and $32$ unvaccinated people became ill. $P(\text{ill}\mid\text{vaccinated})=\tfrac{18}{120}=0.15$ and $P(\text{vaccinated}\mid\text{ill})=\tfrac{18}{50}=0.36$.
- $P(A)=0.6$, $P(B\mid A)=0.3$, $P(B\mid A')=0.5$. Then $P(B)=0.18+0.2=0.38$ and $P(A\mid B)=\tfrac{0.18}{0.38}=0.474$.
AQA 7357 · M2–M3
Examined in: Paper 3 (pure and statistics)
- AQA's booklet does not print the conditional probability formula; learn it.
Edexcel 9MA0 · Statistics 3.2–3.3
Examined in: Paper 3, Section A (statistics)
- Pearson's booklet prints the conditional probability formula.
OCR A H240 · 2.03d, e
Examined in: Paper 2 (pure and statistics)
- OCR prints the formula, but using it to find $P(A\mid B)$ from $P(B\mid A)$ (as in the second example) is excluded on OCR (2.03d).
The normal distribution
The normal distribution $N(\mu,\sigma^2)$ is a symmetric bell-shaped model for continuous data, with points of inflection at $\mu\pm\sigma$. Use your calculator for probabilities and the inverse normal for values. Standardising with $Z$ finds an unknown mean or standard deviation. Choose a model with reasons and recognise when a normal model does not fit.
Key formulae
- $X\sim N(\mu,\sigma^2)$, $Z=\dfrac{X-\mu}\sigma\sim N(0,1)$
- About $68\%$ of values lie within $\mu\pm\sigma$, $95\%$ within $\mu\pm2\sigma$, $99.7\%$ within $\mu\pm3\sigma$
- Approximation: $B(n,p)\approx N\big(np,\ np(1-p)\big)$ when $n$ is large and $p$ is near $0.5$, with a continuity correction
Worked examples
- $X\sim N(50,\ 8^2)$. $P(X>60)=0.106$, $P(45<X<55)=0.468$, and $P(X<a)=0.9$ gives $a=60.3$.
- $X\sim N(50,\ \sigma^2)$ and $P(X<40)=0.2$. Then $\dfrac{40-50}\sigma=-0.8416$, so $\sigma=11.9$.
- $X\sim B(80,\ 0.4)$ is approximated by $Y\sim N(32,\ 19.2)$. $P(X\le28)\approx P(Y<28.5)=0.212$.
AQA 7357 · N2–N3
Examined in: Paper 3 (pure and statistics)
- AQA requires a calculator with an inverse normal function. AQA's specification links the normal distribution to the binomial (N2) without naming continuity corrections.
Edexcel 9MA0 · Statistics 4.2–4.3
Examined in: Paper 3, Section A (statistics)
- Edexcel expects the normal approximation to the binomial, with a continuity correction (as in the third example).
OCR A H240 · 2.04e–h
Examined in: Paper 2 (pure and statistics)
- OCR excludes calculations with the normal approximation to the binomial: you need only know that it exists. OCR also excludes the equation of the normal curve; tables of percentage points are provided.
Hypothesis test for the mean of a normal distribution
If $X\sim N(\mu,\sigma^2)$, the mean of a sample of size $n$ is normally distributed with the same mean and a smaller variance $\dfrac{\sigma^2}n$. Use this to test a claim about $\mu$ when $\sigma$ is known: find the $p$-value of the sample mean, or compare a $z$ statistic with a critical value.
Key formulae
- $\bar X\sim N\!\left(\mu,\dfrac{\sigma^2}n\right)$, $\quad Z=\dfrac{\bar X-\mu}{\sigma/\sqrt n}$
- 5% critical values: $1.645$ (one-tailed), $1.96$ (two-tailed)
Worked examples
- Bags are filled with mean $500$ g, $\sigma=6$ g. A sample of $16$ has mean $497.2$ g. Test at $5\%$ whether the mean has fallen. $z=\dfrac{497.2-500}{6/4}=-1.87<-1.645$ ($p=0.031$), so reject $H_0$: there is evidence the mean has fallen.
- Find the critical region for $\bar X$ in a two-tailed $5\%$ test of $\mu=20$, $\sigma=4$, $n=25$: $\bar X<20-1.96(0.8)=18.43$ or $\bar X>21.57$.
AQA 7357 · O3
Examined in: Paper 3 (pure and statistics)
- AQA's booklet prints the distribution of the sample mean.
Edexcel 9MA0 · Statistics 5.3
Examined in: Paper 3, Section A (statistics)
- Pearson's booklet prints the distribution of the sample mean and the percentage points of the normal distribution.
OCR A H240 · 2.05d, e
Examined in: Paper 2 (pure and statistics)
- OCR's printed formulae do not include the sample-mean result, so learn it; tables of percentage points are provided. Tests for the mean of a non-normal population are excluded.
Hypothesis test for correlation
The product moment correlation coefficient $r$ measures how close bivariate data lie to a straight line, from $-1$ to $1$. To test whether there is correlation in the population, set $H_0:\rho=0$ and compare $r$ with a critical value for the sample size, or use a given $p$-value.
Key formulae
- $H_0:\rho=0$; $\ H_1:\rho>0$, $\rho<0$ or $\rho\ne0$
- Reject $H_0$ if $|r|$ is greater than the critical value (in the right direction for a one-tailed test)
Worked examples
- For $12$ pairs, $r=0.58$. Test at $5\%$ for positive correlation. The critical value is $0.4973$; $0.58>0.4973$, so reject $H_0$: there is evidence of positive correlation.
- For $20$ pairs, $r=-0.41$. Two-tailed test at $5\%$: the critical value is $0.4438$ and $0.41<0.4438$, so there is not enough evidence of correlation.
AQA 7357 · O1
Examined in: Paper 3 (pure and statistics)
- AQA gives the correlation coefficient and a $p$-value or critical value: calculating $r$ is excluded.
Edexcel 9MA0 · Statistics 5.1
Examined in: Paper 3, Section A (statistics)
- Edexcel expects you to calculate $r$ on your calculator; tables of critical values or a $p$-value are given.
OCR A H240 · 2.05f, g
Examined in: Paper 2 (pure and statistics)
- OCR excludes calculating correlation coefficients and provides a table of critical values; the data are assumed to come from a bivariate normal distribution.
Mechanics (5)
Kinematics in two dimensions with vectors
In two dimensions, position, velocity and acceleration are vectors. With constant acceleration the suvat equations work in vector form. With variable acceleration, differentiate or integrate each component separately. Speed is the magnitude of the velocity.
Key formulae
- $\mathbf v=\mathbf u+\mathbf at$, $\quad\mathbf r=\mathbf ut+\tfrac12\mathbf at^2$
- $\mathbf v=\dfrac{d\mathbf r}{dt}=\dot{\mathbf r}$, $\quad\mathbf a=\dfrac{d\mathbf v}{dt}=\ddot{\mathbf r}$; integrate componentwise
- Speed $=|\mathbf v|$; moving parallel to $\mathbf j$ when the $\mathbf i$-component of $\mathbf v$ is $0$
Worked examples
- A particle starts with velocity $(2\mathbf i-\mathbf j)$ m s$^{-1}$ and acceleration $(\mathbf i+2\mathbf j)$ m s$^{-2}$. At $t=3$: $\mathbf v=5\mathbf i+5\mathbf j$ and $\mathbf r=(6\mathbf i-3\mathbf j)+(4.5\mathbf i+9\mathbf j)=10.5\mathbf i+6\mathbf j$.
- $\mathbf r=(t^3-3t)\mathbf i+(2t^2+1)\mathbf j$. $\mathbf v=(3t^2-3)\mathbf i+4t\mathbf j$; at $t=2$, $\mathbf v=9\mathbf i+8\mathbf j$ with speed $\sqrt{145}=12.0$ m s$^{-1}$. It moves parallel to $\mathbf j$ when $t=1$.
AQA 7357 · Q3–Q4
Examined in: Paper 2 (pure and mechanics)
- AQA's booklet prints the vector forms $\mathbf v=\mathbf u+\mathbf at$ and $\mathbf r=\mathbf ut+\tfrac12\mathbf at^2$.
Edexcel 9MA0 · Mechanics 7.3–7.4
Examined in: Paper 3, Section B (mechanics)
- Pearson's booklet does not print the vector forms; Edexcel's guidance includes differentiating and integrating vectors with respect to time.
OCR A H240 · 3.02g (with 3.02d in 2D)
Examined in: Paper 3 (pure and mechanics)
- OCR prints the vector forms and may use dot notation $\dot{\mathbf x}$, $\ddot{\mathbf x}$ and column vectors. The vector form of $v^2=u^2+2as$ is excluded.
Projectiles
A projectile is modelled as a particle moving under gravity alone: no air resistance, no spin and no size. Split the motion into a horizontal part with constant velocity and a vertical part with constant acceleration $g$ downwards. Time $t$ links the two.
Key formulae
- Initial velocity components, speed $U$ at angle $\alpha$ above the horizontal: $u_x=U\cos\alpha$, $u_y=U\sin\alpha$
- Horizontal: $x=(U\cos\alpha)t$. Vertical (up positive): $y=(U\sin\alpha)t-\tfrac12gt^2$, $v_y=U\sin\alpha-gt$
- In vector form: $\mathbf r=\mathbf ut+\tfrac12\mathbf at^2$ with $\mathbf a=-g\mathbf j$
- Landing at the launch height: $T=\dfrac{2U\sin\alpha}g$, $R=\dfrac{U^2\sin2\alpha}g$, $H=\dfrac{U^2\sin^2\alpha}{2g}$. Derive these; don't just quote them
- Path: $y=x\tan\alpha-\dfrac{gx^2}{2U^2\cos^2\alpha}=x\tan\alpha-\dfrac{gx^2}{2U^2}\left(1+\tan^2\alpha\right)$
Worked examples
- A ball is kicked from level ground at $20$ m s$^{-1}$, $30^\circ$ above the horizontal. $u_y=10$, $u_x=10\sqrt3$. It lands when $10t-4.9t^2=0$, so $T=2.04$ s. Range $=10\sqrt3\times2.04=35.3$ m. At the top $v_y=0$: $H=\tfrac{100}{19.6}=5.10$ m.
- A particle is projected from the origin with velocity $(12\mathbf i+14.7\mathbf j)$ m s$^{-1}$. It is back at ground level when $t(14.7-4.9t)=0$, $t=3$ s, $36$ m away. At $t=2$, $\mathbf v=12\mathbf i-4.9\mathbf j$: speed $13.0$ m s$^{-1}$ at $22.2^\circ$ below the horizontal.
- A stone is thrown from a cliff $25$ m above the sea at $14$ m s$^{-1}$, with $\tan\alpha=\tfrac34$. $u_x=11.2$, $u_y=8.4$. It reaches the sea when $4.9t^2-8.4t-25=0$: $t=3.27$ s, $36.7$ m out. At impact $v_y=8.4-9.8\times3.273=-23.68$, so the speed is $\sqrt{11.2^2+23.68^2}=26.2$ m s$^{-1}$.
- A ball is projected from ground level at $15$ m s$^{-1}$ towards a wall $2$ m high and $12$ m away. With $T=\tan\alpha$ the path gives $12T-3.136(1+T^2)>2$, i.e. $3.136T^2-12T+5.136<0$, so $0.491<T<3.336$: the ball clears the wall for $26.2^\circ<\alpha<73.3^\circ$.
AQA 7357 · Q5
Examined in: Paper 2 (pure and mechanics)
- AQA's specification simply says 'projectiles' (Q5); questions are in Paper 2 with pure and vectors.
Edexcel 9MA0 · Mechanics 7.5
Examined in: Paper 3, Section B (mechanics)
- Edexcel (7.5) says the derivations of time of flight, range, greatest height and the equation of the path may be required.
OCR A H240 · 3.02h, i
Examined in: Paper 3 (pure and mechanics)
- OCR (3.02i) lists the range, greatest height, velocity at a given time or position and the Cartesian path. Projectiles on inclined planes or with air resistance are excluded. Velocities may be given as column vectors.
Resolving forces, equilibrium and inclined planes
In two dimensions, resolve each force into components along two perpendicular directions, usually horizontal and vertical, or parallel and perpendicular to a slope. A particle is in equilibrium when the components balance in both directions. The resultant of several forces is their vector sum.
Key formulae
- A force $F$ at angle $\theta$ to a direction has component $F\cos\theta$ along it and $F\sin\theta$ perpendicular to it
- On a slope at angle $\alpha$: weight components $mg\sin\alpha$ down the slope and $mg\cos\alpha$ into it
- Equilibrium: resultant $=\mathbf 0$ (resolve in two directions)
Worked examples
- A $4$ kg particle slides down a smooth plane inclined at $30^\circ$. $a=g\sin30^\circ=4.9$ m s$^{-2}$ and $R=4g\cos30^\circ=33.9$ N.
- Forces of $10$ N (bearing $090^\circ$) and $8$ N (bearing $000^\circ$) act on a particle. A third force keeps it in equilibrium: magnitude $\sqrt{164}=12.8$ N, opposite to the resultant, which makes $38.7^\circ$ with the $10$ N force.
- A $2$ kg particle is pulled up a smooth slope at $20^\circ$ by a force of $15$ N parallel to the slope. $a=\dfrac{15-2g\sin20^\circ}2=4.15$ m s$^{-2}$.
AQA 7357 · R2, R5
Examined in: Paper 2 (pure and mechanics)
- AQA restricts resolving to two dimensions (R2).
Edexcel 9MA0 · Mechanics 8.2, 8.5
Examined in: Paper 3, Section B (mechanics)
- Edexcel's guidance includes forces given in magnitude–direction form and vector diagrams with two or more forces.
OCR A H240 · 3.03p, q
Examined in: Paper 3 (pure and mechanics)
- On OCR, motion on inclined planes is only along a line of greatest slope, so no vector treatment of the motion is needed.
Friction
Friction opposes motion, or the tendency to move, along a rough surface. It can take any value up to a maximum $\mu R$. When the object is moving, or on the point of moving (limiting equilibrium), $F=\mu R$. Find $R$ by resolving perpendicular to the surface first, because other forces often change it.
Key formulae
- $F\le\mu R$; $\ F=\mu R$ when sliding or in limiting equilibrium
- On the point of slipping on a slope with only weight, reaction and friction: $\mu=\tan\alpha$
Worked examples
- A $5$ kg block on rough horizontal ground ($\mu=0.3$) is pulled by $25$ N at $20^\circ$ above the horizontal. $R=49-25\sin20^\circ=40.4$ N, $F=0.3R=12.1$ N, and $a=\dfrac{25\cos20^\circ-12.1}5=2.27$ m s$^{-2}$.
- A box slides down a plane at $30^\circ$ with $\mu=0.2$. $a=g(\sin30^\circ-0.2\cos30^\circ)=3.20$ m s$^{-2}$.
- A block rests on a rough plane and is about to slip when the angle is $25^\circ$. Then $\mu=\tan25^\circ=0.466$.
AQA 7357 · R6
Examined in: Paper 2 (pure and mechanics)
- Friction is A Level only on AQA (R6), including limiting friction and statics.
Edexcel 9MA0 · Mechanics 8.6
Examined in: Paper 3, Section B (mechanics)
- Edexcel uses $F=\mu R$ when moving and $F\le\mu R$ in equilibrium (8.6).
OCR A H240 · 3.03s–v
Examined in: Paper 3 (pure and mechanics)
- OCR excludes the angle of friction and uses the terms normal contact force, frictional force and magnitude of the contact force explicitly (3.03s).
Moments
The moment of a force about a point is the force times the perpendicular distance from the point to its line of action. A rigid body is in equilibrium when the resultant force is zero and the total moment about any point is zero. Take moments about a point where an unknown force acts to eliminate it.
Key formulae
- Moment $=F\times d$ (perpendicular distance); clockwise and anticlockwise moments balance in equilibrium
- A uniform rod's weight acts at its midpoint
- About to tilt about a support: the reaction at the other support is $0$
Worked examples
- A uniform $4$ m rod of mass $10$ kg rests on supports at end $A$ and at $C$, $3$ m from $A$. Moments about $A$: $3R_C=98\times2$, so $R_C=65.3$ N and $R_A=32.7$ N.
- A particle of mass $m$ kg is hung from the other end $B$ of the same rod. The rod is about to tilt about $C$ when $R_A=0$: $98\times1=mg\times1$, so $m=10$.
- A uniform ladder rests against a smooth wall at $60^\circ$ to rough ground. Moments about the foot give the friction needed, $F=\tfrac W2\cot60^\circ$, so it doesn't slip if $\mu\ge0.289$.
AQA 7357 · S1
Examined in: Paper 2 (pure and mechanics)
- AQA's moments content is 'simple static contexts' (S1), in Paper 2.
Edexcel 9MA0 · Mechanics 9.1
Examined in: Paper 3, Section B (mechanics)
- Edexcel's guidance includes parallel and non-parallel coplanar forces, such as ladder problems.
OCR A H240 · 3.04a–c
Examined in: Paper 3 (pure and mechanics)
- OCR models bodies as uniform or non-uniform rods and rectangular laminas only, and gives the ladder as an example; the vector nature of moments is excluded.
Every worked answer is checked before publishing. Notes are written independently from the AQA, Edexcel and OCR specifications; if anything differs from your board's current specification, the board is right.
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AQA (7357), Edexcel (9MA0) and OCR A (H240). The maths is the same for every board, so each note has one core explanation and a short box for each board: its specification reference, which paper examines the topic and anything that differs, such as formulae given in the exam or content left out.
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