One handout per topic, in plain English. Read the handout before the textbook, not after it — each one takes about five minutes and is designed to make the idea land first, so the formal version has somewhere to stick.
12 handoutsCambridge AS & A LevelPrintableFree to copy and share
Completing the square turns a quadratic from something you solve into something you can see.
Picture ity = x² − 6x + 5 tells you almost nothing at a glance. Rewrite it as y = (x − 3)² − 4 and the whole shape appears: a parabola with its lowest point at (3, −4), symmetric about x = 3, minimum value −4. Same equation, all the information now visible.
Completing the square
x² + bx + c = (x + b/2)² − (b/2)² + c. The vertex is at (−b/2, c − b²/4), which gives the maximum or minimum value and the line of symmetry directly. It also derives the quadratic formula and solves any quadratic without factorising.
The discriminant
For ax² + bx + c = 0, the discriminant is b² − 4ac. Positive gives two distinct real roots, zero gives one repeated root, negative gives no real roots. Questions about a line being tangent to a curve are almost always discriminant = 0 in disguise.
Functions, domain and range
The domain is the set of allowed inputs; the range is the set of resulting outputs. A function is one-to-one if no output repeats, and only a one-to-one function has an inverse. Restricting the domain is often what makes an inverse exist.
Inverse and composite functions
To find f⁻¹, write y = f(x), swap x and y, and solve for y. The graph of f⁻¹ is the reflection of f in y = x, and the domain of f⁻¹ is the range of f. For composites, fg(x) means do g first — the order is genuinely counter-intuitive and worth checking every time.
Transformations
y = f(x) + a translates up a. y = f(x + a) translates left a. y = af(x) stretches by factor a vertically. y = f(ax) stretches by factor 1/a horizontally. The horizontal ones are the opposite of what the notation suggests, which is exactly why they are examined.
The bit that catches people outy = f(x + 3) moves the graph three to the left, not right. The reason: to get the same output the curve had at x, you now need an input three smaller. Every student meets this and most get it wrong at least once.
The grown-up words
What it means
What it is called
Note
Rewriting as a squared bracket plus a constant
completing the square
Reveals the vertex
b squared minus 4ac
discriminant
Sign gives the number of roots
Set of allowed input values
domain
Range is the set of outputs
Function where no output repeats
one-to-one function
Required for an inverse
Reflection of f in the line y = x
inverse function
Domain and range swap
f(x + a)
horizontal translation
Moves left by a
Check you have got it
Write x² − 6x + 5 in completed square form and state the minimum point.
(x − 3)² − 4. Minimum at (3, −4), since (x − 3)² is smallest when x = 3.
For what values of k does x² + kx + 9 = 0 have two distinct real roots?
Discriminant k² − 36 > 0, so k > 6 or k < −6.
Edvia Free Resources · Mathematics 9709 · Topic 1 — free to copy and share
Topic 2
Coordinate geometry and circles (P1)
Geometry becomes algebra once you put coordinates on it, and then every geometric fact turns into an equation.
Picture itTwo lines are perpendicular. In geometry that is a picture. In coordinates it is m₁m₂ = −1 — a single equation you can solve. Descartes' idea was that geometry and algebra are the same subject in different clothes, and this topic is where that pays off.
Lines
Gradient m = (y₂ − y₁)/(x₂ − x₁). Equation y − y₁ = m(x − x₁). Parallel lines have equal gradients; perpendicular lines have m₁m₂ = −1. Midpoint is the average of the coordinates; distance is √((x₂−x₁)² + (y₂−y₁)²).
The circle
(x − a)² + (y − b)² = r² has centre (a, b) and radius r. Given the expanded form x² + y² + 2gx + 2fy + c = 0, complete the square in x and in y to recover the centre and radius — this is the same technique as for quadratics, applied twice.
Circle properties worth remembering
The tangent is perpendicular to the radius at the point of contact. The perpendicular from the centre bisects a chord. The angle in a semicircle is a right angle. Each of these turns a geometric question into a gradient or Pythagoras calculation.
Intersections
To find where a line meets a curve, substitute and solve. The discriminant of the resulting quadratic tells you whether they meet twice, touch once (tangent) or miss entirely — the same tool as in the previous topic, used geometrically.
Parametric equations
A curve can be given as x = f(t), y = g(t). Eliminate t to get the Cartesian equation, or differentiate parametrically using dy/dx = (dy/dt) ÷ (dx/dt). Parametric form handles curves that are not functions of x at all.
The bit that catches people outWhen completing the square for a circle, the constant on the right is r², not r. Reading off r = 25 from (x−1)² + (y+2)² = 25 instead of r = 5 is a very fast way to lose the rest of the question.
The grown-up words
What it means
What it is called
Note
Change in y over change in x
gradient
Equal for parallel lines
m1 times m2 equals -1
perpendicular condition
Product of gradients
(x - a)^2 + (y - b)^2 = r^2
equation of a circle
Centre (a, b), radius r
Line touching a circle at one point
tangent
Perpendicular to the radius
Line joining two points on a circle
chord
Bisected by the perpendicular from the centre
Curve defined through a third variable
parametric equations
x and y both functions of t
Check you have got it
Find the centre and radius of x² + y² − 4x + 6y − 12 = 0.
A line has gradient 3/4. What is the gradient of a perpendicular line?
−4/3, since the product of the gradients must be −1.
Edvia Free Resources · Mathematics 9709 · Topic 2 — free to copy and share
Topic 3
Trigonometry (P1)
Trigonometric functions repeat, so trigonometric equations have infinitely many solutions — and finding them all in a given range is the skill being tested.
Picture itsin x = 0.5 has the obvious answer 30°. It also has 150°, and 390°, and 510°, and so on forever. The calculator gives you one. Finding the rest is not a trick — it is reading the symmetry of the graph, and it is where nearly all the marks are.
Exact values and the unit circle
sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3. sin 45° = cos 45° = 1/√2. sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3. These come from the 1-1-√2 and 1-2-√3 triangles and should be recalled instantly, since calculators are sometimes not permitted to give exact surds.
Graphs and their symmetries
sin and cos have period 360° and amplitude 1; tan has period 180° and asymptotes at 90° + 180n. Solving sin x = k over a range means finding the principal value then using sin(180° − x) = sin x, and adding 360° repeatedly. For cos, use cos(360° − x) = cos x.
The two identities that generate everything
sin²x + cos²x = 1 and tan x = sin x ÷ cos x. Almost every P1 trig identity question reduces to substituting one of these and simplifying. If an equation contains both sin and cos to different powers, one of these is the way in.
Radians, arcs and sectors
π radians = 180°. Arc length s = rθ and sector area A = ½r²θ, both with θ in radians. Using degrees in these formulae is the single most common error in the topic, and it produces answers wrong by a factor of about 57.
Solving quadratic-type equations
2sin²x − sin x − 1 = 0 factorises as (2sin x + 1)(sin x − 1) = 0. Treat sin x as the variable, solve the quadratic, then solve each trig equation separately across the full range.
The bit that catches people outSquaring both sides of a trigonometric equation can introduce solutions that do not satisfy the original. Always substitute your answers back — an extra 'solution' that fails the check costs marks even though the algebra was correct.
The grown-up words
What it means
What it is called
Note
sin squared x plus cos squared x
Pythagorean identity
Equals 1
sin x over cos x
tan x
Undefined where cos x = 0
Length of the repeating interval
period
360 degrees for sin and cos
Angle measure where pi equals 180 degrees
radian
Required for arc and sector formulae
s = r theta
arc length
Theta in radians
Half r squared theta
sector area
Theta in radians
Check you have got it
Solve sin x = 0.5 for 0° ≤ x ≤ 360°.
x = 30° and x = 150°. The calculator gives 30°; the second comes from sin(180° − 30°) = sin 30°.
A sector has radius 6 cm and angle 0.8 radians. Find its area and arc length.
Area = ½ × 6² × 0.8 = 14.4 cm². Arc length = 6 × 0.8 = 4.8 cm.
Edvia Free Resources · Mathematics 9709 · Topic 3 — free to copy and share
Topic 4
Series: binomial, arithmetic and geometric (P1)
Some sums have shortcuts, and knowing which shortcut applies is most of the work.
Picture itAdding 1 + 2 + 3 + ... + 100 by hand takes a while. Noticing that it pairs into fifty lots of 101 takes a moment. Series formulae are that observation, generalised — and the hard part of the topic is recognising which kind of sequence you are looking at.
Arithmetic progressions
A common difference d. The nth term is a + (n − 1)d. The sum is Sn = n/2 [2a + (n − 1)d], or equivalently n/2 (a + l) where l is the last term. The second form is quicker whenever you already know the final term.
Geometric progressions
A common ratio r. The nth term is arn−1. The sum is Sn = a(1 − rⁿ)/(1 − r). The sum to infinity exists only when |r| < 1, and equals a/(1 − r) — because the terms shrink fast enough for the total to settle.
Identifying which is which
Check whether consecutive terms have a constant difference (arithmetic) or a constant ratio (geometric). A sequence can be neither. Spending ten seconds on this check prevents applying the wrong formula to the whole question.
The binomial expansion for positive integer n
(a + b)ⁿ = Σ nCr a^(n−r) b^r. The coefficients are the rows of Pascal's triangle, or nCr = n!/(r!(n−r)!). There are n + 1 terms, and questions usually ask for one specific term rather than the whole expansion.
Finding a particular term
The term in x^k is nCr a^(n−r) b^r where the powers work out to give x^k. Set up the power equation first, solve for r, then compute — rather than expanding everything and hunting.
The bit that catches people outThe sum to infinity of a geometric series only exists when |r| < 1. If r = 1.5, the terms grow and the sum diverges — writing a/(1 − r) anyway gives a negative number that looks like an answer and is meaningless.
The grown-up words
What it means
What it is called
Note
Sequence with a constant difference
arithmetic progression
nth term a + (n-1)d
Sequence with a constant ratio
geometric progression
nth term ar^(n-1)
Sum of a geometric series as n grows without limit
sum to infinity
Only if |r| < 1
Number of ways of choosing r from n
nCr
Binomial coefficient
Triangle of binomial coefficients
Pascal's triangle
Each entry is the sum of two above
a(1 - r^n)/(1 - r)
sum of a geometric series
For the first n terms
Check you have got it
A GP has first term 12 and common ratio 1/3. Find the sum to infinity.
Term is 5C3 × 2² × x³ = 10 × 4 × x³ = 40x³, so the coefficient is 40.
Edvia Free Resources · Mathematics 9709 · Topic 4 — free to copy and share
Topic 5
Differentiation (P1)
Differentiation finds the gradient at a point, which turns questions about steepness, rates and maxima into algebra.
Picture itA curve's gradient changes everywhere along it, so 'the gradient' is not a single number — it is a function. dy/dx is that function, and once you have it, every question about where the curve is steepest, flattest, rising or falling becomes a matter of substituting a value.
The rules
d/dx(xⁿ) = nx^(n−1). Sums differentiate term by term. The chain rule: d/dx f(g(x)) = f'(g(x))·g'(x). Products and quotients have their own rules at P1 level for simple cases; always identify which structure you have before starting.
Tangents and normals
The gradient of the tangent at a point is dy/dx evaluated there. The normal is perpendicular, so its gradient is −1 ÷ (dy/dx). Then use y − y₁ = m(x − x₁). Forgetting to take the negative reciprocal is the routine slip.
Stationary points
Set dy/dx = 0 and solve. Classify using the second derivative: d²y/dx² > 0 means a minimum, < 0 means a maximum, = 0 is inconclusive and needs a sign check either side. Stating the test you used is part of the answer.
Increasing and decreasing functions
A function is increasing where dy/dx > 0 and decreasing where dy/dx < 0. Questions asking for the range of x where a curve rises are asking you to solve an inequality in the derivative.
Connected rates of change
If two quantities are related and one is changing with time, the chain rule connects them: dV/dt = (dV/dr)(dr/dt). Write down what you are given and what you want, and the chain of derivatives needed usually becomes obvious.
The bit that catches people outd²y/dx² = 0 does not prove a point of inflection. It is inconclusive — the point could still be a maximum or minimum, as x⁴ at the origin shows. When the second derivative vanishes, you must test the sign of dy/dx either side.
The grown-up words
What it means
What it is called
Note
Gradient function of a curve
derivative
Written dy/dx
Rule for differentiating a function of a function
chain rule
Multiply by the inner derivative
Line perpendicular to the tangent
normal
Gradient is -1 over dy/dx
Point where dy/dx equals zero
stationary point
Maximum, minimum or inflection
Test using the sign of the second derivative
second derivative test
Inconclusive if zero
Linking two rates through a shared variable
connected rates of change
Uses the chain rule
Check you have got it
Find the stationary points of y = x³ − 3x and classify them.
dy/dx = 3x² − 3 = 0, so x = ±1. d²y/dx² = 6x: at x = 1 this is 6 > 0, a minimum at (1, −2); at x = −1 it is −6 < 0, a maximum at (−1, 2).
A curve has dy/dx = 2 at the point (1, 5). Find the equation of the normal there.
Normal gradient = −1/2, so y − 5 = −½(x − 1), giving y = −½x + 5.5.
Edvia Free Resources · Mathematics 9709 · Topic 5 — free to copy and share
Topic 6
Integration (P1)
Integration undoes differentiation, and its geometric meaning is the area under a curve.
Picture itDifferentiation asks how fast something is changing. Integration asks how much has accumulated. They are inverse operations, which is not obvious at all — it is the fundamental theorem of calculus, and it is why an area problem can be solved by reversing a derivative.
The basic rule and the constant
∫xⁿ dx = x^(n+1)/(n+1) + c, valid for all n except −1. The + c is not decoration: infinitely many curves have the same gradient function, and c selects which one. Definite integrals do not need it because it cancels.
Definite integrals and area
∫ from a to b of f(x) dx = F(b) − F(a). This gives the signed area: regions below the x-axis count as negative. For a total physical area, split the integral at the roots and take the modulus of each part.
Area between two curves
Integrate (upper curve − lower curve) between their points of intersection. Find the intersections first by setting the equations equal — attempting the integral before knowing the limits is a common way to waste time.
Volumes of revolution
Rotating about the x-axis: V = π∫y² dx. About the y-axis: V = π∫x² dy, which requires rearranging the equation to give x in terms of y first. Forgetting the π, or forgetting to square, are the two standard errors.
Integration of composite linear functions
∫(ax + b)ⁿ dx = (ax + b)^(n+1) ÷ (a(n+1)) + c. The division by a is the reverse of the chain rule, and omitting it is the most common single mistake in P1 integration.
The bit that catches people outA definite integral that comes out as zero does not mean there is no area. It can mean equal areas above and below the axis cancelling. If the question asks for area, check where the curve crosses the axis before integrating.
The grown-up words
What it means
What it is called
Note
Reverse of differentiation
integration
Needs + c when indefinite
Integral with limits
definite integral
Gives a number, not a function
Area counting below-axis regions as negative
signed area
Split at the roots for total area
pi times the integral of y squared dx
volume of revolution
Rotation about the x-axis
Extra division by a when integrating (ax+b)^n
chain rule adjustment
Most common omission
Integrating upper minus lower between intersections
area between curves
Find limits first
Check you have got it
Evaluate the definite integral of 3x² from 1 to 2.
∫3x² dx = x³, so [x³] from 1 to 2 = 8 − 1 = 7.
Find the volume when y = x is rotated about the x-axis from x = 0 to x = 3.
V = π∫₀³ x² dx = π[x³/3]₀³ = π × 9 = 9π.
Edvia Free Resources · Mathematics 9709 · Topic 6 — free to copy and share
Topic 7
Algebra: modulus, polynomials and partial fractions (P2/P3)
These three techniques exist to turn awkward expressions into ones the rest of your toolkit can handle.
Picture itYou cannot integrate (5x − 7)/((x − 1)(x − 2)) as it stands. Split it into 2/(x − 1) + 3/(x − 2) and both pieces integrate to logarithms in one line. Partial fractions is not a topic in itself so much as a doorway into other topics.
The modulus function
|x| is the distance from zero, always non-negative. To solve |f(x)| = g(x), solve f(x) = g(x) and f(x) = −g(x), then check both in the original — the second branch often produces an invalid solution. Graphically, |f(x)| reflects the negative parts of f above the axis.
Modulus inequalities
|x − a| < b means a − b < x < a + b, which is the interval within distance b of a. Reading it as a distance statement is far more reliable than manipulating the inequality algebraically.
The factor and remainder theorems
If f(a) = 0 then (x − a) is a factor. If you divide f(x) by (x − a), the remainder is f(a). For a divisor (ax − b), evaluate at x = b/a. These turn cubic factorisation from trial division into a few substitutions.
Polynomial division
Long division or comparing coefficients both work; comparing coefficients is usually faster for a known quotient form. Once one root is found by the factor theorem, divide out and the remaining quadratic factorises or yields to the formula.
Partial fractions
Three cases: distinct linear factors → A/(x−a) + B/(x−b). A repeated factor → A/(x−a) + B/(x−a)². An irreducible quadratic → (Ax + B)/(x² + c). If the numerator's degree is not lower than the denominator's, divide first.
The bit that catches people outA repeated linear factor needs both A/(x−a) and B/(x−a)². Using only the squared term gives a system with no solution, and the algebra will not tell you why — you have to know the rule before you start.
The grown-up words
What it means
What it is called
Note
Distance from zero, always non-negative
modulus
|x| written with vertical bars
If f(a) = 0 then (x - a) is a factor
factor theorem
Finds roots of polynomials
Remainder on dividing f(x) by (x - a)
remainder theorem
Equals f(a)
Splitting a fraction into simpler ones
partial fractions
Enables integration
Factor appearing to a power
repeated factor
Needs terms for each power
Numerator degree at least the denominator's
improper fraction
Divide out first
Check you have got it
Given f(x) = x³ − 2x² − 5x + 6, show that (x − 1) is a factor.
f(1) = 1 − 2 − 5 + 6 = 0, so by the factor theorem (x − 1) is a factor.
A/(x−1) + B/(x−2) with 5x − 7 = A(x−2) + B(x−1). At x = 1: −2 = −A so A = 2. At x = 2: 3 = B. So 2/(x−1) + 3/(x−2).
Edvia Free Resources · Mathematics 9709 · Topic 7 — free to copy and share
Topic 8
Logarithms, exponentials and numerical methods (P2/P3)
Logarithms turn multiplication into addition and powers into multiplication, which is why they solve equations with the unknown in the exponent.
Picture itYou cannot solve 3^x = 20 by any amount of rearranging — x is stuck upstairs. Taking logs of both sides brings it down: x ln 3 = ln 20. That single move is the entire reason logarithms exist in this course.
Laws of logarithms
log a + log b = log(ab). log a − log b = log(a/b). log(aⁿ) = n log a. Change of base: log_b a = ln a ÷ ln b. Every log equation is solved by using these to collapse the expression to a single logarithm.
e and natural logarithms
e ≈ 2.71828 is the base for which the derivative of eˣ is eˣ itself. ln x means log to base e, and ln and exp are inverses: e^(ln x) = x. Growth and decay problems use y = Ae^(kt), with k > 0 for growth and k < 0 for decay.
Linearising to find a relationship
If y = kxⁿ, then log y = log k + n log x — a straight line of gradient n and intercept log k when log y is plotted against log x. If y = kaˣ, plot log y against x instead. Deciding which to plot is the whole skill.
Locating roots
If f is continuous and f(a) and f(b) have opposite signs, a root lies between a and b. Stating continuity and showing both signs is what earns the mark — an unsupported claim that 'there is a root here' does not.
Iteration
Rearrange f(x) = 0 into x = g(x) and iterate xn+1 = g(xn). Convergence requires |g'(x)| < 1 near the root, which is why some rearrangements work and others diverge. A cobweb or staircase diagram shows what is happening.
The bit that catches people outln(a + b) is not ln a + ln b. The addition law applies to log of a product, not a sum, and there is no simplification for the log of a sum. This one error appears in almost every set of scripts.
The grown-up words
What it means
What it is called
Note
log a plus log b
log of a product
Equals log(ab)
Logarithm to base e
natural logarithm
Written ln
Base where the derivative of e^x is itself
e
About 2.71828
Plotting logs to get a straight line
linearisation
Gradient gives the power
Opposite signs of f at two points
sign change
Indicates a root between them
x(n+1) = g(x(n))
iterative formula
Converges if |g'(x)| < 1
Check you have got it
Solve 3^x = 20, giving the answer to 3 significant figures.
x ln 3 = ln 20, so x = ln 20 ÷ ln 3 = 2.9957 ÷ 1.0986 = 2.73.
Why does showing f(2) < 0 and f(3) > 0 demonstrate a root between 2 and 3?
Because f is continuous, it must take every value between f(2) and f(3), including zero, somewhere in that interval.
Edvia Free Resources · Mathematics 9709 · Topic 8 — free to copy and share
Topic 9
Further calculus and differential equations (P3)
P3 calculus is a toolkit: for each new form of function or product, there is one technique that unlocks it.
Picture itThe hard part is not performing the technique — it is recognising which one applies. A product of two functions signals the product rule; a function inside another signals the chain rule; an integral of the form f'(x)/f(x) signals a logarithm. Diagnosis before treatment.
Differentiating the standard functions
d/dx(eˣ) = eˣ. d/dx(ln x) = 1/x. d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(tan x) = sec²x. Product rule: (uv)' = u'v + uv'. Quotient rule: (u/v)' = (u'v − uv')/v². Implicit differentiation for equations not solved for y.
Integration techniques
Substitution reverses the chain rule — change variable and remember to change the limits too. Integration by parts: ∫u dv = uv − ∫v du, reverses the product rule; choose u to be the part that simplifies when differentiated. Partial fractions then logs for rational functions.
The standard integrals
∫1/x dx = ln|x| + c. ∫eˣ dx = eˣ + c. ∫f'(x)/f(x) dx = ln|f(x)| + c — recognising this pattern saves an enormous amount of work. ∫sin x dx = −cos x + c, and note the minus sign.
Forming differential equations
'The rate of decay is proportional to the amount present' becomes dN/dt = −kN. Translating a sentence into an equation is usually worth more marks than solving it, and it is where most of the difficulty actually lies.
Solving by separating variables
Rearrange to g(y) dy = f(x) dx, integrate both sides, include a single constant, then use the initial condition to find it. Applying the boundary condition before rearranging into the final form is far less error-prone than after.
The bit that catches people outWhen you substitute in a definite integral, you must change the limits to match the new variable, or convert back to the original variable before substituting. Mixing new-variable working with old-variable limits produces a confident wrong answer.
The grown-up words
What it means
What it is called
Note
(uv)' = u'v + uv'
product rule
For a product of functions
Differentiating an equation not solved for y
implicit differentiation
Gives dy/dx in terms of x and y
Changing variable to reverse the chain rule
integration by substitution
Change the limits too
Reverses the product rule
integration by parts
Integral of u dv = uv minus integral of v du
Integral of f'(x)/f(x)
logarithmic integral
Equals ln|f(x)| + c
Rearranging so each variable is on its own side
separation of variables
Then integrate both sides
Check you have got it
Differentiate y = x²eˣ.
Product rule with u = x², v = eˣ: dy/dx = 2xeˣ + x²eˣ = xeˣ(2 + x).
Solve dy/dx = 2xy given y = 3 when x = 0.
Separate: (1/y)dy = 2x dx. Integrating gives ln|y| = x² + c. At x = 0, y = 3, so c = ln 3. Hence y = 3e^(x²).
Edvia Free Resources · Mathematics 9709 · Topic 9 — free to copy and share
Topic 10
Vectors and complex numbers (P3)
Both topics extend what a number can be — vectors add direction to magnitude, complex numbers add a second dimension to the number line.
Picture it√(−1) has no answer among the real numbers, so mathematicians invented one and called it i. That looks like cheating, but the resulting system is consistent, complete, and turns out to describe electrical circuits and quantum mechanics. Inventing i was the most productive act of stubbornness in mathematics.
Vector basics
A vector has magnitude and direction. Magnitude |a| = √(x² + y² + z²). A unit vector is a ÷ |a|. Position vectors locate points; the vector from A to B is b − a, in that order.
The scalar product
a·b = |a||b| cos θ = a₁b₁ + a₂b₂ + a₃b₃. So cos θ = (a·b) ÷ (|a||b|), and two vectors are perpendicular exactly when their scalar product is zero — a test that requires no angles at all.
Lines in three dimensions
r = a + td, where a is a point on the line and d its direction. Two lines are parallel if their directions are multiples; they intersect if a consistent value of the parameters exists; otherwise they are skew — a possibility with no two-dimensional analogue.
Complex numbers: algebra
z = x + iy with i² = −1. Add and multiply as normal, replacing i² by −1. To divide, multiply top and bottom by the conjugate of the denominator, which makes the denominator real. Roots of real polynomials occur in conjugate pairs.
Modulus-argument form and the Argand diagram
|z| = √(x² + y²) and arg z = arctan(y/x), adjusted for the quadrant. In modulus-argument form, multiplying multiplies the moduli and adds the arguments — which is why complex multiplication is a rotation and a stretch. Loci such as |z − a| = r are circles.
The bit that catches people outThe argument of a complex number depends on the quadrant, and a calculator's arctan only ever returns a value between −90° and 90°. Sketching the point on an Argand diagram before quoting the argument prevents an answer that is exactly 180° wrong.
The grown-up words
What it means
What it is called
Note
Quantity with magnitude and direction
vector
Magnitude by Pythagoras
Vector of magnitude 1
unit vector
Vector divided by its magnitude
a1b1 + a2b2 + a3b3
scalar product
Zero if perpendicular
Lines that neither meet nor are parallel
skew lines
Only in three dimensions
x - iy for z = x + iy
complex conjugate
Used to divide
Distance of z from the origin
modulus
Square root of x squared plus y squared
Angle from the positive real axis
argument
Depends on the quadrant
Check you have got it
Find the angle between a = (1, 2, 2) and b = (2, 0, 1).
Edvia Free Resources · Mathematics 9709 · Topic 10 — free to copy and share
Topic 11
Mechanics (P4)
Mechanics is applied calculus with a physical picture attached, and the picture is always a force diagram.
Picture itEvery mechanics problem starts the same way: draw the object, draw every force acting on it, resolve into two perpendicular directions, and apply F = ma in each. Students who skip the diagram and start writing equations get stuck; students who draw it carefully usually do not.
Forces and equilibrium
Resolve forces into perpendicular components. In equilibrium the components balance in both directions. On an inclined plane, resolve along and perpendicular to the slope: the weight component along the slope is mg sin θ and perpendicular is mg cos θ.
Friction
Friction opposes motion or attempted motion, with F ≤ μR, reaching F = μR at the point of slipping (limiting friction). The normal reaction R is not always mg — on a slope it is mg cos θ, and an applied force with a vertical component changes it too.
Kinematics
For constant acceleration: v = u + at, s = ut + ½at², v² = u² + 2as, s = ½(u + v)t. For variable acceleration use calculus: v = ds/dt, a = dv/dt, and integrate to reverse. Choosing between the two methods depends entirely on whether a is constant.
Newton's laws and connected particles
F = ma applied to each body separately, with the tension in a connecting string the same throughout a light inextensible string over a smooth pulley. Write one equation per body and solve simultaneously — trying to treat the system as one body only works for finding the acceleration.
Energy, work and power
Work done = Fs cos θ. Kinetic energy = ½mv². Gravitational potential energy = mgh. The work-energy principle says the work done by the net force equals the change in kinetic energy. Power = Fv, useful for vehicle problems at constant speed.
The bit that catches people outOn a slope, the normal reaction is mg cos θ, not mg. Using mg gives a friction force that is too large and an answer that is wrong in a way that looks entirely reasonable — which is why it survives checking.
The grown-up words
What it means
What it is called
Note
Force opposing relative motion
friction
F is at most mu R
Friction at the point of slipping
limiting friction
F = mu R
Force perpendicular to the surface
normal reaction
mg cos theta on a slope
v = u + at and related formulae
constant acceleration equations
Only when a is constant
Force times distance in the direction of motion
work done
Measured in joules
Half m v squared
kinetic energy
Changes by the work done
Force times velocity
power
Watts
Check you have got it
A 5 kg block rests on a slope at 30°. Find the component of weight along the slope, using g = 10.
mg sin θ = 5 × 10 × sin 30° = 50 × 0.5 = 25 N.
A car of mass 1200 kg accelerates from rest to 20 m s⁻¹. Find the work done against no resistance.
Work done = change in KE = ½ × 1200 × 20² − 0 = 240 000 J = 240 kJ.
Edvia Free Resources · Mathematics 9709 · Topic 11 — free to copy and share
Topic 12
Probability and statistics (P5/P6)
Statistics is about drawing a conclusion from incomplete information, and being honest about how confident you are entitled to be.
Picture itYou toss a coin ten times and get seven heads. Is the coin biased? Probably not — but 'probably' needs a number. Hypothesis testing is the machinery that turns 'that seems unlikely' into a defensible statement with a stated risk of being wrong.
Representing and summarising data
Histograms use frequency density = frequency ÷ class width, so area represents frequency. Cumulative frequency curves give medians and quartiles. Box plots show spread and outliers. The mean uses every value and is affected by outliers; the median is not.
Probability
P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Conditional probability P(A|B) = P(A ∩ B) ÷ P(B). Events are independent if P(A ∩ B) = P(A)P(B). Tree diagrams handle sequential events; without-replacement problems change the denominator at each stage.
Discrete random variables
A probability distribution lists each value with its probability, and the probabilities sum to 1. E(X) = Σ x P(X = x). Var(X) = Σ x² P(X = x) − [E(X)]². The binomial distribution B(n, p) applies to a fixed number of independent trials with constant probability.
The normal distribution
Continuous, symmetric, defined by mean μ and variance σ². Standardise with z = (x − μ) ÷ σ and read tables. The normal can approximate the binomial when n is large and p is not extreme, using a continuity correction of ±0.5.
Sampling and hypothesis testing
The sample mean of a large sample is approximately normal with mean μ and variance σ²/n — the central limit theorem. A hypothesis test states H₀ and H₁, calculates a test statistic, compares with a critical value at a stated significance level, and concludes in context.
The bit that catches people outA hypothesis test never proves H₀ true. You either reject it or fail to reject it — 'insufficient evidence to reject' is not the same as 'H₀ is correct'. Writing 'this proves the coin is fair' loses the final mark on essentially every test question.
The grown-up words
What it means
What it is called
Note
Frequency divided by class width
frequency density
Height of a histogram bar
Probability of A given B has happened
conditional probability
P(A and B) over P(B)
Events where one does not affect the other
independent events
P(A and B) = P(A)P(B)
Fixed trials, constant probability, independent
binomial distribution
B(n, p)
Subtracting the mean and dividing by sigma
standardising
Gives the z value
Sample means are approximately normal
central limit theorem
For large samples
The statement being tested
null hypothesis
Rejected or not rejected
Check you have got it
X ~ B(10, 0.3). Find P(X = 3), to 3 significant figures.
A class has 40 students in a bar of width 5. What is the frequency density?
40 ÷ 5 = 8. The bar is drawn to a height of 8, so its area represents the frequency of 40.
Edvia Free Resources · Mathematics 9709 · Topic 12 — free to copy and share
Like how this is taught?
Every handout starts with the idea in plain English and only then the formal version. That is how every class at Edvia College works — for two full years of Cambridge A Levels.