BIG IDEAA proof is a chain of statements, each with a reason, ending with exactly what you were asked to show.
alternate segment: ∠TAC = ∠ABC
midpoint theorem: MN ∥ BC, MN = ½BC
Use
Reason to write
angles with parallel lines
alt. ∠s / corr. ∠s / int. ∠s, AB ∥ CD
tangent and chord
∠s in alt. segment (tangent-chord theorem)
joining midpoints
midpoint theorem
circle angles
∠ in semicircle, ∠ at centre, ∠s in same segment, ∠s in opp. segment
tangent
tan ⊥ rad · tangents from ext. pt
triangles
congruent (SSS, SAS, ASA, RHS) / similar (AA, SSS, SAS)
🧠 HOOKProve a quadrilateral is cyclic: opposite ∠s add to 180°, or ∠s in same segment equal. Prove lines parallel: show alt./corr. ∠s equal.
⚠ TRAPNever assume what you are proving. Don't use a diagram's "look" — only given facts and theorems.
🔎 WHY IT WORKSWhy proofs need reasons: a proof must convince someone who cannot see your diagram. Each statement needs a fact or theorem behind it, so the chain cannot break. Work backwards from what you must prove: "to show PQ ∥ BC I need alternate angles equal" — then look for those angles.
🧠 Worked example — follow the thinking
PQ is the tangent at A to a circle through A, B, C, with AB = AC. Prove PQ ∥ BC.
Step
Why I do this
Goal: show ∠QAC = ∠ACB (alternate angles)
💭 Plan first: parallel lines ⇔ alternate angles equal. So I hunt for those two angles.
∠QAC = ∠ABC (∠s in alt. segment)
💭 Angle between tangent and chord AC equals the angle in the opposite segment.
∠ABC = ∠ACB (base ∠s of isos. △, AB = AC)
💭 Given AB = AC, so the base angles are equal.
∴ ∠QAC = ∠ACB → PQ ∥ BC (alt. ∠s equal) ∎
💭 Link the two facts — the chain is complete.
A Math · Sec 4 · SEC K341 (G3)
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C1
Differentiation Rules
standard derivatives, chain, product, quotient
BIG IDEAdy/dx = the gradient of the tangent = the rate of change of y with x.
d(uv)/dx = u v′ + v u′ x² sin x → 2x sin x + x² cos x
➗ Quotient rule
d(u/v)/dx = (v u′ − u v′)/v² ln x / x → (x · 1/x − ln x · 1)/x² = (1 − ln x)/x²
⚠ TRAPAngles must be in radians for sin/cos/tan derivatives. Rewrite roots and fractions as powers first: √x = x^½, 3/x² = 3x⁻².
🧠 HOOKIncreasing function: dy/dx > 0 for all x · decreasing: dy/dx < 0. Show it by writing dy/dx as a square + positive e.g. 3(x − 1)² + 2 > 0.
🔎 WHY IT WORKSWhy dy/dx is the gradient: take two points on the curve very close together. The gradient of the line joining them is (change in y)/(change in x). As the gap shrinks to zero, that line becomes the tangent — and its gradient is dy/dx. For y = x² this limit always comes out as 2x, which is where nxⁿ⁻¹ comes from.
🧠 Worked example — follow the thinking
Differentiate y = e²ˣ cos 3x.
Step
Why I do this
Spot a PRODUCT: u = e²ˣ, v = cos 3x
💭 Two functions multiplied together → product rule. Name them first.
u′ = 2e²ˣ, v′ = −3 sin 3x
💭 Chain rule inside each: the 2 and the 3 come out in front.
dy/dx = uv′ + vu′ = e²ˣ(−3 sin 3x) + cos 3x(2e²ˣ)
💭 Product rule: first × derivative of second + second × derivative of first.
= e²ˣ(2 cos 3x − 3 sin 3x)
💭 Factor out the common e²ˣ to tidy up.
A Math · Sec 4 · SEC K341 (G3)
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C1
Stationary Points, Tangents & Normals
max, min, inflexion, second derivative test
BIG IDEAStationary point: dy/dx = 0. Then use d²y/dx²: > 0 minimum, < 0 maximum.
y = x³ − 3x² − 9x + 5
Steps
① dy/dx = 3x² − 6x − 9 = 3(x − 3)(x + 1) = 0 ② x = 3 → y = −22 · x = −1 → y = 10 ③ d²y/dx² = 6x − 6 x = 3: 12 > 0 → min x = −1: −12 < 0 → max
Sign of dy/dx: left → right
Type
+ → 0 → −
maximum ∩
− → 0 → +
minimum ∪
+ → 0 → + or − → 0 → −
point of inflexion (e.g. y = x³ at 0)
Tangent & normal at x = a
gradient of tangent m = dy/dx at a gradient of normal = −1/m y = x³ − 3x² − 9x + 5 at x = 0: m = −9 tangent y = −9x + 5 · normal y = x/9 + 5
⚠ TRAPIf d²y/dx² = 0 the test fails — use the first derivative sign test instead.
🔎 WHY IT WORKSWhy the second derivative tells max or min: d²y/dx² is the rate of change of the gradient. At a minimum the gradient goes from − to 0 to +, so it is INCREASING → d²y/dx² > 0. At a maximum the gradient goes + → 0 → −, decreasing → d²y/dx² < 0.
A Math · Sec 4 · SEC K341 (G3)
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C1
Rates of Change & Max/Min Problems
connected rates, optimisation
BIG IDEAConnected rates: link with the chain rule dA/dt = dA/dr × dr/dt. Max/min problems: write a formula in one variable, then differentiate.
Connected rates — steps
① write the formula linking the 2 quantities ② differentiate it ③ chain rule with the given rate ④ substitute the instant
Example
radius grows 0.2 cm/s; dA/dt when r = 5? A = πr² → dA/dr = 2πr = 10π dA/dt = 10π × 0.2 = 2π ≈ 6.28 cm²/s
40 m of fence, one side is a wall
Max/min problem
A = x(40 − 2x) = 40x − 2x² dA/dx = 40 − 4x = 0 → x = 10 d²A/dx² = −4 < 0 → max max area = 10 × 20 = 200 m²
⚠ TRAPUnits: rate answers need "per second" etc. A decreasing quantity has a negative rate.
🔎 WHY IT WORKSWhy connected rates use the chain rule: A depends on r, and r depends on t. So when t changes, r changes, which changes A. dA/dt = (how fast A changes per unit r) × (how fast r changes per second) = dA/dr × dr/dt — the "dr" units cancel like fractions.
🧠 Worked example — follow the thinking
Water fills a cone (vertex down) where r = ½h, at 3 cm³/s. Find dh/dt when h = 6 cm.
Step
Why I do this
V = ⅓πr²h = ⅓π(h/2)²h = πh³/12
💭 Write V in terms of ONE variable (h), because the question asks about h. Use r = ½h to remove r.
dV/dh = πh²/4
💭 Differentiate with respect to h.
dV/dt = dV/dh × dh/dt → 3 = (π × 36/4) × dh/dt
💭 Chain rule links the given rate (dV/dt = 3) to the one we want.
dh/dt = 3 ÷ 9π = 0.106 cm/s
💭 Substitute h = 6 only AFTER differentiating.
A Math · Sec 4 · SEC K341 (G3)
∫
C1
Integration
standard integrals, finding the curve, "hence"
BIG IDEAIntegration reverses differentiation. Indefinite integrals always need + c.
∫ f(x) dx
Result
∫ f(x) dx
Result
xⁿ
xⁿ⁺¹/(n + 1) + c
sin x
−cos x + c
(ax + b)ⁿ
(ax + b)ⁿ⁺¹ / a(n + 1) + c
cos x
sin x + c
eˣ
eˣ + c
sec²x
tan x + c
eᵃˣ⁺ᵇ
(1/a)eᵃˣ⁺ᵇ + c
sin(ax + b)
−(1/a)cos(ax + b) + c
cos(ax + b)
(1/a)sin(ax + b) + c
Examples
∫(6x² − 4/x²)dx = 2x³ + 4/x + c ∫(2x + 1)⁴dx = (2x + 1)⁵/10 + c ∫cos 3x dx = ⅓ sin 3x + c
Find the curve
dy/dx = 6x − 2, passes (1, 4) y = 3x² − 2x + c → 4 = 1 + c → c = 3 y = 3x² − 2x + 3
🧠 HOOK"Hence" questions: if d/dx[f(x)] = g(x) was found, then ∫g(x)dx = f(x) + c. Use the earlier part!
⚠ TRAPExpand or split before integrating — ∫(x + 1)² dx: expand first. ∫ of a product is NOT the product of integrals. Not tested: ∫1/x.
🔎 WHY IT WORKSWhy integration needs + c: integration asks "what did I differentiate to get this?". But x², x² + 5 and x² − 7 all differentiate to 2x — the constant disappears. So the answer is x² + c, and you find c from a given point.
🧠 Worked example — follow the thinking
Given d/dx(x ln x) = ln x + 1, find ∫ ln x dx.
Step
Why I do this
∫(ln x + 1) dx = x ln x + c
💭 Integration reverses differentiation, so the given result integrates straight back.
∫ln x dx + ∫1 dx = x ln x + c
💭 Split the integral into two parts.
∫ln x dx + x = x ln x + c
💭 ∫1 dx = x.
∫ln x dx = x ln x − x + c
💭 Rearrange. This "hence" trick is how you integrate things not on your list.
A Math · Sec 4 · SEC K341 (G3)
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C1
Definite Integrals & Area
area under a curve, regions below the axis
BIG IDEA∫ₐᵇ f(x) dx = area between the curve and the x-axis. Area below the axis comes out negative.
∫₀³(x² + 1)dx = [x³/3 + x]₀³ = 12
∫₀⁴(x² − 4x)dx = −32/3 → area 32/3
Sketch and find where the curve cuts the x-axis
Split the integral at each crossing
Integrate each part separately
Add the absolute values
Area with a line
area between curve and a line = area of trapezium/triangle ± ∫ curve (split into shapes you know) area between two curves is not tested
Area with the y-axis
∫ x dy between y = c and y = d (rearrange to x = g(y) first)
🔎 WHY IT WORKSWhy the integral gives area: split the region into thin strips of width δx and height y. Area ≈ Σ y δx. As the strips get infinitely thin this sum becomes ∫ y dx. Below the x-axis y is negative, so those strips count as negative area — that is why you must split at the roots.
🧠 Worked example — follow the thinking
Find the total area enclosed by y = x(x − 2) and the x-axis from x = 0 to x = 3.
Step
Why I do this
Roots: x = 0 and x = 2
💭 Sketch first. The curve crosses the axis at 2, so part is below and part is above.
∫₀²(x² − 2x)dx = [x³/3 − x²]₀² = −4/3
💭 This part is below the axis, so the answer is negative — area is 4/3.
∫₂³(x² − 2x)dx = [x³/3 − x²]₂³ = 0 − (−4/3) = 4/3
💭 Above the axis, positive.
Total = 4/3 + 4/3 = 8/3 units²
💭 Add the SIZES. Doing ∫₀³ in one go gives 0 — wrong!
A Math · Sec 4 · SEC K341 (G3)
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C1
Kinematics
displacement, velocity, acceleration, distance
BIG IDEADifferentiate to go s → v → a. Integrate to go a → v → s (find c from the start conditions).
displacement · velocity · acceleration
Phrase
Means
at rest / instantaneously at rest
v = 0
returns to O
s = 0
max / min velocity
a = 0
initially
t = 0
moving to the left
v < 0
s = t³ − 6t² + 9t
Total distance
find when v = 0 (direction changes) add the distances of each part s(0)=0, s(1)=4, s(3)=0, s(4)=4 distance in 4 s = 4 + 4 + 4 = 12 m displacement at t = 4 is only 4 m
⚠ TRAPDistance ≠ displacement. Always check if the particle turns around (v = 0) inside the time interval.
🔎 WHY IT WORKSWhy differentiate s to get v: velocity is how fast displacement changes = ds/dt. Acceleration is how fast velocity changes = dv/dt. Going backwards (a → v → s) you integrate and use the starting values to find c.
🧠 Worked example — follow the thinking
v = 3t² − 12t + 9 m/s, s = 0 at t = 0. When is it at rest, and what is its initial acceleration?
Step
Why I do this
At rest → v = 0: 3(t² − 4t + 3) = 0
💭 "At rest" always means velocity = 0.
3(t − 1)(t − 3) = 0 → t = 1 s, t = 3 s
💭 Factorise.
a = dv/dt = 6t − 12
💭 Acceleration is the derivative of velocity.
Initially t = 0 → a = −12 m/s²
💭 "Initially" means t = 0. Negative a = slowing down (if moving forward).
A Math · Sec 4 · SEC K341 (G3)
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★
A Math Formula Sheet & Exam Habits
K341 G3
BIG IDEAKnow what the paper gives you. Everything else must be memorised.
📄 GIVEN in the exam
quadratic formula binomial (a + b)ⁿ, ⁿCᵣ = n!/(r!(n − r)!) sin²A + cos²A = 1 · sec²A = 1 + tan²A · cosec²A = 1 + cot²A sin(A ± B), cos(A ± B), tan(A ± B) sin 2A, cos 2A (3 forms), tan 2A sine rule, cosine rule, area = ½bc sin A
🧠 MEMORISE
discriminant conditions · surd rules remainder/factor theorem · a³ ± b³ partial fraction forms · log laws, change of base R-formula · exact values · ASTC · principal values midpoint, gradient, ⊥ m₁m₂ = −1, circle forms linear law conversions all derivatives and integrals · chain/product/quotient kinematics links · alternate segment, midpoint theorems