Every distinction begins with one topic at a time.- start now
| Paper | Core topics | Total marks | Duration |
|---|---|---|---|
| Paper 1 | Algebra, equations & patterns; Functions & graphs; Finance, growth & decay; Differential calculus; Probability | 150 | 3 hours |
| Paper 2 | Statistics; Analytical geometry; Trigonometry (identities, equations, graphs, 2D & 3D); Euclidean geometry | 150 | 3 hours |
| Level | Typical weighting | What earns these marks | Command words to expect |
|---|---|---|---|
| Knowledge | 20% | Stating a definition, formula or fact with no calculation | write down, state, define |
| Routine procedures | 35% | A standard method applied to a familiar, mostly one-step question | calculate, solve, determine |
| Complex procedures | 30% | Several steps or concepts linked together in a less familiar setting | hence, show that, prove |
| Problem solving | 15% | An unseen context where you must choose your own strategy | investigate, hence or otherwise |
Paper 1: 150 marks, 3 hours, covers algebra & patterns, functions, finance, calculus and probability.
Paper 2: 150 marks, 3 hours, covers statistics, analytical geometry, trigonometry and Euclidean geometry.
Cognitive split: 20% knowledge, 35% routine procedures, 30% complex procedures, 15% problem solving.
Round off to two decimal places unless a question states otherwise.
Only a non-programmable, non-graphical scientific calculator is allowed.
| Paper 1, typical order | Usual topic |
|---|---|
| Q1 | Algebraic equations, inequalities and simultaneous equations |
| Q2 | Number patterns (arithmetic, geometric or quadratic) |
| Q3 | Series, sigma notation and convergence |
| Q4-Q5 | Functions and graphs, including the inverse function |
| Q6 | Finance: growth, decay and annuities |
| Q7-Q8 | Differential calculus, including optimisation and sketching |
| Q9 | Probability |
| Paper 2, typical order | Usual topic |
|---|---|
| Q1-Q2 | Statistics: central tendency, spread, and bivariate data |
| Q3 | Analytical geometry |
| Q4-Q5 | Trigonometric identities, equations and graphs |
| Q6-Q7 | 2D and 3D trigonometry (sine, cosine and area rules) |
| Q8-Q9 | Euclidean geometry: circle theorem riders and proportion |
These are the typical TOPIC orders, not a fixed question count. Real papers run longer than the rows above: November 2025 set eleven questions in each paper and November 2024 set twelve in Paper 1 and eleven in Paper 2, so a topic can arrive one or two numbers later than shown. Plan by block and by mark weighting, never by question number.
The order and exact split above shifts slightly from year to year, so treat it as a planning guide, not a fixed rule.
Q: A calculus question on a Paper 1 paper is worth 20 marks. Using the standard cognitive level weighting for the whole paper, estimate how many of those 20 marks would typically sit at each level.
Knowledge: 20% of 20 = 4 marks
Routine procedures: 35% of 20 = 7 marks
Complex procedures: 30% of 20 = 6 marks
Problem solving: 15% of 20 = 3 marks
∴ roughly 4 + 7 + 6 + 3 = 20 marks, so more than half the question (11 out of 20 marks) should be earnable through direct recall and a well-rehearsed method alone.
Read every instruction on the cover page as carefully as you read the questions themselves: a good share of the marks are hiding in how you were told to answer.- on reading the paper properly
Paper 1 and Paper 2 are each 150 marks over 3 hours, split across a fixed set of CAPS topics. Marks are also split by cognitive level: roughly 20% knowledge, 35% routine procedures, 30% complex procedures and 15% problem solving, so most marks reward a well-drilled method, not exotic insight. Round final answers to two decimal places unless told otherwise, use only a non-programmable, non-graphical calculator, and complete any detachable diagram sheet fully before handing it in.
This block opens almost every Paper 1 and it is the biggest single allocation in the paper: 25 marks of equations and inequalities, then 25 marks of patterns and series. None of it is new content. All of it is technique under time pressure, so revise the routine for each question type rather than the full theory behind it.
| Question type | Typical marks | What the marker is looking for |
|---|---|---|
| Factorisable quadratic equation | 2 to 3 | Standard form first, then the factors, then BOTH roots |
| Quadratic formula, two decimal places | 3 to 4 | Correct a, b and c visible in the substitution; rounding only at the very end |
| Quadratic inequality | 3 to 4 | Critical values, a sign test, and the answer written as an interval |
| Simultaneous equations, linear plus quadratic | 5 to 6 | Substitution, the quadratic solved, and the answers given as PAIRS |
| Exponential equation | 3 to 4 | Every term rewritten to the same base before the exponents are equated |
| Surd equation | 4 to 5 | Isolate, square, and the written test that rejects any extraneous root |
| Discriminant and the nature of the roots | 3 to 5 | Δ computed correctly, then the nature stated in words |
| General term of an arithmetic or geometric sequence | 3 to 4 | a and d (or r) established first, then the formula |
| Quadratic pattern, second differences | 4 to 5 | The difference rows shown, then a, b and c solved |
| Sigma notation, expand or evaluate | 4 to 5 | The number of terms counted correctly, then the right sum formula |
| Convergence and the sum to infinity | 4 to 6 | The condition −1 < r < 1 written down explicitly |
Work in this order every single time and you never waste a minute deciding. First: move everything to one side so the equation reads "expression = 0". Second: try to factorise, because it is the fastest route and it carries full marks. Third: if it does not factorise neatly, reach for the quadratic formula. The phrase "correct to TWO decimal places" in the question is a direct instruction that the formula is expected, because an equation that factorises always produces exact roots that need no rounding at all.
Q: Solve for x: (a) 2x² − x − 15 = 0, and (b) 3x² − 8x + 2 = 0, correct to two decimal places.
(a) Already in standard form. Look for two numbers whose product is 2 × (−15) = −30 and whose sum is −1: those are −6 and +5.
2x² − 6x + 5x − 15 = 0
2x(x − 3) + 5(x − 3) = 0
(2x + 5)(x − 3) = 0
∴ x = −5⁄2 or x = 3
(b) This does not factorise, and the wording asks for two decimal places, so use the formula with a = 3, b = −8, c = 2.
Δ = b² − 4ac = (−8)² − 4(3)(2) = 64 − 24 = 40
x = [8 ± √40] ÷ 6, and √40 ≈ 6,3246
x = (8 + 6,3246) ÷ 6 = 14,3246 ÷ 6 = 2,3874…
x = (8 − 6,3246) ÷ 6 = 1,6754 ÷ 6 = 0,2792…
∴ x ≈ 2,39 or x ≈ 0,28
Mark logic: one mark for correct a, b and c inside the formula, one for the value of the discriminant, one for each rounded root. Keep full accuracy in the calculator and round only on the final line.
The discriminant Δ = b² − 4ac answers the question "what kind of roots does this equation have?" without you solving anything. The question almost always wants the answer back in WORDS, so learn the wording alongside the signs. A perfect-square discriminant means the roots are rational; a positive discriminant that is not a perfect square means they are irrational.
| Value of Δ = b² − 4ac | Nature of the roots | What the parabola does |
|---|---|---|
| Δ > 0 and a perfect square | Two real, rational, unequal roots | Cuts the x-axis at two tidy points |
| Δ > 0 but not a perfect square | Two real, irrational, unequal roots | Cuts the x-axis at two points that carry surds |
| Δ = 0 | Two equal real roots (one repeated root) | Touches the x-axis at exactly one point |
| Δ < 0 | No real roots (the roots are non-real) | Never reaches the x-axis at all |
Q: Determine the value(s) of k for which x² + (k − 2)x + 9 = 0 has two equal real roots.
"Two equal real roots" translates directly into Δ = 0.
Here a = 1, b = (k − 2), c = 9
Δ = b² − 4ac = (k − 2)² − 4(1)(9) = (k − 2)² − 36
Set it equal to zero: (k − 2)² − 36 = 0
(k − 2)² = 36
k − 2 = 6 or k − 2 = −6
∴ k = 8 or k = −4
Check with k = 8: the equation becomes x² + 6x + 9 = (x + 3)², which does have the single repeated root x = −3.
An inequality is not an equation. It asks for an INTERVAL, not a pair of numbers. The routine is fixed: move everything to one side so the other side is zero, factorise, read off the critical values (where each factor is zero), then test the sign of the whole expression in each region those critical values carve out.
Q: Solve for x: 3x² − 2x − 8 ≤ 0
The inequality is already in standard form with zero on the right.
Factorise: 3x² − 6x + 4x − 8 = 3x(x − 2) + 4(x − 2) = (3x + 4)(x − 2)
Critical values: 3x + 4 = 0 gives x = −4⁄3; and x − 2 = 0 gives x = 2
Now test one value in each of the three regions:
At x = −2: (3(−2) + 4)(−2 − 2) = (−2)(−4) = +8, so POSITIVE
At x = 0: (0 + 4)(0 − 2) = (4)(−2) = −8, so NEGATIVE
At x = 3: (9 + 4)(3 − 2) = (13)(1) = +13, so POSITIVE
We want where the expression is less than or equal to zero, which is the middle region, endpoints included.
∴ −4⁄3 ≤ x ≤ 2
Mark logic: one mark for correct factorising, one for both critical values, one for the sign reasoning, one for the interval in correct notation.
When one equation is linear and the other is quadratic, make one variable the subject of the LINEAR equation and substitute it into the quadratic one. What you are left with is an ordinary quadratic in a single variable. The marks are lost at the end, not the start: each x-value has its OWN matching y-value, and the answer must be written as pairs.
Q: Solve simultaneously for x and y: y − x = 3 and xy = 10
Make y the subject of the linear equation: y = x + 3
Substitute into the second equation: x(x + 3) = 10
x² + 3x = 10
x² + 3x − 10 = 0
Factorise: (x + 5)(x − 2) = 0, so x = −5 or x = 2
If x = −5, then y = −5 + 3 = −2
If x = 2, then y = 2 + 3 = 5
∴ (x ; y) = (−5 ; −2) or (x ; y) = (2 ; 5)
Check: (−5)(−2) = 10 and (2)(5) = 10, so both pairs satisfy xy = 10.
In an exponential equation the first move is always to rewrite every term as a power of the SAME base. Once the bases match, the exponents can be equated. Where a common factor is hiding, take it out first: an expression like 5x+2 − 5x factorises to 5x(5² − 1) and collapses instantly. In a surd equation you isolate the surd term, square both sides, and solve whatever ordinary equation remains. Squaring can invent roots that do not actually work, so testing the answers back in the original equation is not optional housekeeping, it is a mark.
Q: Solve for x: 5x+2 − 5x = 120
Split the first term using the exponent law: 5x+2 = 5x × 5²
5x(25) − 5x(1) = 120
5x(25 − 1) = 120
24 × 5x = 120
5x = 5
∴ x = 1
Check: 5³ − 5¹ = 125 − 5 = 120, correct.
Q: Solve for x: √(2x + 7) = x + 2
The surd is already isolated, so square both sides:
2x + 7 = (x + 2)² = x² + 4x + 4
0 = x² + 2x − 3
(x + 3)(x − 1) = 0, so x = −3 or x = 1
Test x = 1: LHS = √9 = 3, RHS = 3, so it works
Test x = −3: LHS = √1 = 1, RHS = −1, so it fails
∴ x = 1 only
Why: a square root sign returns a non-negative value, so it can never equal −1.
| Pattern type | How to test it | General term Tn | Sum of n terms, Sn |
|---|---|---|---|
| Linear (arithmetic) | T2−T1 = T3−T2 = d, a fixed value | Tn = a + (n−1)d | Sn = n⁄2[2a+(n−1)d] or n⁄2(a+l) |
| Geometric | T2÷T1 = T3÷T2 = r, a fixed value | Tn = a × rn−1 | Sn = a(rn−1) ÷ (r−1) |
| Quadratic | 1st differences change, but 2nd differences are fixed | Tn = an2 + bn + c | no single formula, add the individual terms once Tn is known |
Tn = a + (n−1)dSn = n⁄2[2a+(n−1)d]Tn = a × rn−1Sn = a(rn−1) ÷ (r−1)S∞ = a ÷ (1−r), only when −1 < r < 1
Q: A community fun run has added a fixed number of new entrants every year since it started. Across its first 14 years, the running total of entrants is 406. In its 20th year, exactly 79 people entered. Determine the number of entrants in year 1 (a) and the yearly increase (d), then state the general term Tn.
Clue 1 (the sum): S14 = 14⁄2[2a+13d] = 406, so 2a + 13d = 58 … (1)
Clue 2 (the term): T20 = a + 19d = 79, so a = 79 − 19d … (2)
Substitute (2) into (1): 2(79−19d) + 13d = 58, giving 158 − 38d + 13d = 58, so −25d = −100, meaning d = 4
Then a = 79 − 19(4) = 79 − 76 = 3
∴ a = 3, d = 4, and Tn = 3 + 4(n−1) = 4n − 1
Q: A workshop machine shaves off one quarter of a steel rod's remaining length with every pass. The rod is 80 cm long before the first pass. Determine the general term for the length shaved off on pass n, calculate the amount shaved off on the 5th pass, and determine whether the total length shaved off over infinitely many passes converges, if so, to what value.
Pass 1 shaves off 1⁄4(80) = 20 cm, leaving 60 cm. Pass 2 shaves off 1⁄4(60) = 15 cm
a = 20, r = 15⁄20 = 3⁄4
Tn = 20 × (3⁄4)n−1
T5 = 20 × (3⁄4)4 = 20 × 81⁄256 ≈ 6,33 cm
Since r = 3⁄4 and −1 < 3⁄4 < 1, the series converges
S∞ = a ÷ (1−r) = 20 ÷ (1 − 3⁄4) = 20 ÷ 1⁄4 = 80 cm
∴ T5 ≈ 6,33 cm; the series converges to S∞ = 80 cm (sensibly, exactly the rod's starting length)
Q: Determine the general term of the quadratic pattern 5 ; 9 ; 15 ; 23 ; 33 ; …
First differences: 4 ; 6 ; 8 ; 10, not constant
Second differences: 2 ; 2 ; 2, constant, so this is quadratic
a = 2 ÷ 2 = 1
b = (first of the 1st differences) − 3a = 4 − 3(1) = 1
c = T1 − a − b = 5 − 1 − 1 = 3
∴ Tn = n2 + n + 3
Q: Evaluate ∑k=620 (2k + 3).
Number of terms = 20 − 6 + 1 = 15
First term (k=6): 2(6) + 3 = 15. Last term (k=20): 2(20) + 3 = 43
This is a linear (arithmetic) series with a = 15, l = 43, n = 15
S15 = 15⁄2(a+l) = 7,5(15+43) = 7,5(58) = 435
∴ ∑k=620 (2k+3) = 435
The classic version of this question hands you two terms that do not sit next to each other and asks for a and d, or a and r. Write each given term out using the general formula and you immediately have two equations in two unknowns. The two families are handled differently at the crunch point: for an ARITHMETIC sequence you SUBTRACT the two equations, because that eliminates a; for a GEOMETRIC sequence you DIVIDE the two equations, because dividing cancels a and leaves a single power of r.
| Arithmetic sequence | Geometric sequence | Quadratic pattern | |
|---|---|---|---|
| Test that identifies it | Constant first difference d | Constant ratio r | Constant second difference |
| General term | Tn = a + (n−1)d | Tn = arn−1 | Tn = an² + bn + c |
| Sum of n terms | Sn = n⁄2[2a + (n−1)d] | Sn = a(rn−1) ÷ (r−1) | Not in the curriculum |
| How you combine two given terms | Subtract the two equations | Divide the two equations | Build three equations from the differences |
| Typical trap sub-question | "Which term equals 97?" | "Does the series converge?" | "Which term is the smallest?" |
Q: The second term of a geometric sequence is 18 and the fifth term is 486. Determine a and r, then calculate the sum of the first 7 terms.
T2 = ar = 18 … (1)
T5 = ar4 = 486 … (2)
Divide (2) by (1): ar4 ÷ ar = 486 ÷ 18
r³ = 27, so r = 3
Substitute r = 3 into (1): a(3) = 18, so a = 6
S7 = a(r7 − 1) ÷ (r − 1) = 6(37 − 1) ÷ (3 − 1)
37 = 2 187, so S7 = 6(2 186) ÷ 2 = 3(2 186) = 6 558
∴ a = 6, r = 3 and S7 = 6 558
Check: the sequence runs 6 ; 18 ; 54 ; 162 ; 486 ; 1 458 ; 4 374, and the second and fifth terms are indeed 18 and 486.
An infinite geometric series only settles on a finite total when −1 < r < 1. Three separate question styles grow out of that single condition. The first simply asks you to test it and then compute S∞ = a ÷ (1 − r). The second gives you a ratio written as an expression in x and asks for the values of x that keep the series convergent, which turns the condition itself into the double inequality you must solve. The third runs the formula backwards: it hands you S∞ plus one other fact and asks you to recover a or r, which usually produces a quadratic in r and therefore possibly TWO valid answers.
Q: Determine the values of x for which the infinite series 3 + 3(2x − 1) + 3(2x − 1)² + … converges.
Each term is multiplied by (2x − 1) to give the next, so r = 2x − 1
Convergence requires −1 < r < 1, so −1 < 2x − 1 < 1
Add 1 to every part: 0 < 2x < 2
Divide every part by 2: 0 < x < 1
∴ the series converges for 0 < x < 1
Q: A convergent geometric series has a sum to infinity of 45, and its second term is 10. Determine the first term and the common ratio.
From the sum to infinity: a ÷ (1 − r) = 45, so a = 45(1 − r) … (1)
From the second term: ar = 10 … (2)
Substitute (1) into (2): 45(1 − r)r = 10
45r − 45r² = 10
45r² − 45r + 10 = 0, and dividing throughout by 5 gives 9r² − 9r + 2 = 0
Factorise: (3r − 1)(3r − 2) = 0, so r = 1⁄3 or r = 2⁄3
If r = 1⁄3: a = 45(1 − 1⁄3) = 45 × 2⁄3 = 30
If r = 2⁄3: a = 45(1 − 2⁄3) = 45 × 1⁄3 = 15
Both ratios satisfy −1 < r < 1, so neither may be discarded.
∴ a = 30 with r = 1⁄3, or a = 15 with r = 2⁄3
Check: 30 × 1⁄3 = 10 and 15 × 2⁄3 = 10, so both give the required second term.
A number pattern only fools you once. Check a second gap or a second ratio, and it hands you its rule for free.- on number patterns
The equations half runs to a fixed script: standard form, then factorise, then the formula if "two decimal places" appears. The discriminant Δ = b² − 4ac names the roots without solving, with Δ = 0 for equal roots and Δ < 0 for none. Inequalities want critical values, a sign test and an interval, and the sign flips whenever you multiply or divide by a negative. Simultaneous systems substitute out of the linear equation and answer in pairs. Exponentials want a common base, surds want a test back in the original equation.
Test first differences for a linear pattern, ratios for geometric, second differences for quadratic. Memorise all five boxed formulas cold. Count sigma terms as (top − bottom + 1) and read Tk straight off the expression next to the ∑ sign. When a question gives two separate clues about a sequence, write each as its own equation and solve them together, substitution is usually fastest. A geometric series only converges, and only then has an S∞, when −1 < r < 1. Always reject a solution for n that isn't a positive whole number.
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