Revision Notes
OxfordAQA A Level Mathematics: Pure Maths 1 — Revision Notes
Condensed recall notes on algebra, quadratics, coordinate geometry and calculus for OxfordAQA 9660 Unit P1, with circle and trigonometry sections marked as Unit PSM1 content.
- Subject
- Mathematics
- Level
- A LEVELS
- Topic
- Unit P1: Pure Maths (International AS)
- Author
- Marlbridge Academic Team
- Updated
- Reviewed by
- Sajawal Zahid (what this means)
Aligned to OxfordAQA A Level Mathematics (9660), Version 5.2 (International AS exams from May/June 2018, A-level from May/June 2019). Official specification .
Syllabus page (what it covers and how it is assessed): OxfordAQA A Level Mathematics.
Syllabus points this page covers
9660
- 1 Unit P1: Pure Maths (International AS) (whole topic)
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Need help with this topic? Request a free trial class for A Level Mathematics (9660).
Condensed for the final weeks. For the full explanation, use the Pure Maths 1 study guide.
Algebra and surds
a^m x a^n = a^(m+n) a^-n = 1/a^n a^(m/n) = (n-th root of a)^m
Rationalise a single-term denominator by multiplying by the surd; for a + √b, multiply by the conjugate a − √b, since (a+√b)(a−√b) = a² − b.
Quadratics
completed square: a(x + p)^2 + q vertex at (-p, q)
discriminant: b^2 - 4ac
| Discriminant | Meaning |
|---|---|
| > 0 | Two distinct roots — crosses the axis twice |
| = 0 | Repeated root — tangent to the axis |
| < 0 | No real roots — never meets the axis |
Almost every “find the values of k” question is a discriminant question. Tangency means set it to zero; two intersections means greater than zero; no intersection means less than zero.
“Always positive” needs two conditions: a > 0 and discriminant < 0. Giving only one loses a mark.
Disguised quadratics — substitute for the repeated term, then substitute back and reject impossible values (√x cannot be negative; aˣ is always positive).
Coordinate geometry (P1)
m = (y2-y1)/(x2-x1) y - y1 = m(x - x1)
distance = sqrt((x2-x1)^2 + (y2-y1)^2)
Perpendicular gradients multiply to −1. P1’s coordinate geometry is lines only (equation of a line, parallel/perpendicular conditions, intersection of a line and a curve) – the circle is not P1 content; see below.
Circle and trigonometry — Unit PSM1 content, not P1
Neither of these two sections is P1 content. Both belong to the other AS unit, PSM1’s pure maths strand (PP1.1 Circle, PP1.2 Trigonometry). They’re included here because every P1 candidate also sits PSM1, but don’t revise them expecting a P1 paper to test them.
circle: (x-a)^2 + (y-b)^2 = r^2
Three circle facts solve most problems: the tangent is perpendicular to the radius at the point of contact; the perpendicular bisector of a chord passes through the centre; the angle in a semicircle is a right angle.
From an expanded circle equation, complete the square in both x and y to find the centre and radius.
sin^2 x + cos^2 x = 1 tan x = sin x / cos x
sine rule: a/sin A = b/sin B
cosine rule: a^2 = b^2 + c^2 - 2bc cos A
area = (1/2)ab sin C
The ambiguous case: when the sine rule gives an angle, there may be a second solution, since sin(180° − θ) = sin θ. Always check whether the obtuse alternative is consistent with the triangle.
When solving trigonometric equations, find all solutions in the stated interval using the graph’s symmetry, and check both ends of the range.
Differentiation (P1)
y = ax^n -> dy/dx = anx^(n-1)
Applications: gradient at a point; tangent with gradient m; normal with gradient −1/m; stationary points where dy/dx = 0, classified by d²y/dx² — positive gives a minimum, negative a maximum.
If d²y/dx² = 0 the test is inconclusive and you must check the sign of the gradient either side. That case appears precisely because the shortcut fails.
The limit definition of the derivative — the gradient of the tangent as the limit of the gradient of a chord — is required knowledge. Carrying out differentiation from first principles itself is not tested; routine questions use the shortcut power rule directly.
Worked example. Find the coordinates and nature of the stationary point of y = x² − 8x + 3.
dy/dx = 2x - 8
2x - 8 = 0 -> x = 4
y = 16 - 32 + 3 = -13 -> (4, -13)
d2y/dx2 = 2, positive -> minimum
Completing the square gives (x − 4)² − 13, confirming the same turning point — a quick, independent check worth doing when time allows.
Integration
integral of ax^n = ax^(n+1)/(n+1) + c n != -1
The + c is a mark. It cancels in a definite integral but is required in an indefinite one.
Area: the integral gives a signed value, so regions below the x-axis integrate negative. For a region entirely below the x-axis, the area is simply the magnitude of the (negative) signed integral. P1 will not set questions on regions that are partially above and partially below the x-axis, so the “split at the roots and take absolute values” technique needed for that mixed case is background knowledge only, not something a P1 paper will require.
Worked example. Find the area enclosed between y = x² − 4 and the x-axis, between x = −2 and x = 2.
integral of (x^2 - 4) dx = x^3/3 - 4x
[8/3 - 8] - [-8/3 + 8] = -32/3
signed integral = -32/3, so area = 32/3
The curve lies entirely below the axis on this interval, so the signed integral comes out negative — the area is its magnitude, 32/3.
For the area between two curves, find the intersections first: they are the limits, and you integrate (upper − lower).
Exam traps
- Giving one condition for “always positive”.
- Omitting
+ c. - Forgetting that the signed integral over a region entirely below the x-axis is negative, and that the area is its magnitude.
- Missing the second solution in the ambiguous case.
- Confusing tangent and normal gradients.
- Forgetting to reject invalid roots after a substitution.
Self-test
- What does
b² − 4ac = 0mean geometrically? - Give both conditions for a quadratic to be positive for all x.
- State three circle theorems useful in coordinate geometry. (PSM1 content, not P1 – see above.)
- When does the second derivative test fail, and what do you do?
- A curve lies entirely below the x-axis between x = a and x = b. Why doesn’t the signed integral directly give the area, and how do you find the area?
Answers: 1. The curve is tangent to the x-axis — one repeated root. 2. a > 0 and b² − 4ac < 0. 3. The tangent is perpendicular to the radius at the point of contact; the perpendicular bisector of a chord passes through the centre; the angle in a semicircle is 90°. 4. When d²y/dx² = 0; you must then examine the sign of the gradient on either side of the stationary point. 5. The signed integral comes out negative because the curve is below the axis throughout, so the area is the magnitude (absolute value) of that negative value. (Splitting a region at the roots and taking absolute values of each part is needed only when a region is partially above and partially below the axis – P1 will not set that case.)
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