OxfordAQA A Level Mathematics: Pure Maths 1 — Revision Notes

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

m = (y2-y1)/(x2-x1)        y - y1 = m(x - x1)
distance = sqrt((x2-x1)^2 + (y2-y1)^2)
circle:  (x-a)^2 + (y-b)^2 = r^2

Perpendicular gradients multiply to −1.

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.

Trigonometry

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

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.

Differentiation from first principles — using the limit definition of the derivative rather than the power rule directly — is examinable, though routine questions use the shortcut rule.

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 total area, split at the roots and take absolute values — simply adding the signed integrals gives the wrong answer.

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.
• Adding signed integrals when a total area is required.
• Missing the second solution in the ambiguous case.
• Confusing tangent and normal gradients.
• Forgetting to reject invalid roots after a substitution.

Self-test

1. What does b² − 4ac = 0 mean geometrically?
2. Give both conditions for a quadratic to be positive for all x.
3. State three circle theorems useful in coordinate geometry.
4. When does the second derivative test fail, and what do you do?
5. Why can adding definite integrals give the wrong 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. Regions below the x-axis give negative values that cancel against positive regions, so the total area must be found by splitting at the roots and taking absolute values.