Edexcel A Level Mathematics: Pure Mathematics 1 — Revision Notes

Condensed for the final weeks. For the full explanation, use the Pure Mathematics 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

Rationalising: multiply by the surd, or by the conjugate for two terms. (a + √b)(a − √b) = a² − b, which removes the surd.

Quadratics

completed square:  a(x + p)^2 + q     vertex at (-p, q)
discriminant:      b^2 - 4ac

Discriminant	Roots	Geometry
> 0	Two distinct	Crosses the x-axis twice
= 0	One repeated	Tangent to the x-axis
< 0	None real	Never meets the x-axis

Almost every “find the values of k” question is a discriminant question. Tangency → set it to zero; two intersections → greater than zero; no intersection → less than zero.

“Always positive” requires two conditions: a > 0 and discriminant < 0. Giving only the discriminant loses a mark.

Worked example. Find the set of values of k for which kx² + 4x + 1 is positive for all real x. Condition 1: k > 0. Condition 2: discriminant < 0, so 4² − 4(k)(1) < 0, giving 16 − 4k < 0, so k > 4. Since k > 4 already forces k > 0, the answer is simply k > 4 — stating both conditions and seeing that the stronger one absorbs the weaker is exactly what full marks require.

Coordinate geometry (straight lines)

gradient   m = (y2 - y1)/(x2 - x1)
line       y - y1 = m(x - x1)
distance   sqrt((x2-x1)^2 + (y2-y1)^2)
midpoint   ((x1+x2)/2, (y1+y2)/2)

Perpendicular gradients multiply to −1; parallel lines have equal gradients. Unit P1 covers only the straight line — before setting up an equation, identify which of these three situations the question describes:

1. Line through two given points — find the gradient first, then substitute one point into y − y1 = m(x − x1).
2. Line parallel to a given line, through a given point — the new line has the same gradient as the given line.
3. Line perpendicular to a given line, through a given point — the new line’s gradient is the negative reciprocal of the given line’s gradient.

The coordinate geometry of circles — equations of the form (x − a)² + (y − b)² = r² and the circle theorems built on them — is not introduced until Pure Mathematics 2. A P1 question that looks like it needs a circle almost always turns out to be a straight-line condition in disguise.

Differentiation

y = ax^n   ->   dy/dx = anx^(n-1)

From first principles:

f'(x) = lim(h->0) [f(x+h) - f(x)] / h

The proof is examinable, and the mark is for writing the limit notation, not just the algebra.

Applications (P1):

• Gradient of a curve at a point.
• Tangent — gradient m at that point. Normal — gradient −1/m, found using the same perpendicular-gradient rule as in the coordinate geometry section above.

Stationary points, and classifying them as maxima or minima using d²y/dx², are introduced in Pure Mathematics 2 — P1 differentiation questions stop at finding a gradient, tangent or normal.

Integration

integral of ax^n = ax^(n+1)/(n+1) + c        n != -1

The + c is a mark — P1 integration is entirely indefinite (there is no n = −1 case either, since that needs logarithms, not introduced until Pure Mathematics 2). Given a gradient function and one point on the curve, substitute that point into the integrated expression to find c and hence the full equation of the curve.

Definite integration, and using it to find the area under a curve or between two curves, is introduced in Pure Mathematics 2 — do not evaluate [F(x)] between two limits on a P1 paper; if a question gives two x-values, check whether it is really asking you to substitute each into a curve or tangent equation instead.

Trigonometry

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
radians:      s = r*theta            (arc length)
              A = (1/2) r^2 * theta  (area of sector)

The ambiguous case: when using the sine rule to find an angle, there may be a second solution, since sin(180° − θ) = sin θ. Always check whether the obtuse alternative — not just the calculator’s default acute answer — is the one consistent with the triangle described.

Radians: a common trap is mixing degree and radian mode on a calculator mid-question — check the angle unit the question uses before substituting into either radian formula above.

The identity sin²x + cos²x = 1 (and tan x = sin x / cos x), and solving trigonometric equations with them, are introduced in Pure Mathematics 2 — P1 trigonometry stays within triangles (sine rule, cosine rule, area) and radian-measure calculations.

Exam traps

• Giving only the discriminant condition for “always positive” — both conditions are needed.
• Omitting + c on an indefinite integral.
• Forgetting the second solution in the sine rule’s ambiguous case.
• Using the normal’s gradient where the tangent’s is needed, or vice versa.
• Mixing up the parallel condition (equal gradients) with the perpendicular condition (gradients multiply to −1) when a question gives one line and a point.
• Working in the wrong angle mode (degrees vs radians) in an arc-length or sector-area calculation.
• Not showing method — method marks are available even with a wrong final answer.

Self-test

1. What does b² − 4ac = 0 mean geometrically?
2. Give the two conditions for ax² + bx + c to be positive for all x.
3. What is the gradient of a line perpendicular to one with gradient 2/3?
4. Give the formulas for arc length and area of a sector in terms of radius r and angle θ (in radians).
5. Why might a P1 question that gives you two x-values be asking for something other than a definite integral?

Answers: 1. The curve is tangent to the x-axis — there is one repeated root. 2. a > 0 and b² − 4ac < 0. 3. −3/2 (the negative reciprocal of 2/3). 4. Arc length s = rθ; area of sector A = ½r²θ. 5. Because definite integration and areas are not introduced until Pure Mathematics 2 — on a P1 paper, two x-values are more likely to be points for substitution into a tangent, normal or curve equation.