Pearson Edexcel International Advanced Level Mathematics: Inequalities, polynomials and graphs (YMA01) – Revision Notes

These notes are aligned to the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (YMA01), Issue 3, April 2019, Unit P1: Pure Mathematics 1, topics 1.6 to 1.12. The specification makes P1 and P2 compulsory for IAS Mathematics, IAL Mathematics and IAL Pure Mathematics. For full explanations read the study guide, then test yourself with the practice questions.

Course pages: the Edexcel A Level Mathematics hub, the printable checklist, the free 10-minute diagnostics, Pure Mathematics 1 revision notes and Pure Mathematics 2 revision notes.

1.6 Simultaneous equations by substitution

Method in steps:

1. Rearrange so that x or y is alone on one side of the linear equation.
2. Put that expression into the non-linear equation. Expand and collect.
3. Solve the quadratic.
4. Put each root back into the linear equation.
5. Write the answers as pairs.

Reminder. Solve x + y = 8 and xy = 15. Then y = 8 − x, so x(8 − x) = 15 and x² − 8x + 15 = 0, so (x − 3)(x − 5) = 0. The solutions are (3, 5) and (5, 3).

If the quadratic has no real roots, the line and curve do not meet. If it has one repeated root, the line touches the curve.

1.7 and 1.9 Inequalities

Situation	Rule
Multiply or divide by a negative	Reverse the sign
Bracket in the inequality	Expand first, then collect
Fraction with x below	Multiply by x² (positive), state x ≠ 0
Quadratic, x² coefficient positive, “< 0”	Between the roots
Quadratic, x² coefficient positive, “> 0”	Outside the roots: two ranges joined by “or”

Method for quadratics:

1. Rearrange to have 0 on one side.
2. Factorise (or use the formula) to find the roots.
3. Sketch the curve with its roots.
4. Read off the region asked for.

Reminders.

5 − 2x ≤ 11        →  −2x ≤ 6  →  x ≥ −3
x² + 4x − 21 > 0   →  (x + 7)(x − 3) > 0  →  x < −7 or x > 3
x² + 3 < 4x        →  x² − 4x + 3 < 0  →  1 < x < 3
4/x > 1            →  4x > x²  →  x² − 4x < 0  →  0 < x < 4

The last line needs a check, because x ≠ 0 was assumed: x = 2 gives 2 > 1 (true), x = 5 gives 0.8 > 1 (false), x = −1 gives −4 > 1 (false).

Graphical reading. px² + qx + r < ax + b means “curve below line”. Find the intersection x-values, sketch both, then choose the interval where the curve is lower. For “curve above line” choose the outer or inner region the sketch shows.

1.8 Regions

Inequality	Boundary line	Shade
y > f(x)	Dotted	Above the boundary
y ≥ f(x)	Solid	Above, and the boundary is included
y < f(x)	Dotted	Below the boundary
y ≤ f(x)	Solid	Below, and the boundary is included

Check every region with a test point such as (0, 0), unless it lies on a boundary. For a region bounded by a curve and a line, find where they meet by solving simultaneously, then mark those points.

Reminder. Region y ≥ x² − 2x and y < 3. The boundaries meet where x² − 2x = 3, so x = −1 or 3, at (−1, 3) and (3, 3). The parabola is solid and the line is dotted. The test point (1, 0) gives 0 ≥ −1 and 0 < 3, both true, so shade above the parabola and below the line.

1.10 Polynomials and f(x)

Result	Use
Difference of two squares	9x² − 16 = (3x − 4)(3x + 4)
Common factor first	x³ − 6x² + 8x = x(x² − 6x + 8) = x(x − 2)(x − 4)
Grouping	Pair terms, then take out the shared bracket
Degree	Highest power of x

Reminder. (x + 3)² − (x − 2)² = (x² + 6x + 9) − (x² − 4x + 4) = 10x + 5.

For f(x) = x² − 2x: f(3) = 3, f(x − 1) = (x − 1)² − 2(x − 1) = x² − 4x + 3, and f(2x) = 4x² − 4x. Replace every x, including the ones inside brackets.

1.11 Sketching graphs

Cubic checklist: sign of the x³ coefficient (end behaviour), roots from factors, a double root touches the axis, y-intercept from x = 0, and label all intercepts.

Reminder. y = x(3 − x)(x + 2): the roots are −2, 0 and 3. The coefficient of x³ is −1, so the curve falls to the right and rises to the left. It is positive for x < −2, negative between −2 and 0, positive between 0 and 3, and negative after 3. Expanded, it is −x³ + x² + 6x. The y-intercept is the origin.

Graph	Asymptotes	Position
y = k/x, k > 0	x = 0, y = 0	Quadrants 1 and 3
y = k/x, k < 0	x = 0, y = 0	Quadrants 2 and 4
y = k/x², k > 0	x = 0, y = 0	Above the x-axis, symmetric
y = k/x², k < 0	x = 0, y = 0	Below the x-axis, symmetric

Graph	Period	Features
y = sin x	360° (2π)	Through the origin; first peak at 90°
y = cos x	360° (2π)	Begins at its peak of 1; first zero at 90°
y = tan x	180° (π)	Through the origin; asymptotes at 90° and 270°

Solving by intersection. Rewrite the equation as two simple graphs, sketch both on one set of axes, and count the crossings. Each crossing is a real solution. The x-coordinates are the solutions of the equation.

1.12 Transformations

Equation	Move	Memory aid
y = af(x)	y-values times a	Outside: as written
y = f(x) + a	Up a	Outside: as written
y = f(x + a)	Left a	Inside: opposite
y = f(ax)	x-values divided by a	Inside: opposite

Asymptotes move with the curve. For y = 3/(x − 4) + 1 start from y = 3/x, move 4 right and 1 up, so the asymptotes are x = 4 and y = 1.

For trigonometric curves: y = 4 cos x has amplitude 4. y = cos 3x has period 120°. y = sin(x − 30°) is sin x moved 30° to the right. y = tan 2x has period 90°.

Given any graph, apply the rule to each labelled point: maximum, minimum, intercepts, asymptotes. Mark the new coordinates and join with the same shape.

Reminder. The curve y = f(x) has a minimum at (2, −5) and crosses the y-axis at (0, 3). For y = 2f(x): minimum (2, −10), y-intercept (0, 6). For y = f(x + 1): minimum (1, −5), the new y-intercept is f(1), which the given points do not show. For y = f(x/2): minimum (4, −5), y-intercept (0, 3), because x = 0 does not move.

Must-know distinctions

• Roots versus intercepts: roots are x-values where y = 0; the y-intercept is found by putting x = 0.
• Strict versus inclusive: < and > use a dotted line and exclude the boundary; ≤ and ≥ use a solid line.
• “Or” versus “and”: two separate ranges use “or”; a single interval is written a < x < b.
• Inside versus outside the bracket: inside changes x and acts in reverse; outside changes y and acts as written.
• k/x versus k/x²: the first has two branches in opposite quadrants; the second has two branches on the same side of the x-axis.

Quick self-test

1. Solve x + 2y = 5 and xy = 3.
2. Solve 4 − 3x > 16.
3. Solve x² − x − 20 < 0.
4. Solve x² − 11x + 24 ≥ 0.
5. Factorise 4x³ − 9x.
6. Expand (x − 2)(x² + 2x + 4).
7. For f(x) = x³ − x, find f(2) and expand f(x − 1).
8. State the asymptotes of y = 3/x² − 4 and where it meets the x-axis.
9. The point (2, −5) lies on y = f(x). Where is its image on y = f(x − 3) + 1? (Apply the two changes in turn.)
10. State the period of y = tan 2x in degrees.
11. Describe the transformation from y = x³ to y = (x − 4)³.
12. Is the boundary of y ≥ x − 2 solid or dotted?

Answers

1. x = 5 − 2y, so (5 − 2y)y = 3 and 2y² − 5y + 3 = 0, giving y = 1 or 3/2. Solutions: (3, 1) and (2, 3/2).
2. −3x > 12, so x < −4.
3. (x + 4)(x − 5) < 0, so −4 < x < 5.
4. (x − 3)(x − 8) ≥ 0, so x ≤ 3 or x ≥ 8.
5. x(2x − 3)(2x + 3).
6. x³ − 8.
7. f(2) = 6; f(x − 1) = (x − 1)³ − (x − 1) = x³ − 3x² + 2x.
8. Asymptotes x = 0 and y = −4; 3/x² = 4 gives x = ±√3/2 (about ±0.866).
9. f(x − 3) moves it right 3 to (5, −5); the +1 raises it to (5, −4).
10. 90°.
11. Translation 4 units to the right (vector (4, 0)).
12. Solid, because ≥ includes the line.

Where marks are usually lost

• Writing x-values only, without the matching y-values, in simultaneous equations.
• Not reversing the inequality sign after dividing by a negative.
• Cancelling or multiplying by x in a fraction inequality, which can lose solutions or reverse the sign.
• Giving a quadratic inequality answer as one long range when the region is two separate parts.
• Leaving out intercepts or asymptote equations from a sketch that asks for them.
• Moving f(x + 3) to the right.
• Applying f(2x) as a doubling of x-coordinates rather than halving.
• Forgetting that a double root touches rather than crosses the axis.
• Treating a dotted boundary as part of the region.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification, Issue 3, April 2019, published by Pearson Education Limited. Unit P1, topics 1.6 to 1.12.