Cambridge IGCSE™ Maths
Chris Pearce, Andrew Milne
Section 1: Number
Section 2: Algebra and graphs
Section 3: Coordinate geometry
Linear graphs 1
graphs 2
Section 4: Geometry Geometric terms and constructions
Section 5: Mensuration
Section 6: Trigonometry
Section 7: Transformations and vectors
Section 8: Probability
Introduction to probability
Probability of combined events
Section 9: Statistics
Mixed exam-style questions: Core
Mixed exam-style questions: Extended
Practice Paper 1: Non-calculator (Core)
Practice Paper 2: Non-calculator (Extended)
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179
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195
Practice Paper 3: Calculator (Core) p. 204
Practice Paper 4: Calculator (Extended) p. 213
List of formulas (Core) p. 224
List of formulas (Extended) p. 225
Answers
Glossary/Index
Proportion
The mass of a coin is directly proportional to its diameter.
A coin with a diameter of 24 mm has a mass of 7.2 g.
Find the mass of a coin with a diameter of 20 mm.
y is directly proportional to √2x + 5 When x = 2, y = 60.
Find y when x = 10.
Show me
Write the inequality as an equation: y = k√
Substitute values of x and y: 60 = k√
So k = and the equation is y =
When x = 10, y = ............
y ∝ x 3 Find the values of a and b in this table. x 45 b
y 960 a 7680
y is inversely proportional to x 2 . When x = 0 . 2, y = 2 . 5.
a Find y when x = 0 4.
b Find x when y = 0 4.
y ∝ 1 5x 1 . When x = 1, y = 10.
Find y when x = 5.
y is inversely proportional to the square root of x.
When x = 2, y = 12.
Find the exac t value of y when x = 16.
Graphs in practical situations
This graph can be used to convert from km/h to m/s.
a Use the graph to convert 54 km/h to m/s.
b Use the graph to convert 10 m/s to km/h.
c Convert 60 m/s to km/h.
This graph shows the journey of a cyclist from Acton to Corlon via Byfield.
a Write down the time the journey started.
b Find the average speed of the journey from Acton to Byfield.
c How long did the cyclist stop at Byfield?
d Work out the average speed from Byfield to Corlon.
e Work out the average speed for the whole journey.
This graph shows the speed of a car over a 15-second interval.
b Work out the acceleration for the last 10 seconds.
c Calculate the distance travelled.
Graphs of functions
a Copy and complete this table of values of y = 8 x 2
x –4–3–2–101234
y –1 7
b Use your table to draw a graph of y = 8 x 2 .
Use a scale of 1 cm to 1 unit on the x-axis and 1 cm to 2 units on the y-axis.
c On the same axes draw the line y = x
d Use the graph to solve the equation 8 x 2 = x
a Copy and complete this table of values of y = 10 x x 12345678910 y 2.51.671.43
b Draw a graph of y = 10 x for 1 ≤ x ≤ 10. Use a scale of 1 cm to 1 unit on each axis.
c On the same axes draw a graph of y = 5 x .
a Copy and complete this table of values of y = x 3 x 2 6x x –3–2–101234 y 4 –6
b Use your table to draw a graph of y = x 3 x 2 6x.
Use a scale of 2 cm to 1 unit on the x-axis and 1 cm to 5 units on the y-axis.
c On the same axes draw the line y = 3x + 6.
d Use your graph to solve the equation x 3 = x 2 + 9x + 6.
a Copy and complete this table of values of y = x + 2 √x x 012345 y 4.83 9.47
b Use your table to draw a graph of y = x + 2 √x .
Use a scale of 2 cm to 1 unit on the x-axis and 1 cm to 1 unit on the y-axis.
c Use the graph to solve the equation √x = 3 . 5 0 . 5x.
Sketching graphs
a Sketch the graph of y = 10 2x.
b Find the equation of this graph.
y
a Solve the equation x 2 + 2x 15 = 0.
b Write x 2 + 2x 15 in completed square form.
c Sketch the graph of y = x 2 + 2x 15. 0 A
The equation of this curve is y = x 2 8x + 7.
Find the coordinates of A, B, C and D.
a Factorise x 3 4x as much as possible.
b Sketch the graph of y = x 3 4x
a (10, a) and (20, b) are on the graph of y = 3 + 4 x . Find the values of a and b
b Find the asymptotes of y = 3 + 4 x .
c Sketch the graph of y = 3 + 4 x . Sketch the graph of y = 2 x for 2 ≤ x ≤ 2.
Differentiation
a Draw the curve y = 6 x for 1 ≤ x ≤ 6.
Use a scale of 1 cm to 1 unit on each axis.
b Draw the tangent to the curve at (3, 2).
c Use the tangent to estimate the gradient of the curve at (3, 2).
The equation of a curve is y = x 3 2 x 2 + 4x.
a Find dy
dx .
b Find the gradient of the curve where it crosses the y-axis.
The equation of a curve is y = 5x x 2 .
Find the equation of the tangent to the curve at the point where it crosses the positive x-axis.
Where the curve crosses the x-axis 5x x 2 = 0
Factorise, x( ) = 0 so x = …. or ….
The point required is (… , …)
dy
dx =
The gradient at (… , …) is …..
The equation of the tangent is …….
The equation of a curve is y = x 3 6 x 2 + 10x.
a Find dy
dx
b Find the coordinates of the two points on the graph where the gradient is 1.
The equation of a curve is y = x 3 3 x 2 + 3.
a Find dy dx
b Find the gradient at (1, 2)
c Find the coordinates of the stationary points.
d State whether each stationary point is a maximum of a minimum. Give reasons for your answers.
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Functions
f(x) = 12 x + 2
a Find f(4).
b Find ff(3).
c Find f −1(x).
d When the domain of f(x) is {x : x ≥ 1} find the range of f(x)
f(x) = 10 2x and g(x) = (x 3)2 .
a Find f(3).
b Find fg(6)
c Find f −1(x)
d Solve the equation g(x) = 4.
f(x) = 3x + 4 and g(x) = 5x 2.
a Find gg(2).
b Find fg(x). Write your answer as simply as possible.
c Find fg(x) + gf(x).
d Solve the equation f −1(x) = 3.
h(x) = 10 x 3
a Find h( 7).
b Find h −1(x).
c Solve the equation h(x) = x
f(x) = x 2 5 and g(x) = 2x + 3.
a Find gf(4)
b Solve the equation f(x) = 4.
c Solve the equation f(x) = g(x)
Linear graphs 1
Learning aims:
• Draw straight-line graphs
• Find the gradient and the equation of a straight-line graph
• Find equations of parallel lines
The equation of a straight line
y = 2x 1 is the equation of a straight line.
Here is a table of values:
x –2–1012345
y –5–3–113579
Here is a graph:
The gradient of the line is 4 2 = 2.
The y-intercept where the line crosses the y-axis is –1.
Both these numbers are in the equation y = 2x 1
The gradient is 2.
The y-intercept is –1.
Syllabus links: C3.1; E3.1; C3.2; E3.2; C3.3; E3.3; C3.5; E3.5; C3.6; E3.6
Example 1
a) Find the gradient of this straight line.
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b) Write down the y-intercept.
c) Find the equation of the line.
a) gradient = 3 6 = 1 2
Use the triangle formed by the axes.
b) The y-intercept is 3. Where the line crosses the y-axis.
c) y = 1 2 x + 3 y = (gradient) x + (y-intercept)
Key Point
y = mx + c where m and c are numbers is the equation of a straight line. The gradient is m and the y-intercept is c.
Parallel lines
Parallel lines have the same gradient.
= 2 3
All these lines have a gradient of 2 3 . They are parallel.
Example 2 Find the equation of the line parallel to y =3x +5 that passes through (4,8).
The gradient is 3. The coefficient of x
The equation is y =3x + c. The equation always has this form.
8=3×4+ c Substitute the coordinates (4,8).
c =8−12=−4
The equation is y =3x −4. Solve the equation to find c.
Quick Test
1. Find the gradient of the line y =6x −8. 2.a) Use this table of coordinates to draw a straight line.
b) Find the gradient of the line.
c) Find the equation of the line parallel to y =− x that passes through (4,2).
E Linear graphs 2
Learning aims:
• E Straight-line graphs and their equations in various forms
• E Gradient, length and midpoint of a line segment
• E Parallel and perpendicular lines
Equation of a straight line graph
y = 2 3 x + 4 is the equation of a straight line with a gradient of 2 3 and a y-intercept of 4.
The equation can be written in different ways such as:
y = 4 2 3 x or 3y = 2x + 12 or 2x + 3y = 12
Example 1 The equation of a line is 4x + 5y = 20.
a) Draw a graph of the line.
b) Find the gradient of the line.
a) If x = 0 then 5y = 20 and y = 4 The easiest way is to find where the line crosses each axis.
Syllabus links: E3.1; E3.2; E3.3; E3.4; E3.5; E3.6; E3.7
(0, 4) is on the line.
If y = 0 then 4x = 20 and x = 5 (5, 0) is on the line.
b) 4x + 5y = 20
5y = 4x + 20 y = 4 5 x + 20
Line segments
The gradient of the line through A and B is:
You can find the gradient from the graph but you can also find it by rearranging the equation to make y the subjec t.
0
a b = difference between y-coordinates difference between x-coordinates = 2 6 9 3 = 8 6 = 4 3
Using Pythagoras’s theorem, the length of the line segment is √a 2 + b 2 = √8 2 + 6 2 = 10
The midpoint of the line segment = ( 3 + 9 2 , 6 + 2 2 ) = (6, 2)
• The gradient of the line through A and B is y 2 y 1 x 2 x 1
• The length of the line segment AB is √(x 2 x 1)2 + (y 2 y 1)2
• The midpoint of AB is ( x 1 + x 2 2 , y 1 + y 2 2 )
Parallel and perpendicular lines
The line L has the equation y = mx + c. The gradient of the line is m.
A line parallel to L has the same gradient, m
A line perpendicular to L has the gradient 1 m
Example 2 A line L has the equation 2x + 5y = 40.
Find the equation of the line perpendicular to L passing through (3, 4)
Key Point
If A(x 1 , y 1) and B(x 2 , y 2) are two points:
5y = 40 2x ⇒ y = 8 2 5 x Rewrite the equation to make y the subject Here are the two lines in the worked example.
Gradient of L = 2 5
Gradient of perpendicular line = 5 2 = 2 . 5 The coefficient of x. y = 2 . 5x + c
4 = 2 . 5 × 3 + c The equation of the perpendicular line c = 4 7 . 5 = 3 . 5 Substitute the coordinates and find c.
The perpendicular line is y = 2 . 5x 3 . 5 This could be written as 2y = 5x 7
Quick Test
1. Find the gradient of the line through ( 6, 0) and (0, 9).
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2. Find the gradient of the line with equation 6x + 5y = 12.
3. Find the equation of a line perpendicular to y = 10 2x that crosses the x-axis at (6, 0).
4. P is ( 4, 2) and Q is (1, 14) a) Find the length of PQ. b) Find the midpoint of PQ.
Linear graphs 1
a Find the gradient of this line.
b Find the equation of the line.
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Two points on the line are (0, 1) and . . . . .
The gradient is . . - 1 .
- 0 = . . .
c Find the equation of the parallel line through (0, 4). x 137 y 53–1
a Use this table of coordinates to draw a straight line.
b Find the gradient of the line.
c Find the y-intercept.
d Write down the equation of the line.
The equation of a straight line is y = 4x + 10.
a Find the gradient of the line.
b Find the coordinates of the point where it crosses the y-axis.
c Find the equation of a parallel line through the origin (0, 0).
d Find the equation of a parallel line through (2, 0).
a Find the equation of a straight line with a gradient of –2 and a y-intercept of 8.
b Find the equation of a parallel line with a y-intercept of –4.
c Find the equation of a parallel line through (1, 2).
a Write down the y-intercept of this line.
b Find the gradient.
c When the line is extended it passes through (10, a). Find the value of a
5
Linear graphs 2
A is ( 10, 2) and B is ( 1, 14).
a Find the gradient of AB.
b Find the length of AB.
Use Pythagoras’ theorem to find the length.
c Find the midpoint of AB.
1 1 0 23 3 2 1 1 2 3 y x
a Find the gradient of this line.
b Find the equation of this line.
c Find the equation of a perpendicular line through (2, 3)
The equation of a line is 2x + 3y = 15.
a The point (a, 8) is on the line. Find the value of a
b Find the equation of a parallel line through (6, 4)
c Find the equation of a perpendicular line through (6, 2)
The line 3x + 4y = 18 crosses the x-axis at A and the y-axis at B.
a Find the coordinates of A and B.
b Find the length of AB.
c Find the gradient of AB.
d Find the equation of the line perpendicular to AB and passing through B.
A line segment joins A (3, -5) and B (6, 7)
a Find the midpoint of the line segment AB
b Find the equation of the straight line through A and B.
6
The coordinates of P are (2, 5)
The coordinates of Q are (8, 1)
Work out the equation of the perpendicular bisector of PQ.
Similarity
Learning aims:
• Calculate length in similar shapes
• E Understand the relationship between lengths and areas in similar shapes
• E Understand the relationship between lengths and volumes in similar solids
Similar shapes
These two shapes are similar
Corresponding angles are the same.
The ratio between corresponding sides is always the same:
A 2 B 2
A 1 B 1 = 12 8 = 1 5
B 2 C 2
B 1 C 1 = 21 14 = 1 . 5
C 2 D 2
C 1 D 1 = 15 10 = 1 . 5
A 2 B 2 C 2 D 2 is an enlargement of A 1 B 1 C 1 D 1 with a scale factor of 1.5
Example 1
a) Explain why these triangles are similar.
b) Find x.
c) Find y.
a) The angles of the triangles are the same. This is sufficient to show that to triangles are similar.
b) XY AB = 25 20 = 1 . 25 XY and AB are corresponding sides because they are both opposite 42°.
Syllabus links: C4.4; E4.4
x = 1 25 × 30 = 37 5
Solve the equation for x
c) YZ BC = 30 y = 1 . 25 YZ and BC are corresponding sides.
y = 30 1 . 25 = 24
Solve the equation for y
E Similar areas
If two shapes are similar with a length scale factor of k then the area scale fac tor is k 2 .
The area of A 1 B 1 C 1 D 1 in the first diagram is 88 cm 2 .
The linear scale fac tor is 1.25 so the area scale factor is 1. 25 2
The area of A 2 B 2 C 2 D 2 is 88 × 1. 25 2 = 198 cm 2
The area of triangle ABC in the first worked example is 240 cm 2 .
The linear scale fac tor is 1.25
The area of DEF is 240 × 1. 25 2 = 375 cm 2 .
E Similar volumes
Two solids are similar when the ratio of corresponding sides are the same. One is an enlargement of the other.
Example 2
These two solids are similar. Complete this table.
The linear scale factor = 24 ÷ 20 = 1 . 2 The heights are lengths. Area scale factor = 1. 2 2 = 1 . 44
Area of base of B = 300 × 1 44 = 432 cm 2 Multiply by the area factor.
the volume of A.
Quick Test
1. The shor test side of a triangle is 24 mm and the longest side is 38 mm. The shor test side of a similar triangle is 84 mm. Find the longest side.
E 2. A trapezium 32 cm wide has an area of 1440 cm 2 Find the area of a similar trapezium 18 cm wide.
E 3. A pyramid 15 cm high has a volume of 800 cm 3
Find the volume of a similar pyramid with a height of 45 cm.
Key Point
If the ratio of corresponding sides in two similar solids is k then the ratio of corresponding areas is k 2 and the ratio of the volumes is k 3 .
Symmetry
Learning aims:
• Recognise lines of symmetry and rotational symmetry in two dimensions
• E Recognise planes of symmetry and rotational symmetry in three dimensions
Symmetry of shapes
Shapes can have reflection symmetry or rotation symmetry.
Here are the symmetries of some common shapes.
Isosceles triangle
1 line of symmetry no rotational symmetry
Square 4 lines of symmetry rotational symmetry of order 4
Kite
1 line of symmetry no rotational symmetry
E Symmetry of solids
Solids can have:
• reflection symmetry with a plane of symmetry
• rotation symmetry with an axis of symmetry
Syllabus links: C4.5; E4.5
Equilateral triangle 3 lines of symmetry rotational symmetry of order 3
Rhombus
2 lines of symmetry rotational symmetry of order 2
Regular pentagon
5 lines of symmetry rotational symmetry of order 5
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Rectangle 2 lines of symmetry rotational symmetry of order 2
Parallelogram no lines of symmetry rotational symmetry of order 2
Circle every diameter is a line of symmetry
Circle theorems
Learning aims:
• Use geometrical properties of circles to find angles
• E Use symmetry properties of circles
Semicircles and tangents
O is the centre of the circle.
AB is a diameter.
ST is the tangent at A.
C is an angle in a semicircle.
• The angle in a semicircle at C is 90°.
• The angle between the tangent AT and the diameter AB is 90°.
E Geometrical properties
O is the centre of the circle.
POQ is the angle at the centre.
PRQ is the angle at the circumference.
The angle at the centre is twice the angle at the circumference:
y = 2x
ABCD is a cyclic quadrilateral.
The vertices are all on a circle.
The sum of opposite angles of a cyclic quadrilateral is 180°:
Syllabus links: C4.7; E4.7; E4.8
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a + c = 180 and b + d = 180 S
A, B and C are angles at the circumference from P and Q
The angles at A, B and C are all equal.
Angles at the circumference in the same segment are equal.
ST is a tangent to the circle at P
Angles TPQ and PRQ are angles in the same segment.
Angles in the same segment are equal.
