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Title: maths olevel notes with solutions
Description: maths notes with worked examples
Description: maths notes with worked examples
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Mathematics IGCSE notes
Index
1
...
Accuracy and Error
3
...
Ratio & proportion
click on a topic to
visit the notes
5
...
Percentages
7
...
Algebra: simplifying and factorising
9
...
Rearranging formulae
11
...
Parallel lines, bearings, polygons
13
...
Trigonometry
15
...
Similar triangles, congruent triangles
17
...
Loci and ruler and compass constructions
19
...
Straight line graphs
21
...
Distance, velocity graphs
23
...
Graphical transformations
25
...
Statistical calculations, diagrams, data collection
27
...
Calculus
29
...
Decimals and standard form
top
(a) multiplying and dividing
(i) 2
...
36 Move the decimal points to the right until each is a whole
number, noting the total number of moves, perform the multiplication, then
move the decimal point back by the previous total:
→ 25 × 136 = 3400 , so the answer is 3
...
00175 ÷ 0
...
→ 1
...
In this case, both numbers share a 7, so divide this out: → 0
...
0416
6 0
...
0416
(iii) decimal places
To round a number to n d
...
, count n digits to the right of the decimal point
...
e
...
round 3
...
p
...
012678 → 3
...
013 to 3 d
...
(iv) significant figures
To round a number to n s
...
, count digits from the left starting with the first
non-zero digit, then proceed as for decimal places
...
g
...
85 to 3 s
...
, 3109
...
85 so 3110 to 3 s
...
e
...
round 0
...
f
...
0030162 → 0
...
00302 to 3 s
...
(b) standard form
(iii) Convert the following to standard form: (a) 25 000
(b) 0
...
So 25000 = 2
...
0000123 = 1
...
4 × 105 ) × ( 3
...
4 × 3
...
4 × 1011 = 1
...
2 ÷ 2
...
2 × 1012
2
...
28 × 10 9
Again, rearrange the calculation to
(vi) adding/subtracting in standard form: (2
...
75 × 107 ) The
hardest of the calculations
...
e
...
= (0
...
75 × 107 ) = 4 × 107
Questions
(a) 2
...
5
(b) 2
...
015
(c) Convert into standard form and multiply: 25 000 000 × 0
...
6 × 103 ) ÷ (2 × 10−2 )
(e) (1
...
5 × 10−4 )
Answers
(a) → 254 × 15 = 3810 , so 2
...
5 = 3
...
55 ÷ 0
...
Notice a factor of 5, so let’s cancel it first:
= 510 ÷ 3 = 170
(c) = (2
...
4 × 10−10 ) = 6 × 10−3
(d) = (2
...
3 × 105
(e) = (1
...
25 × 10−3 ) = 1
...
Accuracy and Error
top
To see how error can accumulate when using rounded values in a calculation,
take the worst case each way: e
...
this rectangular space is
3m
measured as 5m by 3m, each measurement being to the nearest
metre
...
5m and the width as low as
2
...
So the smallest possible area is 4
...
5 = 11
...
Now, the length
could be anything up to 5
...
5m itself (which
would be rounded up to 6m) So the best way to deal with this is to use the
(unattainable) upper bounds and get a ceiling for the area as
5
...
5 = 19
...
Then these two facts can be expressed as 11
...
25m2
...
5cm by 5
...
0cm, each to the nearest 0
...
What is the volume of the block?
(b) A runner runs 100m, measured to the nearest metre, in 12s, measured to
the nearest second
...
0, b = 2
...
f
...
45 × 4
...
95 = 241
...
upper bound volume = 2
...
05 × 20
...
193875 cm3
...
943625cm3 ≤ volume<258
...
99
...
5
So
, i
...
7
...
739…ms-1
< speed <
12
...
5
(b) Since speed =
(c) for the smallest value of a − b , we need to take the smallest value of a
together with the biggest value of b, etc
...
4 < a − b < 0
...
95 − 2
...
05 − 2
...
e
...
Powers and roots
top
1) x a × x b = x a + b
2) x a ÷ x b = x a −b
3) ( x a )b = x ab
1
4) x − a = a
x
0
5) x = 1
(a) whole number powers
Note that the base numbers (x’s) have to be the same;
25 × 32 cannot be simplified any further
...
g
...
001
3
10 1000
5) 100 = 1
...
g
...
6
Q
...
1
(a) 16 2 =
16 = 4
1
(b) 64 3 =
3
64 = 4
3
(c) 4 2 = ( 4)3 = (2)3 = 8
3
(d) 81 4 = ( 4 81)3 = (3)3 = 27
2
(e) easier to use power law (3) above: ( x 6 ) 3 = x
7
6×
2
3
= x4
top
4
...
If quantities are linearly related,
either directly or inversely, (like number of workers and time taken to do a job),
calculate by multiplying by a ratio:
e
...
If 8 workers can together do a job in 6 days, how long would the same job take
with 12 workers?
8
ans: it will take less time, so we multiply by the ratio 12
...
e
...
If a workforce of 20 can produce 12 cars in 15 days, how many workers
should be used if 15 cars are needed in 10 days?
15
15
ans:
no
...
2
(b) Proportion
Where quantities are related not necessarily linearly
...
y ∝ x 2 , and you are given that y is 7
...
eg
Rewrite as
y = kx 2 , and substitute the given values to find k:
7
...
2
...
2 x 2 ,
and any problems solved
...
e
...
If y is inversely proportional to the cube of x,
1
then
y∝ 3
x
k
Rewrite as
y = 3 , and proceed as usual
...
e
...
Radiation varies inversely as the square of the distance away from the source
...
What is the radiation
at 50m away?
ans: as distance increases by a factor of 5, radiation must decrease by a factor of 52 ,
so the radiation is 75 ÷ 25 = 3
...
(a) Water needs to be removed from an underground chamber before work can
commence
...
If the water is now at a depth of 5m (same crosssection), and you want to empty the chamber in 10 hours time, how many pipes need
to be used?
(b) y is proportional to x 2 and when x is 5 y is 6
...
64
(c) The time t seconds taken for an object to travel a certain distance from rest is
inversely proportional to the square root of the acceleration a
...
What is the value of a if the time taken is 5 seconds?
Answers
5 4
(a) No
...
(b)
y ∝ x2
y = kx 2
and we know when x is 5, y is 6, so
6 = k × 52 , so k =
y=
6
25
6
25
, and we can write the relationship as
x2
...
6
(ii) When y is 8
...
64 = 25 x 2 , so
25 × 8
...
44 , so x = 5 ×1
...
1
,
a
k
So t =
...
a
4
4
16
or 0
...
When t = 5, 5 =
, so a = , and a =
5
25
a
(c) t ∝
9
5
...
g
...
Convert to vulgar form first:
− ,
6 3
6 3
then find the lowest common denominator, in this case 6
...
6 3
6
6
2
1 7
16 7
(ii) Multiplying/dividing: e
...
5 ×
...
3 8
3
3 ×1
3
8
/
To divide, turn the ÷ into a × and invert the second fraction
...
g
...
075
3
is 0
...
40 3
...
075 as a fraction? 0
...
1000
40
(b) Ratios
(iv) To divide a quantity into 3 parts in the ratio 3: 4:5, call the divisions 3
parts, 4 parts and 5 parts
...
(v) To find the ratio of several quantities, express in the same units then cancel
or multiply up until in lowest terms e
...
what is the ratio of 3
...
25m to
75cm?
Perhaps metres is the best unit to use here, so the ratio is 3 :2
...
75
...
So the ratio is 12 : 9 : 3, and we can now cancel down to 4:3:1
10
Questions
3
5
(a) (2 ) 2 × 1
4
11
1 3
1
(b) (1 − ) ÷ 2
3 5
5
(c) What is 0
...
Each part is £5000 ÷ 8 = £625 , so the
£5000
£5000 splits into £625, £1250, and £3125
...
Percentages
top
(i) What is 75g as a percentage of 6kg? Express as a fraction, then multiply by
75
100 to covert to a percentage
...
6000
60
4
(ii) Find 23% of 3
...
This is
23
23
× 3
...
736kg
...
The original amount is always regarded as 100%,
and this problem wants to find 112%
...
This can be
112
accomplished in one go, however, by multiplying by
, i
...
1
...
100
So the answer is £20 × 1
...
40
...
This means we are trying to find 88% of the
original, so the answer is £20 × 0
...
60
...
The 20% refers to 20% of the original amount which
we don’t know, not 20% of £6000
...
The information says that
6000
£ x × 1
...
1
...
e
...
e
...
A coke can advertises 15% extra free, and contains 368ml
...
2 ml
...
2 = 48 ml
...
6 million
...
My account has £27000 this
year
...
Big Ears can buy 24 toadstools
for £1 this year
...
38
100
1000
(c) (i) £27000 × 1
...
0810 = £58290
...
(iii) £ x × 1
...
08
(d) Inflation at 4% per year means that if you pay £100 for some goods this
year, the same goods will cost you £104 in next years’ money
...
045 = £1
...
in 5 years’ time, and so £1 will
1
buy him 24 ×
, i
...
19
...
2166529
...
Rational and irrational numbers
top
a
where a and b are
b
integers
...
Fractions, integers, and
2
recurring decimals are rational
...
25, 3 8
...
1234
...
A rational number is one which can be expressed as
a
(to confirm they really are rational)
b
125
1
A terminating decimal: 0
...
123
...
123123123
...
123123
...
123123
...
=
999
333
(i) Converting rationals to the form
(ii) rationalising a denominator:
2
3
has a 3 in the denominator, so multiply top and bottom by
3
does not change the value of the expression, only the shape):
2
3
6
3
×
= 3
= 6
...
To simplify expressions using these:
200 = 100 × 2 = 100 × 2 = 10 2
18
18
=
= 9 = 3
2
2
(iv) Finding irrational numbers in a given area:
e
...
find an irrational number between 5 and 6
...
g
...
(Or say π + 2 for example)
...
375
6
(ii)
(iii) 72
2
2
(c) Find an irrational number between 1 and 1
...
36
(iv)
3
250
Answers
(a) (i) 0
...
36363636
...
363636
...
363636 = 36 + x
...
g
...
3,
101
etc
10
15
36
4
=
99
11
8
...
g
...
These are −4
and +1
...
(multiply out the answer to check!)
(d) 2 x 2 + 9 x + 4 (full quadratic with more than one x2): multiply the 2 by the 4
to get 8, and repeat the previous process i
...
look for two numbers which
multiply to 8 and add up to 9
...
Now split the middle term
accordingly and group into 2 pairs:
2 x 2 + 9 x + 4 = 2 x 2 + 8 x + x + 4 = (2 x 2 + 8 x) + ( x + 4) Then factorise each
group, = 2 x ( x + 4) + ( x + 4) , and notice the bracket factor which you now
extract: = ( x + 4)(2 x + 1)
...
= (ab + ac) − (b 2 + bc) = a(b + c) − b(b + c) ,
and there just happens to be a big factor: = (b + c )( a − b)
16
Questions
(a) Simplify a (b − c ) + b(c − a ) + c ( a − b)
(b) Factorise (i) 4 p 2 q − 6 pq (ii) 2 x 2 + 6 x
(v) 3 x 2 + 11x + 6
(iv) x 2 + 10 x + 21
(iii) 4 x 2 − 1
(vi) 2ab − 6ac + b − 3c
Answers
(a) = ab − ac + bc − ab + ac − bc = 0
(b) (i) = 2 pq (2 p − 3) (ii) = 2 x ( x + 3)
(iii) = (2 x − 1)(2 x + 1)
(iv) = ( x + 7)( x + 3) (v) 3 × 6 gives 18, so 9 and 2 are the required numbers:
3x 2 + 9 x + 2 x + 6 = (3x 2 + 9 x) + (2 x + 6) = 3 x( x + 3) + 2( x + 3) and finally
= (3 x + 2)( x + 3)
...
Equations: linear, quadratic, simultaneous
top
(a) Linear
Perform the same operation on both sides to isolate x:
2x
1
[ ×6 ]
+x=
3
2
4x + 6x = 3
10 x = 3
[ ÷10 ]
3
...
If it can be factorised, do so (see 8
...
Then:
(2 x − 1)( x + 3) = 0 means one of the brackets must be 0, so
2 x − 1 = 0 or x + 3 = 0 , which can be solved to give
1
x = , −3
2
If not, use x =
−b ± b 2 − 4ac
and round the answers suitably
...
2x − y = 7
multiply equation 1 by 2, then add:
3x + 2 y = 5
4 x − 2 y = 14
3x + 2 y = 5
7 x = 19
solve and substitute back in to equation 1 to find y
...
18
one linear, one quadratic:
x 2 + y 2 = 25
x + y = 0
...
8 − x , so x 2 + (0
...
Multiply out, and solve the quadratic in x
...
5
(a)
[ ×6 ]
→ 2 x − 3 + 3x = 6
(b) x 2 + 2 x − 15 = 0
( x + 5)( x − 3) = 0
x = −5, 3
...
These are 4, -3
...
2
19
1
= 2 [ ×x ]
x
x2 − 1 = 2x
x2 − 2x − 1 = 0
(d) x −
2 ± (−2) 2 − 4 × 1× −1
2± 8
x=
{Note that 8 = 4 × 2 = 2 2
=
2
2 ×1
and so 2 can be cancelled}: = 1 ± 2 , so x = -0
...
41 to 2 d
...
(e)
x + 2y = 5
x 2 − y 2 = −3
rearrange equation1 : x = 5 − 2 y , and substitute:
∴ (5 − 2 y ) 2 − y 2 = −3
∴ 25 − 20 y + 4 y 2 − y 2 = −3
∴ 3 y 2 − 20 y + 28 = 0
∴ (3 y − 14)( y − 2) = 0
14
13
∴ y = , 2
...
3 3
20
10
...
g
...
[ −b ] gives ax = cd − b , and finally [ ÷a ] gives x =
a
(ii) with a variable which appears more than once, gather together and
factorise: e
...
ax = bx + c [ −bx ] gives ax − bx = c , factorising
c
gives ( a − b) x = c , then [ ÷(a − b) ] gives x =
...
Rearrange it to make F the subject
...
∴x =
1− b
−a
would also be correct, as top and bottom
{ Note that x =
b −1
are multiplied by –1}
(b)
21
11
...
Avoid it!
(b) quadratic
e
...
x 2 < 4 First treat like an equation and factorise if possible (formula
otherwise): x 2 − 4 < 0 , then ( x − 2)( x + 2) < 0
...
Draw a number line, and a sketch of the function (in this case a
“happy” parabola) which reveals the region in which x 2 − 4 < 0 :
−2 < x < 2
...
2
(c) 2 variable linear inequalities
e
...
3 x − 2 y ≥ 6
...
g
...
2
1
– 2–
–
–
–
y
1 2 3 x
1
2
3
4
The origin’s coordinates make 3 × 0 − 2 × 0 which is not ≥ 6 , so that side is
rejected:
2
1
– 2–
–
–
–
1
2
3
4
y
1 2 3 x
22
y
Questions
5
4
3
2
1
(a) Solve 2(1 − x ) < 6
(b) Solve 12 − x ≤ x 2
– 4 3 – 1
– – 2 1
–
– 2
(c) Find the 3 inequalities which identify this region:
– 3
– 4
1 2 3 4 5 x
Answers
[ ÷2 ]
[ + x, − 3 ]
(a) 2(1 − x ) < 6
∴1 − x < 3
x > −2
(b) 12 − x ≤ x 2
[rearrange]
2
x + x − 12 ≥ 0
∴ ( x + 4)( x − 3) ≥ 0 , giving critical values of –4 and +3
...
2
By considering a point (e
...
origin) in the shaded region, the inequalities are
1
y < 2 x + 1 , y > x − 2 , and x + y < 4
...
Parallel lines, bearings, polygons
top
(a) Parallel lines
alternate angles equal
corresponding angles equal
allied or interior add up to 180 °
N
(b) bearings
Bearings are measured clockwise from North:
bearing of B from A is 135º
A
45°
B
(c) polygons
for a polygon with n sides,
sum of interior angles = (n − 2)180 º
sum of exteriors = 360º
24
exterior angle
interior angle
Questions
a
(a) In the diagram opposite, find
the value of θ in terms of a and b
...
Given also that AB = BC, find the
bearing of C from A
...
Make a sketch of the pentagon, marking in the angles
...
ADE = θ (opposite)
...
A rearrangement gives θ = b − a
...
The bearing of C from A is therefore 105º
...
A line of symmetry means
the set up is like this:
The only way of allocating 100º and 120º to a, b, c
and make a total of 540º is to have three 100º’s
and two 120º’s
...
Areas and volumes, similarity
top
(a) Areas of plane figures
CIRCLE
TRIANGLE
h
r
C
b
2
1
bh
2
πr
B
a
A
b
1
absinC
2
or
TRAPEZIUM
PARALLELOGRAM
b
h
h
a
b
bh
1
(a + b)h
2
(b) Surface area and volume
Shape
surface area
volume
PRISM
Prism
A
A× l
p×l
Cylinder
2π rh
πr h
Cone
π rl
4π r
p
l
1 2
πr h
3
Sphere
r
2
h
CONE
SPHERE
1
× base area × h
3
Pyramid
Pipe flow: number of m3/s flowing through (or out of) a pipe
= cross-sectional area × speed
v
26
l
h
4 3
πr
3
2
CYLINDER
r
PYRAMID h
r
(b) Similarity
Enlargement scale factor =
k
Area scale factor =
k
2
Volume scale factor =
3
k
Questions
(a) A cylinder has volume 100cm3, and height 5cm
...
What is the volume of the base part? (frustum)
(c) The empty swimming pool shown opposite
is to be filled with water
...
How long will the pool take to fill?
25m
10m
1m
(d) Two blocks are geometrically similar,
and the big blocks weighs 20 times the
small block
...
52 cm
...
05cm to 3sf
{Note the pre-corrected value was doubled resulting in 5
...
04}
27
3m
(b) The upper small cone has base radius 5cm and height 10cm
...
1
So volume of pool = (1 + 3)25 × 10 = 500 m3
...
s
...
× speed = π 52 × 200 = 5000π cm3, which
is 5000π ÷ 106 m3/s
...
e
...
500 ÷ (5000π ÷ 106 ) =
π
(d) assuming same density material, weight is directly proportional to volume
...
e
...
∴ k = 3 20 , and so the ratio of surface
areas,1 : k 2 , is 1 : 7
...
Trigonometry
top
O
H
A
cosθ =
H
O
tan θ =
A
sin θ =
H
O
θ
A
Sine rule:
a
b
c
=
=
sin A sin B sin C
Cosine rule: a 2 = b 2 + c 2 − 2bc cos A
y
x
Two opposite pairs: use sine rule
y
x
Three sides and one angle: use cosine rule
z
Angle between line and plane is the angle between the line and its
projection on the plane: e
...
for the angle between this diagonal and
the base, draw the projection, and the angle is shown here:
Trigonometric functions for all angles:
1
y
1
y
90 180 270 360x
– 1
5
90 180 270 360x
90 180 270 360x
– 5
– 1
sinx
cosx
29
y
tanx
C
Questions
(a) ADB is a straight line of length 20cm
...
6cm
40°
A
D
θ
(b) In triangle ABC, AB = 5cm, BC = 8cm, and BCA = 30 º
...
How far and at what
bearing is it from its original point?
(d) Is an internal diagonal of a cube at 45º elevation from the base?
(e) Find two values of x in the range 0º to 360º for which sin x = −0
...
No, a decent diagram!
θ lies in the triangle on the right, and all the information
we have is in the left triangle
...
CD
, ∴ CD = 6sin 40º = 3
...
sin 40º =
6
AD
, ∴ AD = 6 cos 40º = 4
...
, so BD = 20 − 4
...
= 15
...
cos 40º =
6
3
...
Then tan θ =
= 0
...
, so θ = 14
...
f
...
4
...
1
...
1
...
9º
B
5 cm
A
N
(c) Using cosine rule,
A
N
OB 2 = 52 + 62 − 2 × 5 × 6 × cos135º
5
so OB = 10
...
f
...
7 º
=
6
10
...
The bearing of B from O is therefore 069
...
(Enlargement won’t make any difference to the
angles)
...
1
...
3º to 3 s
...
So tan θ =
2
(e) First find the principal value from
the calculator: -30º
...
5
– 1
y
-30
– 90
90 180 270 360 x
-0
...
So x = 210º, 330º
31
15
...
Find the area of the sector
...
How much oil is
contained in the cylinder?
(c) Find θ in the following diagrams:
θ
120°
(a)
(b)
(c)
35°
θ
θ
40°
Answers
(a) Arc length =
gives θ =
90
θ
360
× 2π 20 and this is given as 10cm
...
Therefore sector area =
π
which simplifies nicely to 100cm2
...
Above the oil is an isosceles triangle, so
split it down the line of symmetry:
50 cm
OIL
25 cm
This gives an angle of cos −1 0
...
3…cm by Pythagoras
...
Therefore the area of the segment
120
is
× π 50 2 − 43
...
× 25 = 1535cm2
...
307m3
...
θ
ˆ
C = 35 º (angles in the same segment)
ˆ
CAB = 90 º (angle in a semicircle)
so θ = 180 − 35 − 90 = 55 º (angle sum
of a triangle)
ˆ
ABT = 40 º (alternate segment theorem)
ˆ
Isosceles triangle gives BAT = 70 º, and
ˆ
so C = 70 º (angles in same segment)
C
35°
A
θ B
C
θ
A
B
T 40°
34
16
...
e
...
to prove congruent:
SSS
SAS
AAS
ASA
RHS
but not ASS – there are sometimes two different triangles with the same
ASS
35
Questions
(a) (i) Prove that triangles BCD and
ACE are similar
...
(iii)If the
A
area of triangle BCD is 12 what is the
ABDE?
D
E
6
B
4
6
8
C
BD and
area of the trapezium
(b) Use congruent triangles to prove that the diagonals of a
parallelogram bisect each other
...
The third angle is
shared, so AAA is established and they are similar
...
So BD = 6 ÷ , or
8 2
2
2
3
CE is 6 × = 9 , so DE is 9 – 6 = 3
6 × = 4
...
So the trapezium has area 27 – 12 = 15
...
So we have two
congruent triangles ABX and DCX by
D
C
ASA
...
So AX = XC and DX = XB, i
...
the diagonals bisect each other
...
Transformations
top
⎛ a⎞
(i) translation by vector ⎜ ⎟ shifts a to the right and b up
...
[Note e
...
+90º means 90º anticlockwise]
perform a rotation using compasses,
or if a multiple of 90º, use the L shape:
To find the centre of a rotation already performed, perpendicularly bisect
a line joining any point with its image
...
Alternatively, if it’s a 90º rotation, find the centre by trial and error then
confirm by using L shapes
...
To find the mirror line of a given
reflection, join a point to its image and
mark the mid-point
...
(iv) enlargement from P with a scale factor k
...
37
To find a centre of enlargement, join a point to its image and extend
the line back
...
Questions
y
(a) What single transformation will carry
triangle A onto (i) B (ii) C?
(b) A “glide reflection” is a reflection
followed by a translation
...
What is the vector
of the subsequent translation?
(c) E is transformed onto F: state the
single transformation which accomplishes
this
...
(Check with L shapes)
(ii) reflection through the line y = − x − 1
(b) The diagram shows A reflected to
A’
...
⎝1⎠
D
y
4
3
2
1
– 4– 3– 2– 1
– 1
– 2
– 3
– 4
(c) Draw lines joining points with
their images, and extend them downwards
...
So it’s an enlargement, centre (-2, -4) with scale factor ½
...
Loci and ruler and compass constructions
top
(a) In 2-D: locus of points equidistant from:
1 fixed point is a circle
2 fixed points A and B is the perpendicular bisector of AB
A
Circumcentre
3 fixed points A, B and C the circumcentre of ABC
B
C
(b) locus of points equidistant from:
2 fixed lines is the angle bisector
incentre
3 fixed lines is the incentre of the
triangle
Questions
(a) Construct the triangle ABC where AB = 8cm, BC = 5cm and
CA = 6cm
...
(b) In 3-D, describe the locus of points exactly 1cm away from the
nearest point on a line segment AB
...
Then note that the boundary
lines for the two requirements are the
ˆ
angle bisector of BAC and the perpendicular
bisector of AB, and the intersection of the
two regions must be selected
...
(or a hollow sausage, as we say in the trade)
...
Vectors
Vectors are most easily seen as journeys for a particular distance
in a particular direction
...
g
...
a + b : join the arrows nose to tail:
{Note in books and exam papers vectors
will be bold lower case letters without bars
...
3a
To get from A to B via given vectors, the route chosen
doesn’t matter - the expressions will all simplify down to
the same answer
...
The vectors
a , b , and c are defined as shown
...
a
B
A
b
c
C
D
X
→
(i) Find two different expressions for AX in terms of a , b , and c
...
→
B
→
(b) In triangle OAB, OA = a and OB = b
...
Find in terms of a and b :
→
→
b
O
a
A
→
→
(ii) OX
(iii) OK
(iv) AK
(i) AB
What does the final answer tell you geometrically?
Answers
→
(a) (i) Going via B and C we get AX = − a + c + 1 (2 a ) = − 1 a + c
4
2
→
Via D, however, we get AX = − b − 3 (2 a ) = − b − 3 a
...
So − 2 a + c = − b − 2 a ,
which simplifies to a + b + c = 0
...
This relationship can be seen
easily if we join A to the mid-point of CD and observe that there is a closed
triangle illustrating that a + b + c = 0 :
a
B
A
c
C
→
→
(b) (i) AB = − a + b
2
3
a+ 1b
3
b
c
Y
a
(ii) OX = a + 1 (− a + b ) , which simplifies to
3
→
(iii) OK = 3 ( 2 a + 1 b ) = a + 1 b
...
2
2
That AK is parallel to OB and half as long
...
Straight line graphs
top
y
gradient m =
∆y
∆x
∆y
∆x
x
y
Equation of a straight line through the origin,
gradient m, is y = mx
y = mx
x
y
Equation of a straight line gradient m and
y-intercept c is y = mx + c
c
y = mx + c
x
Equation of a straight line gradient m and
passing through (x1,y1) is y − y1 = m( x − x1 )
y
Intersecting lines: solve their equations
simultaneously to find the intersection
...
(c) Where do the lines y = 3 x − 5 and
3 x + 2 y = 6 intersect?
3
2
1
y
–– 11 1 2 3 x
(d) A is (2,3), B is (5,6) and C is (4,0)
...
42
Answers
(a) Rearrange: 2 x + 6 y + 12 = 0
[-2x, -12]
[ ÷6 ]
6 y = − 2 x − 12
12
1
y = −3 x−2
so the gradient is − 1 and the y-intercept is –2
...
3
The equation is y = − 2 x + 2
[+ 2 x ]
3
3
2
[ ×3 ]
3 x+ y = 2
2x + 3y = 6
(c) Solve simultaneously
...
Sub back into
9
1
16 1
(1) gives y =
...
3
9 3
6−3
1
= 1
...
5− 2
1
So the required equation is y = − x + c but what is c? Get this by
substituting the coordinates of a point on the line, i
...
C
...
More graphs
top
1
graphs of x, x 2 , x 3 , , k x and x 2 + y 2 = r 2
x
(a) x as above, linear
y
2
y =x – x +1
(b) x 2 parabolae
x
y
3
(c) x3 cubics(!)
x
y = x –x
y
(d)
1
hyperbolae
x
y =
x
2
x
4
3
2
1
(e) k x , where k > 0 and x is an integer
y
x
y =2
– 2 – 1
2 x
1
y
(f) x 2 + y 2 = r 2 circle radius r, centre origin
2
x
2
x + y = 1
Solving equations using graphs:
(i) Draw the graph of y = x 2 and on the same grid
y = 3
...
e
...
e
...
If y = x + 2 is drawn, the x-values at the
intersections are solutions to x 2 = x + 2 ,
i
...
x 2 − x − 2 = 0 , which could be factorised
to ( x − 2)( x + 1) = 0 , giving x = −1, 2 , which
can be seen on the graph
...
Questions
(a) Plot y = x 2 and y = 4 − x 2 on the same grid and find the x-values
of their intersections
...
State the number
of bacteria after (i) 1 day (ii) 2 days (ii) 3 days (iv) 4 days (v) x days
Sketch the graph of the number of bacteria against x, the number of days
after the start, for 0 ≤ x ≤ 5
...
Answers
(a) Intersection x-values are approx
...
4 and 1
...
At intersection, y = x 2 and y = 4 − x 2
...
(which means these two values
of x are actually ± 2 )
5
4
3
2
1
– 2– 1
– 1
y
2
y = x
y =4 –x
1
(b) At the two intersections, x = -0
...
6 (approx)
...
Therefore the equation representing x-values at intersection
is x 2 = x + 1 , i
...
x 2 − x − 1 = 0
...
7 days, (b) the gradient of the tangent
at x = 3 shows the rate of growth at that moment, and is about 550
bacteria/day
...
Distance, velocity graphs
top
{We are really dealing with displacement, i
...
how far along a certain route,
usually a straight line, from an origin
...
g
...
(b) Velocity – time
v
B
Gradient measured between A and B
= average acceleration
A
t
Gradient of a tangent
= acceleration at that point
Area between the curve and x-axis = displacement
{note: under the x-axis, area counts negative}
(c) Trapezium rule
an approximate method for counting
area under a curve:
y
d
Area ≈ { y0 + 2 y1 +
...
2
This replaces each strip with a trapezium, i
...
the top becomes a straight line segment, and will
under- or over- estimate the true area
...
Find
(i) its average acceleration over the first second
(ii) its instantaneous acceleration at t = 1
2
(iii) the distance it covers using the trapezium rule with 4 strips
...
Find the values of V and T
...
5
= 1ms-2
...
e
...
0
...
25
≈
{0 + 2 × 0
...
52 + 2 × 0
...
34375, or
2
0
...
f
...
34375 is an overestimate due to the
concave curve}
(b) Splitting into two trapezia and a triangle, area under curve
1
1
1 V
= (V + 2V )T + (2V + 1 V )T + T which = 3VT
...
Substituting gives
Acceleration on first part =
T
5T 2 = 20 which leads to T = 2, and V = 10
...
Sequences; trial and improvement
top
(a) Sequences
Numbers in a sequence u receive the names u1 , u2 , u3 ,
...
A sequence may be defined directly: un = 3n + 1 (that is 4, 7, 10, …
...
where u1 = 1, u2 = 1, and un = un −1 + un − 2
(b) trial and improvement
to solve an equation or maximise/minimise a quantity
e
...
Solve x 3 − x − 1 = 0 correct to 1
...
p
...
5
1
...
4
1
...
3 and 1
...
e
...
35 to indicate
...
3 and 1
...
3
48
Questions
(a) un = 3n − 7
...
(c) Find the number of straight lines joining n dots, and prove your formula
...
How many different ways are there
for the counter to get from the 1st square to the 10th square?
(e) Find, to 1 d
...
the value of x which minimises the function x 2 + 2 x
Answers
(a) (i) u10 = 3 × 10 − 7 = 23
...
e
...
So the number of lines between n dots is 1 n(n − 1)
2
(d) We are having to advance the counter 9 places
...
)
The first move is either a 1 or a 2, after which the number of ways
remaining to get to the end is un−1 or un − 2 respectively
...
Noting that u1 = 1
and u2 = 2 , the sequence must go 1, 2, 3, 5, 8, 13, 21, 34, 55, …
...
3
3
y
2
2
x
y = x + 2
1
1 x
– 1
x2 + 2x
0
...
0
...
0
...
0
...
x
-0
...
4
-0
...
2
So far, we are assuming there is a simple minimum, but all we know
is that it’s somewhere in the vicinity of –0
...
p
...
29
-0
...
27
x2 + 2x
0
...
0
...
0
...
We now know it’s between –0
...
29, so rounded to 1 d
...
the value of x is indeed –0
...
Graphical transformations
top
For any graph y = f ( x ) ,
y = f ( x − a)
→ +a
a translation of a steps in the + x direction
y − a = f ( x)
{i
...
y = f ( x ) + a }
↑ +a
a translation of a steps in the + y direction
x
y= f( )
a
↔ ×a
a stretch by factor a in the + x direction
×a
a stretch by factor a in the + y direction
y
= f ( x)
a
{i
...
y = af ( x ) }
e
...
1
– 1
y
1
180
360x
y
– 1
1
180
360x
x
y = sin
2
y = sinx
stretch by 2 in x
– 1
y
180
360x
y = sin 2x
stretch by 1/2 in x
multiple transformations:
this is truly tricky
...
g
...
So the original parabola must be moved 1 step
to the right then 3 steps up
...
Thus the transformations are: a
2
2
3
stretch in the x direction by factor 3, then a translation by +1 in the x direction,
and finally a stretch by factor 2 in the y direction, illustrated here:
2
1
y
2
1
–– 1 123x
2
y
2
1
––– 1123 x
32
⎛ x ⎞
⎟
⎝ 3 ⎠
2
y= ⎜
y =x
y
2
1
–– 1 123x
2
⎛ x – 1 ⎞
⎟
⎝ 3 ⎠
2
y = ⎜
y
–– 1 123x
2
2
⎛ x – 1 ⎞
⎟
⎝ 3 ⎠
y =2⎜
2
Questions
(a) The graph of y = cos x is shown
...
(i) Perform a stretch on this
by a factor of 2 in the y direction,
drawing the result on the empty grid
...
5
4
3
2
1
y
5
4
3
2
1
–– 1 1 2 3 4 x
2 1
–
y
–– 1 1 2 3 4 x
2 1
–
The ensuing line will have equation
y
= x + 1 , i
...
y = 2 x + 2 , and this
2
is confirmed by the diagram
...
Suppress the urge to divide x by 3 first (as you would do in
a calculation):
x
x
replace x by x +1: → y = ( x + 1) 2
...
So the transformations are:
3
translate by –1 in x direction, then stretch by factor 3 in the x direction, then
stretch by factor ½ in the y direction
...
Probability
top
One definition of the probability of an event is the limit to which
the relative frequency (no
...
of trials) tends as the
no
...
So 7 tails from 10 flips of a fair coin gives 0
...
However, one would expect by n = 1000 to have
converged more closely to 0
...
That would cast doubt on the fairness of the
coin
...
g
...
Pick one ball at random, keep
it out, then pick another ball
...
2 5
1 10
10
The probability is ×
which is
...
54
Questions
(a) Draw a table of results for the rolling of two dice
...
What is the probability that
(i) they are both boys (ii) there is at least 1 girl ?
(c) A game consists of three turns of an arrow which lands randomly
between 1 and 5, with the scores are added together
...
(i) P(difference = 2) =
1
1
B
2
3
4
5 6
1
2
3
8
2
...
(there’s an overlap)
...
gives =
36
(ii) P(total = 6) =
10 9
15
(ii) P(at least 1 girl) =
×
=
22 21 77
62
1 – P(both boys) =
...
) = ( )3 + ( )3 +
...
(ii) P(prize) = P(554 or 545 or 455 or 555) = ( )3 + ( )3 +
...
Statistical calculations, diagrams, data collection
top
(a) calculations
(i) averages:
mean =
∑x
i
n
median = value of the middle item when listed in order
mode = most commonly occurring value
(ii) measures of spread:
range = max – min
Interquartile range = Upper quartile – lower quartile
Quartiles in small data sets: fiddly and pointless, but here we go
...
If the number of data was even, split the data into two sets; if the
number of data was odd, ignore the median and consider the remaining
values as two sets
...
(b) diagrams
others
(i) pie chart
for categoric data (non-numerical) e
...
modes of
transport used to school
(ii) frequency diagram
gas
frequency
8
6
4
2
5 152535455565758595 no of runs
in a time sequence, the mean of the last 10
(say) values is calculated then plotted
...
(iv) scatter graphs
to see correlation between two variables
Latin %
(iii) moving average
90
85
80
75
70
65
60
56
60 65 70 75 80 85 90 95
Maths %
nuclear
oil
(v) stem and leaf diagrams
the data is transcribed straight from a
table onto the stems: this is a back-toback stem and leaf
...
5
8
12
……
x
0
...
freq
5
13
25
……
and the cum freq’s plotted at the right end of the interval
...
q
...
q
...
If the data is integer valued, the class
boundaries will be between integers
...
e
...
a random sample
of size 50 to be selected from the school population of
830: take a Ran# from calculator, ×1000 , discard if
over 830, otherwise choose that member of the list
...
stratified sampling – when the population is divided into
strata and you wish each stratum to be represented in
the sample proportionately to its size
...
e
...
in a prison in the age groups 18 - 25, 26 - 40, 41 –
60, and 61- 100 there are 100, 300, 250 and 150
inmates
...
From the 18 – 25 group you must take a random sample
100
of size
× 50 , i
...
6 (nearest integer), and so on
...
No
questions which could have a variety of possible
responses, better yes/no or tick boxes, or score on a
scale of 1 to 5, say
...
Functions
top
Functions are rules which require an input, x, and give a single output, f ( x) , (also
called y)
...
Domain
This is the set of input values
...
Tha natural domain of f ( x) = x − 3 is x ≥ 3 , since any values of x
below 3 do not give a real output
...
For f ( x) = x − 3 , the range is all the numbers between 0 and infinity,
i
...
f ( x) ≥ 0 (or y ≥ 0 )
...
For example, if f ( x) = 2 x + 1 and g ( x) = 3x − 2 ,
then g ( f (1)) is g (3) which = 7
...
Inverse
The inverse of a function, called f −1 ( x) , reverses the action of the
function
...
g
...
To find a formula for the inverse of f ( x) , call this y, and rearrange the
formula so that x is the subject
...
Check with
the above example,
So f −1 ( y ) =
59
9 10
x
f −1 (5) =
5 +1
= 3 , which is correct!
2
Questions
(a) Find the natural domains of :
(i) f ( x) = 3 x − 2
(iv) f ( x) =
1
2x − 2
(ii) f ( x) =
1
x
(iii) f ( x) = 5 − x
(v) f ( x) = x(4 − x)
(b) Find the ranges of all the functions in q
...
(d) Find the inverse of (i) f ( x) =
2x + 3
5
60
(ii) h( x) =
x−3
x +1
Answers
(a) (i) the entire real line
(ii) all real numbers except 0
(iii) x ≤ 5
(iv) x > 1
(v) the inside bit needs to be ≥ 0
...
-1
1
2
3
4 x
(b) (i) the entire real line
(ii) all real numbers except 0
(iii) y ≥ 0
(iv) y > 0
(v) 0 ≤ y ≤ 2
(c) g ( f ( x)) =
(5 x − 3) − 3
5x − 6
=
(5 x − 3) + 2
5x −1
2x + 3
5
5 y = 2x + 3
5 y − 3 = 2x
5y − 3
5x − 3
...
Why all these minuses? Let’s multiply top and bottom by -1, and
y −1
3+ x
...
Calculus
Differentiation
If y = f ( x) , then
dy
(or f '( x ) ) is the name of the gradient
dx
function of y
...
So, for
dx
example:
dy
(i) on the graph of y against x,
represents the gradient
...
dt
dP
is the rate of change of the
(iii) if P is the price of a share,
dt
share price
...
dx
At a max or min the gradient will be 0
...
g
...
dy
= 3x 2 − 3
dx
dy
when x = 1 ,
= 3 × 12 − 3 = 0 , so there is a stationary point
...
dx
dx
x
dy
dx
62
1-
1
1+
-
0
+
The diagram shows that we have a minimum
...
5
(iii) 5x 2
2
x +1
(v)
(vi)
x
x
(iv) 3 x ( x + 1)
(b) Show that the curve y = x3 + x has no stationary points
...
(d) The displacement of a toy car during the first 10 seconds after
t3
release is given by s = t 2 −
...
63
Answers
(iv) f ( x) = 3 x 2 + x so
−2
(v) f ( x) = 2 x −1 so f '( x) = 2 × −1x −2 , i
...
2
...
So
x
x
−1
f '( x) = −1x −2 , i
...
2
...
= 3 x 2 + 1
...
(b)
dy
= 3 x 2 + 6 x − 9
...
Therefore
dx
2
x + 2x − 3 = 0
( x + 3)( x − 1) = 0
∴ x = −3 or 1
...
Is (1,0) a max or min?
dy
Using
= 3( x + 3)( x − 1) ,
dx
11
1+
x
dy
+
0
dx
(c)
so we have a minimum
...
5
dt
22
= 3
...
5
(ii) To find the maximum value of v, we need to differentiate the
expression for v
...
dt
5
Putting this = 0 solves to t = 5
...
5
(i) After 2 seconds, v = 2 × 2 −
64
top
29
...
A small set may be shown on a Venn diagram, e
...
given the universal set is positive
integers less than 12, and A is the set of primes:
ξ
A
2 11
3
5
7
1
8
4 9
10
6
A∩ B :
the intersection of A and B
A∪ B :
the union of A and B
A' :
A⊂ B:
the complement of A
A
B
A is a subset of B
6∈ A:
6 is a member of the set A
∅:
the empty set
Intersections are overlaps, unions are all elements in one or the other or both
...
(i) Translate into normal English: B ∩ C ≠ ∅
(ii) Describe the set B ∩ A '
(iii) Is a white mouse a member of the set A ∩ ( B ∪ C ) ' ?
A
B
(c)
...
How many play both football and water polo?
(d) ξ is the set of all employed people in England
...
B is the set of those with a building society account
...
(i) Shade the set of those in catering with a bank account but no building
society account, and describe this in set notation
...
66
Answers
(a)
A
B
(b) (i) there do exist black cats
(ii) black inanimate objects
(iii) the white mouse is certainly not a member of black objects or cats, and it is an
animal, so yes!
(c) let’s call the number in the intersection x
...
e
...
(Note also that there are none in the left hand compartment, i
...
n( F ∩ W ') = 0 , which
means that in this case F ⊂ W
...
)
(d)
B
A
(i) this is A ∩ C ∩ B '
C
(ii)
B
A
C
The members of C ∩ ( A ∪ B ) ' are those in the catering industry without a bank or
building society account
Title: maths olevel notes with solutions
Description: maths notes with worked examples
Description: maths notes with worked examples