higher maths - conditional offer workbook · 2019-02-21 · in europe, this means multiply. for...
TRANSCRIPT
Perth College
Higher Maths
Conditional offer workbook
Applicant name: ______________________________________
Please complete this workbook within one calendar month of receipt. Place the workbook in an A4 envelope and post it first class recorded mail (cost £2·48 at time of writing) to FAO G Fyfe Core Skills dept Perth College Crieff Road Perth PH1 2NX Please note a normal stamp on your envelope is not enough! Underpaid postage may lead to a delay in your application and your envelope may even be refused. If you live locally you can just hand it in personally at the main college reception.
FAQ
Why do I have to do this workbook when I already have the necessary qualification given on your website/prospectus?
We only give out unconditional offers to applicants who are resitting or whose “gap” since achieving their qualification does not exceed one academic year. It is very easy to forget important background skills if it has been a while since you studied maths. Doing this workbook will prepare you very well for Higher, and increase your chances of coping with a demanding course.
Something has happened and I don’t think I will be able to return the workbook within one month. What should I do?
Contact us immediately. Places on the course are limited, and if you don’t tell us you risk losing your place.
What happens when I send my workbook back?
It is carefully marked by one of the maths tutors. You then receive written feedback by email on your answers. Depending on how you have done, you will either be offered either an unconditional place or your application will be rejected.
What is the pass mark?
We go by the overall standard of your answers. You probably won’t get everything perfectly correct, and there may be topics you struggle with. However almost everybody does a good enough job to get on the course. Do your best and enjoy getting back into maths!
Introduction
We are pleased that you want to study Higher maths. It is a very interesting course which is needed for careers in Science, Engineering, Medicine, Veterinary Medicine, Architecture, and Actuarial work In any case you want to pass, so it is important you know what to expect and that you are well prepared for starting the course. Here is a list of the current maths courses offered in Scotland:
National 3
Pretty basic, most people could probably cope with this without special preparation. Old equivalents were Access 3 maths and Foundation level maths (standard grade 5 or 6).
National 4
Starts getting more mathematical, for example contains
equations, Pythagoras, and Trigonometry. Old equivalents were Intermediate 1 maths and General level maths (standard grade 3 or 4). This is the normal entry requirement for Nat 5. Nat 4 Lifeskills is a different course which is not suitable for progression.
National 5
Much more mathematical. Most of the general public would only understand a little of this. Old equivalents were O-Grade maths, Intermediate 2 maths and Credit level maths (standard grade 1 or 2). This is the usual entry requirement for the careers mentioned earlier.
Higher
Proper specialised maths. Most people would probably just shake their heads at this, which is a pity because this is where it starts to get really interesting!
Advanced Higher
Absolute hieroglyphics, and absolutely fascinating. Old equivalent was CSYS. English A-level is similar.
The point of this workbook is to help you brush up on the most important parts of the Nat 5 course – the parts that will help you do well at Higher.
,
·
:
For European applicants
In Britain we write some mathematical things differently to the way you write them in Europe. Commas can be confusing. In Europe, a comma separates the units from the tenths; the whole number from the decimal part. For example, you maybe weigh 70,2 kg. In Britain, commas are sometimes used to make large numbers easier to read. They come after the thousands or the millions. For example, the population of Perth is 50,000 and the number of people living in Scotland was 5,295,403 according to the 2011 census. Using commas like that is a bit old-fashioned. We don’t use them like that in this workbook, we leave a small space instead.
In Europe, this means multiply. For example 2·7 means 14
In Britain, it is a decimal point. So 2·7 here means the same as the European 2,7
When we write 22 5 it means 22,5 2,5 2,5 6,25 and does not mean 2 25 50 !
If we want to show we are multiplying the 2 by the 7, we write 2 7 instead. In Europe, 20 : 4 means divide and get the answer 5. We do use the : sign in Britain, but only in things called ratios which compare two amounts. For example, the ratio of men to women might be 5 : 6. That means there are five men for every six women, it’s not a sum where you need to do anything.
In Britain if we want to divide we use a divide sign ÷ instead, so we write 20 4 5
Calculator
This is a Casio fx-85 calculator. You will need one (an fx-82 or 83 is almost as good) for Higher maths and for some of the work in this book. There are other makes of good calculators, but they don’t all have the special features the fx-85 possesses. If you have not used one of these before, please read through the owner’s manual. Before you go any further in this book, make sure you know
how to save time using the cursor, the delete o and last answer M buttons.
how to change the shape of “weird” answers using the n button.
how to change between Line mode and Math mode.
how to use the fraction button a.
that 0 3 means endless 3s and 0 72 means 0·72727272 … repeating endlessly.
the button for entering negative numbers is z, not p.
how to enter - it’s in yellow writing above your K button. (The “3” at the end here is just the calculator rounding things off.)
Contents
Expanding brackets
Common factors
Difference of two squares
Trinomials
Fractions
Simultaneous equations
Surds
Indices
Rearranging
Gradient
Trig equations and graphs
Expanding brackets
You should know how to multiply a simple bracket by a number
5 2 5 10x x
2 3 8 6 16w w
or by a letter
23 6 3 6x x x x
2 3 25 3 5 3a a a a a a
or by both
22 5 2 10p p p p
23 2 5 6 15r r r r
and be able to tidy up expressions with a negative sign before a bracket
15 4 3 2 15 4 3 6
12 10
w w w w
w
1) 5 4d
3) 1k
5) 8x x
7) 2 6c c
9) 7 2 4y y
11) 6 3f f f
2) 2 4 3v
4) 1a a
6) 4 8m m n
8) 22 3 5a a a
10) 4 9 4h h
12) 28 3 7 2b b b
F L
I
O
(x + 2)(x + 3)
You should know how to multiply out pairs of brackets. One method uses what we call the Distributive Law:
2 3
3 2 3
x x
x x x
2
2
3 2 6
5 6
x x x
x x
The second method uses the word FOIL. We perform four multiplication sums.
F tells us to multiply the first terms in each bracket together 2x x x O tells us to multiply the two outermost terms together 3 3x x I tells us to multiply the two innermost terms together 2 2x x L tells us to multiply the last terms in each bracket together 2 3 6
So
2
2
2 3 3 2 6
5 6
x x x x x
x x
The first bracket gets split up. Repeat the second bracket.
Two more examples:
2
2
5 4
4 5 4
4 5 20
20
x x
x x x
x x x
x x
or
2
F O I L
2
5 4
4 5 20
20
x x
x x x
x x
2
2
3 6
6 3 6
6 3 18
9 18
t t
t t t
t t t
t t
or
2
F O I L
2
3 6
6 3 18
9 18
t t
t t t
t t
13) 4 7x x
15) 9 6y y
17) 2 3 4g g
19) 2 3 2 3x x
21) 3 2 2 1r r
14) 9 6c c
16) 2
3x
18) 3 1 2 1d d
20) 5 6p p
22) 7 2 3 2t t
FOIL is quicker but you need to know the Distributive Law for dealing with longer brackets.
2 2 2
3 2 2
3 2
6 4 4 6 4
4 6 6 24
7 2 24
x x x x x x x x
x x x x x
x x x
23) 23 2 4x x x
24) Begin by multiplying the last pair of brackets here.
1 7 3x x x
Factorising is a huge part of Higher Maths and it is vital you can cope with it. There are three kinds of factorising:
Common Factor
Difference of Two Squares (DoTS)
Trinomials Common factors
Simple examples might involve dividing every term by a number
2 6 2 3x x
21 14 7 3 2p p
Sometimes we divide every term by a letter or a letter and number
5 5km k k m
2 4 4u u u u
230 10 10 3cd de d cd e
Note in the last example we divide by 10, the biggest possible number, rather than 2 or 5.
25) 6 9w
27) 2ab a
29) 2f f
31) 26 9x x
33) 218 3m mn
35) 26 12x x
26) 35 10r p
28) 212c c
30) 24 6h h
32) 220 8xy y
34) 212 18v v
36) 2 215 9a b ab
Difference of two squares
These are easy to spot – look for square numbers like 1, 4, 9, 16, 25 … and squared letters.
2 9 3 3w w w
21 1 1y y y
2 216 4 4s t s t s t
37) 2 4x
39) 2 1a
41) 29 4x
43) 236 1b
38) 2 49v
40) 216 p
42) 225 9u
44) 2 216 81c d
and add together to make this number?
What two numbers multiply together to make this number?
2 6 8x x
Trinomials
This is really important.
In general terms a trinomial quadratic has the shape 2ax bx c . To begin with we’ll look at ones with a single x2.
To factorise a trinomial like 2 6 8x x , ask yourself these two questions:
In other words, ? ? 8
? ? 6
This one is easy, the numbers must be 2 and 4. So the brackets are 2x and 4x .
2 6 8 2 4x x x x
It is harder when negatives are involved.
To factorise 2 12x x , we ask ourselves
? ? 12
? ? 1
Note the “1” comes from the single x. The numbers must be normal 4 and −3, so
2 12 4 3x x x x
To factorise 2 11 28x x , we ask ourselves
? ? 28
? ? 11
This time the numbers must both be negative: −4 and −7.
2 11 28 4 7x x x x
45) 2 3 2x x
47) 2 8 7y y
49) 2 2 15c c
51) 2 3 18m m
53) 2 9 20p p
46) 2 2 1x x
48) 2 10 24w w
50) 2 5 14k k
52) 2 20r r
54) 2 10 9b b
and add together to make this number?
What two numbers multiply together to make this number?
22 7 3x x
2 × 3 = 6
2
Same bracket!
2 6 3
2 3 1 3
x x x
x x x
When the trinomial has more than one x2, it can be done using trial and error. Some people prefer using a method which reinforces the Distributive Law. This method is explained about 13:30 minutes into the video at http://www.mathcentre.ac.uk/video/292.
Consider factorising 22 7 3x x .
Using the video method, first you must multiply the 3 units at the end by the 2 at the start.
So our questions are ? ?
? ? 7
6
and the numbers must be 6 and 1.
Now we do something which seems strange, we split up the 7x into 6x and 1x.
2 22 7 3 2 6 3x x x x x
This allows us to find a common factor for the first pair of terms …
2
Find a CF
2 6 3
2 3 3
x x x
x x x
… and repeat for the second pair. The beauty of this method is the bracket you find for the first pair is always repeated for the second pair!
Finally the bracket itself is now a common factor:
2 3 1 3
2 1 3
x x x
x x
Consider factorising 26 5 6x x .
Again, if you are using the video method, begin by multiplying the 6 units at the end by the −6 at the start.
Our questions are ? ? 36
? ? 5
Obviously this needs more thought, but the numbers must be −9 and 4. Split up the −5x into −9x and 4x
2 26 5 6 6 9 4 6x x x x x
and look for common factors in each pair of terms.
26 9 4 6
3 2 3 2 2 3
x x x
x x x
Again there’s a repeated bracket.
3 2 3 2 2 3
3 2 2 3
x x x
x x
and add together to make this number?
What two numbers multiply together to make this number?
6 × (−6) = −36
26 5 6x x
Factorise these expressions using either the video method or trial and error:
55) 22 5 3x x
57) 26 7 2b b
59) 23 5 2a a
61) 24 11 6x x
63) 23 14 15n n
56) 210 17 3m m
58) 23 2 8d d
60) 24 7 2s s
62) 212 23 10e e
64) 210 11 6w w
Casio in Math mode Other calculators
Fractions
You need to be confident adding, subtracting, multiplying and dividing with fractions. You need to be able to cancel fractions down and be able to do all this without a calculator. Multiplying fractions is easy – multiply the tops and multiply the bottoms.
Do these without a calculator:
65) 3 3
4 8
67) 4 4
5 3
66) 2 5
3 9
68) 1 5
of 4 6
(“of” means ×)
Try it with letters.
69) 5 2
a c
71) 1 c
b d
70) 4 2
u v
72) 3 4x
y y
Often the answers will cancel down. It’s not always easy to spot the cancelling at the end of a sum like this though. A better idea is to cancel before multiplying. You can cancel anything on the top with anything on the bottom:
2 3
3 5
3 into 3 goes once. The numbers reduce and the sum gets easier:
1
2
3
3
1
2
5 5
Here’s another example.
1
5
4 12
9
3
10 2
3
8
5 in to 5 goes once, 5 into 10 goes twice. 3 into 12 goes four times, 3 into 9 goes three.
6a b
1
2
5
8
r
r b
1
2
6
r
a
r
5 r
1
3
8
6 5
8
6
a
r
5
8
a
r
4
3 5
4
15
4
a
r
a
r
Cancel before multiplying:
73) 8 5
9 8
76) 2 35
of 7 12
74) 6 1
of 7 6
77) 3 14
4 15
75) 2 3
5 4
78) 8 15
of 21 16
Letters can be cancelled too.
For example, in the sum 3
10
t
t the t’s obviously cancel and you’re left with
3
10.
It is only a small step to mastering something more complicated like
2
6 5
8
ab r
r b
There’s a lot will cancel here. The b’s will go. The r on the top will cancel one of the r’s in the r2 underneath. And don’t forget the 6 and the 8 can both be divided by 2:
Have a go at these:
79) 1 3
4 2
y
y
82) 25
15
w
w
80) 3 2
8
m
q m
83) 2
2 2r h
h r
81) 2
3 4
de
d
84) 2
1 2
6
ab
a b
To divide a quantity by a fraction, turn the fraction upside-down and multiply instead. So
12
4 21 1
81
4
or 8
Try the next lot without a calculator. Remember to look for cancelling when you reach the multiplication stage.
85) 1
23
2
1
88) 2 5
3 9
86) 4
45
4
1
89) 34
2
2
1
87) 1 1
2 3
1
2
90)
52
52
52
x
x
These ones have letters.
91)
a c
b d
a
b
94)
t r
r t
92)
ff
e
95) 2
mw
n
93) 2
2
2
a a
b b
96) 2 2
r r
v a
97) 2 bottles 3 bottles
2 3
2 sevenths 3 sevenths
2 3
7 7
a a
Note the answer to the last sum is 57
, not 514
!
So adding and subtracting fractions is easy if the denominators are the same:
98) 1 3
5 5
99) 5 2
6 6
100) 11 6
12 12
The same idea works if letters are involved. If the denominators are the same, just add or subtract the tops.
4 2 6
c c c
8 7 1
v v v
2 2
5 5 5
x x
101) 1 4
b b
104) 6
5 5
t
102) 9 2
w w
105) 8 8
c d
103) 10 7
2 2r r
106) 2
6 6
u v
Usually though the denominators are different. When this happens you must find a common denominator. Once the bottoms are the same, it’s easy. Try these:
107) 1 1
2 5
10 10
110) 2 1
5 4
108) 2 3
3 8
111) 1 1
10 3
109) 2 1
3 5
112) 7 1
6 4
12
The same idea applies when you have algebraic fractions.
2
3
6
3 3
6
3
c
w
cw
w w
cw
w
113) 5
3
3 3
3
a
t
t t
t
116) 4
4
n m
d
d
114) 6 2
m p
mp mp
mp
117) a c
b d
115) 1 5
x y
xy
118) u v
v u
uv uv
The whole number “1” can be written as a fraction in lots of different ways, for example 2
2
2 3 4, , , ,
2 3 4
d m
d m. As long as the top and bottom are the same, the fraction is equal to 1.
119) 2
13
2 3
3 3
122) 3
15
120) 2
13
2
3
123) 3
14
121) 1
14
1
4 4
124) 5
12
A topic you may see later in Higher involves starting with a power of x and doing two things to it:
raise the power by 1
divide by the new power For example,
32 3
45 4
3
4
xx x
xx x
If the power is a fraction, it gets more complicated. Try the two steps with these, using the method of questions 90) to write your answer as neatly as possible:
125) 12x
127) 25x
129) 34x
126) 13x
128) 12x
130) 53x
Simultaneous equations
It is possible to add and subtract equations. Suppose an elephant weighs the same as a car
e = c
while a rabbit weighs the same as a handbag
r = h
Then it is certainly true that the elephant and rabbit combined weigh the same as the car and handbag together:
So we can add two equations to get a new equation.
Add:
e c
r h
e r c h
We can also subtract two equations to get a new equation, though it’s not so easy to draw pictures to represent this.
To solve the simultaneous equations
20
4
a b
a b
we can just add in columns : the a’s, the b’s then the units. Something special happens when we add the b’s. One “b” is positive, the other negative so they cancel. It’s just like adding 7 and −7. So we get
20
4
Add: 2 24
a b
a b
a
which means a = 12. Now if a = 12 it’s always been 12. So the original equations were really saying
12 20
12 4
b
b
If we pick either of these, we can now calculate what b stands for. Let’s pick the first equation. (We could also pick the second, it doesn’t matter.) First equation becomes
12 20b
so we know both letters: a = 12 and b = 8
Solve these pairs of simultaneous equations by adding. Work out both letters. 131)
2 17
2 7
x y
x y
132)
4 3 58
2 3 2
w f
w f
Adding equations together doesn’t always help us though.
4 23
2 15
Add:
p q
p q
Here we get a new equation, 6 2 38p q , but it’s no use as it still contains two unknown
letters. Perhaps we could subtract the equations instead?! That should make the q’s vanish, then you can solve the equations.
4 23
2 15
Subtract:
p q
p q
Use subtracting to work out both letters in these simultaneous equations. 133)
5 8
4
j k
j k
134)
7 3 34
3 10
a c
a c
Decide whether to add or subtract, then solve each pair of equations. 135)
18
2 15
r s
r s
137)
2 5 8
4 5 1
a n
a n
139)
7 11
5
x y
x y
136)
2 8
8
t u
t u
138)
3 2 21
3 6
e k
e k
140)
4 5 13
4 2 6
f g
f g
Solve the following after first multiplying one of the equations by a suitable number. 141)
3 22
5 2 33
a b
a b
143)
3 4
2 5 6
c d
c d
142)
2 7 5
3 2
x y
x y
144)
2 9 17
8
f g
f g
Sometimes simultaneous equations can be solved using the method of substitution. This is a nice method when the equations have different “shapes”.
2 3 32
5 12
x y
y x
Here the second equation tells us that y is equivalent to the expression 5 12x , so we can
replace y in the first equation by 5 12x and play with that:
2 3 5 12 32
2 15 36 32
17 68
4
5 4 12
8
x x
x x
x
x
y
Solve these using the method of substitution: 145)
2 3 7
3
h k
k h
146)
7 2 24
2 9
a b
b a
Surds
These do not play a big part at Higher, but you should know how to simplify a surd and how to rationalise a denominator. Nowadays many calculators will do this automatically.
But of course Paper 1 is a non-calculator exam … The square root of a product is the same as the product of the individual square roots:
ab = a × b
In the same way, rewrite the following surds as products of surds.
147) 35 = ×
149) 14 = ×
148) 15 = ×
150) 8 = ×
Let’s look more closely at that last answer : 8 = 4 × 2 . One of the two parts is not
actually a surd, because 4 is of course a whole number, namely 2.
So we can take our working a little further.
8 = 4 × 2
= 2 × 2
or just 2 2
When we rewrite 8 as 2 2 , we say we have simplified 8 . That’s because we are using
a smaller surd.
Surds can often be simplified in this manner. Try rewriting the following without a calculator. It is helpful if you write the non-surd first when splitting it up.
151) 45 5
___ 5
152) 12 3
153) 125
Sometimes there are several ways in which we can split up a surd. Consider 50 . This can
be split up in three different ways:
50 1 50
This is a bit pointless. It’s just saying
50 1 50
which isn’t telling us anything we don’t already know!
50 10 5
This is better, but both 10
and 5 are surds so we
cannot tidy it up any further.
50 25 2
This is much more useful,
because 25 is a whole
number not a surd! We can continue
50 5 2
The third way is best here because 25 is a square number. When splitting up a surd, look for the biggest possible square number that goes into it.
If you have a choice between picking 4 and 16 , pick the larger one.
Avoid using the number “1”.
50
Simplify each surd.
154) 18
156) 20
158) 75
160) 32
162) 1000
155) 200
157) 108
159) 63
161) 98
163) 54
If you have to multiply surds together, there’s usually a quick way. Remember that
a b ab
So for example,
6 24 6 24
Use this method to simplify
164) 2 8
165) 3 12
166) 20 5
It would be nice if there was a simple way of adding surds together, or subtracting them.
167a) Is it true that 25 9 16 ?
b) Is it true that 48 3 51 ? Use a calculator to check.
No, we can’t add or subtract surds like that.
Fractions such as 2
5, where the denominator is a surd, are considered ugly in Maths.
There is however a way we can rewrite such fractions so that we get a whole number on the bottom and a surd on the top instead. That’s considered a “nicer” shape!
So
Once we have got rid of the surd on the bottom, we are said to have expressed the fraction with a rational denominator. Express these with rational denominators.
168) 7
3
170) 2
11
172) 1
10
174) 15
5
169) 2
2
171) 6
3
173) 3
6
175) 20
5
Multiply top and bottom by the surd in the denominator.
Note this is like multiplying by 1 2 2 5
= × 5 5 5
2 5=
5
Indices
In Maths we often see things like 72 or 43 or c5 where a number or letter is raised to a power. The power or index is the small number up top, while we call the bottom number the base.
176) 72 stands for
178) 6f stands for
177) 3 3 3 3 can be shortened to
179) m m m can be shortened to
We now start combining numbers written in index form, and look for shortcuts to the answer. Consider the multiplying sum
5 34 4
Is there a way of simplifying this? Perhaps it might tidy up to 154 or something similar? Let’s write it out in longhand first:
5 3
5 3
4 4
4 4 4 4 4 4 4 4 4 4
We just have eight copies of the number 4 all multiplied together. In other words
5 3 84 4 4
Likewise
3 2
3 2
10 10
5
10 10 10 10 10 10 10
10
When multiplying two numbers written in index form, we add the powers. Simplify
180) 2 45 5
182) 7 42 2
184) 8 6r r
181) 83 3
183) 2 5w w
185) 4y y
43 base index
mm n
n
aa
a
186) 7 42 2
188) 3 6x x
187) 21210 10
189) 2 73 3f f
The rule does not work though if the bases are different. For example, something like 2 54 3
or 4 3y z cannot be simplified.
The rule we’ve been using is often called the First Law of Indices. In symbols,
So we add the powers when multiplying. And, not surprisingly, we subtract the powers when dividing numbers written in index form.
This is demonstrated if we write a sum like 5
3
d
d out in longhand :
5
3
d d d d d d
d d d d
Now we get lots of cancelling and end up with
d d d d d
d
d d
2d
So
52
3
dd
d
This is the Second Law of Indices :
m n m na a a
Simplify
190) 6
4
k
k
193) 3
7
v
v
191) 5m
m
194) 2
3
t
t
192)
72
12
w
w
195)
43
13
u
u
To simplify something like 6 5
3
s s
s
, tidy up the top line first.
6 5 11
3 3
s s s
s s
Simplify
196) 6 5
4
x x
x
197) 2
4
r r
r
198) 3
5 4
p
p p
We can explain how to simplify something like 2 74 2p p by writing the sum out in
“longhand”:
2 7
2 7
4 2
4 2 4 2
p p
p p p p p p p p p p p
On the right-hand side we are multiplying a whole lot of things together. But we can multiply things in any order we like, so let’s rewrite it with the numbers first:
98
4 2
p
p p p p p p p p p
You can now see it will simplify to 98p
So if there are any whole numbers in front of the bases, multiply them together and then add the powers. For example,
3 2 58 5 40v v v
Negative powers mean fractions. They mean 1
the positive version
Simplify
199) 2 58 2n n
201) 35 2e e
203) 1 12 26 3p p
200) 3 212 3a a
202) 2 39 4y y
204) 31
2 215 2w w
Before we look at what negative powers and fractional powers mean, it is worth learning a few everyday powers off by heart. Fill these in.
22 =
32 =
42 =
52 =
62 =
23 =
33 =
43 =
24 =
34 =
25 =
35 =
26 =
27 =
28 =
29 =
-5
5
1 means x
x
1
1
1 14 means or
4 4
-2 meansx
32 means or
Rewrite these negative powers as fractions. Show both steps.
205) 2 13
207) 2 12
209) 2 17
211) 4 12
213) 3 15
215) 5 12
206) 2 15
208) 1 14
210) 2 16
212) 2 14
214) 1 17
216) 6 12
217) 2007 Int 2
Simplify the expression below, giving your answer with a positive power.
5 8m m
Now consider the sum 2
310 . This stands for
2
3
6
10 10 10 10 10 10 10
10 10 10 10 10 10
10
The fact suggests the so-called Third Law of Indices:
5
2 104 4
44 21 12 27 7 7 or 49
33 2
2233f f f
55 4
4455g g g
Simplify these:
218) 3
2c
220) 12
6
d
222) 525a
219) 1
4t
221) 313p
223) 434y
Use the Third Law to evaluate
224) 2124
226) 3233
225) 6562
227) 4345
When we raise a power to a new power, we multiply the powers.
n
m mna a
228) 2002 Credit
Express in its simplest form
2
8 3y y
.
Consider the following sum which involves the third law:
22
1
11227 7
7
7
Think what this is actually telling us. It’s saying, “If we square the thing called 127 , we get 7.”
But that’s also true of the surd 7 :
2
7 7
In other words 127 and 7 mean the same.
1216 16 or 4
Evaluate
229) 129
230) 12100
231) 1264
Likewise a power of 13
means the same as a cube root ( 3 ).
3138 8 or 2
The list after q 204) may come in handy for the following.
232) 31327
233) 13125
234) 1364
12x means the same as x .
Often the fractional powers are more complicated, for example 23125 .
If you have to evaluate something like this, it will usually be in the non-calculator paper and the whole question revolves around recognising 125 as a common power. Use the following method:
3
2
2233125 5
5
25
235) 2 25 532
236) 3 34 416
In a similar way, evaluate these without a calculator.
237) 238
239) 3532
241) 329
243) 3481
238) 13125
240) 238
242) 23125
244) 5664
245) 2006 Int 2
Evaluate 3416 .
246) 2009 Int 2 Prelim
Evaluate 23125
.
247) 2002 Int 2 Winter diet
Evaluate 234a when 27a .
248) 2004 Prelim
Simplify 1 12 43x x
and state its value when 81x .
The Fourth law of Indices is less important : a product raised to a power can be split up.
A good example of this law in action is when you are trying to square something like 3x, maybe as part of a FOIL sum.
It’s not just 3x2 of course, you have to square that 3 as well!
2 2 2
2
3 3
9
x x
x
m m mab a b
F
3 3x x
249) 2
5r
251) 2
23ab
250) 3 3
2 3 22 2
n n
252) 2
3 14p q
Now a surprising result. Try raising a number to the power of 0 on your calculator. For example, 70
7^0=
Try it a few times with different bases. What do you notice?
253) 0
5
255) 0
m
254) 0
14
256) 0
2a
257) Evaluate 0 3t when 4t
258) Evaluate 0
2w when 5w
These are straightforward enough, but be careful with an expression like 02w when 5w .
The difference is subtle. Remember a power in Maths only refers to the thing it is sitting beside.
With 02w you are raising w to the power of zero, getting 1 of course, and then doubling the answer.
259) Evaluate 03y when 4y
260) Evaluate 010 8r when 2r
Any number, or piece of algebra, raised to the power of 0 is always equal to 1.
261) Simplify 3 32x x
262) 2003 Int 2
Express 2 2 23 3 3a a a
in its simplest form.
× 5
+ 2
5t 5t + 2 t
Rearranging
In our working in Maths we have to move things about. There are two vital ideas to understand:
When you move something over to the other side of an equation it has to become the opposite thing
adding things become subtracts multiplying things become _________________ squares become ________________
and so on. Sometimes at school this is mistakenly called “change side, change sign”.
Expressions are “created” in a certain order and must be rearranged in a certain order. An expression in Maths is just a mixture of numbers and letters, such as
4w 2 5y 3 2 1f 3
dc
Expressions always involve some combination of the four basic ideas of arithmetic
adding
subtracting
multiplying
dividing
and perhaps some more advanced ideas such as
squaring
square rooting
taking the Sine
cubing
Starting with the letter “t”, if we multiply it by 5 then add 2 to the answer we create the simple expression
5 2t
We can represent the above process by a diagram:
Multiplying t by 5 first gives us the intermediate answer of 5t, then we add 2 to get the final answer.
+ 2
× 5
t + 2 5(t + 2) t
sq
+ 7
m
sq
× 2
p
+ 3
sq
u
× 7
sq
r
× 4
× 4
− 12
y
+ 3
× 10
f
− 6
× 4
r
÷ 2
+ 8
w
+ 4
÷ 3
c
− 5
÷ 9
a
÷ 8
− 8
g
Note if we change the order of the two steps so that we are adding 2 first then multiplying the answer by 5. Does it make any difference to the final answer? Yes! If we want to show that we’re taking 5 lots of the whole of “t + 2”, we must wrap the (t + 2) in a bracket.
Fill in the intermediate and final expressions for these sets of instructions. 263) 264) 265) 266) 267) 268) 269)
For squares and square roots we will use these shorthands:
sq = square
= take the square root
Fill in the following.
270) 271) 272) 273)
+ 8
n
− 4
k
× 5
b
?? ??
6h − 5 h
× 6 − 5
6h 6h − 5 h
2x + 4 x
3(f + 1) f
5e − 22 e
5(p − 13) p
n2 + 4 n
(s + 8)2 s
12w w
1q q
274) 275) 276)
An important skill which textbooks often ignore is to be able to look at an expression and work out how it has been formed. We’re working backwards, trying to find out what the steps were and get them in the correct order. For example, consider the expression 6h − 5. What happened to “h” to form this expression?
We must have multiplied by 6 first, then added 5. We can fill everything in:
Try the same for these. 277) 278) 279) 280) 281) 282) 283) 284)
3
4
c c
42
a a
2
3
h
h
2
7
d d
× n
+ t
m
÷ e
− w
c
+ 3v
× d
u
× q
− f2
p
jk + a j
b(k − d) k
4y
wz y
2r s
t
r
285) 286) 287) 288)
Sometimes instead of using numbers we use other letters. Instead of multiplying x by 3 then subtracting 1 we might multiply x by y then subtract something called 5z. Fill in the following.
289) 290) 291) 292) 293) 294) 295) 296)
× 6
6r r
× 6
Q r
Here are three similar sums.
10 7 3 3 10 7 7 10 3
Each expresses the connection between 10, 3 and 7 in a different way. It’s a similar idea with the “Speed – Distance – Time” formulas from school.
D D
S T D S TT S
Basically these are all just different versions of the one relationship. Each version has a different letter on the left-hand side. We call the letter on its own the subject of the formula.
We often have to take a formula like 3a y
fm
and rearrange it so that one of the other
letters is the subject, for example
a =
The bubble will contain all the other original letters and numbers. There will be an “f”, a “3”, a “y” and an “m” kicking about. But they may be in different positions. The “−” may become a “+”, and there may not even be a fraction any more! The key idea behind rearranging a formula is to understand how it has been formed and then reverse the steps involved. Let’s look at a very simple example:
6Q r
Here, to get Q, we have simply multiplied r by 6.
or even
If we want to make “r” the subject we need to undo the multiplying. The opposite of multiplying by 6 is of course dividing by 6. Divide each side of the equation by 6:
6
6
Q r
Qr
r is now the subject of the formula. If you prefer, swap sides and write 6
Qr instead.
× 6
÷ 6
morning put on socks
put on shoes
take off socks
take off shoes
evening
× 3
÷ 2f
+ 7w
− 1
cube
Reverse each process, for example 297) 299) 301)
298) 300) 302)
If we have two or more steps in an expression the last step gets undone first. Suppose that in the morning, just before you go out, you put your socks on and then put your shoes on. In the evening when you come home you start by dealing with your shoes first, then you deal with your socks!
× 2
+ 7
× 6
− 2
sq
× 3
+ 9
- 4
× 9
+ 1
sq
÷ 2
+ h
× m
− 6
− 3e
If we want to reverse the two-step process we need to deal with the 7 first, then deal with the 2. Fill in the order we deal with things when reversing these processes:
303)
deal with the _____ first
then deal with the _____ 305)
deal with the _____ first
then deal with the _____ 307)
deal with the _____ first
then deal with the _____
304)
deal with the _____ first
then deal with the _____ 306)
deal with the _____ first
then deal with the _____ 308)
deal with the _____ first
then deal with the _____
The same idea works if there are more steps involved. Reverse this process: 309)
deal with the _____ first
then deal with the _____
then deal with the _____
d w
dealing with the c
dealing with the 4
4
4
4
d w c
d c w
d cw
310) Consider the formula 4d w c
Suppose we want to rewrite it with “w” as the subject. Our first task is to work out what has happened to w to create the thing called d. Fill in the two steps involved:
Now when we reverse the process we must
deal with the _____ first
then deal with the _____ “Dealing” with something just means moving it to the other side. So the full working would be
k y
g m
2
2k y d
2
2
2
2
k y d
k y d
k d y
311) Change the subject of the formula below to y.
Fill in what happened to y on the right-hand side to make k: Reverse the process:
deal with the _____ first
then deal with the _____ Now the working, a line at a time:
312) Change the subject of the formula
g b m x
to m.
deal with the _____ first
then deal with the _____
g b m x
m
dealing with the square, it becomes a square root
dealing with the 2d
Change the subject of these formulas to the required letter. 313) 4 to ?m s t s
315) 5 to ?p r w r
317) 2 to ?d e f e
319) to ?y mx c m
314) 5
to ?v
u va
316) to ?h
m r hn
318) 7 to ?R ac x c
320) 6 to ?Q x x
321) 2
3 to ?H T f T
323) 3 to ?p d y d
325) 24 to ?b w w
322) 3 to ?F mr z r
324) to ?R r
L R
326) 2
4 to ?b w w
Gradient
This is a measure of the steepness of a line. If you travel from one point to another, the basic definition of gradient is
number of squares travelled up
number of squares travelled to the rightm
Always read the distances off from left to right, the same direction we read text in. Travelling from left to right (F to P),
FP
76
7 up
6 to right m
Lines which slope downhill from as we read from left to right have negative gradients. This is something you should be able to spot instantly when you see a line.
CW
310
310
3 up (3 down)
10 to rightm
F
P
7
6
3
10
C
W
Write down the gradient of each sloping line in its simplest form. 327) 329)
328) 330)
Usually the coordinates of two points on the line are supplied. In this case we can use a formula from standard grade
B AAB
B A
y ym
x x
The formula is usually written as 2 1AB
2 1
y ym
x x
but I prefer to use whatever the names of the
points are called. So for the gradient of a line called PQ, we would write
Q PPQ
Q P
y ym
x x
Write down the correct version of this formula for finding:
331) CDm
332) VFm
8
2
7
21
4
8
12
5
H(−2, 6)
N(3, 14)
333) TLm
334) EKm
To find HNm in this diagram we do
N HHN
N H
85
14 6
3 2
y ym
x x
The answer is positive which agrees with the diagram. You need to be able to do gradients like this quickly without a diagram in front of you.
335) G is (1, −5) and H is (8, 9). Find GHm .
336) R is (12, 1) and A is (−2, 8). Find ARm .
337) N is (−6, −3) and Z is (−11, 2). Find NZm .
338) Y is (0, 6) and D is (6, 15). Find DYm .
339) J is (−1, 1) and G is (17, −5). Find GJm .
340) B is (10, −15) and P is (−25, 15). Find BPm .
341) V is (7, 6) and X is (−1, 10). Find VXm .
342) T is (−4, −15) and A is (3, 6). Find TAm .
343) K is (−5, −4) and L is (−1, 3). Find KLm .
35°
0o, 360o
90o
180o
270o
1
1
-1
-1
quadrant 1 quadrant 2
quadrant 4 quadrant 3
Trig equations and graphs
Trig is an area Higher students are often weak at. You need to know about the four
quadrants; how to find “partners” of an angle; how the signs of j, k and l change in the quadrants; how to solve simple trig equations; exact values and trig graphs.
This is a four quadrants diagram.
In trig we measure angles by turning anticlockwise from the positive x-axis. 35°lies in quadrant 1. Which quadrant are these angles in?
344) 212° 346) 69°
345) 299° 347) 108°
Make sure your calculator is set in degree mode (a little “D” at the top of the screen).
For any given angle, j, k and l are decimals associated with that angle. They can be positive or negative depending on which quadrant your angle is in.
348) Complete this table by writing + or – in the boxes depending on the sign of the decimals:
qd 1 (eg 50°) qd 2 (eg 164°) qd 3 (eg 237°) qd 4 (eg 303°)
j +
k
l
C
A S
T
360 x
x 180 x
180 x
To learn the signs in the four quadrants you just need to know where the positive ones are (and all the others must be negative). Some people learn this diagram:
Others learn a sentence like “All Sin Tan Cos” or “All Students Take Chemistry”. Without a calculator, are these positive or negative?
349) k 18°
353) l 224°
357) k 345°
350) l 100°
354) l 333°
358) j 353°
351) j175°
355) j 254°
359) k 129°
352) j 38°
356) k 225°
360) l 80°
361) Type these in to your calculator:
j 8° j 188° j 172° j 352°
What do you notice? Angles which behave the same way as each other like this are called partners. They form a family of four, one in each quadrant. To find the partners of an angle in quadrant 1, learn these rules:
362) Write down all the partners of 20° 363) Write down all the partners of 60° 364) Write down all the partners of 45° 365) Write down all the partners of 87° 366) Write down all the partners of 155°
(The same rules work, you just get the answers in a different order) Trig equations are like puzzles where the mystery numbers are angles. Most of the time these angles must be between 0° and 360°, and this will be written in the question as 0 360x , possibly with ≤ signs depending on whether they want to include 0
and 360 in the interval.
To solve a simple trig equation like sin 0 448, 0 360x x , first decide which
quadrants the angle might lie in. Use your calculator to find one angle, then find its partner in the other quadrant.
1
(qd 1)(qd 2)
sin 0 448, 0 360
Angles lie in qd 1 or 2
sin 0 488
29 2 or 180 29 2
29 2 or 150 8
x x
x
Solve these trig equations for 0 360x :
367) sin 0 205x
368) 310
cosx
369) tan 3x
370) 27
cosx
If the decimal or fraction is negative, ignore the negative sign when you type it in. This will then give you a quadrant 1 angle which you can use to get the two partners you really want.
1
(qd 1)
cos 0 15, 0 360
Angles lie in qd 2 or 3
cos 0 15
81 4
x x
x
(qd 2) (qd 3)
or 180 81 4 or 180 81 4
98 6 or 261 4
Solve these trig equations for 0 360x :
371) sin 0 921x
372) tan 0 577x
373) 12
cosx
374) 53
tanx
Certain special angles come up a lot in maths diagrams and we need to learn their j, k
and l off by heart. We learn them as exact fractions though, not decimals.
375) If you chop a square of side 1 in half, you get a 45° angle.
By Pythagoras the hypotenuse must be 2 . Now use
SOHCAHTOA to make fractions:
sin45
cos45
tan45
376) Cut an equilateral triangle of side 2 in half and you get a 30, 60, 90 triangle.
By Pythagoras the height must be 3 , so again using SOHCAHTOA
sin30
cos30
tan30
sin60
cos60
tan60
You must learn these nine answers off by heart for Higher, or be in a position to quickly work them out from the triangles.
377) Use the above to solve 32
sin , 0 360x x
45°
1
2 1
30°
60°
2
3
1
90 180 270 360
-1
1
x
y
90 180 270 360
-1
1
x
y
You should have met the trig graphs before. The tan graph is seldom seen at Higher, but you need to memorise the sin and cos graphs.
378) Here are the waves of the sin and cos graphs, but which is which?
Label them, using the signs throughout the four quadrants to help you.