8 sundayherald 2 April 2006
Photograph: Rex Features
EXAM GUIDE
MATHEMATICS
BY ANDREW STEWART
Andrew Stewart is principal teacher of
mathematics at Alness Academy,
Ross-shire, a post he has held for 25 years.
He has taught mathematics at all levels
including evening classes and Easter study
courses. Andrew has been a marker for
SQA for 25 years and examiner for 15. For
the past 10 years he has been joint setter
for Standard Grade Maths.
T
HIS guide is intended to help you
prepare for your Maths exams at
Standard Grade and Higher. Like all
guides, it offers advice and hints; it
does not do the work for you. Only you can
fully prepare yourself for the exam. This
involves hard work and time. Remember
too that the biggest asset you have is your
teacher. In the run-up to the exams, you
must ask when you are not sure of
something. Help your teacher to help you.
When you are studying on your own,
focus on certain topics at a time and try to
master them. Make sure you also do as
many past papers as possible and
concentrate on the most recent ones.
TOOL BOX
In some ways maths is like a box full of
tools. These tools represent the skills you
have worked hard to master.
If someone tells you to use a particular
tool, then you have no difficulty. You can
add, you can subtract, you can factorise,
you can simplify, you can differentiate and
so on. But when it comes to the exam,
nobody is there to tell you what tool to use.
One of the difficulties in maths is
knowing how to start a question. Not all
questions are simply skills-based ones,
which tell you what to do clearly. Many
questions are context-based and require
you to cut your way through the verbiage
and get to the core maths of the question.
Then you have to decide how you are going
to solve the problem – ie, decide which tool
to use.
Maths exams are perhaps different from
any other in that we all have that fear that
we will not be able to do anything at all, that
our mind will go blank and we will freeze.
Everyone sitting a maths exam gets that
Helping more Scottish students to achieve their goals
2 April 2006 sundayherald 9
MATHEMATICS
deep breath and return to the question
at hand.
Perhaps if a question contains two or
more parts then it is possible to do part (b)
without part (a). If you still cannot start it,
have a look at one of the other questions,
but resist the temptation to flit from
question to question too quickly. You have
plenty of time, so use it. Once you have
started writing, then it will become easier.
It is important to get off to a good start
and getting the first question under your
belt is the best way. In Standard Grade
Paper 1, this is not too difficult as the first
few questions will be short numerical ones.
In the second paper and in both Higher
papers it is important to get yourself
settled. The effect of knowing (rightly or
wrongly) that you have the first question
correct boosts your confidence for the rest
of the paper.
If you do not like the first question and
feel it is not your favourite topic, then have
a look at the next two or three, they may be
better bets for you. However, the examiners
will make the first few questions the easier,
more straightforward ones and so it would
not make sense to start at the end and work
your way back.
A maths paper can usually be divided
into three parts. The earlier questions are
usually the standard ones, easy to recognise and hopefully easy to do. The final two
or three are likely to be the testing ones; the
last one usually designed for the top candidates. They are not impossible though, just
more difficult.
The middle group of questions then form
the “battleground” of your exam. They are
the ones you need to get correct to give you
your pass. You should take care and try to
answer each part. Give yourself time in this
section and do not panic if they seem
impossible – they are not, just think of your
toolbox and decide what the crucial aspects
of the question are and what tool to take
from the box. You have probably done
similar questions before, you just need to
access the correct part of your brain to
recall the method.
MARKS AND CLUES
Be guided by the number of marks
allocated to a question. A question worth
one mark is likely to be one in which you
are expected simply to write down the
answer (no working needed here).
Questions worth two, three or four marks
are likely to involve a few steps of working
but should be fairly straightforward.
When you come to questions worth five,
six, or seven marks, they may well be
divided up into smaller parts or require
quite sustained thinking and reasoning.
These higher mark questions are not so
common at Standard Grade.
feeling from time to time. It is true that
maths exams do not lend themselves to
waffle: facts and figures are absolutes. So it
is important to relax and read the questions
carefully. Do not jump in and answer what
you think or hope they are asking: answer
what is being asked.
DON’T PANIC!
If you do get a feeling of panic then
stop, put your pen down and look around
you. Tell yourself that you are as good as
anyone else in the room and if they can
do the questions then so can you. Take a
Revise
Watch out for clues in the wording:
n Any word in bold is making a point! It is
trying to draw your attention to something
important.
n “Hence” means you must (notice the
bold?) use the answer you have just found
for the next part of the question.
n “Evaluate” simply means find the
value of.
n “Show” means you must be very
precise and put down all the steps of your
reasoning.
Being allowed the use of a calculator
does not mean that you can just write down
the answer. Even correct answers could
receive no credit if there is no working out
shown.
Markers are looking for the steps
involved and want to know how you got
your answer. Marks are allocated to this
/ Practise
part of your solution – it is not the case
that all the marks are given for the final
answer. If you write down only the answer
and it is wrong, then you will receive
no marks. The same wrong answer with
working could gain you three of the four
marks.
Percentage profit
= 40/160 x 100%
= 1/4 x 100%
= 25%
2005 Paper II Question 1
Question:
A night train from London to Edinburgh
leaves at 23:21 and arrives at 06:51.
STANDARD GRADE
(a) How long does the train journey take?
GENERAL LEVEL
The following worked examples illustrate
some of the commonly asked questions at
this level:
(b) The distance from London to
Edinburgh is 644 kilometres. Find the
average speed of the train in kilometres per
hour. Give your answer correct to one
decimal place.
2005 Paper I Question 1(a)
Question:
Carry out the following calculations:
(a) 209.3 – 175.48
Answer:
You should set the answer out like this:
209.30 (Notice the insertion of 0)
– 175.48
= 33.82
The important bit here is writing 209.3 as
209.30 to fit in with the second number.
2005 Paper I Question 7
Question:
(a) Graham goes into a shop and buys a
bottle of water and a cheese roll for £1.38.
In the same shop, Alan pays £1.77 for two
bottles of water and a cheese roll. How
much does a bottle of water cost?
Answer:
Beware! You cannot use your calculator for
time calculations. You must also be careful
how you do time calculations without a
calculator – time is not decimal. The
following method is advised:
(a) We have to work out how long it is
between 23:21 and 06:51:
So, 23:21 to 24:00, 24:00 to 06:00, 06:00 to
06:51 = 39 minutes plus 6 hours plus 51
minutes
(= 6 hours 90 minutes)
= 7 hours 30 minutes
(b) Speed =
distance
time
= 644/7.5 (Note the change from
7 hours 30 minutes to
7.5 hours)
(b) Craig goes into the shop and buys four
bottles of water and three cheese rolls. How
much will this cost?
= 85.9 km/hour
Answer:
This is a reasoning question so you have to
think your way through it and show your
working. Remember your working shows
how you are thinking and can gain you
marks even if you don’t finish the question.
(b) Solve the inequality: 3x – 5 ) 13
(a) One water + cheese roll = £1.38
Two waters and cheese roll = £1.77
£1.77 – £1.38 = 39 pence
So one water alone costs 39 pence.
Answer:
(a) 5 + 3 (2x – 5)
= 5 + 6x – 15
= 6x – 10
(b) There are two ways we can do this
From (a) we know that a bottle of water
costs 39 pence, so we can work out that a
cheese roll costs:
£1.38 – £0.39 = £0.99
(b) 3x – 5 ) 13
3x ) 18
x)6
So four bottles of water and three cheese
rolls costs:
4 x 39 pence + 3 x 99 pence = £4.53
OR
From the information in part (a) we see that
we can work out the cost of four bottles of
water and three cheese rolls by adding:
2 x £1.38 + £1.77 = £4.53
That is: 2 x (bottle of water + cheese roll) +
(2 bottles of water + cheese roll).
2005 Paper I Question 8
Question:
John buys a football programme for £1.60
and sells it for £2.00. Calculate his percentage profit.
Answer:
Profit = selling price minus buying price
= £2.00 – £1.60 = 40 pence
The percentage profit is 40 pence
expressed as a percentage of the buying
price of £1.60:
Test
2005 Paper II Question 5
Question:
(a) Remove the brackets and simplify:
5 + 3 (2x – 5).
2005 paper II Question 9
Question:
Serge drives from his home in Paris
to Madrid, a journey of 1280 kilometres.
His car has a 60-litre petrol tank and
travels 13 kilometres per litre. Serge
starts his journey with a full tank of
petrol. What is the least number of times he
has to stop to refuel? Give a reason for your
answer.
Answer:
60 litre tank
13 kilometres per litre
Distance car can travel
= 60 x 13 = 780 kilometres
The car starts with a full tank so it can
travel 780 kilometres and then refuel. The
car will then be able to travel another 780
kilometres. But the distance left to travel is
1280 – 780 = 500 kilometres, so Serge only
needs to re fuel once.
Note: Working must be shown in
questions like this. An answer of “he refuels
once” with no working will receive no
marks.
Review
Turn to next page
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10 sundayherald 2 April 2006
MATHEMATICS
From previous page
2005 Paper II Question 11
Question:
A rectanglar shop sign is supported by a
metal bar, AB. The length of the shop sign is
90 centmetres and the bar AB is atached to
the wall 35 centimetres above the sign.
Calculate the size of the shaded angle AB.
Do not use a scale drawing.
side and you want to find the angle, so
using SohCahToa gives you the tan ratio.
Tan ABC = 35/90
Angle ABC = 21.30
CREDIT LEVEL
The following worked examples illustrate
some of the commonly asked questions at
this level.
2005 Paper I Question 3
Question:
Evaluate 12.5% of £140
Answer:
Since this is a non-calculator paper, we
have to recognise that
12.5% = 1/8
(If we do not then the question becomes
more complicated.)
Answer:
When you see a right-angled triangle you
should automatically think of two things:
Pythagoras and trigonometry. You have to
decide which has to be used. Here, because
you are trying to connect angles and sides,
it must be trigonometry.
You have to choose which ratio to use:
sine, cosine or tangent. In this case, you
have the opposite side and the adjacent
So, 12.5% of £140
= 1/8 of £140
= £17.50
An alternative way of working out the
percentages is:
10% of £140
2% of £140
1% of £140
0.5% of £140
= £14
= £2.80
= £1.40
= £0.70
So 12.5% of £140
= 10% + 2% + 0.5%
= £14 + £2.80 + £0.70
= £17.50
2005 Paper I Question 6
Question:
Solve the equation 2/x + 1 = 6
Answer:
In the last question we used a fraction to
make things easier. Here we want to get rid
of the fraction to make things easier. You
have to make a judgement about how to
solve a maths question – you have to use
the appropriate tool for the job.
To get rid of fractions in an equation, we
multiply throughout by the number or letter that will eliminate the denominator.
So here we multiply by x, giving:
2 + x = 6x
which is a much easier equation to solve
6x = 2 + x
5x = 2
x = 2/5
If you do not eliminate the fraction at the
beginning, the solution would be:
2/ + 1 = 6
x
2/ = 5
x
Answer:
This type of question is not normally well
done, so practice is necessary.
We need to have the same denominator
before we can add or subtract fractions.
Here the common denominator is found by
multiplying the two denominators m and
(m+1). Remember to multiply the top of the
fraction as well.
= 3(m+1) +
m (m+1)
4m
m (m+1)
=
3(m+1) + 4m
m(m + 1)
=
3m + 3 + 4m
m(m + 1)
=
7m + 3
m(m+1)
2005 Paper II Question 8
Question:
The side length of a cube is 2x centimetres
The expression for the volume in cubic
centimetres is equal to the expression for
the surface area in square centimetres.
Calculate the side length of the cube.
2/ = 5/
x
1
Now we can cross-multiply to get:
5x = 2
x = 2/5
2005 Paper I Question 9
Question:
(a) Emma puts £30 worth of petrol into the
empty fuel tank of her car. Petrol costs 75
pence per litre. Her car uses 5 litres of petrol
per hour, when she drives at a particular
constant speed. At this constant speed, how
many litres of petrol will remain in the car
after 3 hours?
(b) The next week, Emma puts £20 worth of
petrol into the empty fuel tank of her car.
Petrol costs c pence per litre. Her car uses k
litres of petrol per hour, when she drives at
another constant speed. Find a formula for
R, the amount of petrol remaining in the
car after t hours
Answer:
This question is in two parts. The first part
is knowledge and the second part reasoning. The first part is a lead-in – you should
use your method in part (a) to help you with
part (b). Always show all your working,
especially in reasoning questions.
(a) Emma buys £30 of petrol. The cost of
petrol is 75p per litre. So we can find how
many litres she bought by dividing £30 by
75p. Note: watch the units.
Answer:
This is a reasoning question so we have to
think our way through the steps required
and write down all our working.
The volume of a cube is the length of the
side cubed.
So V = 2x x 2x x 2x = 8x3
Surface area is the area of all six sides.
The area of one side is 2x x 2x
So surface area is 6 x 2x x 2x = 24x2
So
8x3 = 24x2
8x3 – 24x2 = 0
8x2(x – 3) = 0
So since x cannot be zero,
x=3
and so the side length is 6cm.
2005 Paper II Question 9
Question:
The monthly bill for a mobile phone is made
up of a fixed rental plus call charges. Call
charges vary as the time used.
The relationship between the monthly
bill, y (pounds), and the time used, x
(minutes) is represented in the graph below.
Litres = 3000/75
[note we have changed £30 into 3000 pence
to keep the units the same]
= 40
In 3 hours Emma used 5 x 3 litres = 15 litres.
So (40 – 15) litres remains
25 litres remains
(b) Here we use the same method as in part
(a) but with some letters this time:
Emma buys £20 of petrol
The cost of petrol is c pence per litre
Number of litres bought is 2000/c
Number of litres used is k x t
So R = 2000/c – kt
2004 Paper I Question 4
Question:
+
4
Simplify
3
m
(m+1)
(a) Write down the fixed rental.
(b) Find the call charge per minute.
Answer:
This is another reasoning question.
(a) The fixed charge is £10, which is found
from where the line crosses the y axis.
(b) From the graph we see that the cost for
60 minutes is £13, which includes the fixed
charge of £10.
So the call charge for 60 minutes is £3.
Find the cost per minute by dividing £3 by
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2 April 2006 sundayherald 11
MATHEMATICS
60 minutes. (Change £3 to 300 pence for
ease of working.)
Charge per minute = 300/60 = 5 pence.
The cosine rule is given in the formula
sheet as:
a2 = b2 + c2 – 2bc cosA
2004 paper II Question 7
Question:
A square trapdoor of side 80cm is held open
by a rod as shown.
In this question we will take the side we are
wanting to calculate as a
a2 = 802 + 402 – 2 x 80 x 40 x cos 76º
So
(Be careful to work out the whole of
2 x 80 x 40 x cos 76º before subtracting it.)
a2 = 8000 – 1548.3
a2 = 6451.7
(Remember to take the square root!)
a = 80.3
The rod is attached to a corner of the
trapdoor and placed 40cm along the
edge of the opening. The angle between the
trapdoor and the opening is 76o. Calculate
the length of the rod to two significant
figures.
Answer:
There is always a question on non-rightangled trigonometry in the Credit exam.
This work covers the sine rule, the cosine
rule and the area of a triangle. All of the
necessary formulae are given inside the
front cover of the question paper.
Since you are being asked to find the
length of a side in this triangle, you use
either the sine rule or cosine rule. Because
the information you are given consists of
two sides and the angle between them, you
could move straight to the cosine rule as
the sine rule will not work (neither will it
work if you are given only three sides or
three angles).
•
The question asked for the answer to be
given correct to two significant figures,
remember to round:
a = 80
HIGHER GRADE
Remember that at Higher Level 60% of the
marks are allocated at grade C level, so
there will be questions that you can do, and
do well.
The questions in Paper I are likely to be
shorter than those in Paper II and more to
the point. It should be your aim to build up
a good mark in Paper I to make Paper II less
daunting. The early questions in Paper I
should be reasonably straightforward and
explicit. As you work through the paper, the
questions will get trickier, but they should
still be fairly direct.
In Paper II you will find the questions
longer and more problem-based. This is
particularly true of the later ones. If you
have scored well in Paper I, then a couple of
good questions early on in Paper II and you
could be over the magic 50%.
WORKED EXAMPLES
The following examples illustrate the different types of questions you can expect in
the exam. The first six are graded C so you
should all be able to attempt them with an
expectation of success. There is then a B
and an A graded question.
While the last two are more difficult it is
important that you should try to get as much
as possible out of every question and realise
that you can gain partial marks. Don’t disregard questions because they look daunting,
parts may be quite straightforward.
2005 Paper I Question 5
Question:
Differentiate (1 + 2 sin x)4 with respect to x.
Answer:
Let f(x) = (1 + 2sin x)4
Using the chain rule to differentiate, we get:
f '(x) = 4 (1 + 2sin x)3 x 2 cos x
= 8 cos x (1 + 2sin x)3
Remember that you must first differentiate
the bracket, then differentiate what is
inside the bracket.
2005 Paper I Question 9
Question:
If cos 2x = 7/25 and 0 < x < W/2 , find the exact
values of cos x and sin x.
Answer:
cos 2x
= 7/25
cos 2x
= 2cos2 x – 1
2cos2 x – 1 = 7/25
2
2cos x = 1 + 7/25
2cos2 x = 32/25
cos2 x
= 16/25
cos x
= 4/5
From the 3,4,5 right-angled triangle
sin x
= 3/5
FOR PUPILS
web links for homework and revision
•
FOR PARENTS
an essential guide to
•
National Qualifications
FOR TEACHERS
free classroom resources
visit www.LTScotland.org.uk/NQ
to discover more and to download this exam guide
2005 paper II Question 3
Question:
(a) A chord joins the points A (1,0) and B
(5,4) on the circle as shown on the diagram.
Show that the equation of the perpendicular
bisector of chord AB is x + y = 5.
y
y
B (5,4)
C
A (1,0)
O A (1,0)
x
O
x
(b) The point C is the centre of this circle.
The tangent at point A on the circle has
equation x + 3y = 1.
Find the equation of the radius CA.
Answer:
(a) Mid point of AB is ((1+5)/2,(0+4)/2) = (3,2)
Gradient of AB = 4-0/5-1 = 4/4 = 1
Gradient of perpendicular bisector
=-1{m1 xm2=-1)
So we need the equation of the line with
gradient -1 which passes through (3,2).
We n ow u s e o n e o f t h e m o s t u s e f u l
equations of the straight line:
y – b = m(x – a)
so:
y – 2 = -1(x – 3)
which simplifies to y = 5 – x
(b)
x + 3y = 1
3y = 1 – x
y = 1/3 – x/3
y = –1/3 x + 1/3
Turn to next page
12 sundayherald 2 April 2006
MATHEMATICS
From previous page
so the gradient of the radius is -1/3 meaning
that the gradient of the tangent is 3 as the
tangent is perpendicular to the radius.
In the dark, Andrew and Bob locate Tracy
using heat-seeking beams.
B(–12, 0, 9)
A(23, 0, 8)
Again: y – b = m(x – a)
A is (1,0)
So
y – 0 = 3(x – 1)
Which simplifies to
y = 3x – 3
y=5–x
and
y = 3x – 3
Taking these equations together we get:
5 – x = 3x – 3
-x -3x = -3 -5
-4x = -8
x=2
substituting gives
y=3
So C is the point (2,3)
2005 Paper II Question 4
Question:
The sketch shows the positions of Andrew
(A), Bob (B) and Tracy (T) on three hill-tops.
Relative to a suitable origin, the
co-ordinates (in hundreds of metres) of the
three people are A (23, 0, 8), B (-12, 0, 9) and
T (28, -15, 7).
c
TA = `(-5)2 + 152 + 12 =
c
TB = `402 + 152 + 22 =
`251
`1829
Then we evaluate the scalar product by
multiplying the corresponding components and adding:
c c
TA.TB = (-5) x (-40) + 15 x 15 + 1 Op{
= 200 + 225 + 2
= 427
(c) (i)
(c) (ii) To find the equation of the circle we
use:
(x –a)2 + (y – b)2 = r2
a = 2 and b = 3
the value of r is found by using the distance
formula:
r2 = (2-1)2 + (3 -0)2 = 12 + 32 = 10
So the equation of the circle is:
(x –2)2 + (y – 3)2 = 10
The lengths of the two vectors are found:
T(28, –15, 7)
c
c
(a) Express the vectors TA and TB in component form.
(b) Calculate the angle between these two
beams.
Answer:
(a) c
TA =
c
TB =
( )( )
( )( )
23 – 28
0 + 15 =
8–7
-5
15
1
12 – 28
-40
0 + 15 = 15
9–7
2
To find the angle between two lines, you
consider the vectors represented by the
lines and use the scalar product. The scalar
product can be evaluated in two ways.
By equating the two approaches it is
possible to find the angle between the two
vectors and hence the angle between the
two lines.
Now we apply the other method of evaluating the product. This method involves
multiplying the lengths of the two vectors
and the cosine of the angle between them.
We apply the product and equate it to 427
c c
TA x TB x cos A = 427
So cos A =
and A
427
` 251 x Hw{>p
= 0.6302…
= 50.9
The diagram shows the graph of
y = 24 , x > 0.
`x
Find the equation of the tangent at P, where
x = 4.
Answer:
To find the equation of a straight line we
need to know a point on the line and the
gradient of the line. Most often we are given
a point, or it is straightforward to find. There
are several ways of finding the gradient. Use
whichever method is appropriate to the
information given in the question.
Here we are asked to find the equation of a
tangent to a curve and we are given the equation of the curve. We need to remember that
the gradient of a curve, or the gradient of the
tangent to the curve, is found by differentiating the equation of the curve and substituting the value of the x co-ordinate of the point.
So y = 24 = 24x-1/2
`x
dy = – 1 x 24x -3/2
2
dx
x=4
2005 Paper II Question 6
Question:
dy = – 1 x 24 x 4 -3/2
dx
2
y
= – 12 x 1
`43
P
= – 12 x 1
23
y = 24
`x
= – 12 x 1/8
O
x
4
= -3/2
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2 April 2006 sundayherald 13
MATHEMATICS
Calculators used to look like this – it sure was hard to replace the batteries …
So the gradient of the tangent is: -3/2
Photograph: Rex Features
coefficients of the quadratic factor left
when (x +1) is removed.
f(x) = (x + 1)(x2 - 2x – 3)
= (x + 1)(x + 1)(x – 3)
x =4
y = 24
`4
So the point P is (4,12)
(b) Two of the factors of f(x) are the same,
so this indicates that the turning point is
found by equating the double factor to
zero and so the answer is: (-1,0)
The equation of the tangent is now found
by applying the formula:
y – b = m(x – a)
y – 12 = – 3 (x – 4)
2
2y – 24 = -3(x – 4)
2y – 24 = -3x + 12
2y + 3x = 36
2005 Paper I Question 4
Question:
Functions f(x) = 3x – 1 and g(x) = x2 + 7 are
defined on the set of real numbers.
(a) Find h(x) where h(x) = g(f(x)).
(b) (i) Write down the co-ordinates of the
minimum turning point of y = h(x).
(b) (ii) Hence state the range of the function h.
2004 Paper I Question 2
Question:
f (x) = x3 – x2 – 5x – 3.
(a) (i) Show that (x + 1) is a factor of f (x).
(ii) Hence or otherwise factorise f (x) fully.
(b) One of the turning points of the graph of
y = f (x) lies on the x-axis. Write down the
co-ordinates of this turning point.
Answer:
(a) f(x) = 3x – 1
g(x) = x2 + 7
h(x) = g(f(x))
= g(3x – 1)
Answer:
To show that (x + 1) is a factor of f (x), we
need to show that f(-1) = 0. While this can
be done fairly easily by substitution, it is
better to use synthetic division as it helps
with the rest of the question.
(a) (i)
(b) (i) We are asked to write down the coordinates. This means there should be no
extra working to do, we should be able to
look at our answer to (a) and see the answer
to (b). But the simplified answer to (a) is not
helpful, as it does not allow us to write down
what we need. If we look two lines before,
we should recognise the format (x + a)2 + b.
This is the completed square format.
Completing the square is a fairly routine
task and we should not have a problem if
asked to do it. But recognising the completed square format is not so easy. If we do
recognise it, then we can indeed just write
down the answer.
One of the uses of the completing the
square is that it allows us to write down the
turning point of a quadratic function. If
we complete the square to get the form
(x + a)2 + b, then the turning point is (-a,b).
= 12
–1
1
1
–1
–5
–3
–1
2
3
–2
–3
0
The 0 at the end shows that f(-1) = 0 and so
(x + 1) is a factor
(ii) To factorise fully, we use the bottom line
entries in the table. The 1,-2 and -3 from
the bottom line of the table are the
Revise
This is using the fact that log A – log B = log
A for same base logs.
B
(ii) Since this is a minimum turning point,
the lowest value y can take is 7. So the range
of the function (ie, the possible values of y)
must be:
y )7
Now we use the fact that logay = x is the
same as y = ax:
2005 Paper II Question 7
Question:
Solve the equation
log4(5–x) – log4 (3–x) = 2, x<3
42 = 5 – x
3–x
16(3 – x) = 5 – x
48 – 16x = 5 – x
-16x + x = 5 – 48
Answer:
log4(5 – x) – log4(3 – x) = 2
-15x = -43
log4 5 – x = 2
3–x
x = 43
15
MATHEMATICS
EXAM TIMETABLE
= (3x – 1)2 + 7
= 9x2 -6x +1 + 7
= 9x2 -6x + 8
/ Practise
So from (3x – 1)2 + 7 we can write down that
the turning point is
(1/3, 7)
Level/Paper
Time
Friday May 5
Foundation
Paper 1 (Non-calculator) 9am-9.20am
Foundation
Paper 2
9.40am-10.20am
General
Paper 1 (Non-calculator)
10.40am11.15am
General
Paper 2
11.35am-12.30pm
Credit
Paper 1 (Non-calculator)
1.30pm2.25pm
Credit
Paper 2
2.45pm-4.05pm
Test
Level/Paper
Time
Friday May 19
Higher
Paper 1 (Non-calculator)9am-10.10am
Higher
Paper 2
10.30am-noon
Intermediate 1
Paper 1 (Non-calculator) 1pm-1.35pm
Intermediate 1
Paper 2
1.55pm-2.50pm
Intermediate 2
Paper 1 (Non-calculator)
1pm-1.45
Intermediate 2
Paper 2
2.05pm-3.35pm
Advanced Higher
1pm-4pm
Review
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