YEAR 11 INTO YEAR 12 BRIDGING UNITSfluencycontent2-schoolwebsite.netdna-ssl.com/FileCluster/...1 A burger contains 4 500 000 J of energy. Write this in: a kilojoules b megajoules

YEAR 11 INTO YEAR 12

BRIDGING UNITS Key contacts: LBTS6

Any questions about entry requirements - please email;

Mr. Souch (Head of Sixth Form) [email protected]

Any questions about Bridging Units, ‘Insight Into’ Days or how to prepare for LBTS6 - please email;

Miss Akowe (LBTS6 Achievement Coordinator) [email protected]

If you would like to change your first choice courses- please e-mail;

Ms.Deniran (LBTS6 Pastoral Officer) [email protected]

Why do you have to complete ‘Bridging Units’?

Around the UK, many students have struggled with the transition between Year 11 and Year 12. This is primarily due to the increase in difficulty between GCSE and A-level/BTec Level 3 as well as students’ misunderstanding about the level of academic independence required to become an ‘outstanding’ sixth form student. As a result, LBTS6 has put together ‘Bridging Units’ for every A-level and BTec course to ensure that every student is well prepared for starting Year 12 in September 2020. What subjects do LBTS offer? Every prospective LBTS6 student should have chosen three subjects from different option blocks below. If you would like to change your first choice subjects or you have seen a clash (e.g. two of your first choice subjects are in the same option block), please can you e-mail Ms. Deniran at [email protected] as soon as possible.

Option 1 Option 2 Option 3 Option 4 Option 5

A‐level Chemistry A‐level Sociology A‐level Economics A‐level Law A‐level Psychology

A‐level Computer Science A‐level History A‐level Biology A‐level Maths A‐level Further Maths

A‐level Religious Studies A‐level Physics A‐level English

BTEC Media

(equivalent to 1 A‐level)

BTec Performing Arts

(Equivalent to 1 A‐level)

BTec Applied Science


BTec Sport


BTec Business Diploma

(Equivalent to 2 A‐

levels)

BTec Health and Social

Care (equivalent to

1 A‐level)

BTec ICT

(Equivalent to 1 A‐level) BTec Business Diploma

(Equivalent to 2 A‐levels)

BTec Business Ext

Cert

(Equivalent to 1 A‐

level)

Languages: If you are a native speaker of French or Spanish, LBTS6 also offers both as an A-level course. This must be taken as an additional fourth subject. Expectations for student completion of ‘Bridging Units’ LBTS6 expects EVERY student to complete the Bridging Units in ALL the subjects they have applied for. For instance, if you have applied for A-level Economics, Psychology and Law - you ONLY complete the bridging units for those subjects. If you want to change your subjects, then please complete the subjects in which you WANT TO STUDY not the ones you have applied for. You should email Ms. Deniran at [email protected] if you wish to change your first choice subjects. The bridging unit for each subject must be handed in (PRINTED) on GCSE Results Day on the 20th August. If this is not complete, LBTS6 reserves the right to not allow you onto your first choice courses. Entry Requirements: This information is available on the website at www.lilianbaylis.com/lbts6. If you want to ask any questions, you can e-mail Mr. Souch at [email protected]

BRIDGING UNIT: A-level BIOLOGY Contact Email: [email protected]

On the following pages there is some of the content and skills which you will need to be able to do in order for you to be successful on this course.

You will need to spend time learning and memorising the content on the pages you have been given. You need to hand in answers to all of the questions on a sheet of paper.

You will be tested on this knowledge in your first lesson back.

GCSE → A Level transition

Student sheet

AQA Biology

© Oxford University Press 2019 http://www.oxfordsecondary.co.uk/acknowledgements

This resource sheet may have been changed from the original 1

Transition from GCSE to A Level

Moving from GCSE Science to A Level can be a daunting leap. You’ll be expected to remember a lot more facts, equations, and definitions, and you will need to learn new maths skills and develop confidence in applying what you already know to unfamiliar situations.

This worksheet aims to give you a head start by helping you:

to pre-learn some useful knowledge from the first chapters of your A Level course

understand and practice of some of the maths skills you’ll need.

Learning objectives

After completing the worksheet you should be able to:

define practical science key terms

recall the answers to the retrieval questions

perform maths skills including:

o converting between units, standard form, and prefixes

o using significant figures

o rearranging formulae

o magnification calculations

o calculating percentages, errors, and uncertainties

o drawing and interpreting line graphs.

http://www.oxfordsecondary.co.uk/acknowledgements


Student sheet

AQA Biology



Retrieval questions You need to be confident about the definitions of terms that describe measurements and results in A Level Biology.

Learn the answers to the questions below then cover the answers column with a piece of paper and write as many answers as you can. Check and repeat.

Practical science key terms

When is a measurement valid? when it measures what it is supposed to be measuring

When is a result accurate? when it is close to the true value

What are precise results? when repeat measurements are consistent/agree closely with each

other

What is repeatability? how precise repeated measurements are when they are taken by

the same person, using the same equipment, under the same

conditions

What is reproducibility? how precise repeated measurements are when they are taken by

different people, using different equipment

What is the uncertainty of a measurement? the interval within which the true value is expected to lie

Define measurement error the difference between a measured value and the true value

What type of error is caused by results varying

around the true value in an unpredictable way?

random error

What is a systematic error? a consistent difference between the measured values and true

values

What does zero error mean? a measuring instrument gives a false reading when the true value

should be zero

Which variable is changed or selected by the

investigator?

independent variable

What is a dependent variable? a variable that is measured every time the independent variable is

changed

Define a fair test a test in which only the independent variable is allowed to affect the

dependent variable

What are control variables? variables that should be kept constant to avoid them affecting the

dependent variable



Student sheet

AQA Biology



Biological molecules


What are monomers? smaller units from which larger molecules are made

What are polymers? molecules made from a large number of monomers joined together

What is a condensation reaction? a reaction that joins two molecules together to form a chemical

bond whilst eliminating of a molecule of water

What is a hydrolysis reaction? a reaction that breaks a chemical bond between two molecules and

involves the use of a water molecule

What is a monosaccharide? monomers from which larger carbohydrates are made

How is a glycosidic bond formed? a condensation reaction between two monosaccharides

Name the three main examples of

polysaccharides.

glycogen, starch, cellulose

Describe Benedict’s test for reducing sugars gently heat a solution of a food sample with an equal volume of

Benedict’s solution for five minutes, the solution turns orange/brown

if reducing sugar is present

Name the two main groups of lipids phospholipids, triglycerides (fats and oils)

Give four roles of lipids source of energy, waterproofing, insulation, protection

What is an ester bond? a bond formed by a condensation reaction between glycerol and a

fatty acid

Describe the emulsion test for lipids mix the sample with ethanol in a clean test tube, shake the sample,

add water, shake the sample again, a cloudy white colour indicates

that lipid is present

What are the monomers that make up proteins? amino acids

Draw the structure of an amino acid

How is a peptide bond formed? a condensation reaction between two amino acids

What is a polypeptide? many amino acids joined together

Describe the biuret test for proteins mix the sample with sodium hydroxide solution at room

temperature, add very dilute copper(II) sulfate solution, mix gently,

a purple colour indicates that peptide bonds are present

How does an enzyme affect a reaction? it lowers the activation energy

Give five factors which can affect enzyme action. temperature, pH, enzyme concentration, substrate concentration,

inhibitor concentration

What is a competitive inhibitor? a molecule with a similar shape to the substrate, allowing it to

occupy the active site of the enzyme

What is a non-competitive inhibitor? a molecule that changes the shape of the enzyme by binding

somewhere other than the active site.



Student sheet

AQA Biology



Maths skills

1 Numbers and units

1.1 Units and prefixes

A key criterion for success in biological maths lies in the use of correct units and the management of numbers. The units scientists use are from the Système Internationale – the SI units. In biology, the most commonly used SI base units are metre (m), kilogram (kg), second (s), and mole (mol). Biologists also use SI derived units, such as square metre (m

2), cubic metre

(m3), degree Celsius (°C), and litre (l).

To accommodate the huge range of dimensions in our measurements they may be further modified using appropriate prefixes. For example, one thousandth of a second is a millisecond (ms). Some of these prefixes are illustrated in the table below.

Multiplication factor Prefix Symbol

109 giga G

106

mega M

103

kilo k

10–2

centi c

10–3

milli m

10–6

micro µ

10–9

nano n

Practice questions

1 A burger contains 4 500 000 J of energy. Write this in:

a kilojoules b megajoules.

2 HIV is a virus with a diameter of between 9.0×10−8

m and 1.20×10−7

m.

Write this range in nanometres.

1.2 Powers and indices

Ten squared = 10 × 10 = 100 and can be written as 102. This is also called ‘ten to the

power of 2’.

Ten cubed is ‘ten to the power of three’ and can be written as 103 = 1000.

The power is also called the index.

Fractions have negative indices:

one tenth = 10−1

= 1/10 = 0.1

one hundredth = 10−2

= 1/100 = 0.01

Any number to the power of 0 is equal to 1, for example, 290 = 1.

If the index is 1, the value is unchanged, for example, 171 = 17.

When multiplying powers of ten, you must add the indices.

So 100 × 1000 = 100 000 is the same as 102 × 10

3 = 10

2 + 3 = 10

5



Student sheet

AQA Biology



When dividing powers of ten, you must subtract the indices.

So 100/1000 = 1/10 = 10−1

is the same as 102/10

3 = 10

2 − 3 = 10

−1

But you can only do this when the numbers with the indices are the same.

So 102 × 2

3 = 100 × 8 = 800

And you can’t do this when adding or subtracting.

102 + 10

3 = 100 + 1000 = 1100

102 − 10

3 = 100 − 1000 = −900

Remember: You can only add and subtract the indices when you are multiplying or dividing the numbers, not adding or subtracting them.

Practice questions

3 Calculate the following values. Give your answers using indices.

a 108 × 10

3 b 10

7 × 10

2 × 10

3

c 103 + 10

3 d 10

2 − 10

−2

4 Calculate the following values. Give your answers with and without using indices.

a 105 ÷ 10

4 b 10

3 ÷ 10

6

c 102 ÷ 10

−4 d 100

2 ÷ 10

2

1.3 Converting units

When doing calculations, it is important to express your answer using sensible numbers. For example, an answer of 6230 μm would have been more meaningful expressed as 6.2 mm.

If you convert between units and round numbers properly, it allows quoted measurements to be understood within the scale of the observations.

To convert 488 889 m into km:

A kilo is 103 so you need to divide by this number, or move the decimal point three places to the

left.

488 889 ÷ 103 = 488.889 km

However, suppose you are converting from mm to km: you need to go from 103 to 10

−3, or move

the decimal point six places to the left.

333 mm is 0.000 333 km

Alternatively, if you want to convert from 333 mm to nm, you would have to go from 10−9

to 10−3

, or move the decimal point six places to the right.

333 mm is 333 000 000 nm

Practice question

5 Calculate the following conversions:

a 0.004 m into mm b 130 000 ms into s

c 31.3 ml into μl d 104 ng into mg

6 Give the following values in a different unit so they make more sense to the reader.

Choose the final units yourself. (Hint: make the final number as close in magnitude to zero as you can. For example, you would convert 1000 m into 1 km.)

a 0.000 057 m b 8 600 000 μl c 68 000 ms d 0.009 cm



Student sheet

AQA Biology



2 Decimals, standard form, and significant figures

2.1 Decimal numbers

A decimal number has a decimal point. Each figure before the point is a whole number, and the figures after the point represent fractions.

The number of decimal places is the number of figures after the decimal point. For example, the number 47.38 has 2 decimal places, and 47.380 is the same number to 3 decimal places.

In science, you must write your answer to a sensible number of decimal places.

Practice questions

1 New antibiotics are being tested. A student calculates the area of clear zones in Petri dishes in which the antibiotics have been used. List these in order from smallest to largest.

0.0214 cm2 0.03 cm

2 0.0218 cm

2 0.034 cm

2

2 A student measures the heights of a number of different plants. List these in order from smallest to largest.

22.003 cm 22.25 cm 12.901 cm 12.03 cm 22 cm

2.2 Standard form

Sometimes biologists need to work with numbers that are very small, such as dimensions of organelles, or very large, such as populations of bacteria. In such cases, the use of scientific notation or standard form is very useful, because it allows the numbers to be written easily.

Standard form is expressing numbers in powers of ten, for example, 1.5×107 microorganisms.

Look at this worked example. The number of cells in the human body is approximately 37 200 000 000 000. To write this in standard form, follow these steps:

Step 1: Write down the smallest number between 1 and 10 that can be derived from the number to be converted. In this case it would be 3.72

Step 2: Write the number of times the decimal place will have to shift to expand this to the original number as powers of ten. On paper this can be done by hopping the decimal over each number like this:

until the end of the number is reached.

In this example that requires 13 shifts, so the standard form should be written as 3.72×10

13.

For very small numbers the same rules apply, except that the decimal point has to hop backwards. For example, 0.000 000 45 would be written as 4.5×10

−7.

Practice questions

3 Change the following values to standard form.

a 3060 kJ b 140 000 kg c 0.000 18 m d 0.000 004 m

4 Give the following numbers in standard form.

a 100 b 10 000 c 0.01 d 21 000 000



Student sheet

AQA Biology



5 Give the following as decimals.

a 106

b 4.7×109

c 1.2×1012

d 7.96×10−4

2.3 Significant figures

When you use a calculator to work out a numerical answer, you know that this often results in a large number of decimal places and, in most cases, the final few digits are ‘not significant’. It is important to record your data and your answers to calculations to a reasonable number of significant figures. Too many and your answer is claiming an accuracy that it does not have, too few and you are not showing the precision and care required in scientific analysis.

Numbers to 3 significant figures (3 s.f.):

7.88 25.4 741

Bigger and smaller numbers with 3 significant figures:

0.000 147 0.0147 0.245 39 400 96 200 000 (notice that the zeros before the figures and after the figures are not significant – they just show you how large the number is by the position of the decimal point).

Numbers to 3 significant figures where the zeros are significant:

207 4050 1.01 (any zeros between the other significant figures are significant).

Standard form numbers with 3 significant figures:

9.42×10−5

1.56×108

If the value you wanted to write to 3.s.f. was 590, then to show the zero was significant you would have to write:

590 (to 3.s.f.) or 5.90 × 102

Remember: For calculations, use the same number of figures as the data in the question with the lowest number of significant figures. It is not possible for the answer to be more accurate than the data in the question.

Practice question

6 Write the following numbers to i 2 s.f. and ii 3 s.f.

a 7644 g

b 27.54 m

c 4.3333 g

d 5.995×102

cm3

7 The average mass of oxygen produced by an oak tree is 11800 g per year.

Give this mass in standard form and quote your answer to 2 significant figures.

3 Working with formulae

It is often necessary to use a mathematical formula to calculate quantities. You may be tested on your ability to substitute numbers into formulae or to rearrange formulae to find specific values.

3.1 Substituting into formulae

Think about the data you are given in the question. Write down the equation and then think about how to get the data to substitute into the equation. Look at this worked example.



Student sheet

AQA Biology



A cheek cell has a 0.06 mm diameter. Under a microscope it has a diameter 12 mm. What is the magnification?

magnification = image size (mm) ÷ object size (mm) or M = O

I

Substitute the values and calculate the answer:

M = 12 mm/0.06 mm = 12/0.06 = 200

Answer: magnification = ×200 (magnification has no units)

Sometimes an equation is more complicated and the steps need to be carried out in a certain order to succeed. A general principle applies here, usually known by the mnemonic BIDMAS. This stands for Brackets, Indices (functions such as squaring or powers), Division, Multiplication, Addition, Subtraction.

Practice questions

1 Calculate the magnification of a hair that has a width of 6.6 mm on a photograph. The hair is 165 µm wide.

2 Estimate the area of a leaf by treating it as a triangle with base 2 cm and height 9 cm.

3 Estimate the area of a cell by treating it as a circle with a diameter of 0.7 µm. Give your answer in µm

2.

4 An Amoeba population starts with 24 cells. Calculate how many Amoeba cells would be present in the culture after 7 days if each cell divides once every 20 hours. Use the equation Nt = N0 × 2

n where Nt = number after time t, N0 = initial population, n = number of divisions in

the given time t.

5 In a quadrat sample, an area was found to contain 96 aphids, 4 ladybirds, 22 grasshoppers,

and 3 ground beetles. Calculate the diversity of the site using the equation D =

2

1

N

n

where n = number of each species, N = grand total of all species, and D = diversity.

Remember: In this equation there is a part that needs to be done several times then summed, shown by the symbol Σ.

3.2 Rearranging formulae

Sometimes you will need to rearrange an equation to calculate the answer to a question. For example, the relationship between magnification, image size, and actual size of specimens in

micrographs usually uses the equation M = O

I, where M is magnification, I is size of the image,

and O = actual size of the object.

You can use the algebra you have learnt in Maths to rearrange equations, or you can use a triangle like the one shown.

Cover the quantity you want to find. This leaves you with either a fraction or a multiplication:

M = I ÷ O O = I ÷ M I = M × O



Student sheet

AQA Biology



Practice questions

6 A fat cell is 0.1 mm in diameter. Calculate the size of the diameter seen through a microscope with a magnification of ×50.

7 A Petri dish shows a circular colony of bacteria with a cross-sectional area of 5.3 cm2.

Calculate the radius of this area.

8 In a photograph, a red blood cell is 14.5 mm in diameter. The magnification stated on the image is ×2000. Calculate the real diameter of the red blood cell.

9 Rearrange the equation 34 = 2a/135 × 100 and find the value of a.

10 The cardiac output of a patient was found to be 2.5 dm3

min−1

and their heart rate was 77 bpm. Calculate the stroke volume of the patient.

Use the equation: cardiac output = stroke volume × heart rate.

11 In a food chain, efficiency = in taken biomass

dtransferre biomass × 100

A farmer fed 25 kg of grain to his chicken. The chicken gained weight with an efficiency of 0.84. Calculate the weight gained by the chicken.

4 Magnification

To look at small biological specimens you use a microscope to magnify the image that is observed. The microscope was developed in the 17th century. Anton van Leeuwenhoek used a single lens and Robert Hooke used two lenses. The lenses focus light from the specimen onto your retina to produce a magnified virtual image. The magnification at which observations are made depends on the lenses used.

4.1 Calculating the magnifying power of lenses

Lenses each have a magnifying power, defined as the number of times the image is larger than the real object. The magnifying power is written on the lens.

To find the magnification of the virtual image that you are observing, multiply the magnification powers of each lens used. For example, if the eyepiece lens is ×10 and the objective lens is ×40 the total magnification of the virtual image is 10 × 40 = 400.

Practice questions

1 Calculate the magnification of the virtual image produced by the following combinations of lenses:

a objective ×10 and eyepiece ×12 b objective ×40 and eyepiece ×15

4.2 Calculating the magnification of images

Drawings and photographs of biological specimens should always have a magnification factor stated. This indicates how much larger or smaller the image is compared with the real specimen.

The magnification is calculated by comparing the sizes of the image and the real specimen. Look at this worked example.

The image shows a flea which is 1.3 mm long. To calculate the magnification of the image, measure the image (or the scale bar if given) on the paper (in this example, the body length as indicated by the line A–B).



Student sheet

AQA Biology



For this image, the length of the image is 42 mm and the length of the real specimen is 1.3 mm.

magnification = specimen real of length

image of length = 42/1.3 = 32.31

The magnification factor should therefore be written as ×32.31

Remember: Use the same units. A common error is to mix units when performing these calculations. Begin each time by converting measurements to the same units for both the real specimen and the image.

Practice question

2 Calculate the magnification factor of a mitochondrion that is 1.5 µm long.

4.3 Calculating real dimensions

Magnification factors on images can be used to calculate the actual size of features shown on drawings and photographs of biological specimens. For example, in a photomicrograph of a cell, individual features can be measured if the magnification is stated. Look at this worked example.

The magnification factor for the image of the open stoma is ×5000.

This can be used to find out the actual size of any part of the cell, for example, the length of one guard cell, measured from A to B.

Step 1: Measure the length of the guard cell as precisely as possible. In this example the image of the guard cell is 52 mm long.

Step 2: Convert this measurement to units appropriate to the image. In this case you should use µm because it is a cell.

So the magnified image is 52 × 1000 = 52 000 µm

Step 3: Rearrange the magnification equation (see Topic 3.2) to get:

real size = size of image/magnification = 52 000/5000 = 10.4

So the real length of the guard cell is 10.4 µm.



Student sheet

AQA Biology



Practice question

3 Use the magnification factor to determine the actual size of a bacterial cell.

5 Percentages and uncertainty

A percentage is simply a fraction expressed as a decimal. It is important to be able to calculate routinely, but is often incorrectly calculated in exams. These pages should allow you to practise this skill.

5.1 Calculating percentages as proportions

To work out a percentage, you must identify or calculate the total number using the equation:

percentage = number total

number total of percentage a as want younumber× 100%

For example, in a population, the number of people who have brown hair was counted.

The results showed that in the total population of 4600 people, 1800 people had brown hair.

The percentage of people with brown hair is found by calculating:

people of number total

hair brown withpeople of number × 100

= 4600

1800 × 100 = 39.1%

Practice questions

1 The table below shows some data about energy absorbed by a tree in a year and how some of it is transferred.

Energy absorbed by the tree in a year 3 600 000 kJ/m2

Energy transferred to primary consumers 2240 kJ/m2

Energy transferred to secondary consumers 480 kJ/m2

Calculate the percentage of energy absorbed by the tree that is transferred to

a primary consumers b secondary consumers.

2 One in 17 people in the UK has diabetes.

Calculate the percentage of the UK population that have diabetes.



Student sheet

AQA Biology



5.2 Calculating the percentage change

When you work out an increase or a decrease as a percentage change, you must identify, or calculate, the total original amount:

% increase = amount original

increase × 100

% decrease = amount original

decrease × 100

Remember: When you calculate a percentage change, use the total before the increase or decrease, not the final total.

Practice questions

3 Convert the following mass changes as percentage changes.

Sucrose conc. / mol dm

−3

Initial mass / g Final mass / g Mass change /

g Percentage

change in mass

0.9 1.79 1.06

0.7 1.86 1.30

0.5 1.95 1.70

0.3 1.63 1.76

0.1 1.82 2.55

5.3 Measurement uncertainties

When you measure something, there will always be a small difference between the measured value and the true value. This may be because of the size of the scale divisions on your measuring equipment, or the difficulty of taking the measurement. This is called an uncertainty.

To estimate the uncertainty of a measurement with an instrument with a marked scale such as a ruler, a good rule of thumb is to let the uncertainty be equal to half the smallest division on the scale being used.

Using a ruler with a mm scale, the length of the leaf seems to be 74 mm. The smallest division is 1 mm, so the uncertainty is 0.5 mm.

The true length is therefore 74 mm +/− 0.5 mm.



Student sheet

AQA Biology



Practice question

4 Give the uncertainty for the following pieces of equipment:

a large measuring cylinder with 2 cm3

divisions

b digital stopwatch timer measuring to the nearest hundredth of a second

c thermometer with 0.1 °C divisions.

5.4 Calculating percentage uncertainties

The uncertainty is the range of possible error either side of the true value due to the scale being used, so the value recorded for the measurement = closest estimate +/− uncertainty.

The difference between the true value and the maximum or minimum value is called the absolute error.

Once the absolute error has been established for a particular measurement, it is possible to express this as a percentage uncertainty or relative error. The calculation to use is:

relative error = value measured

error absolute× 100%

In the leaf example above, the absolute error is +/−0.5 mm.

The relative error is therefore:

0.5/74 × 100% = 0.7%

Practice questions

5 Complete the table to show the missing values in the last two columns.

Measurement made Equipment used Absolute error Relative error

Length of a fluid column in a respirometer is 6 mm

mm scale 0.5 mm

Volume of a syringe is 12 cm3 of liquid 0.5 cm

3 divisions

Change in mass of 1.6 g balance with 2 d.p.

6 Scatter graphs and lines of best fit

The purpose of a scatter graph with a line of best fit is to allow visualisation of a trend in a set of data. The graph can be used to make calculations, such as rates, and also to judge the correlation between variables. It is easy to draw such a graph but also quite easy to make simple mistakes.

6.1 Plotting scatter graphs

The rules when plotting graphs are:

Ensure that the graph occupies the majority of the space available:

o In exams, this means more than half the space

o Look for the largest number to help you decide the best scale

o The scale should be based on 1, 2, or 5, or multiples of those numbers



Student sheet

AQA Biology



Ensure that the dependent variable that you measured is on the y-axis and the independent variable that you varied is on the x-axis

Mark axes using a ruler and divide them clearly and equidistantly (i.e. 10, 20, 30, 40 not 10, 15, 20, 30, 45)

Ensure that both axes have full titles and units are clearly labelled

Plot the points accurately using sharp pencil ‘x’ marks so the exact position of the point is obvious

Draw a neat best fit line, either a smooth curve or a ruled line. It does not have to pass through all the points. Move the ruler around aiming for:

o as many points as possible on the line

o the same number of points above and below the line

If the line starts linear and then curves, be careful not to have a sharp corner where the two lines join. Your curve should be smooth

Confine your line to the range of the points. Never extrapolate the line beyond the range within which you measured

Add a clear, concise title.

Remember: Take care, use only pencil and check the positions of your points.

Practice questions

1 Use your calculated data in Topic 5.2 question 3 to plot a graph of % mass change against sucrose concentration.

2 For each of the tables of data:

a Plot a scatter graph

b Draw a line of best fit

c Describe the correlation

Sodium bicarbonate

concentration / %

Rate of oxygen production by

pondweed / mm3 s

−1

6.5 1.6

5.0 2.1

3.5 1.2

2.0 0.8

1.0 0.5

0.5 0.2

Turbidity of casein samples at different pH

pH % transmission

(blue light)

9.00 99

8.00 99

6.00 87

5.00 67

4.75 26

4.50 30

4.00 24

3.75 43

3.50 64



Student sheet

AQA Biology




BRIDGING UNIT: A-level CHEMISTRY

Contact Email: [email protected] On the following pages there is some of the content and skills which you will need to be able to do in order for you to be successful on this course.




Student sheet

AQA Chemistry








Learning objectives





o converting between units and standard form and decimals

o balancing chemical equations

o rearranging equations

o calculating moles and masses

o calculating percentage yield and percentage error

o interpreting graphs of reactions.



Student sheet

AQA Chemistry



Retrieval questions

You need to be confident about the definitions of terms that describe measurements and results in A Level Chemistry.






other

What is repeatability? how precise repeated measurements are when they are taken by


conditions






around the true value in an unpredictable way?

random error


values


should be zero


investigator?

independent variable


changed


dependent variable


dependent variable



Student sheet

AQA Chemistry



Atomic structure


What does an atom consist of? a nucleus containing protons and neutrons, surrounded by electrons

What are the relative masses of a proton,

neutron, and electron? 1, 1, and

1840

1 respectively

What are the relative charges of a proton,

neutron, and electron?

+1, 0, and -1 respectively

How do the number of protons and electrons

differ in an atom?

they are the same because atoms have neutral charge

What force holds an atomic nucleus together? strong nuclear force

What is the atomic number of an element? the number of protons in the nucleus of a single atom of an element

What is the mass number of an element? number of protons + number of neutrons

What is an isotope? an atom with the same number of protons but different number of

neutrons

What is an ion? an atom, or group of atoms, with a charge

What is the function of a mass spectrometer? it accurately determines the mass and abundance of separate

atoms or molecules, to help us identify them

What is a mass spectrum? the output from a mass spectrometer that shows the different

isotopes that make up an element

What is the total number of electrons that each

electron shell (main energy level) can contain?

2n2 electrons, where n is the number of the shell

How many electrons can the first three electron

shells hold each?

2 electrons (first shell), 8 electrons (second shell), 18 electrons

(third shell)

What are the first four electron sub-shells

(orbitals) called?

s, p, d, and f (in order)

How many electrons can each orbital hold? a maximum of 2 electrons

Define the term ionisation energy, and give its

unit

the energy it takes to remove a mole of electrons from a mole of

atoms in the gaseous state, unit kJ mol-1

What is the equation for relative atomic mass

(Ar)? relative atomic mass

Cof atom 1 of mass 12

1

atom 1 of mass average

12th

What is the equation for relative molecular mass

(Mr)? relative molecular mass

Cof atom 1 of mass 12

1

molecule 1 of mass average

12th



Student sheet

AQA Chemistry



Maths skills

1 Core mathematical skills

A practical chemist must be proficient in standard form, significant figures, decimal places, SI units, and unit conversion.

1.1 Standard form

In science, very large and very small numbers are usually written in standard form. Standard form is writing a number in the format A × 10

x where A is a number from 1 to 10 and x is the

number of places you move the decimal place.

For example, to express a large number such as 50 000 mol dm−3

in standard form, A = 5 and x = 4 as there are four numbers after the initial 5.

Therefore, it would be written as 5×104 mol dm

−3.

To give a small number such as 0.000 02 Nm2 in standard form, A = 2 and there are five

numbers before it so x = −5.

So it is written as 2×10−5

Nm2.

Practice questions

1 Change the following values to standard form.

a boiling point of sodium chloride: 1413 °C

b largest nanoparticles: 0.0 001×10−3

m

c number of atoms in 1 mol of water: 1806×1021

2 Change the following values to ordinary numbers.

a 5.5×10−6

b 2.9×102 c 1.115×10

4 d 1.412×10

−3 e 7.2×10

1

1.2 Significant figures and decimal places

In chemistry, you are often asked to express numbers to either three or four significant figures. The word significant means to ‘have meaning’. A number that is expressed in significant figures will only have digits that are important to the number’s precision.

It is important to record your data and your answers to calculations to a reasonable number of significant figures. Too many and your answer is claiming an accuracy that it does not have, too few and you are not showing the precision and care required in scientific analysis.

For example, 6.9301 becomes 6.93 if written to three significant figures.

Likewise, 0.000 434 56 is 0.000 435 to three significant figures.

Notice that the zeros before the figure are not significant – they just show you how large the number is by the position of the decimal point. Here, a 5 follows the last significant digit, so just as with decimals, it must be rounded up.

Any zeros between the other significant figures are significant. For example, 0.003 018 is 0.003 02 to three significant figures.

Sometimes numbers are expressed to a number of decimal places. The decimal point is a place holder and the number of digits afterwards is the number of decimal places.

For example, the mathematical number pi is 3 to zero decimal places, 3.1 to one decimal place, 3.14 to two decimal places, and 3.142 to three decimal places.



Student sheet

AQA Chemistry



Practice questions

3 Give the following values in the stated number of significant figures (s.f.).

a 36.937 (3 s.f.) b 258 (2 s.f.) c 0.043 19 (2 s.f.) d 7 999 032 (1 s.f.)

4 Use the equation:

number of molecules = number of moles × 6.02 × 1023

molecules per mole

to calculate the number of molecules in 0.5 moles of oxygen. Write your answer in standard form to 3 s.f.

5 Give the following values in the stated number of decimal places (d.p.).

a 4.763 (1 d.p.) b 0.543 (2 d.p.) c 1.005 (2 d.p.) d 1.9996 (3 d.p.)

1.3 Converting units

Units are defined so that, for example, every scientist who measures a mass in kilograms uses the same size for the kilogram and gets the same value for the mass. Scientific measurement depends on standard units – most are Système International (SI) units.

If you convert between units and round numbers properly it allows quoted measurements to be understood within the scale of the observations.

Multiplication factor Prefix Symbol

109 giga G

106

mega M

103

kilo k

10–2

centi c

10–3

milli m

10–6

micro µ

10–9

nano n

Unit conversions are common. For instance, you could be converting an enthalpy change of 488 889 J mol

−1 into kJ mol

−1. A kilo is 10

3 so you need to divide by this number or move the

decimal point three places to the left.

488 889 ÷ 103 kJ mol

−1 = 488.889 kJ mol

−1

Converting from mJ mol−1

to kJ mol−1

, you need to go from 103 to 10

−3, or move the decimal point

six places to the left.

333 mJ mol−1

is 0.000 333 kJ mol−1

If you want to convert from 333 mJ mol−1

to nJ mol−1

, you would have to go from 10−9

to 10−3

, or move the decimal point six places to the right.

333 mJ mol−1

is 333 000 000 nJ mol−1

Practice question

6 Calculate the following unit conversions.

a 300 µm to m

b 5 MJ to mJ

c 10 GW to kW



Student sheet

AQA Chemistry



2 Balancing chemical equations

2.1 Conservation of mass

When new substances are made during chemical reactions, atoms are not created or destroyed – they just become rearranged in new ways. So, there is always the same number of each type of atom before and after the reaction, and the total mass before the reaction is the same as the total mass after the reaction. This is known as the conservation of mass.

You need to be able to use the principle of conservation of mass to write formulae, and balanced chemical equations and half equations.

2.2 Balancing an equation

The equation below shows the correct formulae but it is not balanced.

H2 + O2 → H2O

While there are two hydrogen atoms on both sides of the equation, there is only one oxygen atom on the right-hand side of the equation against two oxygen atoms on the left-hand side. Therefore, a two must be placed before the H2O.

H2 + O2 → 2H2O

Now the oxygen atoms are balanced but the hydrogen atoms are no longer balanced. A two must be placed in front of the H2.

2H2 + O2 → 2H2O

The number of hydrogen and oxygen atoms is the same on both sides, so the equation is balanced.

Practice question

1 Balance the following equations.

a C + O2 → CO

b N2 + H2 → NH3

c C2H4 + O2 → H2O + CO2

2.3 Balancing an equation with fractions

To balance the equation below:

C2H6 + O2 → CO2 + H2O

Place a two before the CO2 to balance the carbon atoms.

Place a three in front of the H2O to balance the hydrogen atoms.

C2H6 + O2 → 2CO2 + 3H2O

There are now four oxygen atoms in the carbon dioxide molecules plus three oxygen atoms in the water molecules, giving a total of seven oxygen atoms on the product side.

To balance the equation, place three and a half in front of the O2.

C2H6 + 3½O2 → 2CO2 + 3H2O

Finally, multiply the equation by 2 to get whole numbers.

2C2H6 + 7O2 → 4CO2 + 6H2O



Student sheet

AQA Chemistry



Practice question

2 Balance the equations below.

a C6H14 + O2 → CO2 + H2O

b NH2CH2COOH + O2 → CO2 + H2O + N2

2.4 Balancing an equation with brackets

Ca(OH)2 + HCl → CaCl2 + H2O

Here the brackets around the hydroxide (OH−) group show that the Ca(OH)2 unit contains one

calcium atom, two oxygen atoms, and two hydrogen atoms.

To balance the equation, place a two before the HCl and another before the H2O.

Ca(OH)2 + 2HCl → CaCl2 + 2H2O

Practice question

3 Balance the equations below.

a Mg(OH)2 + HNO3 → Mg(NO3)2 + H2O

b Fe(NO3)2 + Na3PO4 → Fe3(PO4)2 + NaNO3

3 Rearranging equations and calculating concentrations

3.1 Rearranging equations

In chemistry, you sometimes need to rearrange an equation to find the desired values.

For example, you may know the amount of a substance (n) and the mass of it you have (m), and need to find its molar mass (M).

The amount of substance (n) is equal to the mass you have (m) divided by the molar mass (M):

M

mn

You need to rearrange the equation to make the molar mass (M) the subject.

Multiply both sides by the molar mass (M):

M × n = m

Then divide both sides by the amount of substance (n):

N

mm

Practice questions

1 Rearrange the equation V

nc to make:

a n the subject of the equation

b V the subject of the equation.

2 Rearrange the equation PV = nRT to make:

a n the subject of the equation



Student sheet

AQA Chemistry



b T the subject of the equation.

3.2 Calculating concentration

The concentration of a solution (a solute dissolved in a solvent) is a way of saying how much solute, in moles, is dissolved in 1 dm

3 or 1 litre of solution.

Concentration is usually measured using units of mol dm−3

. (It can also be measured in g dm3.)

The concentration of the amount of substance dissolved in a given volume of a solution is given by the equation:

V

nc

where n is the amount of substance in moles, c is the concentration, and V is the volume in dm3.

The equation can be rearranged to calculate:

the amount of substance n, in moles, from a known volume and concentration of solution

the volume V of a solution from a known amount of substance, in moles, and the concentration of the solution.

Practice questions

3 Calculate the concentration, in mol dm−3

, of a solution formed when 0.2 moles of a solute is dissolved in 50 cm

3 of solution.

4 Calculate the concentration, in mol dm−3

, of a solution formed when 0.05 moles of a solute is dissolved in 2.0 dm

3 of solution.

5 Calculate the number of moles of NaOH in an aqueous solution of 36 cm3 of 0.1 mol dm

−3.

4 Molar calculations

4.1 Calculating masses and gas volumes

The balanced equation for a reaction shows how many moles of each reactant and product are involved in a chemical reaction.

If the amount, in moles, of one of the reactants or products is known, the number of moles of any other reactants or products can be calculated.

The number of moles (n), the mass of the substance (m), and the molar mass (M) are linked by:

M

mn

Note: The molar mass of a substance is the mass per mole of the substance. For CaCO3, for example, the atomic mass of calcium is 40.1, carbon is 12, and oxygen is 16. So the molar mass of CaCO3 is:

40.1 + 12 + (16 × 3) = 100.1. The units are g mol−1

.



Student sheet

AQA Chemistry



Look at this worked example. A student heated 2.50 g of calcium carbonate, which decomposed as shown in the equation:

CaCO3(s) → CaO(s) + CO2(g)

The molar mass of calcium carbonate is 100.1 g mol−1

.

a Calculate the amount, in moles, of calcium carbonate that decomposes.

M

mn 2.50/100.1 0.025 mol

b Calculate the amount, in moles, of carbon dioxide that forms.

From the balanced equation, the number of moles of calcium carbonate number of moles

of carbon dioxide 0.025 mol

Practice questions

1 In a reaction, 0.486 g of magnesium was added to oxygen to produce magnesium oxide.

2Mg(s) + O2(g) → 2MgO(s)

a Calculate the amount, in moles, of magnesium that reacted.

b Calculate the amount, in moles, of magnesium oxide made.

c Calculate the mass, in grams, of magnesium oxide made.

2 Oscar heated 4.25 g of sodium nitrate. The equation for the decomposition of

sodium nitrate is:

2NaNO3(s) → 2NaNO2(s) + O2(g)

a Calculate the amount, in moles, of sodium nitrate that reacted.

b Calculate the amount, in moles, of oxygen made.

3 0.500 kg of magnesium carbonate decomposes on heating to form magnesium oxide and carbon dioxide. Give your answers to 3 significant figures.

MgCO3(s) → MgO(s) + CO2(g)

a Calculate the amount, in moles, of magnesium carbonate used.

b Calculate the amount, in moles, of carbon dioxide produced.

5 Percentage yields and percentage errors

5.1 Calculating percentage yield

Chemists often find that an experiment makes a smaller amount of product than expected. They can predict the amount of product made in a reaction by calculating the percentage yield.

The percentage yield links the actual amount of product made, in moles, and the theoretical yield, in moles:

percentage yield 100product of moles) (in amount ltheoretica

product of moles) (in amount actual

Look at this worked example. A student added ethanol to propanoic acid to make the ester, ethyl propanoate, and water.

C2H5OH + C2H5COOH → C2H5COOC2H5 + H2O

The experiment has a theoretical yield of 5.00 g.

The actual yield is 4.50 g.



Student sheet

AQA Chemistry



The molar mass of C2H5COOC2H5 = 102.0 g mol−1

Calculate the percentage yield of the reaction.

Actual amount of ethyl propanoate: M

mn = 4.5/102 0.0441 mol

Theoretical amount of ethyl propanoate: M

mn = 5.0/102 0.0490 mol

percentage yield (0.0441/0.0490) × 100% 90%

Practice questions

1 Calculate the percentage yield of a reaction with a theoretical yield of 4.75 moles of product and an actual yield of 3.19 moles of product. Give your answer to 3 significant figures.

2 Calculate the percentage yield of a reaction with a theoretical yield of 12.00 moles of product and an actual yield of 6.25 moles of product. Give your answer to 3 significant figures.

5.3 Calculating percentage error in apparatus

The percentage error of a measurement is calculated from the maximum error for the piece of apparatus being used and the value measured:

percentage error value measured

error maximum× 100%

Look at this worked example. In an experiment to measure temperature changes, an excess of zinc powder was added to 50 cm

3 of copper(II) sulfate solution to produce zinc sulfate and

copper.

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

The measuring cylinder used to measure the copper(II) sulfate solution has a maximum error of ±2 cm

3.

a Calculate the percentage error.

percentage error (2/50) × 100% 4%

b A thermometer has a maximum error of ±0.05 °C.

Calculate the percentage error when the thermometer is used to record a temperature rise of 3.9 °C. Give your answer to 3 significant figures.

percentage error (2 × 0.05)/3.9 × 100% 2.56%

(Notice that two measurements of temperature are required to calculate the temperature change so the maximum error is doubled.)

Practice questions

3 A gas syringe has a maximum error of ±0.5 cm3. Calculate the maximum percentage error

when recording these values. Give your answers to 3 significant figures.

a 21.0 cm3 b 43.0 cm

3

4 A thermometer has a maximum error of ±0.5 °C. Calculate the maximum percentage error when recording these temperature rises. Give your answers to 3 significant figures.

a 12.0 °C b 37.6 °C



Student sheet

AQA Chemistry



6 Graphs and tangents

6.1 Deducing reaction rates

To investigate the reaction rate during a reaction, you can measure the volume of the product formed, such as a gas, or the colour change to work out the concentration of a reactant during the experiment. By measuring this concentration at repeated intervals, you can plot a concentration–time graph.

Note: When a chemical is listed in square brackets, it just means ‘the concentration of’ that chemical. For example, [O2] is just shorthand for the concentration of oxygen molecules.

By measuring the gradient (slope) of the graph, you can calculate the rate of the reaction. In the graph above, you can see that the gradient changes as the graph is a curve. If you want to know the rate of reaction when the graph is curved, you need to determine the gradient of the curve. So, you need to plot a tangent.

The tangent is the straight line that just touches the curve. The gradient of the tangent is the gradient of the curve at the point where it touches the curve.

Looking at the graph above. When the concentration of A has halved to 1.0 mol dm−3

, the tangent intercepts the y-axis at 1.75 and the x-axis at 48.

The gradient is 48

751. = −0.0365 (3 s.f.).

So the rate is 0.0365 mol dm−3

s−1

.

Practice question

1 Using the graph above, calculate the rate of reaction when the concentration of A halves again to 0.5 mol dm

−3.

6.2 Deducing the half-life of a reactant

In chemistry, half-life can also be used to describe the decrease in concentration of a reactant in a reaction. In other words, the half-life of a reactant is the time taken for the concentration of the reactant to fall by half.



Student sheet

AQA Chemistry



Practice question

2 The table below shows the change in concentration of bromine during the course of a reaction.

Time / s [Br2] / mol dm−3

0 0.0100

60 0.0090

120 0.0066

180 0.0053

240 0.0044

360 0.0028

a Plot a concentration–time graph for the data in the table.

b Calculate the rate of decrease of Br2 concentration by drawing tangents.

c Find the half-life at two points and deduce the order of the reaction.


BRIDGING UNIT: A-level COMPUTER SCIENCE

Contact Email: [email protected]

Browse through the specification so you are aware of the course contents: https://www.aqa.org.uk/subjects/computer-science-and-it/as-and-a-level/computer-science-7516-7517/specification-at-a-glance

1. Finite state machines

Watch these videos then answer the question below:

A. https://www.youtube.com/watch?v=4rNYAvsSkwk

B. https://www.youtube.com/watch?v=LuGs7WhlHWA

Look at the diagram below. Circle the sequence of characters that are acceptable:

a) caabb

b) bac

c) aaabbccc

d) abc

e) aabbca

2. Graphs and tree structures

Watch this video then answer the question below:

https://www.youtube.com/watch?time_continue=3&v=PMPMDZqeipw&feature=emb_title

Draw a directed graph on the right using the

following adjacency matrix.

A B C D

A 0 1 0 1

B 0 0 0 0

C 0 0 0 0

D 0 1 1 0

Populate this adjacency matrix using the

undirected graph on the left.

A B C D

A

B

C

D

C

A

D

B

C

A

D

B

3. Binary multiplication

Watch this video then answer the questions below:

https://www.youtube.com/watch?v=Va_UvwJULcI

Multiply the following binary values

a) Multiply the following binary values

1 1 1 X

0 1

b) Multiply the following binary values

1 0 0 X

1 0

c) Multiply the following binary values

0 0 1 X

1 1

d) Multiply the following binary values

1 1 1 X

1 0

BRIDGING UNIT: A-level ECONOMICS

Contact Email: [email protected] and [email protected]

Task: Presentation on the impact of Coronavirus to the UK Economy

The Coronavirus (COVID-19) pandemic has had an enormous impact on the economic performance of the UK. The impact of this virus will have an impact on individuals and businesses for many years to come.

In preparation for your A-level Economics course, you should use the readings below (AND YOUR OWN READING) to create a presentation entitled ‘The Impact of Coronavirus on the UK Economy’. The presentation on ‘Google Slides’ or ‘Microsoft Powerpoint’ which should include the following bullet points;

1) Introduction: How is the performance of the UK economy measured? (Key words: Economic growth, unemployment, inflation and balance of trade in goods and services)

2) Using evidence to support you, explain five ways in which the Coronavirus pandemic has had on the economic performance of the UK.

3) Using evidence to support you, explain the different strategies the UK government has used to minimise the impact of the Coronavirus pandemic on individuals and firms within the UK

4) Conclusion: Write a conclusion (with justification) about whether you think the government has done enough to support firms and individuals in the UK during the Coronavirus pandemic.

You should send your presentation to [email protected] and [email protected].

Prize 1: £20 Amazon voucher for the best presentation - this will be judged by Mr. Souch and Mr. Elton.

Prize 2: The first five students to finish a complete presentation and send to both email addresses above will win a £10 Amazon voucher.

Wider reading on the economic impact of Coronavirus on the UK

1. Coronavirus: A visual guide to the economic impact https://www.bbc.co.uk/news/business-51706225

2. Coronavirus will have a significant impact on the economy https://www.bbc.co.uk/news/av/uk-politics-52282814/coronavirus-will-have-significant-impact-on-economy

3. Research on the Economic Impact of the Coronavirus on the UK https://www.oxfordeconomics.com/coronavirus

4. Coronavirus: What could the long-term impact be for you? https://www.youtube.com/watch?v=LhzQi1kty-Y

5. UK unemployment rate set to surge in Coronavirus lockdown https://www.cityam.com/ons-job-vacancies-sank-as-uk-entered-coronavirus-lockdown/

6. Coronavirus lockdown tips UK economy into biggest slump on record

https://www.theguardian.com/business/2020/apr/23/coronavirus-lockdown-tips-uk-economy-into-biggest-slump-on-record

7. Coronavirus: More than 140,000 firms claim wage bill help https://www.bbc.co.uk/news/business‐52346685

8. More than 9 million expected to be furloughed https://www.bbc.co.uk/news/business-52209790

9. Coronavirus: Four ways the economy has been affected https://www.bbc.co.uk/news/52138535

10. Impact of Coronavirus on the exports and imports within the UK https://www.statista.com/statistics/1110268/impact-of-coronavirus-on-exports-and-imports-uk/

BRIDGING UNIT: A-level ENGLISH LITERATURE


Read at least two of the following novels:

The Handmaid’s Tale by Margaret Atwood

The Kite Runner by Khaled Hosseini

Life After Life by Kate Atkinson

I Capture the Castle by Dodie Smith

The Picture of Dorian Gray by Oscar Wilde

Half of a Yellow Sun by Chimamanda Ngozi Adichie

1984 by George Orwell

Enduring Love by Ian McEwan

The God of Small Things by Arundhati Roy

The Other Hand by Chris Cleave*

White Tiger by Aravind Adiga*

My sister the serial killer by Oyinkan Braithwaite

We need to talk about Kevin by Lionel Shriver*

The Color Purple by Alice Walker

The books with a * are real page turners (and still high quality literature) so I would start with these.

Task: Select two of the novels and compare and contrast them in any ways that seem interesting to you, paying particular attention to narrative voice.

You might also consider structure, language and style.

2 sides of A4 minimum

BRIDGING UNIT: A-level FRENCH Contact Email: [email protected]

Google Classroom Code : mmpnxpa

In A level French, you will see the following topics :

Students must study the following themes and sub-themes in relation to at least one French-speaking country. Students must study the themes and sub-themes using a range of sources, including material from online media.

1. Aspects of French-speaking society: current trends

1. The changing nature of family (La famille en voie de changement) Grands-parents, parents et enfants – soucis et problèmes Monoparentalité, homoparentalité, familles recomposée La vie de couple – nouvelles tendances

2. The 'cyber-society' (La « cyber-société ») Qui sont les cybernautes ? Comment la technologie facilite la vie quotidienne Quels dangers la « cyber-société » pose-t-elle ?

3. The place of voluntary work (Le rôle du bénévolat) Qui sont et que font les bénévoles ? Le bénévolat – quelle valeur pour ceux qui sont aidés ? Le bénévolat – quelle valeur pour ceux qui aident ?

2. Artistic culture in the French-speaking world

1. A culture proud of its heritage (Une culture fière de son patrimoine) Le patrimoine sur le plan national, régional et local Comment le patrimoine reflète la culture Le patrimoine et le tourisme

2. Contemporary francophone music (La musique francophone contemporaine) La diversité de la musique francophone contemporaine Qui écoute et apprécie cette musique ? Comment sauvegarder cette musique ?

3. Cinema: the 7th art form (Cinéma : le septième art) Pourquoi le septième art ? Le cinéma – une passion nationale ? Evolution du cinéma – les grandes lignes

In addition, students will study a film. We study La Haine, from Mathieu Kassovitz (1995). You can watch the trailer and other video clips on Youtube and start doing some research on the movie : themes, characters, techniques… You will also find a lot of resources on the Google Classroom :)

BRIDGING UNIT: A-level HISTORY Contact Email: [email protected]

Watch these videos and make a research document tracking the key events of American History between these periods

https://ap.gilderlehrman.org/

Then, create a piece that argues which event or factor between the years 1757 and 1848 changed history for Black people in the US the most.

You must include at least 8 factors.

Eg, Missouri Compromise, Creation of the Cotton Gin, wars, economics, politics, the consitituion.

BRIDGING UNIT: A-level LAW Contact Email: [email protected]

For each of the questions below, use the websites listed (or your own reasearch) to help you find the answers.

Parliament or Government?: http://www.parliament.uk/about/how/role/relations-with-other-institutions/parliament-government/

1) What is the difference between “Parliament” and “Government”?

Creating Laws in Parliament: http://www.parliament.uk/about/how/laws/

1) What is a “Bill”? 2) What are the four different types of Bills? Describe each one. 3) What is an “Act?” 4) What are the stages a Bill goes through to become an Act in:

a. The House of Commons b. The House of Lords

6) What is “Delegated Legislation”?

Criminal Courts: https://www.gov.uk/courts

7) What are the different types of criminal courts? What sorts of things do they deal with?

Criminal Offences: http://e-lawresources.co.uk/Criminal.php

8) Describe what is meant by ACTUS REUS & MENS REA

9) How are the following offences committed? Think about the ACTUS REUS and MENS REA elements

A. GBH B. ABH C. Murder D. Assault E. Theft F. Robbery G. Burglary

10) What are the MAXIMUM penalties given for each of those offences? (you’ll need to do your own internet

research for these, rather than the elawresources website)

Judicial Precedent: http://www.britpolitics.co.uk/judicial-precedent-a-level-uk-politics

11) What is “Judicial Precedent”?

Key Cases:

12) What happened in the case of “Donoghue V Stevenson”? Why is this important legal case? http://www.e-lawresources.co.uk/Donoghue-v-Stevenson.php ;

13) What happened in the case of “British Railways Board V Herrington”? Why is this important legal case? http://e-lawresources.co.uk/British-Railways-Board-v-Herrington.php ;

14) Why was it important to decide whether Jaffa Cakes were biscuits or cakes? https://www.kerseys.co.uk/jaffa-cakes-cakes-biscuits/

15) What happened to poor little Mary? https://www.theguardian.com/uk/2002/feb/05/sarahboseley

16) Why was Kiranjit Ahluwalia not guilty of murder? http://www.e-lawresources.co.uk/R-v-Ahluwalia.php

BRIDGING UNIT: A-level MATHS Contact Email: [email protected]




Welcome to Maths at Lilian Baylis Technology School The aim of this pack is to aid your preparation for an Advanced Mathematics course and make the transition from

GCSE to AS level as smooth as possible. It contains GCSE Higher tier material that you should be confident with before

you begin your AS course. Given the situation with school closures, it is more important this year than ever that you

put the effort in and come prepared.

We want you to be successful and we will do all we can to be supportive but the hard work has to come from you.

You MUST work through this booklet over the summer and bring your completed booklet with you to enrollment.You will be tested on what you have learned at the end of your first week in school.

Please read ‘The Guide to Level 3 Mathematics courses’ on the next page, it will inform you about which maths

course would be most appropriate for you. A decision will need to be made about which course will best fit with your

intended career path and your other subjects.

Should you encounter any problems with the work you can email [email protected] and we will get back

to you as soon as possible.

Good luck with the work. We hope you enjoy your summer break and we look forward to meeting you at enrollment.

The Lilian Baylis Maths Team

https://alevelmathsrevision.com/s6fprep

mailto:[email protected]

mailto:[email protected]

http://www.facebook.com/S6FMaths

A guide to Level 3 Maths Courses at LBTS6F

A Level Mathematics (Edexcel) Minimum entry requirement GCSE grade 7

This is the traditional advanced level mathematics course. It is particularly appropriate for mathematically able

students who show both interest and commitment. Algebraic competency is very important as much of the course

relies heavily on this skill. A high level at GCSE is not always an automatic indicator of this and even students achieving

grade 7 at GCSE may need to work especially hard to succeed.

The course is split into three distinct areas: Pure Maths, Statistics and Mechanics.

• Pure Maths is the most abstract of the three areas. Pure Maths describes maths that is not necessarily appliedto real-life situations and need not be put into a context. To name but a few, Surds, Solving QuadraticEquations, Calculus are all topics covered as part of Pure Maths.

• Statistics is the branch of Maths that involves probability and data handling and is the type of Maths that liesat the heart of how financial, economic and even political decisions are made. You can go barely anywhere inthe modern world without being faced with Statistical Maths in one form or another.

• Mechanics is the Maths behind Physics: you will learn about the laws that govern Forces and Energy, look inmore depth at Newton’s Laws of Motion and much more. To summarise, Mechanics is the Maths behind howeverything works and is at the forefront of technological innovation.

A Level Further Maths (Edexcel) Minimum entry requirement GCSE grade 8

This can only be studied as a 2nd Maths A-level alongside A Level Mathematics and is suitable only for those students

with a passion and flair for the subject. Further maths is available at both AS and A level.

Further maths is an excellent choice for those considering Maths or related courses at university and it combines

well with a wide range of subjects.

Further Maths looks in more detail at Statistics, Mechanics and Pure Maths.

Core Maths (AQA)

Minimum entry requirement GCSE grade 5

Core Maths is the new Level 3 AS level qualification for students who wish to develop their practical maths skills for

the real world, be it in work, study or everyday life. It can be taken alongside A Levels or Level 3 BTEC.

What you study is not purely theoretical or abstract; it can be applied on a day-to-day basis in work, study or life and

will include a financial maths element. It will also help with other A-level subjects – in particular with science,

geography, business studies, psychology and economics.

The skills developed in the study of mathematics are increasingly important in the workplace and in higher education;

studying Core Maths will help you keep up these essential skills. Most students who study maths after GCSE improve

their career choices and increase their earning potential.

Core Maths can’t be taken alongside A-level Maths or Further Maths but provides a valuable opportunity for those

students who want to continue Maths studies but don’t feel confident or have the required GCSE grade to study A-

level Maths.

1

Introduction

Thank you for choosing to study Mathematics in the sixth form.

Over the course, you will study topics in Pure Maths, Mechanics and Statistics. If you have chosen

to study Further Mathematics as well as Maths then, you will study extra topics within these areas.

The Mathematics Department is committed to ensuring that you make good progress throughout

your A level course. In order that you make the best possible start to the course, we have prepared

this booklet. It is vital that you spend time working through the questions in this booklet over the

summer. You need to have a good knowledge of these topics before you commence your course in

September. You should have met all the topics before at GCSE. Work through what you need to

from each chapter, making sure that you understand the examples. Then tackle the exercise to

ensure you understand the topic thoroughly. The answers are at the back of the booklet. You will

need to be very organised so keep your work in a folder & mark any queries to ask at the beginning

of term.

A mock test is provided at the back of this booklet.

Use this introduction to give you a good start to your Year 12 work that will help you to enjoy, and

benefit from, the course. The more effort you put in, right from the start, the better you will do.

Sources for further help are indicated throughout the booklet.

All the topics can be found on resources such as Corbett Maths or DrFrostMaths.

2

Helpful Videos and Resources for Working From Home Topics to

Prepare

for A-level

Maths

Hegarty Maths Videos and

Resources

(Requires a school subscription)

Done

?

Corbett Maths Videos and Resources

(No subscription required)

Done

?

1 Manipulati

ng algebraic

expressions

166 - Expanding Triple Brackets

172 – Expressions with algebraic fractions

229 – Simplify Algebraic Fractions

(involving quadratics)

15 - Algebra: expanding three brackets

21 - Algebraic Fractions: addition

22 - Algebraic Fractions: division

23 - Algebraic Fractions: multiplication 24 - Algebraic Fractions: simplifying

2 Surds 115 - Simplifying Surds

117 - Brackets Involving Surds 2

119 - Rationalising Surds 2

305 - Surds: intro, rules, simplifying

307 - Surds: rationalising denominators

308 - Surds: expanding brackets

3 Rules of

indices

104 – Index form 3 (power of negative

integers)

109 – Index form 8 (powers of non-unit

fractions)

174 - Indices With Algebraic Expressions 2

175 - Indices With Algebraic Expressions 3

173 - Indices: fractional

174 - Indices: laws of

175 - Indices: negative

4 Factorising

expressions

227 - Factorising Quadratic Expressions 5

228 - Factorising Quadratic Expressions 6

119a - Factorisation: splitting the middle

120 - Factorisation: difference of 2 squares

5 Completing

the square

235 - Completing the Square 1



10 - Algebra: completing the square

6 Solving

quadratic

equations

232 - Solving Quadratic Equations (by

Factorising) 3

238 - Solving by Completing the Square 1

242 - Solving using the Quadratic Formula

244 - Quadratic Equations from Algebraic

Fractions

266 - Quadratics: solving (factorising)

267 - Quadratics: formula

267a - Quadratics: solving (completing the

square)

7 Sketching

quadratic

graphs

252 - Finding the y-intercept of a Quadratic

Graph

253 - Finding the x-intercept (Roots) of a

Quadratic Graph

256 - Finding the Turning Point of a

Quadratic Graph 2

257 - Sketch a Fully Labelled Quadratic

Graph

265 - Quadratic graphs: sketching using

key points

371 - Quadratic graph (completing the

square)

8 Solving

linear

simultaneou

s equations

193 - Simultaneous Equations by

Elimination 4

194 - Simultaneous Equations by

Substitution

295 - Simultaneous equations (elimination)

296 - Simultaneous equations (substitution,

both linear)

9 Solving

quadratic

simultaneou

s equations

314 - Equation of a Circle 1

246 - Simultaneous Equations Involving

Quadratics

12 - Algebra: equation of a circle

298 - Simultaneous equations (advanced)

10 Solving

simultaneou

s equations

graphically

259 - Simultaneous Equations Using

Graphs (Quadratic & Linear)

297 - Simultaneous equations (graphical)

11 Linear

inequalities

273 - Linear Inequalities as Graph Regions

1


2


3

180 - Inequalities: graphical y>a or x>a

181 - Inequalities: graphical y>x+a

182 - Inequalities: region

12 Quadratic

inequalities

277 - Solving Quadratic Inequalities 378 - Inequalities: quadratic

13 Sketching

cubic and

298 - Cubic Graphs (from a Table of

Values)

344 – Types of graph: cubics

346 - Types of graph: reciprocal

reciprocal

graphs

299 - Cubic Graphs (Recognising)

300 - Reciprocal Graphs 1

14 Translating

graphs

307 - Graph Transformations 1 f(x)+a

308 - Graph Transformations 2 f(x+a)

323 - Transformations of graphs

15 Straight line

graphs

210 - Straight Lines Graphs 5

213 - Straight Lines Graphs 8

194 - Linear graphs: find equation of a line

195 - Linear graphs: equation through

2 points

16 Parallel and

perpendicul

ar lines

214 - Straight Line Graphs (Parallel)

216 - Straight Line Graphs (Perpendicular)

2

320 - Circle Normals and Tangents

196 - Linear graphs: parallel lines

197 - Linear graphs: perpendicular lines

372 - Equation of a Tangent to a Circle

17 Pythagoras’

Theorem

506 - 3D Pythagoras 2

501 - Pythagoras (Applied) 1

502 - Pythagoras (Applied) 2

259 - Pythagoras: 3D

260 - Pythagoras: rectangles/isosceles

triangles

263 - Pythagoras: distance points

18 Direct and

inverse

proportion

344 – Algebraic Direct Proportion 2

346 – Algebraic Inverse Proportion 1

254 - Proportion: direct

255 - Proportion: inverse

19 Circle

theorems

604 - Circle Theorems (Multi Step) 1



64 - Circle theorems – theorems

65 - Circle theorems – examples

20 Trigonomet

ry

585 - 3D Trigonometry 5

303 - Sine Graph

304 - Cosine Graph

305 - Tangent Graph

521 - Sine Rule (Find Side) 1

523 - Sine Rule (Find Angle) 1

525 - Sine Rule (Ambiguous Case)

527 - Cosine Rule (Find Side) 1

529 – Cosine Rule (Find Angle) 1

332 - Trigonometry: 3D

338 - Trigonometry: Sine graph

339 - Trigonometry: Cosine graph

340 - Trigonometry: Tangent graph

333 - Trigonometry: sine rule (sides)

334 - Trigonometry: sine rule (angles)

334a - Trigonometry: sine rule (ambiguous

case)

335 - Trigonometry: cosine rule (sides)

336 - Trigonometry: cosine rule (angles)

21 Rearrangin

g equations

186 - Solve Equations With x on Both

Sides 3

187 - Solve Equations With Algebraic

Fractions

111 - Equations: involving fractions

112 - Equations: fractional advanced

112 - Equations: cross multiplication

22 Volume and

surface area

of 3D

shapes

576 - Cones (Volume) 1

578 - Frustums (Volume)

579 - Rectangular Based Pyramids

(Volume)

580 - Spheres (Volume) 1

359 - Volume: cone

360 - Volume: pyramid

360a - Volume: Frustum

361 - Volume: sphere

313 - Surface area: sphere

314 - Surface area: cone

315 - Surface area: cylinders

23 Area under

a graph and

gradients

892 - Area Under a Curve 2

893 - Area Under a Curve 3

889 - Gradient at a Point on a Curve

890 – Instantaneous Rate of Change

389 - Area under a Graph

390a - Instantaneous Rate of Change

Contents

Contents ...................................................................................................................................... 3

Reading List ................................................................................................................................ 4

1 Fractions ............................................................................................................................ 5

2 Algebra .............................................................................................................................. 7

2.1 Expanding Brackets .................................................................................................... 7

2.2 Linear Equations ......................................................................................................... 9

2.3 Equations Containing Fractions................................................................................. 11

2.4 Forming Equations .................................................................................................... 13

2.5 Linear Inequalities .................................................................................................... 14

2.6 Simultaneous Equations ............................................................................................ 16

2.7 Factorising Linear Expressions ................................................................................. 18

2.8 Factorising Quadratic Expressions ............................................................................ 19

2.9 Completing the Square .............................................................................................. 21

2.10 Solving Quadratic Equations .................................................................................... 23

2.11 Changing the Subject ............................................................................................... 25

2.12 Indices ..................................................................................................................... 28

2.13 Surds........................................................................................................................ 31

2.14 Algebraic Fractions .................................................................................................. 36

2.15 Linear and Quadratic Simultaneous Equations ......................................................... 39

2.16 Expanding More Than Two Binomials ..................................................................... 41

2.17 Quadratic Inequalities .............................................................................................. 44

2.18 Using Completing the Square to Find Turning Points .............................................. 47

3 Functions ........................................................................................................................... 51

3.1 Function Notation .................................................................................................... 51

3.2 Composite Functions ............................................................................................... 54

3.3 Inverse Functions ..................................................................................................... 59

4 Graphs ................................................................................................................................ 62

4.1 Straight line graphs ................................................................................................... 62

4.2 Basic shapes of curved graphs ................................................................................... 64

4.3 Factors ...................................................................................................................... 68

5 Trigonometry ..................................................................................................................... 71

Trigonometric Equations ................................................................................................... 71

Practice Booklet Test ................................................................................................................. 74

Solutions to the Exercises .......................................................................................................... 75

Solutions to the Practice Booklet Test ........................................................................................ 89

3

Reading List

As a student who is choosing to study Mathematics at A Level, it is logical to assume that you have

an interest in the subject.

With that said, the following books may be of interest to you.

50 Mathematical Ideas You Really Need to Know (Tony Crilly)

Alex’s Adventures in Numberland (Alex Bellos)

Cabinet of Mathematical Curiosities (Ian Stewart)

The Calculus Wars (Jason Socrates Bardi)

The Code Book (Simon Singh)

The Curious Incident of the Dog in the Night-time by Mark Haddon

How Many Socks Make a Pair?: Surprisingly Interesting Maths (Rob Eastway)

Hello World: How to be Human in the Age of the Machine (Hannah Fry)

Humble Pi: A Comedy of Maths Errors (Matt Parker)

The Life-Changing Magic of Numbers (Bobby Seagull)

The Num8er My5teries (Marcus du Sautoy)

4

1 Fractions

To add or subtract fractions, find the lowest common denominator of the two fractions and then

rewrite the fractions accordingly. Ensure that you simplify as far as possible.

Examples

2

3+

3

4=

8

12+

9

12

=17

12

= 15

12

6

7−

3

5=

30

35−

21

35

=9

35

When multiplying fractions, it is far more efficient to “cancel” first; this avoids trying to simplify

fractions with unnecessarily large numerators and/or denominators.

To multiply with fractions, simply multiply the numerators and denominators together.

Example

2

9×

6

7=

2

3×

2

7

=4

21

To divide by a fraction, we simply multiply by the reciprocal of the second fraction (i.e. we “flip the

second fraction over”).

Example

2

3÷

10

9=

2

3×

9

10

=1

1×

3

5

=3

5

5

For addition and subtraction with mixed numbers, add (or subtract) the integer (whole number) parts

first and then work with the fractions.

Examples

21

4+ 3

1

2= 5

1

4+

1

2

= 51

4+

2

4

= 53

4+

2

4

= 55

4

= 61

4

31

5− 1

2

3= 2

1

5−

2

3

= 23

15−

10

15

= 2−7

15

= 2 −7

15

= 18

15

To multiply and divide with mixed, convert the mixed numbers to improper fractions and then calculate

as normal.

21

4× 3

1

5=

9

4×

16

5

=9

1×

4

5

=36

5

= 71

5

51

4÷ 2

1

2=

21

4÷

5

2

=21

4×

2

5

=21

2×

1

5

=21

10

= 21

10

It should also be noted that in the study of A Level Mathematics, answers are preferred as improper

fractions rather than mixed numbers.

EXERCISE

1. 2

3+

1

5

5. 2

2

3+ 1

1

4

9. 2

3

4× 1

1

22. 5

6−

1

4

6. 3

5

8+ 2

3

7

10. 3

1

2× 2

3

73. 7

8×

4

5

7. 7

7

8− 4

2

5

11. 5

3

7÷ 4

2

54. 5

6÷

3

4

8. 5

1

6− 2

3

4

12. 3

1

2÷ 2

1

4

6

2 Algebra

2.1 Expanding Brackets

To remove a single bracket multiply every term in the bracket by the number or expression outside:

Examples

1) 3 (x + 2y) = 3x + 6y

2) -2(2x - 3) = (-2)(2x) + (-2)(-3)

= -4x + 6

To expand two brackets multiply everything in the first bracket by everything in the second bracket.

You may have used

* the smiley face method

* FOIL (First Outside Inside Last)

* using a grid.

Examples:

1) (x + 1)(x + 2) = x(x + 2) + 1(x + 2)

or

(x +1)(x + 2) = x2 + 2 + 2x + x

= x2 + 3x +2

or

x 1

x x2 x

2 2x 2

2) (x - 2)(2x + 3) = x(2x + 3) - 2(2x +3)

= 2x2 + 3x – 4x - 6

= 2x2 – x – 6

or

(x - 2)(2x + 3) = 2x2 – 6 + 3x – 4x = 2x2 – x – 6

or

x -2

2x 2x2 -4x

3 3x -6

(x +1)(x + 2) = x2 + 2x + x + 2

= x2 + 3x +2

(2x +3)(x - 2) = 2x2 + 3x - 4x - 6

= 2x2 - x - 6

7

EXERCISE A Multiply out the following brackets and simplify.

1. 7(4x + 5)

2. -3(5x - 7)

3. 5a – 4(3a - 1)

4. 4y + y(2 + 3y)

5. -3x – (x + 4)

6. 5(2x - 1) – (3x - 4)

7. (x + 2)(x + 3)

8. (t - 5)(t - 2)

9. (2x + 3y)(3x – 4y)

10. 4(x - 2)(x + 3)

11. (2y - 1)(2y + 1)

12. (3 + 5x)(4 – x)

Two Special Cases

Perfect Square: Difference of two squares:

(𝑥 + 𝑎)2 = (𝑥 + 𝑎)(𝑥 + 𝑎) = 𝑥2 + 2𝑎𝑥 + 𝑎2 (𝑥 + 𝑎)(𝑥 − 𝑎) = 𝑥2 − 𝑎2

(2𝑥 − 3)2 = (2𝑥 − 3)(2𝑥 − 3) = 4𝑥2 − 12𝑥 + 9 (2𝑥 + 3)(2𝑥 − 3) = 4𝑥2 − 9

EXERCISE B Expand the following

1. (x - 1)2

2. (3x + 5)2

3. (7x - 2)2

4. (x + 2)(x - 2)

5. (3x + 1)(3x - 1)

6. (5y - 3)(5y + 3

8

2.2 Linear Equations

When solving an equation whatever you do to one side must also be done to the other.

You may

• add the same amount to both side

• subtract the same amount from each side

• multiply the whole of each side by the same amount

• divide the whole of each side by the same amount.

If the equation has unknowns on both sides, collect all the letters onto the same side of the equation.

If the equation contains brackets, you often start by expanding the brackets.

A linear equation contains only numbers and terms in x. (Not 2 3or x x or 1

𝑥 etc)

Example 1: Solve the equation 64 – 3𝑥 = 25

Solution: There are various ways to solve this equation. One approach is as follows:

Step 1: Add 3x to both sides (so that the x term is positive): 64 = 3𝑥 + 25

Step 2: Subtract 25 from both sides: 39 = 3𝑥

Step 3: Divide both sides by 3: 13 = 𝑥

So the solution is x = 13.

Example 2: Solve the equation 6𝑥 + 7 = 5 – 2𝑥.

Solution:

Step 1: Begin by adding 2x to both sides 8𝑥 + 7 = 5

(to ensure that the x terms are together on the same side)

Step 2: Subtract 7 from each side: 8𝑥 = −2

Step 3: Divide each side by 8: 𝑥 = −1

4

EXERCISE A: Solve the following equations, showing each step in your working:

1) 2x + 5 = 19 2) 5x – 2 = 13 3) 11 – 4x = 5

4) 5 – 7x = -9 5) 11 + 3x = 8 – 2x 6) 7x + 2 = 4x – 5

9

Example 3: Solve the equation 2(3𝑥 – 2) = 20 – 3(𝑥 + 2)

Step 1: Multiply out the brackets: 6𝑥 – 4 = 20 – 3𝑥 – 6

(taking care of the negative signs)

Step 2: Simplify the right hand side: 6𝑥 – 4 = 14 – 3𝑥

Step 3: Add 3x to each side: 9𝑥 – 4 = 14

Step 4: Add 4: 9𝑥 = 18

Step 5: Divide by 9: 𝑥 = 2

EXERCISE B: Solve the following equations.

1) 5(2x – 4) = 4 2) 4(2 – x) = 3(x – 9)

3) 8 – (x + 3) = 4 4) 14 – 3(2x + 3) = 2

10

2.3 Equations Containing Fractions

When an equation contains a fraction, the first step is usually to multiply through by the denominator of

the fraction. This ensures that there are no fractions in the equation.

Example 4: Solve the equation 5 112

y+ =

Solution:

Step 1: Multiply through by 2 (the denominator in the fraction): 10 22y + =

Step 2: Subtract 10: y = 12

Example 5: Solve the equation 1

(2 1) 53

x + =

Solution:

Step 1: Multiply by 3 (to remove the fraction) 2 1 15x+ =

Step 2: Subtract 1 from each side 2x = 14

Step 3: Divide by 2 x = 7

When an equation contains two fractions, you need to multiply by the lowest common denominator.

This will then remove both fractions.

Example 6: Solve the equation 1 2

24 5

x x+ ++ =

Solution:

Step 1: Find the lowest common denominator:

Step 2: Multiply both sides by the lowest common

denominator

Step 3: Simplify the left hand side:

Step 4: Multiply out the brackets:

Step 5: Simplify the equation:

Step 6: Subtract 13

Step 7: Divide by 9:

The smallest number that both 4 and 5

divide into is 20.

20( 1) 20( 2)40

4 5

x x+ ++ =

205

( 1)

4

x + 20+

4

( 2)

5

x +40=

5(x + 1) + 4(x + 2) = 40

5x + 5 + 4x + 8 = 40

9x + 13 = 40

9x = 27

x = 3

11

Example 7: Solve the equation 2 3 5

24 6

x xx

− −+ = −

Solution:

The lowest number that 4 and 6 go into is 12. So we multiply every term by 12:

Simplify

Expand brackets

Simplify

Subtract 10x

Add 6

Divide by 5

12( 2) 12(3 5 )12 24

4 6

x xx

− −+ = −

12 3( 2) 24 2(3 5 )x x x+ − = − −

12 3 6 24 6 10x x x+ − = − +

15 6 18 10x x− = +

5 6 18x− =

5x = 24

x = 4.8

Exercise: Solve these equations

1) 1

( 3) 52

x + = 2) 2

1 43 3

x x− = +

3) 3 54 3

y y+ = − 4)

2 32

7 14

x x− −= +

5) 7 1

132

xx

−= − 6)

1 1 2 5

2 3 6

y y y− + ++ =

7) 1 5 3

22 3

x xx

− ++ = 8)

5 102 1

x x− = −

12

2.4 Forming Equations

Example 8: Find three consecutive numbers so that their sum is 96.

Solution: Let the first number be n, then the second is n + 1 and the third is n + 2.

Therefore n + (n + 1) + (n + 2) = 96

3n + 3 = 96

3n = 93

n = 31

So the numbers are 31, 32 and 33.

1) Find 3 consecutive even numbers so that their sum is 108.

2) The perimeter of a rectangle is 79 cm. One side is three times the length of the other. Form an

equation and hence find the length of each side.

3) Two girls have 72 photographs of celebrities between them. One gives 11 to the other and finds

that she now has half the number her friend has.

Form an equation, letting n be the number of photographs one girl had at the beginning.

Hence find how many each has now.

13

2.5 Linear Inequalities

Linear inequalities can be solved using the same techniques as linear equations (for the most part).

We may add and subtract the same numbers on both sides and we can also multiply and divide by

positive numbers; multiplying/dividing both sides by a negative needs further explanation.

Example

2x - 3 < 11

Here we can simply add 3 to both sides: 2x < 14

Next, as with linear equations we divide by 2: x < 7

However, if we were to have 3 – 2x > 6, we would need to adopt a different technique. If we wish to

divide or multiply by a negative number, we must reverse the direction of the inequality.

Example

3 − 2𝑥 > 6

As before, we would subtract 3 from both sides: −2𝑥 > 3

Divide by -2 and subsequently reverse the inequality: 𝑥 < −3

2

We can see this working on a more basic level; it is true to state that 3 < 4 but it is incorrect if we

multiply both sides by a negative and keep the sign as it was: -6 < -8 is not true.

You may find it easier to rearrange the inequality:

Example

3 − 2𝑥 > 6

If we add 2x to both sides, we remove the hassle: 3 > 6 + 2𝑥

We then subtract 6: −3 > 2𝑥

Divide by two as normal: −3

2> 𝑥

Remember that you can change this round to say 𝑥 < −3

2

Both of these techniques are acceptable and is more a matter of preference.

14

Exercise: Solve each inequality

1) 𝑥 + 6 < 10 2) 𝑦 – 7 > 14

3) 2𝑥 + 6 < 8 4) 3𝑦 + 7 ≥ 31

5) −𝑥 + 6 ≤ 9 6) 8 − 𝑦 > 12

7) 7 – 3𝑥 < 8 8) 19 – 4𝑦 ≥ 40

15

2.6 Simultaneous Equations

Example 3x + 2y = 8

5x + y = 11

x and y stand for two numbers. Solve these equations in order to find the values of x and y by

eliminating one of the letters from the equations.

In these equations it is simplest to eliminate y. Make the coefficients of y the same in both equations.

To do this multiply equation by 2, so that both equations contain 2y:

3x + 2y = 8

10x + 2y = 22 2× =

To eliminate the y terms, subtract equation from equation . We get: 7x = 14

i.e. x = 2

To find y substitute x = 2 into one of the original equations. For example put it into :

10 + y = 11

y = 1

Therefore the solution is x = 2, y = 1.

Remember: Check your solutions by substituting both x and y into the original equations.

Example: Solve 2x + 5y = 16

3x – 4y = 1

Solution: Begin by getting the same number of x or y appearing in both equation. Multiply the top

equation by 4 and the bottom equation by 5 to get 20y in both equations:

8x + 20y = 64

15x – 20y = 5

As the SIGNS in front of 20y are DIFFERENT, eliminate the y terms from the equations by ADDING:

23x = 69 +

i.e. x = 3

Substituting this into equation gives:

6 + 5y = 16

5y = 10

So… y = 2

The solution is x = 3, y = 2.

16

Exercise: Solve the pairs of simultaneous equations in the following questions:

1) x + 2y = 7 2) x + 3y = 0

3x + 2y = 9 3x + 2y = -7

3) 3x – 2y = 4 4) 9x – 2y = 25

2x + 3y = -6 4x – 5y = 7

5) 4a + 3b = 22 6) 3p + 3q = 15

5a – 4b = 43 2p + 5q = 14

17

2.7 Factorising Linear Expressions

Example 1: Factorise 12x – 30

Solution: 6 is a common factor to both 12 and 30. Factorise by taking 6 outside a bracket:

12x – 30 = 6(2x – 5)

Example 2: Factorise 6x2 – 2xy

Solution: 2 is a common factor to both 6 and 2. Both terms also contain an x.

Factorise by taking 2x outside a bracket.

6x2 – 2xy = 2x(3x – y)

Example 3: Factorise 9x3y2 – 18x2y

Solution: 9 is a common factor to both 9 and 18.

The highest power of x that is present in both expressions is x2.

There is also a y present in both parts.

So we factorise by taking 9x2y outside a bracket:

9x3y2 – 18x2y = 9x2y(xy – 2)

Example 4: Factorise 3x(2x – 1) – 4(2x – 1)

Solution: There is a common bracket as a factor.

So we factorise by taking (2x – 1) out as a factor.

The expression factorises to (2x – 1)(3x – 4)

EXERCISE A Factorise each of the following

1) 3x + xy

2) 4x2 – 2xy

3) pq2 – p2q

4) 3pq - 9q2

5) 2x3 – 6x2 6) 8a5b2 – 12a3b4

7) 5y(y – 1) + 3(y – 1)

18

2.8 Factorising Quadratic Expressions

Simple quadratics: Factorising quadratics of the form 𝒙𝟐 + 𝒃𝒙 + 𝒄

The method is:

Step 1: Form two brackets (x … )(x … )

Step 2: Find two numbers that multiply to give c and add to make b. Write these two numbers

at the end of the brackets.

Example 1: Factorise x2 – 9x – 10.

Solution: Find two numbers that multiply to make -10 and add to make -9. These numbers are -10 and

1.

Therefore x2 – 9x – 10 = (x – 10)(x + 1).

General quadratics: Factorising quadratics of the form 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄

One method is that of combining factors. There are many more options that you can use.

Another method is:

Step 1: Find two numbers that multiply together to make ac and add to make b.

Step 2: Split up the bx term using the numbers found in step 1.

Step 3: Factorise the front and back pair of expressions as fully as possible.

Step 4: There should be a common bracket. Take this out as a common factor.

Example 2: Factorise 6x2 + x – 12.

Solution: We need to find two numbers that multiply to make 6 × -12 = -72 and add to make 1. These

two numbers are -8 and 9.

Therefore, 6x2 + x – 12 = 6x2 - 8x + 9x – 12

= 2x(3x – 4) + 3(3x – 4) (the two brackets must be identical)

= (3x – 4)(2x + 3)

Difference of two squares: Factorising quadratics of the form 2 2x a−

Remember that 2 2x a− = (x + (a)(x – (a).

Therefore: 2 2 29 3 ( 3)( 3)x x x x− = − = + −

2 2 216 25 (2 ) 5 (2 5)(2 5)x x x x− = − = + −

Also notice that: 2 22 8 2( 4) 2( 4)( 4)x x x x− = − = + −

and 3 2 2 23 48 3 ( 16 ) 3 ( 4 )( 4 )x xy x x y x x y x y− = − = + −

19

Factorising by pairing or grouping

Factorise expressions like 22 2x xy x y+ − − using the method of factorising by pairing:

22 2x xy x y+ − − = x(2x + y) – 1(2x + y)

= (2x + y)(x – 1)

(factorise front and back pairs, both brackets

identical)

EXERCISE B Factorise

1) 𝑥2 − 𝑥 − 6

2) 𝑥2 + 6𝑥 − 16

3) 2𝑥2 + 5𝑥 + 2

4) 2𝑥2 − 3𝑥

5) 3𝑥2 + 5𝑥 − 2

6) 2𝑦2 + 17𝑦 + 21

7) 7𝑦2 − 10𝑦 + 3

8) 10𝑥2 + 5𝑥 − 30

9) 4𝑥2 − 25

10) 𝑥2 − 3𝑥 − 𝑥𝑦 + 3𝑦2

11) 4𝑥2 − 12𝑥 + 8

12) 16𝑚2 − 81𝑛2

13) 4𝑦3 − 9𝑎2𝑦

14) 8(𝑥 + 1)2 − 2(𝑥 + 1) − 10

20

2.9 Completing the Square

A related process is to write a quadratic expression such as 𝑥2 + 6𝑥 + 11 in the form (𝑥 + 𝑎)2 + 𝑏. This is

called completing the square. It is often useful, because 𝑥2 + 6𝑥 + 11 is not a very transparent expression –

it contains x in more than one place, and it’s not easy either to rearrange or to relate its graph to that of 2x .

Completing the square for quadratic expressions in which the coefficient of 𝑥2 is 1 (these are called monic

quadratics) is very easy. The number a inside the brackets is always half of the coefficient of x.

Example 1 Write x 2 + 6x + 4 in the form (x + a)2 + b.

Solution x 2 + 6 x + 4 is a monic quadratic, so a is half of 6, namely 3.

When you multiply out (x + 3)2, you get x2 + 6x + 9.

[The x-term is always twice a, which is why you have to halve it to get a.]

x2 + 6x + 9 isn’t quite right yet; we need 4 at the end, not 9, so we can write

x2 + 6x + 4 = (x + 3)2 – 9 + 4

= (x + 3)2 – 5.

This version immediately gives us several useful pieces of information. For instance, we now know a lot

about the graph of y = x2 + 6x + 4:

• It is a translation of the graph of y = x2 by 3 units to the left and 5 units down

• Its line of symmetry is x = –3

• Its lowest point or vertex is at (–3, –5)

We also know that the smallest value of the function x2 + 6x + 4 is –5 and this occurs when

x = –3.

And we can solve the equation x2 + 6x + 4 = 0 exactly without having to use the quadratic equation formula,

to locate the roots of the function:

x2 + 6x + 4 = 0

(x + 3)2 – 5 = 0

(x + 3)2 = 5

(x + 3) = 5 [don’t forget that there are two possibilities!]

x = –3 5

These are of course the same solutions that would be obtained from the quadratic equation formula – not

very surprisingly, as the formula itself is obtained by completing the square for the general quadratic

equation ax2 + bx + c = 0.

21

Non-monic quadratics

Everyone knows that non-monic quadratic expressions are hard to deal with. Nobody really likes trying to

factorise 6x2 + 5x – 6 (although you should certainly be willing and able to do so for A Level, which is why

some examples are included in the exercises here).

Example 2 Write 2x2 + 12x + 23 in the form a(x + b)2 + c.

Solution First take out the factor of 2:

2x2 + 12x + 23 = 2(x2 + 6x + 11.5) [you can ignore the 11.5 for now]

Now we can use the method for monic quadratics to write

x2 + 6x + 11.5 = (x + 3)2 + (something)

So we have, so far

2x2 + 12x + 23 = 2(x + 3)2 + c [so we already have a = 2 and b = 3]

Now 2(x + 3)2 = 2(x2 + 6x + 9)

= 2x2 + 12x + 18

We want 23 at the end, not 18, so:

2x2 + 12x + 23 = 2(x + 3)2 – 18 + 23

= 2(x + 3)2 + 5.

If the coefficient of x2 is a perfect square you can sometimes get a more useful form.

Example 3 Write 4x2 + 20x + 19 in the form (ax + b)2 + c.

Solution It should be obvious that a = 2 (the coefficient of a2 is 4).

So 4x2 + 20x + 19 = (2x + b)2 + c

If you multiply out the bracket now, the middle term will be 2 2x b = 4bx.

So 4bx must equal 20x and clearly b = 5.

And we know that (2x + 5)2 = 4x2 + 20x + 25.

So 4x2 + 20x + 19 = (2x + 5)2 – 25 + 19

= (2x + 5)2 – 6.

EXERCISE A

1 Write the following in the form (x + a)2 + b.

(a) x2 + 8x + 19 (b) x2 – 10x + 23 (c) x2 + 2x – 4

(d) x2 – 4x – 3 (e) x2 – 3x + 2 (f) x2 – 5x – 6

2 Write the following in the form a(x + b)2 + c.

(a) 3x2 + 6x + 7 (b) 5x2 – 20x + 17 (c) 2x2 + 10x + 13

3 Write the following in the form (ax + b)2 + c.

(a) 4x2 + 12x + 14 (b) 9x2 – 12x – 1 (c) 16x2 + 40x + 22

Half of 6

22

2.10 Solving Quadratic Equations

A quadratic equation has the form 2 0ax bx c+ + = .

There are two methods that are commonly used for solving quadratic equations:

* factorising

* the quadratic formula

Not all quadratic equations can be solved by factorising.

Method 1: Factorising

Make sure that the equation is rearranged so that the right hand side is 0. It usually makes it easier if the

coefficient of x2 is positive.

Example 1 : Solve x2 –3x + 2 = 0

Factorise (x –1)(x – 2) = 0

Either (x – 1) = 0 or (x – 2) = 0

So the solutions are x = 1 or x = 2

Note: The individual values x = 1 and x = 2 are called the roots of the equation.

Example 2: Solve x2 – 2x = 0

Factorise: x(x – 2) = 0

Either x = 0 or (x – 2) = 0

So x = 0 or x = 2

23

Method 2: Using the formula

The roots of the quadratic equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 are given by the formula:

𝑥 =−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎

Example 3: Solve the equation 22 5 7 3x x− = −

Solution: First we rearrange so that the right hand side is 0. We get 22 3 12 0x x+ − =

We can then tell that a = 2, b = 3 and c = -12.

Substituting these into the quadratic formula gives:

𝑥 =−3 ± √32 − 4 × 2 × (−12)

2 × 2=

−3 ± √105

4(this is the surd form for the solutions)

If we have a calculator, we can evaluate these roots to get: x = 1.81 or x = -3.31

Exercise

1) Use factorisation to solve the following equations:

(a) x2 + 3x + 2 = 0 (b) x2 – 3x – 4 = 0 (c) x2 = 15 – 2x

2) Find the roots of the following equations:

(a) x2 + 3x = 0 (b) x2 – 4x = 0 (c) 4 – x2 = 0

3) Solve the following equations either by factorising or by using the formula:

(a) 6x2 - 5x – 4 = 0 (b) 8x2 – 24x + 10 = 0

4) Use the formula to solve the following equations to 3 significant figures where possible

(a) x2 +7x +9 = 0

(b) 6 + 3x = 8x2

(c) 4x2 – x – 7 = 0

(d) x2 – 3x + 18 = 0

(e) 3x2 + 4x + 4 = 0

f) 3x2 = 13x – 16

24

2.11 Changing the Subject

Rearranging a formula is similar to solving an equation –always do the same to both sides.

Example 1: Make x the subject of the formula y = 4x + 3.

Solution: y = 4x + 3

Subtract 3 from both sides: y – 3 = 4x

Divide both sides by 4; 3

4

yx

−=

So 3

4

yx

−= is the same equation but with x the subject.

Example 2: Make x the subject of y = 2 – 5x

Solution: Notice that in this formula the x term is negative.

y = 2 – 5x

Add 5x to both sides y + 5x = 2 (the x term is now positive)

Subtract y from both sides 5x = 2 – y

Divide both sides by 5 2

5

yx

−=

Example 3: The formula 5( 32)

9

FC

−= is used to convert between ° Fahrenheit and ° Celsius.

Rearrange to make F the subject.

5( 32)

9

FC

−=

Multiply by 9 9 5( 32)C F= − (this removes the fraction)

Expand the brackets 9 5 160C F= −

Add 160 to both sides 9 160 5C F+ =

Divide both sides by 5 9 160

5

CF

+=

Therefore the required rearrangement is 9 160

5

CF

+= .

EXERCISE A Make x the subject of each of these formulae:

1) y = 7x – 1

2) 5

4

xy

+=

3) 4 23

xy = −

4) 4(3 5)

9

xy

−=

25

Example 4: Make x the subject of 𝑥2 + 𝑦2 = 𝑤2

Solution: 𝑥2 + 𝑦2 = 𝑤2

Subtract 2y from both sides: 𝑥2 = 𝑤2 − 𝑦2 (this isolates the term involving x)

Square root both sides: 𝑥 = ±√𝑤2 − 𝑦2

Remember the positive & negative square root.

Example 5: Make a the subject of the formula 1 5

4

at

h=

Solution: 1 5

4

at

h=

Multiply by 4 5

4a

th

=

Square both sides 2 5

16a

th

=

Multiply by h: 216 5t h a=

Divide by 5: 216

5

t ha=

EXERCISE B: Make t the subject of each of the following

1) 32

wtP

r=

2) 2

32

wtP

r=

3) 21

3V t h=

4) 2t

Pg

=

5) ( )w v t

Pag

−=

6) 𝑟 = 𝑎 + 𝑏𝑡2

26

When the Subject Appears More Than Once

Sometimes the subject occurs in more than one place in the formula. In these questions collect the

terms involving this variable on one side of the equation, and put the other terms on the opposite side.

Example 6: Make t the subject of the formula 𝑎 − 𝑥𝑡 = 𝑏 + 𝑦𝑡

Solution: 𝑎 − 𝑥𝑡 = 𝑏 + 𝑦𝑡

Start by collecting all the t terms on the right hand side:

Add xt to both sides: 𝑎 = 𝑏 + 𝑦𝑡 + 𝑥𝑡

Now put the terms without a t on the left hand side:

Subtract b from both sides: 𝑎 − 𝑏 = 𝑦𝑡 + 𝑥𝑡

Factorise the RHS: 𝑎 − 𝑏 = 𝑡(𝑦 + 𝑥)

Divide by (y + x): a b

ty x

−=

+

So the required equation is a b

ty x

−=

+

Example 7: Make W the subject of the formula 2

WaT W

b− =

Solution: This formula is complicated by the fractional term. Begin by removing the fraction:

Multiply by 2b: 2𝑏𝑇 − 2𝑏𝑊 = 𝑊𝑎

Add 2bW to both sides: 2𝑏𝑇 = 𝑊𝑎 + 2𝑏𝑊 (this collects the W’s together)

Factorise the RHS: 2𝑏𝑇 = 𝑊(𝑎 + 2𝑏)

Divide both sides by a + 2b: 2

2

bTW

a b=

+

Exercise C Make x the subject of these formulae:

1) 𝑎𝑥 + 3 = 𝑏𝑥 + 𝑐

2) 3(𝑥 + 𝑎) = 𝑘(𝑥 − 2)

3) 2 3

5 2

xy

x

+=

−

4) 1x x

a b= +

27

2.12 Indices

Basic rules of indices

𝑦4 means 𝑦 × 𝑦 × 𝑦 × 𝑦. 4 is called the index (plural: indices), power or exponent of y.

There are 3 basic rules of indices:

1) 𝑎𝑚 × 𝑎𝑛 = 𝑎𝑚+𝑛 e.g. 34 × 35 = 39

2) 𝑎𝑚 ÷ 𝑎𝑛 = 𝑎𝑚−𝑛 e.g. 38 × 36 = 32

3) (𝑎𝑚)𝑛 = 𝑎𝑚𝑛 e.g. (32)5 = 310

Further examples

𝑦4 × 5𝑦3 = 5𝑦7

4𝑎3 × 6𝑎2 = 24𝑎5

2𝑐2 × (−3𝑐6) = −6𝑐8

24𝑑7 ÷ 3𝑑2 =24𝑑7

3𝑑2= 8𝑑5

(multiply the numbers and multiply the a’s)

(multiply the numbers and multiply the c’s)

(divide the numbers and divide the d terms by

subtracting the powers)

EXERCISE A Simplify the following: Remember that 𝑏 = 𝑏1

1) 𝑏 × 5𝑏5

2) 3𝑐2 × 2𝑐5

3) 𝑏2𝑐 × 𝑏𝑐3

4) 2𝑛6 × (−6𝑛2)

5) 8𝑛8 ÷ 2𝑛3

6) 𝑑11 ÷ 𝑑9

7) (𝑎3)2

8) (−𝑑4)3

28

Zero index: Remember 0 1a = For any non-zero number, a.

Therefore ( )0

00 35 1 1 5.2304 1

4

= = − =

Negative powers

A power of -1 corresponds to the reciprocal of a number, i.e. 1 1

aa

− =

Therefore 1 1

55

− =

1 1

0.25 40.25

− = =

14 5

5 4

−

=

(Find the reciprocal of a fraction by turning it upside down)

This result can be extended to more general negative powers: 1n

na

a

− = .

This means:

2

2

1 13

93

− = = 4

4

1 12

162

− = =

22 1 2

1 1 416

4 4 1

− − = = =

There is a particularly nice way of understanding the negative powers.

Consider the following:

31 32 33 34 35

3 9 27 81 243

Every time you move one step to the right you multiply by 3.

Now consider the sequence continuing, right-to-left:

3-2 3-1 30 31 32 33 34 35

19 1

3 1 3 9 27 81 243

Each time you move one step to the left you divide by 3.

Take particular care when there are numbers as well as negative powers.

Example 110

10xx

−= but 11 1

10 10x

x

−= or (10x)–1.

The usual rules of powers and brackets tell you that 10x–1 is not the same as (10x)–1.

3⎯⎯→ 3

⎯⎯→

3⎯⎯ 3

⎯⎯3

⎯⎯

3⎯⎯→ 3

⎯⎯→

3⎯⎯ 3

⎯⎯ 3

⎯⎯ 3

⎯⎯

29

Fractional powers:

Fractional powers correspond to roots: 𝑎1/2 = √𝑎 𝑎1/3 = √𝑎3 𝑎1/4 = √𝑎4

In general: 𝑎1/𝑛 = √𝑎𝑛

Therefore:

81

3 = √83

= 2 251

2 = √25 = 5 100001

4 = √10000

4= 10

A more general fractional power can be dealt with in the following way: 𝑎𝑚

𝑛 = (𝑎1

𝑛)𝑚

So 43

2 = (√4)3

= 23 = 8

(8

27)

23

= ((8

27)

13 )

2

= (2

3)

2

=4

9

(25

36)

−3/2

= (36

25)

3/2

= (√36

25)

3

= (6

5)

3

=216

125

EXERCISE B: Find the value of:

1) 41

2

2) 271

3

3) (1

9)

1

2

4) 5−2

5) 180

6) 7−1

7) 272

3

8)

22

3

−

9) 8−2

3

10) (0.04)1

2

11)

2 / 38

27

12)

3 / 21

16

−

Simplify each of the following:

13) 2𝑎1

2 × 3𝑎5

2

14) 𝑥3 × 𝑥−2

15) (𝑥2𝑦4)1

2

30

2.13 Surds

A surd is a root of a number that cannot be expressed as an integer. Surds are part of the set of irrational

numbers.

Example:

√3 and √8 are surds but √4 is not as it equals 2.

Simplifying Surds

Start to simplify surds by using two rules:

√𝑎𝑏 = √𝑎 × √𝑏 and √𝑎

𝑏=

√𝑎

√𝑏

By using the multiplication rule, simplify surds by finding a factor of the number you are taking a root

of which is a square number. Always try to find the largest square number factor, otherwise you will

have to simplify further.

Example:

√8 = √4 × √2

= 2√2

3√12 = 3 × √4 × √3

= 3 × 2 × √3

= 6√3

√600

√2= √

600

2

= √300

= √100 × √3

= 10√3

EXERCISE A

Simplify

1) √50

2) √72

3) √27

4) √80

5) √360

6) √900

√3

31

Multiplying and Dividing with Surds

The rules of algebra are true for any numeric value; these include surds. To multiply and divide

expressions with surds, deal with any integers together and then deal with any surds.

Examples:

2√3 × √2 = 2√6

3√5 × 6√2 = 18√10

2√5 × 7√8 = 14√40

= 14 × √4 × √10

= 28√10

√2(5 + 2√3) = 5√2 + 2√6

8√14

2√7= 4√2

(1 + √3)(2 − √2) = 2 − 2√2 + 2√3 − √6

(3 + √2)(3 − √2) = 32 − (√2)2

= 1

In this example, you could expand as usual but this is an example of the difference of two squares.

EXERCISE B

Simplify

1) √3 × √7

2) 5√2 × 4√5

3) 3√3 × 2√6

4) √8 × √27

5) 5√20

6√5

6) 8√18

4√2

7) (√2 + 1)(√2 +

5)

8) (5 − √3)(√2 −

8)

32

Addition and Subtraction with Surds

You can only add or subtract with surds if the surd is the same; sometimes if they are not the same, you

may be able to simplify them so that the same surd is present.

Example:

2√3 + 4√3 + 6√5 = 6√3 + 6√5

Here add the 2√3 and 4√3 as the same surd is present but you cannot add the 6√5.

2√5 + √45 = 2√5 + 3√5

= 5√5

By simplifying √45 to 3√5, you can add the two surds together.

These methods also work for subtraction of surds.

Exercise C

Simplify

1) √3 + √7

2) 5√2 + 4√2

3) 3√6 + √24

4) √50 + √8

5) √27 + √75

6) 2√5 − √5

7) √72 − √50

8) 6√3 − √12 + √27

9) √200 + √90 − √98

10) √72 − √75 + √108

33

Rationalising the Denominator

It is far easier to calculate with a fraction if the surd if the denominator is a rational number (i.e. not a

surd). The process of this is known as rationalising the denominator. To do this, multiply by the surd in

the denominator, doing so makes use of the fact that √𝑎 × √𝑎 = (√𝑎)2

= 𝑎

Example: 1

√3

Multiply the denominator by √3 to rationalise it and so multiply the numerator by √3 also:

1

√3×

√3

√3=

√3

3

Example 2:

4

√2=

4

√2×

√2

√2

=4√2

2

= 2√2

Example 3:

2 + √3

√5=

2 + √3

√5×

√5

√5

=√5(2 + √3)

5

=2√5 + √15

5

If there is more than just a surd in the denominator, we make use of the difference of two squares by

multiplying by its conjugate.

Example:

Rationalise 2

3−√7

We multiply the numerator and denominator by its

conjugate: 3 + √7

It’s a difference of two squares so expand as usual

2

3 − √7×

3 + √7

3 + √7=

2(3 + √7)

(3 − √7)(3 + √7)

=2(3 + √7)

32 − (√7)2

=2(3 + √7)

9 − 7

=2(3 + √7)

2

= 3 + √7

34

Exercise D

Rationalise the following:

1

a) 1

√2 b)

3

√5 c)

10

√5

d) 5

2√7 e)

√3

√2 f)

10

√10

g) 4+√7

√3 h)

6+8√5

√2 i)

6−√5

√5

2

a) 1

2 1− b)

2

6 2− c)

6

7 2+

d) 1

3 5+ e)

1

6 5−

35

2.14 Algebraic Fractions

Algebraic fractions can be treated in exactly the same way as numerical fractions.

Example Multiply 3𝑥

7𝑦 by 2.

Solution 3 2 = 6x, so the answer is6𝑥

7𝑦. (Not

6𝑥

14𝑦as this is just an equivalent fraction!)

Example Divide 3𝑦2

4𝑥 by y.

Solution 3𝑦2

4𝑥÷ 𝑦 =

3𝑦2

4𝑥×

1

𝑦

=3𝑦2

4𝑥𝑦

=3𝑦

4𝑥 (Don’t forget to simplify!)

Example Divide𝑃𝑄𝑅

100 by T.

Solution 𝑃𝑄𝑅

100÷ 𝑇 =

𝑃𝑄𝑅

100×

1

𝑇

=𝑃𝑄𝑅

100𝑇

Here it would be wrong to say just

𝑃𝑄𝑅

100

𝑇, which is a mix (as well as a mess!)

Double fractions, or mixtures of fractions and decimals, are always wrong.

For instance, if you want to divide 𝑥𝑦

𝑧 by 2, you should not say

0.5𝑥𝑦

𝑧 but

𝑥𝑦

2𝑧.

This sort of thing is extremely important when it comes to rearranging formulae.

36

Example Simplify 3

𝑥−1−

1

𝑥+1

Solution Use a common denominator. [You must treat (x – 1) and (x + 1) as separate

expressions with no common factor.]

3 1 3( 1) ( 1)

1 1 ( 1)( 1)

x x

x x x x

+ − −− =

− + − +

3 3 1

( 1)( 1)

x x

x x

+ − +=

− +

2 4

( 1)( 1)

x

x x

+=

− + .

Do use brackets, particularly on top – otherwise you are likely to forget the minus at the end

of the numerator (in this example subtracting -1 gives +1).

Don’t multiply out the brackets on the bottom. You will need to see if there is a factor, which

cancels out (although there isn’t one in this case).

Example Simplify 2

3𝑥−3+

5

𝑥2−1.

Solution A common denominator may not be obvious, you should look to see if the

denominator factorises first.

2

2 5 2 5

3 3 3( 1) ( 1)( 1)1x x x xx+ = +

− − + −−

2( 1) 5 3

3( 1)( 1)

x

x x

+ + =

− +

2 2 15

3( 1)( 1)

x

x x

+ +=

− +

2 17

3( 1)( 1)

x

x x

+=

− + If one of the terms is not a fraction already, the best plan is to make it one.

Example Write 3

𝑥+1+ 2 as a single fraction.

Solution

3 3 22

1 1 1x x+ = +

+ +

3 2( 1)

1

x

x

+ +=

+

2 5

1

x

x

+=

+ This method often produces big simplifications when roots are involved.

𝑥 − 1 is a common factor, so the common denominator is 3(𝑥 − 1)(𝑥 + 1).

37

Example Write 𝑥

√𝑥−2 +√𝑥 − 2as a single fraction.

Solution

22

12 2

x x xx

x x

−+ − = +

− −

2( 2)

2

x x

x

+ −=

−

( 2)

2

x x

x

+ −=

−

2 2

2

x

x

−=

−

It is also often useful to reverse this process – that is, to rewrite expressions such as𝑥

𝑥−2. The

problem with this expression is that x appears in more than one place and it is not very easy

to manipulate such expressions (for example, in finding the inverse function, or sketching a

curve). Here is a very useful trick.

Exercise

1 Write as single fractions.

(a) 2

𝑥 − 1+

1

𝑥 + 3(b) 2

𝑥 − 3−

1

𝑥 + 2(c) 1

2𝑥 − 1−

1

3𝑥 + 2(d) 3

𝑥 + 2+ 1

(e) 12

1x−

−

(f) 23

1

x

x−

+

(g) 2

3 1

4(2 1) 4 1x x−

− −

2 Write as single fractions.

(a) 𝑥 + 1

√𝑥+ √𝑥

(b) 2𝑥

√𝑥 + 3+ √𝑥 + 3

(c) 𝑥

√𝑥 − 23 + √(𝑥 − 2)23

38

2.15 Linear and Quadratic Simultaneous Equations

I am sure that you will be very familiar with the standard methods of solving simultaneous

equations (elimination and substitution). You will probably have met the method for solving

simultaneous equations when one equation is linear and one is quadratic. Here you have no

choice; you must use substitution.

Example 1 Solve the simultaneous equations x + 3y = 6

x2 + y2 = 10

Solution Make one letter the subject of the linear equation: x = 6 – 3y

Substitute into the quadratic equation (6 – 3y)2 + y2 = 10

Solve … 10y2 – 36y + 26 = 0

2(y – 1)(5y – 13) = 0

… to get two solutions: y = 1 or 2.6

Substitute both back into the linear equation x = 6 – 3y = 3 or –1.8

Write answers in pairs: (x, y) = (3, 1) or (–1.8, 2.6)

• You can’t just square root the quadratic equation.

• You could have substituted for y instead of x (though in this case that would

have taken longer – try to avoid fractions if you can).

• It is very easy to make mistakes here. Take great care over accuracy.

• It is remarkably difficult to set questions of this sort in such a way that both

pairs of answers are nice numbers. Don’t worry if, as in this example, only one

pair of answers are nice numbers.

Questions like this appear in many GCSE papers. They are often, however, rather simple

(sometimes the quadratic equations are restricted to those of the form x2 + y2 = a) and it is

important to practice less convenient examples.

39

Exercise

Solve the following simultaneous equations.

1 x2 + xy = 12 2 x2 – 4x + y2 = 21

3x + y = 10 y = 3x – 21

3 x2 + xy + y2 = 1 4 x2 – 2xy + y2 = 1

x + 2y = –1 y = 2x

5 c2 + d 2 = 5 6 x + 2y = 15

3c + 4d = 2 xy = 28

7 2x2 + 3xy + y2 = 6 8 2x2 + 4xy + 6y2 = 4

3x + 4y = 1 2x + 3y = 1

9 4x2 + y2 = 17 10 2x2 − 3xy + y2 = 0

2x + y = 5 x + y = 9

11 x2 + 3xy + 5y2 = 15 12 xy + x2 + y2 = 7

x − y = 1 x − 3y = 5

13 x2 + 3xy + 5y2 = 5 14 4x2 − 4xy −3y2 = 20

x − 2y = 1 2x − 3y = 10

15 x2 − y2 = 11 16 12 1

3x y+ =

x − y = 11 x + y = 7

40

2.16 Expanding More Than Two Binomials

You should already be able to expand algebraic expressions of the form (ax + b)(cx + d).

e.g. (2x + 1)(3x − 2) = 6x2 − 4x + 3x − 2 = 6x2 − x − 2

e.g. (5x + 4)(5x − 4) = 25x2 − 20x + 20x − 16 = 25x2 − 16

We are now going to algebraic expressions of the form (ax + b)(cx + d)(ex + f).

To simplify the product of three binomials, first expand any two of the brackets and then

multiply this answer by each term in the third bracket

Example 1:

Expand and simplify (x − 2)(2x + 3)(x + 7)

(x − 2)(2x + 3) = 2x2 + 3x − 4x − 6 First expand two of the brackets

(You may prefer to use the grid method) = 2x2 − x − 6 Simplify

Now (x − 2)(2x + 3)(x + 7) = (x + 7)(2x2 − x − 6)

= x(2x2 − x − 6) + 7(2x2 − x − 6) Multiply your expansion by each term

in the 3rd bracket

= 2x3 − x2 − 6x + 14x2 − 7x − 42

= 2x3 + 13x2 − 13x − 42 Simplify

Example 2:

Show that (2x + 5)( x − 1)(4x − 3) = 8x3 + 6x2 − 29x + 15 for all values of x.

(2x + 5)(x − 1) = 2x2 − 2x + 5x − 5 First expand any two of the brackets.

= 2x2 + 3x − 5 Simplify

Now (2x + 5)(x − 1)(4x − 3) = (4x − 3)( 2x2 + 3x − 5)

= 4x(2x2 + 3x − 5) − 3(2x2 + 3x − 5) Multiply your expansion by each term

in the 3rd bracket

= 8x3 + 12x2 − 20x − 6x2 − 9x + 15 Remember the minus outside the 2nd bracket

changes each sign inside the 2nd bracket

= 8x3 + 6x2 − 29x + 15 Simplify

To simplify the product of four binomials, first expand any two of the brackets and then

expand the other two brackets and then multiply these answers.

41

Example 3:

Expand and simplify (x + 3)(x − 3)(2x + 1)(5x − 6)

(x + 3)(x − 3) = x2 − 9 Expand two of the brackets

(2x + 1)(5x − 6) = 10x2 − 7x − 6 Expand the other two brackets

(x + 3)(x − 3)(2x + 1)(5x − 6)

= (x2 − 9)(10x2 − 7x − 6) Use the two expansions above

= x2(10x2 − 7x − 6) − 9(10x2 − 7x − 6) Multiply the 2nd bracket

by each term in the 1st bracket = 10x4 − 7x3 − 6x2 − 90x2 + 63x + 54

= 10x4 − 7x3 − 96x2 + 63x + 54 Simplify

EXERCISE:

1. Expand and simplify

(a) (x + 1)(x + 4)(x + 5) (b) (2x + 7)(3x + 1)(x + 8)

(c) (x − 3)(x − 1)(2x − 3) (d) (3x + 8)(x − 2)(2x − 5)

(e) (5x − 1)(2x + 5)(3x − 2) (f) (4x + 1)(2x + 7)(4x − 1)

(g) (x + 4)2 (3x − 7) (h) (6x + 5)(2x − 1)2

(i) (x − 1)(x + 1)(4x − 1)(2x − 5) (j) (x + 5)2(x − 2)2

2. Show that (2x + 3)3 = 8x3 + 36x2 + 54x + 27 for all values of x.

3. Show that (x − 4)2(x + 3) simplifies to x3 + ax2 + bx + c where a, b and c are

integers.

4. Express (3x − 1)4 in the form ax4 + bx3 + cx2 + dx + e where a, b, c, d and e are

integers.

5. (3x + 5)(x − 4)(3x − 2) = 9x3 + Ax2 + Bx + 40

Work out the value of A and the value of B.

6. (x − 3)(2x + 1)(Ax + 1) = 8x3 + Bx2 + Cx − 3

Work out the value of A, the value of B and the value of C.

7. Here is a cuboid.

x − 2

x + 5

2x + 1

All measurements are in centimetres.

Show that the volume of the cuboid is (2x3 + 7x2 − 17x − 10) cm3.

42

8. f(x) = 3x3 − 2x2 + 4

Express f(x + 2) in the form ax3 + bx2 + cx + d.

9. The smallest of three consecutive positive odd numbers is (2x − 1).

Work out the product of the three numbers.

Give your answer in the form ax3 + bx2 + cx + d.

43

2.17 Quadratic Inequalities

You should be able to solve quadratic equations of the form ax2 + bx + c = 0

e.g. x2 − 3x − 4 = 0 (x − 4)(x + 1) = 0 x = 4 or x = −1

e.g. 3x2 − 14x + 8 = 0 (3x − 2)(x − 4) = 0 x = 3

2 or x = 4

e.g. x2 = 10 − 3x x2 + 3x − 10 = 0 (x + 5)(x − 2) = 0 x = − 5 or x = 2

You should also know the shape of a quadratic curve.

If the coefficient of x2 is positive, the curve is ‘smiling’.

If the coefficient of x2 is negative, the curve is ‘frowning’.

If f(x) > 0 or f(x) ≥ 0 we want the values of x where f(x) is above the x-axis.

If f(x) < 0 or f(x) ≤ 0 we want the values of x where f(x) is below the x-axis.

Example 1:

Solve x2 + 5x – 24 ≥ 0

(x + 8)(x − 3) ≥ 0 First factorise your quadratic expression

Critical values are x = −8 and x = 3 Solve (x + 8)(x − 3) = 0

−8 3 x Always draw a sketch of your curve

Show where the curve cuts the x-axis

by solving (x + 8)(x − 3) = 0

x ≤ −8 and x ≥ 3 We want the area where y ≥ 0

If you are asked to write the solution set of the inequality x2 + 5x − 24 ≥ 0 then the answer

would be: x : x ≤ −8, x ≥ 3

NOTE: There are TWO regions so we write the answer as TWO inequalities.

44

Example 2:

Find the solution set of the inequality 6(x2 + 2) < 17x

6x2 + 12 < 17x First expand the bracket

6x2 − 17x + 12 < 0 Rearrange to the form ax2 + bx + c < 0

(3x − 4)(2x − 3) < 0 Factorise in order to find where it cuts the x-axis

Critical values are x = 4/3 and 3/2 Solve (3x −4)(2x − 3) = 0

3

4

2

3 x Sketch the curve and shade below the axis

3

4 < x <

2

3We want the region where f(x) is below the x-axis

Solution set = x : 3

4 < x <

2

3 Make sure your answer is given in the correct form

NOTE: There is ONE region so we write the answer as ONE inequality.

Example 3:

Solve x(x + 9) ≤ 0

x(x + 9) ≤ 0 This is already factorised with 0 on one side

so there is no need to expand the brackets

Critical values are x = 0 and x = −9

−9 0 x Sketch the curve and shade below the axis

−9 ≤ x ≤ 0 We want the region where f(x) is below the x-axis

There is only one region so write as one inequality

45

Example 4:

Solve the inequality 14 + 5x < x2

14 + 5x − x2 < 0 Rearrange to the form ax2 + bx + c < 0

(2 + x)(7 − x) < 0 Factorise in order to find where it cuts the x-axis

−2 7 x The curve is 'frowning' as we have −x2

x < −2 and x > 7 We want the region where f(x) is below the x-axis

OR

14 + 5x − x2 < 0 Rearrange to the form ax2 + bx + c < 0

x2 − 5x − 14 > 0 Multiply each term by −1 which changes < to >

(x + 2)(x − 7) > 0 Factorise in order to find where it cuts the x-axis

−2 7 x The curve is 'smiling' as we have +x2

x < −2 and x > 7 This method gives the same answer as the 1st method

EXERCISE

1. Solve each of these inequalities.

(a) x2 + 9x + 18 ≤ 0 (b) x2 − x − 20 < 0

(c) (x − 2)(x + 7) > 0 (d) x2 − 5x ≥ 0

(e) 2x2 − 11x + 12 < 0 (f) (5 + x)( 1 − 2x) ≥ 0

(g) 15 + 2x – x2 ≤ 0 (h) 21 − x – 2x2 > 0

(i) x(5x − 2) > 0 (j) x2 − 2x > 35

2. Find the solution set for each of these inequalities.

(a) x2 − 4x + 3 ≤ 0 (b) x2 + x − 42 < 0

(c) x(x + 2) > 48 (d) 3x2 + 14x − 5 ≥ 0

(e) 2x2 > 11x − 12 (f) 16 – x2 ≤ 6x

(g) 7 + 2(4x2 – 15x) ≤ 0 (h) x2 – 4(x + 6) > 8

(i) 3x(5 − x) > 0 (j) (x + 5)2 ≥ 1

3. Solve2

122 +x ≥ 4x

4. Find the solution set for which 15 + 2x ≤ x2

5. Find the set of values for which 6 + x ≥ x2 and x + 2 < x2

6. Find the solution set for (x − 3)(2x + 3) < 2x(1 − 2x) − 5

46

2.18 Using Completing the Square to Find Turning Points

You should already be able to express a quadratic equation in the form a(x + b)2 + c by

completing the square.

e.g. x2 − 6x + 3 = (x − 3)2 − 9 + 3 = (x − 3)2 − 6

e.g. 3x2 + 6x + 5 = 3[x2 + 2x] + 5 = 3[(x + 1)2 − 1] + 5 = 3(x + 1)2 + 2

We are now going to deduce the turning points of a quadratic function after completing the

square.

Example 1:

Given y = x2 + 6x − 5, by writing it in the form y =(x + a)2 + b, where a and b are

integers, write down the coordinates of the turning point of the curve. Hence sketch the

curve.

y = x2 + 6x − 5

= (x + 3)2 − 9 − 5 Remember to halve the coefficient of x

= (x + 3)2 − 14 and subtract (−3)2 to compensate

The turning point occurs when (x + 3)2 = 0, i.e. when x = −3

When x = −3, y = (−3 + 3)2 − 14 = 0 − 14 = −14

So the coordinates of the turning point is (−3, −14)

The graph y = x2 + 6x − 5 cuts the y-axis when x = 0, i.e. y = −5

Sketch:

y

x

(0, −5)

(−3, −14)

When y = (x + a)2 + b then the coordinates of the turning point is (−a, b).

The minimum or maximum value of y is b.

47

Example 2:

Given that the minimum turning point of a quadratic curve is (1, −6), find an equation of

the curve in the form y = x2 + ax + b. Hence sketch the curve.

y = (x − 1)2 − 6 If the minimum is when x = 1, we know we have (x − 1)2

= (x2 − x − x + 1) − 6 If the minimum is when y = −6, we know we have (...)2 − 6

= x2 − 2x − 5

An equation of the curve is y = x2 − 2x − 5

The graph cuts the y-axis when x = 0, i.e. at y = −5

Sketch: It is a minimum turning point so the shape is

y

x

(0, −5)

(1, −6)

NOTE: There are other possible equations as, for example y = 4(x − 1)2 − 6 also has a turning

point of (1, −6). If it was a maximum turning point then the coefficient of x2 would be negative.

Example 3:

Find the maximum value of −x2 + 4x − 7 and sketch the curve.

− x2 + 4x − 7 = − ( x2 − 4x + 7) First take out the minus sign

= − [(x − 2)2 − 4 + 7] Remember to use square brackets

= − [(x − 2)2 + 3]

= − (x − 2)2 − 3 Multiply (x − 2)2 and +3 by −1

The maximum value is −3

y It is a maximum value so the shape is

x

(2, −3)

(0, −7)

48

Exercise:

1. By writing the following in the form y = (x + a)2 + b, where a and b are integers, write

down the coordinates of the turning point of the curve. Hence sketch the curve.

(a) y = x2 − 8x + 20 (b) y = x2 − 10x − 1

(c) y = x2 + 4x − 6 (d) y = 2x2 − 12x + 8

(e) y = −x2 + 6x +10 (f) y = 5 − 2x − x2

2. Given the following minimum turning points of quadratic curves, find an equation of the

curve in the form y = x2 + ax + b. Hence sketch each curve.

(a) (2, −3) (b) (−4, 1)

(c) (−1, 5) (d) (3, −12)

(e) (1, −7) (f) (−4, −1)

3. Find the maximum or minimum value of the following curves and sketch each curve.

(a) y = x2 + 4x + 2 (b) y = 1 − 6x − x2

(c) y = −x2 + 2x − 3 (d) y = x2 − 8x + 8

(e) y = x2 − 3x − 1 (f) y = −3x2 + 12x − 9

4. The expression x2 – 3x + 8 can be written in the form (x – a)2 + b for all values of x.

(i) Find the value of a and the value of b.

The equation of a curve is y = f(x) where f(x) = x2 – 3x + 8

The diagram shows part of a sketch of the graph of y = f(x).

The minimum point of the curve is M.

(ii) Write down the coordinates of M.

49

5. (i) Sketch the graph of f(x) = x2 – 6x + 10, showing the coordinates of the turning point

and the coordinates of any intercepts with the coordinate axes.

(ii) Hence, or otherwise, determine whether f(x) − 3= 0 has any real roots.

Give reasons for your answer.

*6. The minimum point of a quadratic curve is (1, −4). The curve cuts the y-axis at −1.

Show that the equation of the curve is y = 3x2 − 6x −1

*7. The maximum point of a quadratic curve is (−2, −5). The curve cuts the y-axis at −13.

Find the equation of the curve. Give your answer in the form ax2 + bx + c.

* = extension

50

3 Functions

3.1 Function Notation

In GCSE Mathematics equations are written as shown below:

y = 3x + 4 y = x² + 5

However, we also used function notation.

We often use the letters f and g and we write the above equations as

f(x) = 3x + 4 g(x) = x² + 5

Example 1:

Using the equation y = 3x + 4, find the value of y if

(a) x = 4 (b) x = −6

(a) y = 3(4) + 4 = 12 + 4 = 16 Substitute for x = 4 in the equation

(b) y = 3(–6) + 4 = –18 + 4 = –14 Substitute for x = −6 in the equation

Example 2:

f is a function such that f(x) = 3x + 4

Find the values of

(a) f(4) (b) f(−6)

(a) f(4) = 3(4) + 4 = 12 + 4 = 16 Substitute for x = 4 in the equation

(b) f(−6) = 3(–6) + 4 = –18 + 4 = –14 Substitute for x = −6 in the equation

51

Example 3:

g is a function such that g(x) = 2x² – 5

Find the values of (a) g(3) (b) g(−4)

(a) g(3) = 2(3)² – 5 = 18 – 5 = 13 Substitute for x = 3 in the equation

(b) g(–4) = 2(–4)² – 5 = 32 – 5 = 27 Substitute for x = −4 in the equation

Example 4:

The functions f and g are defined for all real values of x and are such that

f(x) = x² – 4 and g(x) = 4x + 1

Find (a) f(−3) (b) g(0.3)

(c) Find the two values of x for which f(x) = g(x).

(a) f(−3) = (−3)² – 4 = 9 – 4 = 5 Substitute for x = −3 in the equation f(x)

(b) g(0.3) = 4(0.3) + 1 = 1.2 + 1 = 2.2 Substitute for x = 0.3 in the equation g(x)

(c) x² – 4 = 4x + 1 Put f(x) = g(x)

x² – 4 – 4x – 1 = 0 Rearrange the equation as a quadratic = 0

x² – 4x – 5 = 0 Simplify

(x – 5)(x + 1) = 0 Solve the quadratic by factorising

x – 5 = 0 or x + 1 = 0

x = 5 or x = –1

52

Exercise:

1. The function f is such that f(x) = 5x + 2

Find (a) f(3) (b) f(7) (c) f(−4)

(d) f(−2) (e) f(−0.5) (f) f(0.3)

2. The function f is such that f(x) = x2 − 4

Find (a) f(4) (b) f(6) (c) f(−2)

(d) f(−6) (e) f(−0.2) (f) f(0.9)

3. The function g is such that g(x) = x3 − 3x2 – 2x + 1

Find (a) g(0) (b) g(1) (c) g(2)

(d) g(−1) (e) g(−0.4) (f) g(1.5)

4. The function f is such that 𝑓(𝑥) = √2𝑥 + 5

Find (a) f(0) (b) f(1) (c) f(2)

(d) f(−1) (e) f(−0.7) (f) f(1.5)

5. f(x) = 3x2 – 2x – 8

Express f(x + 2) in the form ax2 + bx

6. The functions f and g are such that

f(x) = 3x – 5 and g(x) = 4x + 1

(a) Find (i) f(−1) (ii) g(2)

(b) Find the value of x for which f(x) = g(x).

7. The functions f and g are such that

f(x) = 2x2 – 1 and g(x) = 5x + 2

(a) Find f(−3) and g(−5)

(b) Find the two values of x for which f(x) = g(x).

53

3.2 Composite Functions

A composite function is a function consisting of 2 or more functions.

The term composition is used when one operation is performed after another operation.

For instance:

This function can be written as f(x) = 5(x + 3)

Suppose f(x) = x2 and g(x) = 2 + 3x

What is fg(x)? Now fg(x) = f[g(x)]

This means apply g first and then apply f.

fg(x) = f(2 + 3x) = (2 + 3x)²

What is gf(x)?

This means apply f first and then apply g.

gf(x) = g(x²) = 2 + 3x²

NOTE: The composite function gf(x) means apply f first followed by g.

NOTE: The composite function fg(x) means apply g first followed by f.

NOTE: fg(x) can be written as fg and gf(x) can be written as gf; fg is not the same as gf.

+ 3 × 5 x x + 3 5(x + 3)

g(x) is replaced

with 2 + 3x

It is then substituted

into f(x)

f(x) is replaced

with x2

It is then substituted

into g(x)

54

Example 1:

f and g are functions such that f(x) = x

1 and g(x) = 3 – 2x

Find the composite functions (a) fg (b) gf

(a) fg = fg(x) = f (3 − 2x) Do g first: Put (3 − 2x) instead of g(x)

=x23

1

−Substitute (3 − 2x) for x in

x

1

(b) gf = gf(x) = g

x

1 Do f first: Put

x

1 instead of f(x)

= 3 – 2x

1 = 3 –

x

2 Substitute

x

1 for x in (3 − 2x)

Example 2:

f(x) = 7 – 2x g(x) = 4x – 1 h(x) = 3(x – 1)

Find the following composite functions: (a) gf (b) gg (c) fgh

(a) gf = gf(x) = g(7 − 2x) Do f first: Put (7 − 2x) instead of f(x)

= 4(7 − 2x) − 1 Substitute (7 − 2x) for x in 4x − 1

= 27 − 8x Simplify 28 − 8x − 1

(b) gg = gg(x) = g(4x − 1) Put (4x − 1) instead of g(x)

= 4(4x − 1) − 1 Substitute (4x − 1) for x in 4x − 1

= 16x − 5 Simplify 16x − 4 − 1

(c) fgh = fgh(x) = fg[3(x − 1)] Put 3(x − 1) instead of h(x)

= fg (3x − 3) Expand 3(x − 1)

= f [4(3x − 3) − 1] Substitute (3x − 3) for x in 4x − 1

= f (12x − 13) Simplify 12x − 12 − 1

= 7 − 2(12x − 13) Substitute (12x − 13) for x in 7 − 2x

= 33 − 24x Simplify 7 − 24x + 26

55

Example 3:

f(x) = 7 – 2x g(x) = 4x – 1 h(x) = 3(x – 1)

Evaluate (a) fg (5) (b) ff (−2) (c) ghf (3)

(a) fg(5) = f(20 − 1) = f(19) Substitute for x = 5 in 4x − 1

= 7 − 2(19) = −31 Substitute for x = 19 in 7 − 2x

(b) ff(−2) = f[7 − 2(−2)] = f(11) Substitute for x = −2 in 7 − 2x and simplify

= 7 − 2(11) = −15 Substitute for x = 11 in 7 − 2x and simplify

(c) ghf(3) = gh(7 − 6) = gh(1) Substitute for x = 3 in 7 − 2x and simplify

= g[3(1 − 1)] = g(0) Substitute for x = 1 in 3(x − 1) and simplify

= 4(0) − 1 = −1 Substitute for x = 0 in 4x − 1 and simplify

Example 4:

f(x) = 3x + 2 and g(x) = 7 − x

Solve the equation gf(x) = 2x

gf(x) = g(3x + 2) Put (3x + 2) instead of f(x)

= 7 − (3x + 2) g's rule is subtract from 7

= 5 − 3x Simplify 7 − 3x − 2

5 − 3x = 2x Put gf(x) = 2x and solve

5 = 5x Add 3x to both sides

x = 1

56

Example 5:

(more challenging question)

Functions f, g and h are such that

f: x 4x − 1 g: x 2

1

+x, x ≠ −2 h: x (2 – x)2

Find (a)(i) fg(x) (ii) hh(x) (b) Show that hgf(x) =

2

14

18

+

+

x

x

(a) fg(x) = f

+ 2

1

x

= 4

+ 2

1

x– 1

= 2

)2(14

+

+−

x

x

= 2

2

+

−

x

x

(b) hh(x) = h(2 – x)2 Substitute for h(x)

= [2 – (2 – x)2]2

= [2 – (4 – 2x + x2)]2 (2 − x)2 = (2 − x)(2 − x) = 4 − 2x + x2

= (−2 + 4x − 2x2 )2

(c) hgf(x) = hg(4x – 1)

= h(214

1

+−x)

=

2

14

12

+−

x

=

2

14

1

14

142

+−

+

+

xx

x Simplify using a common denominator of 4x + 1

=

2

14

128

+

−+

x

x

=

2

14

18

+

+

x

x

Substitute for g(x)

Simplify using a common denominator of x + 2 denominator o

Put (4x – 1) for f(x)

Substitute (4x – 1) for x in g(x)

4x – 1 + 2 = 4x + 1, so put 4x + 1 for x in h(x)

f’s operation is × 4 − 1

h’s operation is subtract from 2 and then square

57

Exercise:

1. Find an expression for fg(x) for each of these functions:

(a) f(x) = x − 1 and g(x) = 5 – 2x

(b) f(x) = 2x + 1 and g(x) = 4x + 3

(c) f(x) =x

3and g(x) = 2x − 1

(d) f(x) = 2x2 and g(x) = x + 3

2. Find an expression for gf(x) for each of these functions:

(a) f(x) = x − 1 and g(x) = 5 – 2x

(b) f(x) = 2x + 1 and g(x) = 4x + 3

(c) f(x) =x

3and g(x) = 2x − 1

(d) f(x) = 2x2 and g(x) = x + 3

3. The function f is such that f(x) = 2x − 3

Find (i) ff(2) (ii) Solve the equation ff(a) = a

4. Functions f and g are such that

f(x) = x2 and g(x) = 5 + x

Find (a)(i) fg(x) (ii) gf(x)

(b) Show that there is a single value of x for which fg(x) = gf(x) and find this value of x.

5. Given that f(x) = 3x – 1, g(x) = x2 + 4 and fg(x) = gf(x), show that x2 – x – 1 = 0

6. The function f is defined by f(x) =x

x 1−, x ≠ 0

Solve ff(x) = −2

7. The function g is such that g(x) =x−1

1 for x ≠ 1

(a) Prove that gg(x) =x

x 1−

(b) Find ggg(3)

8. Functions f, g and h are such that f(x) = 3 – x, g(x) = x2 − 14 and h(x) = x − 2

Given that f(x) = gfh(x), find the values of x.

58

3.3 Inverse Functions

The function f(x) = 3x +5 can be thought of as a sequence of operations as shown below

Now reversing the operations

The new function, 3

5−x, is known as the inverse function.

Inverse functions are denoted as f −1(x).

Example 1:

Find the inverse function of f(x) = 3x − 4

y = 3x − 4 Step 1: Write out the function as y = ...

x = 3y − 4 Step 2: Swap the x and y

x + 4 = 3y Step 3: Make y the subject

3

4+x = y

f −1(x) = 3

4+x Step 4: Instead of y = write f −1(x) =

÷ 3 − 5 x x − 5

× 3 + 5 x 3x 3x +5

59

Example 2:

Find the inverse function of f(x) = 7

2−x

y = 7

2−xStep 1: Write out the function as y = ...

x = 7

2−yStep 2: Swap the x and y

7x = y − 2 Step 3: Make y the subject

7x + 2 = y

f −1(x) = 7x + 2 Step 4: Instead of y = write f −1(x) =

Example 3:

Find the inverse function of f(x) = 4+x

y = 4+x Step 1: Write out the function as y = ...

x = 4+y Step 2: Swap the x and y

x2 = y + 4 Step 3: Make y the subject

x2 − 4 = y

f −1(x) = x2 − 4 Step 4: Instead of y= write f −1(x) =

RULES FOR FINDING THE INVERSE f −1(x):

Step 1: Write out the function as y = ...

Step 2: Swap the x and y

Step 3: Make y the subject

Step 4: Instead of y= write f −1(x) =

60

Exercise:

1. Find the inverse function, f −1(x), of the following functions:

(a) f(x) = 3x − 1 (b) f(x) = 2x + 3

(c) f(x) = 1 – 2x (d) f(x) = x2 + 5

(e) f(x) = 6(4x – 1) (f) f(x) = 4 − x

(g) f(x) = 3x2 − 2 (h) f(x) = 2(1 − x)

(i) f(x) =1

2

+x(j) f(x) =

2

1

−

+

x

x

2. The function f is such that f(x) = 7x − 3

(a) Find f −1(x).

(b) Solve the equation f −1(x) = f(x).

3. The function f is such that f(x) =2

8

+x

(a) Find f −1(x).

(b) Solve the equation f −1(x) = f(x).

4. The function f is such that f(x) =4

1

+x, x ≠ −4.

Evaluate f −1(3). [Hint: First find f −1(x) and then substitute for x = −3]

5. f(x) =3+x

x, xR, x ≠ −3

(a) If f −1(x) = −5, find the value of x.

(b) Show that ff −1(x) = x

6. Functions f and g are such that

f(x) = 3x + 2 g(x) = x2 + 1

Find an expression for (fg)−1(x) [Hint: First find fg(x) ]

61

4 Graphs

No doubt you will have plotted many graphs of functions such as y = x2 – 3x + 4 by working out

the coordinates of points and plotting them on graph paper. But it is actually much more useful for

A Level mathematics (and beyond) to be able to sketch the graph of a function. It might sound less

challenging to be asked to draw a rough sketch than to plot an accurate graph, but in fact the

opposite is true. The point is that in order to draw a quick sketch you have to understand the basic

shape and some simple features of the graph, whereas to plot a graph you need very little

understanding. Many professional mathematicians do much of their basic thinking in terms of

shapes of graphs, and you will be more in control of your work, and understand it better, if you

can do this too.

When you sketch a graph you are not looking for exact coordinates or scales. You are simply

conveying the essential features:

• the basic shape

• where the graph hits the axes

• what happens towards the edges of your graph

The actual scale of the graph is irrelevant. For instance, it doesn’t matter what the y-

coordinates are.

4.1 Straight line graphs

I am sure that you are very familiar with the equation of a straight line in the form y = mx + c,

and you have probably practised converting to and from the forms

ax + by + k = 0 or ax + by = k,

usually with a, b and k are integers. You need to be fluent in moving from one form to the

other. The first step is usually to get rid of fractions by multiplying both sides by a common

denominator.

Example 1 Write 35

2y x= − in the form ax + by + k = 0, where a, b and k are integers.

Solution Multiply both sides by 5: 5y = 3x – 10

Subtract 5y from both sides: 0 = 3x – 5y – 10

or 3x – 5y – 10 = 0

In the first line it is a very common mistake to forget to multiply the 2 by 5.

It is a bit easier to get everything on the right instead of on the left of the equals sign, and this

reduces the risk of making sign errors.

In plotting or sketching lines whose equations are written in the form ax + by = k, it is useful

to use the cover-up rule:

62

Example 2 Draw the graph of 3x + 4y = 24.

Solution Put your finger over the “3x”. You see “4y = 24”.

This means that the line hits the y-axis at (0, 6).

Repeat for the “4y”. You see “3x = 24”.

This means that the line hits the x-axis at (8, 0).

[NB: not the point (8, 6)!]

Mark these points in on the axes.

You can now draw the graph.

1 Rearrange the following in the form ax + by + c = 0 or ax + by = c as convenient,

where a, b and c are integers and a > 0.

(a) y = 3x – 2 (b) 12

3y x= +

(c) 34

3y x= − + (d) 7 52 4

y x= −

(e) 323 4

y x= − + (f) 4 27 3

y x= −

2 Rearrange the following in the form y = mx + c. Hence find the gradient and the

y-intercept of each line.

(a) 2x + y = 8 (b) 4x – y + 9 = 0

(c) x + 5y = 10 (d) x – 3y = 15

(e) 2x + 3y + 12 = 0 (f) 5x – 2y = 20

(g) 3x + 5y = 17 (h) 7x – 4y + 18 = 0

3 Sketch the following lines. Show on your sketches the coordinates of the intercepts of

each line with the x-axis and with the y-axis.

(a) 2x + y = 8 (b) x + 5y = 10

(c) 2x + 3y = 12 (d) 3x + 5y = 30

(e) 3x – 2y = 12 (f) 4x + 5y + 20 = 0

(0, 6)

(8, 0)

y

x

63

Version 1 64 © OCR 2017

4.2 Basic shapes of curved graphs

You need to know the names of standard types of expressions, and the graphs associated with

them.

(a) The graph of a quadratic function (e.g. y = 2x2 + 3x + 4) is a parabola:

Notes:

• Parabolas are symmetric about a vertical line.

• They are not U-shaped, so the sides never reach the vertical. Neither do they dip outwards at

the ends.

These are wrong:

(b) The graph of a cubic function (e.g. y = 2x3 – 3x2 + 4x – 5) has no particular name; it’s

usually referred to simply as a cubic graph. It can take several possible shapes:

64

(c) The graph ofa number

yx

= is a hyperbola:

The graph of a hyperbola gets closer and closer to the axes without ever actually touching

them. This is called asymptotic behaviour, and the axes are referred to as the asymptotes of

this graph.

(d) The graph of 2

a numbery

x= is similar (but not identical) to a hyperbola to the right

but is in a different quadrant to the left:

(e) Graphs of higher even powers (f) Graphs of higher odd powers

y = x4 (y = x6 etc. are similar): y = x5 (y = x7 etc. are similar):

65

Which way up? This is determined by the sign of the highest power.

If the sign is positive, the right-hand side is (eventually) above the x-axis.

This is because for big values of x the highest power dominates the expression.

(If x = 1000, x3 is bigger than 50x2).

Examples y = x2 – 3x – 1 y = 10 – x2

These are often referred to (informally!) as happy and sad parabolas respectively .

y = x3 – 3x – 2 y = 2 – x – x5

66

Sketch (do not plot) the general shape of the graphs of the following curves.

Axes are not required but can be included in the questions marked with an asterix.

1 y = x2 – 3x + 2 2 y = –x2 + 5x + 1

3 y = 1 – x2 4 y = (x – 2)(x + 4)

5 y = (3 – x)(2 + x) 6 y = (1 – x)(5 – x)

7 y = x3 8 y = –x3

9* 3

yx

= 10* 2

yx

= −

11 y = (x – 2)(x – 3)(x + 1) 12* 2

2y

x=

13 Sketch on the same axes the general shape of the graphs of 𝑦 = 𝑥2 and 𝑦 = 𝑥4.

14 Sketch on the same axes the general shape of the graphs of 𝑦 = 𝑥3 and 𝑦 = 𝑥5.

67

4.3 Factors

Factors are crucial when curve-sketching.

They tell you where the curve meets the x-axis.

Do not multiply out brackets!

Example Sketch the graph of y = (x – 2)(x + 3).

Solution The graph is a positive (happy!) parabola

so start by drawing the correct shape

with a horizontal axis across it.

The factors tell you that it hits the x-axis

at x = –3 and x = 2.

Mark these on your sketch:

and only now put in the y-axis, which is

clearly slightly nearer 2 than –3:

Note: the lowest point on the graph is

not on the y-axis. (Because the

graph is symmetric, it is at x = 12

− .)

Repeated factors

Suppose you want to sketch y = (x – 1)2(x + 2).

You know there is an intercept at x = –2.

At x = 1 the graph touches the axes, as if it were

the graph of y = (x – 1)2 there.

[More precisely, it is very like y = 3(x – 1)2 there. That is because, close to x = 1, the (x – 1)2 factor

changes rapidly, while (x + 2) remains close to 3.]

Likewise, the graph of y = (x + 2)(x – 1)3

looks like y = (x – 1)3 close to x = 1.

[Again, more precisely, it is

very like y = 3(x – 1)3 there.]

–3 2

–3 2

68

Sketch the curves in questions 1–21. Use a different diagram for each.

Show the x-coordinates of the intersections with the x-axis.

1 y = x2 2 y = (x – 1)(x – 3)

3 y = (x + 2)(x – 4) 4 y = x(x – 3)

5 y = (x + 2)(3x – 2) 6 y = x(4x + 3)

7 y = –x(x – 3) 8 y = (2 – x)(x + 1)

9 y = (3 – x)(2 + x) 10 y = (x + 2)(x – 1)(x – 4)

11 y = x(x – 1)(x + 2) 12 y = –x(x – 1)(x + 2)

13 y = (3 – x)(2 – x)(1 – x) 14 y = (x – 1)2(x – 3)

15 y = (x – 1)(x – 3)2 16 y = (x + 1)3

17 y = (2 – x)(x + 1)3 18 y = (x + 1)(x + 2)(x – 1)(x – 2)

19 y = –(x + 3)(x + 2)(x – 1)(x – 4) 20 y = (x – 2)2(x + 2)2

21 y = (x – 1)(x – 2)2(x – 3)3

22 (a) Sketch the graph of y = x2.

(b) Sketch y = 2x2 on the same axes.

(c) Sketch y = x2 + 1 on the same axes.

23 (a) Sketch the graph of y = x.

(b) Sketch y = 2x on the same axes.

24 (a) Sketch the graph of1

yx

= .

(b) Sketch1

1yx

= + on the same axes.

25 (a) Sketch the graph of2

1y

x= .

(b) Sketch2

2y

x= on the same axes.





69

28 (a) Sketch the graph of y = x3 – 4x.

[Hint: It cuts the x-axis at –2, 0 and 2.]

(b) Sketch y = 2x3 – 8x on the same axes.

29 (a) Sketch the graph of y = x4 – x2.

[Hint: It cuts the x-axis at 1 and –1, and touches the axis at 0.]

(b) Sketch y = –x4 + x2 on the same axes.

30 Sketch, on separate axes, the following graphs. Show the x-coordinates of the intersections with

the x-axis.

(a) y = 4 – x2

(b) y = (x – 2)(x + 1)

(c) y = –(x – 2)(x + 1)

(d) y = x(x + 4)

(e) y = (x – 2)2

(f) y = –(x + 1)2

(g) y = (1 – x)(2 + x)

70

5 Trigonometry The following two aspects are worth emphasising at this stage.

Trigonometric Equations

You can of course get one solution to an equation such as sin x = –0.5 from your calculator. But what about

others?

Example 1 Solve the equation sin x = –0.5 for 0 x < 360.

Solution The calculator gives sin–1(0.5) = –30.

This is usually called the principal value of the function sin–1.

To get a second solution you can either use a graph or a standard rule.

Method 1: Use the graph of y = sin x

By drawing the line y = -0.5 on the same set of axes as the graph of the sine curve, points of

intersection can be identified in the range

0 x < 360.

(The red arrows each indicate 30 to one side or the other.)

Hence the required solutions are 210 or 330.

Method 2: Use an algebraic rule.

To find the second solution you use sin (180 – x) = sin x

tan (180 + x) = tan x

cos (360 – x) = cos x.

Any further solutions are obtained by adding or subtracting 360 from the principal value or

the second solution.

In this example the principal solution is –30.

Therefore, as this equation involves sine, the second solution is:

180 – (–30) = 210

–30 is not in the required range, so add 360 to get:

360 + (–30) = 330.

Hence the required solutions are 210 or 330.

y = sin x

y = –0.5

x = 210 x = 330 x = –30

71

You should decide which method you prefer. The corresponding graphs for cos x and tan x are shown

below.

y = cos x

y = tan x

To solve equations of the form y = sin (kx), you will expect to get 2k solutions in any interval of 360. You

can think of compressing the graphs, or of using a wider initial range.

72

Exercise

1 Solve the following equations for 0 x < 360. Give your answers to the nearest 0.1.

(a) sin x = 0.9 (b) cos x = 0.6 (c) tan x = 2

(d) sin x = –0.4 (e) cos x = –0.5 (f) tan x = –3

2 Solve the following equations for –180 x < 180. Give your answers to the nearest 0.1.

(a) sin x = 0.9 (b) cos x = 0.6 (c) tan x = 2

(d) sin x = –0.4 (e) cos x = –0.5 (f) tan x = –3

73

Practice Booklet Test

You may NOT use a calculator

1. Expand and simplify

(a) (2x + 3)(2x – 1) (b) (a + 3)2 (c) 4x(3x – 2) – x(2x + 5)

2. Factorise

(a) x2 – 7x (b) y2 – 64 (c) 2x2 + 5x – 3 (d) 6t2 – 13t + 5

3. Simplify

(a) 32

3

8

4

yx

yx(b)

3

23 +x+

6

14 −x(c)

2

𝑥−2+

4

𝑥+3

4. Solve the following equations

(a) 4

1−h+

5

3h = 4 (b) x2 – 8x = 0 (c) p2 + 4p = 12

5. If 22𝑥+1 × 4𝑥+1 = 8𝑥+2, find the value of x.

6. a) Solve the simultaneous equations 3x – 5y = -11

5x – 2y = 7

b) Solve the simultaneous equations 𝑥2 + 𝑦2 = 25

𝑥 + 𝑦 = −1

7. Rearrange the following equations to make x the subject

(a) v2 = u2 + 2ax (b) V =3

1πx2h (c) y =

1

2

+

+

x

x

8. Solve 5x2 – x – 1 = 0 giving your solutions in surd form

9. Find the values of x which satisfy the following inequalities

(a) 5𝑥 − 2 < 6 (b) 4 − 2𝑥 > 9 (c) 𝑥2 − 6𝑥 − 16 > 0

10. Given 𝑓(𝑥) = 7𝑥 − 2 and 𝑔(𝑥) = 2𝑥2 + 5𝑥 − 1, find

(a) 𝑓(3) (b) 𝑔(3) (c) 𝑔(−4) (d) 𝑓𝑔(−2)

11. Simplify

(a) √18 × √75 (b) √54 + √27 (c) 5

√2(d)

3−√2

√6(e)

1−√2

3+√2

12. Solve for −360° ≤ 𝜃 ≤ 360°

(a) sin 𝜃 = 0.2 (b) cos 𝜃 = −0.35 (c) tan 𝜃 = 12.6

74

Solutions to the Exercises

1 Fractions

1) 13

15

2) 7

12

3) 7

10

4) 10

9

5) 311

12

6) 63

56

7) 319

40

8) 25

12

9) 41

8

10) 81

2

11) 118

77

12) 15

9

2.1 Expanding Brackets

Ex A

1) 28x + 35

2) -15x + 21

3) -7a + 4

4) 6y + 3y2

5) -4x – 4

6) 7x – 1

7) x2 + 5x + 6

8) t2 – 7t – 10

9) 6x2 + xy – 12y2

10) 4x2 + 4x – 24

11) 4y2 – 1

12) 12 + 17x – 5x2

Ex B

1) x2 – 2x + 1

2) 9x2 + 30x + 25

3) 49x2 – 28x + 4

4) x2 – 4

5) 9x2 -1

6) 25y2 – 9

2.2 Linear Equations

Ex A

1) 7 2) 3 3) 1½ 4) 2 5) −3

56) −

7

3

Ex B

1) 2.4 2) 5 3) 1 4) ½

2.3 Equations Containing Fractions

1) 7

2) 15

3) 24/7

4) 35/3

5) 3

6) 2

7) 9/5

8) 5

2.5 Linear Inequalities

1) 𝑥 < 4

2) 𝑦 > 21

3) 𝑥 < 1

4) 𝑦 ≥ 8

5) 𝑥 ≥ −3

6) 𝑦 < −47) 𝑥 > −

1

3

8) 𝑦 ≤ −5.25

2.6 Simultaneous Equations

1) x = 1, y = 3

2) x = -3, y = 1

3) x = 0, y = -2

4) x = 3, y = 1

5) a = 7, b = -2

6) p = 11/3, q = 4/3

75

2.7 Factorising Linear Expressions

1) x(3 + y)

2) 2x(2x – y)

3) pq(q – p)

4) 3q(p – 3q)

5) 2x2(x - 3)

6) 4a3b2(2a2 – 3b2)

7) (y – 1)(5y + 3)

2.8 Factorising Quadratic Expressions

1) (x – 3)(x + 2)

2) (x + 8)(x – 2)

3) (2x + 1)(x + 2)

4) x(2x – 3)

5) (3x -1 )(x + 2)

6) (2y + 3)(y + 7)

7) (7y – 3)(y – 1)

8) 5(2x – 3)(x + 2)

9) (2x + 5)(2x – 5)

10) (x – 3)(x – y)

11) 4(x – 2)(x – 1)

12) (4m – 9n)(4m + 9n)

13) y(2y – 3(a)(2y + 3(a)

14) 2(4x + 5)(x – 4)

2.9 Completing the Square

1 (a) (x + 4)2 + 3 (b) (x – 5)2 – 2 (c) (x + 1)2 – 5

(d) (x – 2)2 – 7 (e) (x – 1½)2 – ¼ (f) (x – 2½)2 – 12¼

2 (a) 3(x + 1)2 + 4 (b) 5(x – 2)2 – 3 (c) 2(x + 2½)2 + ½

3 (a) (2x + 3)2 + 5 (b) (3x – 2)2 – 5 (c) (4x + 5)2 – 3

2.10 Solving Quadratic Equations

1) (a) -1, -2 (b) -1, 4 (c) -5, 3

2) (a) 0, -3 (b) 0, 4 (c) 2, -2

3) (a) -1/2, 4/3 (b) 0.5, 2.5

4) (a) -5.30, -1.70 (b) 1.07, -0.699 (c) -1.20, 1.45

(d) no solutions (e) no solutions f) no solutions

2.11 Changing the Subject of a Formula

Ex A

1) 1

7

yx

+=

2) 4 5x y= − 3) 3(4 2)x y= + 4)

9 20

12

yx

+=

Ex B

1) 32rP

tw

=

2) 32rP

tw

=

3) 3V

th

=

4) 2

2

P gt =

5) Pag

t vw

= −

6) r a

tb

−=

Ex C

1) 3c

xa b

−=

− 2)

3 2

3

a kx

k

+=

−3)

2 3

5 2

yx

y

+=

− 4)

abx

b a=

−

76

2.12 Indices

Ex A

1) 5b6

2) 6c7

3) b3c4

4) -12n8

5) 4n5

6) d2

7) a6

8) -d12

Ex B

1) 2

2) 3

3) 1/3

4) 1/25

5) 1

6) 1/7

7) 9

8) 9/4

9) ¼

10) 0.2

11) 4/9

12) 64

13) 6a3

14) x

15) xy2

2.13 Surds

Ex A

1) 5√2

2) 6√2

3) 3√3

4) 4√5

5) 6√10

6) 10√3

Ex B

1) √21

2) 20√10

3) 18√2

4) 6√6

5) 5

3

6) 6

7) 7 + 6√2

8) 5√2 − 40 − √6 + 8√3

Ex C

1) √3 + √7

2) 9√2

3) 5√6

4) 7√2

5) 8√3

6) √5

7) √2

8) 7√3

9) 3√2 + 3√10

10) 6√2 + √3

Ex D

1

a) √2

2

b) 3√5

5

c) 2√5

d) 5√7

14

e) √6

2

f) √10

g) 4√3+√21

3

h) 3√2 + 4√5

i) 6√5−5

5

2

a) 2 + 1

b) 6 + 2

c) 2(7 – 2)

d) 14(3 5)−

e) 6 + 5

77

2.14 Algebraic Fractions

1 (a) 3 5

( 1)( 3)

x

x x

+

− +(b)

7

( 3)( 2)

x

x x

+

− + (c)

3

(2 1)(3 2)

x

x x

+

− +

(d) 5

2

x

x

+

+(e)

2 3

1

x

x

−

−(f)

3

1

x

x

+−

+

(g) 6 1

4(2 1)(2 1)

x

x x

−

− +

2 (a) 2 1x

x

+(b)

3 3

3

x

x

+

+(c)

3

2 2

2

x

x

−

−

2.15 Linear and Quadratic Simultaneous Equations

1 (2, 4), (3, 1)

2 (6, –3), (7, 0)

3 (1, –1), (–1, 0)

4 (1, 2), (–1, –2)

5 (2, –1), ( 3825

− , 4125

)

6 (7, 4), (8, 72)

7 (–5, 4), ( 195

, 135

− )

8 (–1, 1), ( 53, 7

9− )

9 (2, 1), ( 12, 4)

10 (3, 6), ( 92, 9

2)

11 (2, 1), ( 59

− , 149

− )

12 (–1, –2), ( 3813

, 913

− )

13 ( 53, 1

3), ( 3

5− , 4

5− )

14 (2, –2) (only)

15 (6, –5) (only)

16 (6, 1), ( 143

, 73)

2.16 Expanding More Than Two Binomials

1. (a) x3 + 10x2 + 29x + 20 (b) 6x3 + 71x2 + 191x + 56

(c) 2x3 − 11x2 + 18x − 9 (d) 6x3 − 11x2 − 42x + 80

(e) 30x3 + 49x2 − 61x + 10 (f) 32x3 + 112x2 − 2x − 7

(g) 3x3 + 17x2 − 8x − 112 (h) 24x3 − 4x2 − 14x + 5

(i) 8x4 − 22x3 − 3x2 + 22x − 5 (j) x4 + 6x3 − 11x2 − 60x + 100

2. Proof

3. x3 − 5x2 − 8x + 48

4. 81x4 − 108x3 + 54x2 − 12x + 1

5. A = −27 B = −46

6. A = 4 B = −18 C = −17.

7. Proof

8. 3x3 + 16x2 + 28x + 20

9. 8x3 + 12x2 − 2x − 3

2.17 Quadratic Inequalities

1. (a) −6 ≤ x ≤ −3 (b) −4 < x < 5

(c) x < −7 or x >2 (d) x ≤ 0 or x ≥ 5

(e) 2

3< x < 4 (f) −5 ≤ x ≤

2

1

(g) x ≤ −3 or x ≥ 5 (h) −2

7< x < 3

(i) x < 0 or x >5

2(j) x < −5 or x > 7

78

2. (a) x : 1 ≤ x ≤ 3 (b) x : −7 < x < 6

(c) x : x < −8, x > 6 (d) x : x ≤ −5, x ≥3

1

(e) x : x <2

3, x > 4 (f) x : x ≤ −8, x ≥ 2

(g) x :4

1 ≤ x ≤

2

7 (h) x : x < −4, x > 8

(i) x : 0 < x < 5 (j) x : x ≤ −6, x ≥ −4

3. x ≤ 2 or x ≥ 6

4. x : x ≤ − 3, x ≥ 5

5. −2 ≤ x < −1 and 2 < x ≤ 3

6. x : − 2

1

< x < 3

4

2.18 Using Completing the Square to Find Turning Points

1. (a) (4, 4) (b) (5, −26)

(c) (−2, −10) (d) (3, −10)

(e) (3, 19) (f) (−1, 6)

2. (a) y = x2 − 4x + 1 (b) y = x2 + 8x +17

(c) y = x2 + 2x + 6 (d) y = x2 − 6x − 3

(e) y = x2 − 2x − 6 (f) y = x2 − 8x + 15

3. (a) minimum (−2, −2) (b) maximum (−3, 10)

(c) maximum (1, −2) (d) minimum (4, −8)

(e) minimum (1.5, −3.25) (f) maximum (−2, 3)

4. (i) a = 1.5 b = 5.75

(ii) (1.5, 5.75)

5. (i) y f(x)

10

(3, 1)

x

(ii) It has 2 real roots as if you move the graph 3 down it will cut the x-axis twice

as the minimum point will be (3, −2)

6. Minimum point is (1, −4) thus y = A(x − 1)2 − 4 = Ax2 − 2Ax + A − 4

Cuts y-axis at −1, thus A − 4 = −1

A = 3 y = 3x2 − 2(3)x + 3 − 4

y = 3x2 − 6x −1

7. y = −2x2 − 8x − 13

79

3.1 Function Notation

1. (a) 17 (b) 37 (c) −18

(d) –8 (e) −0.5 (f) 3.5

2. (a) 12 (b) 32 (c) 0

(d) 32 (e) −3.96 (f) −3.19

3. (a) 1 (b) −3 (c) −7

(d) –1 (e) 1.256 (f) −5.375

4. (a) 5 (b) 7 (c) 3

(d) 3 (e) 5

103(f) 2 2

5. 3x2 + 10x

6. (a) f(− 1) = − 8 and g(2) = 9 (b) x = − 6

7. (a) f(− 3) = 17 and g(− 5) = − 23 (b) x =2

1− and x = 3

3.2 Composite Functions

1. (a) 4 − 2x (b) 8x + 7 (c) 12

3

−x(d) 2(x + 3)2

2. (a) 7 − 2x (b) 8x + 7 (c) 16−

x(d) 2x2 + 3

3. (a) – 1 (b) a = 3

4. (a)(i) (5 + x)2 (ii) 5 + x2 (b) x = –2

5. 3(x2 + 4) – 1 = (3x – 1)2 + 4

3x2 + 12 – 1 = 9x2 – 6x + 1 + 4

3x2 + 11 = 9x2 – 6x + 5

6x2 – 6x – 6 = 0

x2 – x – 1 = 0

6. x =2

3

7. (a) x

x

x

x

x xx

x

111

1

1

1

1g

111

11

−=

−

−==

−=

−−−−

−

(b) 3

8. x = 1 and x = 8

80

3.3 Inverse Functions

1. (a) 3

1+x (b)

2

3−x (c)

2

1 x−(d) 5− x

(e) 24

6+x(f) 4 – x (g)

3

2+

x (h)

2

2 x−

(i) x

x−2(j)

1

12

−

+

x

x

2. (a) 7

3+x(b) x = 0.5

3. (a) 28−

x (b) x = − 4 and x = 2

4. 3

13−

5. (a)2

5(b) x

x

x

xf

xxx

x

x

x

xxx

x

xx

x

x

====+

=

−−

−+

−

−

−+

−

−

−

3

3

31

3

1

333

1

3

1

)1(33

1

3

1

3

1

3

6. 3

5−

x

4.1 Straight Line Graphs

1 (a) 3x – y = 2 (b) x – 2y + 6 = 0

(c) 3x + 4y = 12 (d) 14x – 4y = 5

(e) 8x + 12y = 9 (f) 12x – 21y = 14

2 (a) y = –2x + 8; –2, 8 (b) y = 4x + 9; 4, 9

(c) 15

2y x= − + ; 15

− , 2 (d) 13

5y x= − ; 13, –5

(e) 23

4y x= − − ; 23

− , –4 (f) 52

10y x= − ; 52, –10

(g) 3 175 5

y x= − + ; 35

− , 175

(h) 7 94 2

y x= + ; 74, 9

2

81

3 (a) (b)

(c) (d)

(e) (f)

y

x

8

4

y

x

2

10

y

x 6

4

y

x

–6

4

y

x

6

10

x

y

–5

–4

82

4.2 Basic Shape of Curved Graphs

1 2

3 4

5 6

7 8

9

10

11 12

83

13 14

red: y = x2 blue: y = x4 red: y = x3 blue: y = x5

4.3 Factors

1 2

3 4

5 6

7 8

1 3

–2 4

3 0

–2 ⅔

–¾ 0

0 3

–1 2

84

9 10

11 12

13 14

15 16

17 18

3 –2

0 1

–2

–2

0 1

1 3

–1

–1 2

–1–2 1 2

1 –2 4

1 3

3 2 1

85

19 20

21

[In this graph in particular, do NOT worry about the y-

coordinates of the minimum points.]

22

red: y = x2

blue: y = 2x2

purple: y = x2 + 1

23

red: y = x

blue: y = 2x

red: 1

yx

=24

blue: 1

1yx

= +

blue dotted: y = 1 [horizontal asymptote]

–3 –21 4

2 –2

1 2 3

86

red: 2

1y

x=25

blue: 2

2y

x=

26

red: y = x3

blue: y = 2x3

27

red: y = x4

blue: y = 3x4

28

red: y = x3 – 4x

blue: y = 2x3 – 8x

29

red: y = x4 – x2

blue: y = –x4 + x2

87

30 (a) (b)

(c) (d)

(e) (f)

(g)

NOTE: in parts (b), (c) and (g) in particular, the maximum or minimum point is not on the y-axis.

5 Trigonometry

1 (a) 64.2, 115.8 (b) 53.1, 306.9 (c) 63.4, 243.4

(d) 203.6, 336.4 (e) 120, 240 (f) 108.4, 288.4

2 (a) 64.2, 115.8 (b) 53.1, –53.1 (c) 63.4, –116.6

(d) –23.6, –156.4 (e) 120, –120 (f) –71.5, 108.4

88

Solutions to the Practice Booklet Test

1) (a) 4x2 + 4x – 3 (b) a2 + 6a + 9 (c) 10x2 -13x

2) (a) x(x – 7) (b) (y + 8)(y – 8) (c) (2x - 1)(x + 3) (d) (3t - 5)(2t – 1)

3) (a)22

x

y(b)

10 3

6

x +(c)

6𝑥−2

(𝑥−2)(𝑥+3)

4) (a) h = 5 (b) x = 0 or x = 8 (c) p = -6 or p = 2

5) 𝑥 = 3

6) (a) x = 3, y = 4 (b) 𝑥 = −4. 𝑦 = 3 and 𝑥 = 3, 𝑦 = −4

7) (a)2 2

2

v ux

a

−= (b)

3Vx

h= (c)

2

1

yx

y

−=

−

8) 1 21

10x

=

9) (a) x < 1.6 (b) x < -2.5 (c) x < -2 or x > 8

10) (a) 19 (b) 32 (c) 11 (d) -23

11) (a) 15√6 (b) 9√3 (c) 5√2

2(d)

3√6−2√3

6(e)

5−4√2

9

12) (a) 𝜃 = −348.5, −191.5, 11.5, 168.5 (b) 𝜃 = ±110.5, ±249.5

(c) 𝜃 = −274.5, −94.5, 85.5, 265.5

89

BRIDGING UNIT: A-level PHYSICS Contact Email: [email protected]





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Learning objectives





o unit conversions

o uncertainties

o using standard form and significant figures

o resolving vectors

o rearranging equations

o equations of work, power, and efficiency



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Retrieval questions

You need to be confident about the definitions of terms that describe measurements and results in A Level Physics.






other

What is repeatability?

how precise repeated measurements are when they are taken by


conditions






around the true value in an unpredictable way? random error


values


should be zero


investigator? independent variable


changed


dependent variable


dependent variable



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Matter and radiation


What is an atom made up of? a positively charged nucleus containing protons and neutrons,

surrounded by electrons

Define a nucleon a proton or a neutron in the nucleus

What are the absolute charges of protons,

neutrons, and electrons?

1.601019, 0, and 1.601019

coulombs (C) respectively

What are the relative charges of protons,


1, 0, and 1 respectively (charge relative to proton)

What is the mass, in kilograms, of a proton, a

neutron, and an electron?

1.671027, 1.671027

, and 9.111031 kg respectively

What are the relative masses of protons,


1, 1, and 0.0005 respectively (mass relative to proton)

What is the atomic number of an element? the number of protons

Define an isotope isotopes are atoms with the same number of protons and different

numbers of neutrons

Write what A, Z and X stand for in isotope

notation ( XAZ )?

A: the number of nucleons (protons + neutrons)

Z: the number of protons

X: the chemical symbol

Which term is used for each type of nucleus? nuclide

How do you calculate specific charge? charge divided by mass (for a charged particle)

What is the specific charge of a proton and an

electron?

9.58107 and 1.7610

11 C kg1

respectively

Name the force that holds nuclei together strong nuclear force

What is the range of the strong nuclear force? from 0.5 to 34 femtometres (fm)

Name the three kinds of radiation alpha, beta, and gamma

What particle is released in alpha radiation? an alpha particle, which comprises two protons and two neutrons

Write the symbol of an alpha particle α42

What particle is released in beta radiation? a fast-moving electron (a beta particle)

Write the symbol for a beta particle β01

Define gamma radiation electromagnetic radiation emitted by an unstable nucleus

What particles make up everything in the

universe?

matter and antimatter

Name the antimatter particles for electrons,

protons, neutrons, and neutrinos

positron, antiproton, antineutron, and antineutrino respectively

What happens when corresponding matter and

antimatter particles meet?

they annihilate (destroy each other)

List the seven main parts of the electromagnetic

spectrum from longest wavelength to shortest

radio waves, microwaves, infrared, visible, ultraviolet, X-rays,

gamma rays

Write the equation for calculating the wavelength

of electromagnetic radiation wavelength ()

)(frequency

)( light of speed

f

c

Define a photon a packet of electromagnetic waves

What is the speed of light? 3.00108 m s

1

Write the equation for calculating photon energy photon energy (E) Planck constant (h) frequency (f)

Name the four fundamental interactions gravity, electromagnetic, weak nuclear, strong nuclear



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Maths skills

1 Measurements

1.1 Base and derived SI units

Units are defined so that, for example, every scientist who measures a mass in kilograms uses the same size for the kilogram and gets the same value for the mass. Scientific measurement depends on standard units – most are Système International (SI) units. Every measurement must give the unit to have any meaning. You should know the correct unit for physical quantities.

Base units

Physical quantity Unit Symbol Physical quantity Unit Symbol

length metre m electric current ampere A

mass kilogram kg temperature difference Kelvin K

time second s amount of substance mole mol

Derived units

Example:

taken time

travelled distancespeed

If a car travels 2 metres in 2 seconds:

s

m1

seconds 2

metres 2speed = 1m/s

This defines the SI unit of speed to be 1 metre per second (m/s), or 1 m s−1

(s−1

= s

1).

Practice questions

1 Complete this table by filling in the missing units and symbols.

Physical quantity Equation used to derive unit Unit Symbol and name

(if there is one)

frequency period−1

s−1

Hz, hertz

volume length3 –

density mass ÷ volume –

acceleration velocity ÷ time –

force mass × acceleration

work and energy force × distance



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1.2 Significant figures

When you use a calculator to work out a numerical answer, you know that this often results in a large number of decimal places and, in most cases, the final few digits are ‘not significant’. It is important to record your data and your answers to calculations to a reasonable number of significant figures. Too many and your answer is claiming an accuracy that it does not have, too few and you are not showing the precision and care required in scientific analysis.

Numbers to 3 significant figures (3 s.f.):

3.62 25.4 271 0.0147 0.245 39 400

(notice that the zeros before the figures and after the figures are not significant – they just show you how large the number is by the position of the decimal point).

Numbers to 3 significant figures where the zeros are significant:

207 4050 1.01 (any zeros between the other significant figures are significant).

Standard form numbers with 3 significant figures:

9.42×10−5

1.56×108

If the value you wanted to write to 3.s.f. was 590, then to show the zero was significant you would have to write:

590 (to 3.s.f.) or 5.90 × 102

Practice questions

2 Give these measurements to 2 significant figures:

a 19.47 m b 21.0 s c 1.673×10−27

kg d 5 s

3 Use the equation:

current

difference potentialresistance

to calculate the resistance of a circuit when the potential difference is 12 V and the current is 1.8 mA. Write your answer in kΩ to 3 s.f.

1.3 Uncertainties

When a physical quantity is measured there will always be a small difference between the measured value and the true value. How important the difference is depends on the size of the measurement and the size of the uncertainty, so it is important to know this information when using data.

There are several possible reasons for uncertainty in measurements, including the difficulty of taking the measurement and the resolution of the measuring instrument (i.e. the size of the scale divisions).

For example, a length of 6.5 m measured with great care using a 10 m tape measure marked in mm would have an uncertainty of 2 mm and would be recorded as 6.500 ± 0.002 m.

It is useful to quote these uncertainties as percentages.

For the above length, for example,

100tmeasuremen

yuncertaintyuncertaint percentage

6.500

0.002yuncertaint percentage × 100% = 0.03%. The measurement is 6.500 m ± 0.03%.



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Values may also be quoted with absolute error rather than percentage uncertainty, for example, if the 6.5 m length is measured with a 5% error,

the absolute error = 5/100 × 6.5 m = ±0.325 m.

Practice questions

4 Give these measurements with the uncertainty shown as a percentage (to 1 significant figure):

a 5.7 ± 0.1 cm b 450 ± 2 kg c 10.60 ± 0.05 s d 366 000 ± 1000 J

5 Give these measurements with the error shown as an absolute value:

a 1200 W ± 10% b 330 000 Ω ± 0.5%

6 Identify the measurement with the smallest percentage error. Show your working.

A 9 ± 5 mm B 26 ± 5 mm C 516 ± 5 mm D 1400 ± 5 mm

2 Standard form and prefixes

When describing the structure of the Universe you have to use very large numbers. There are billions of galaxies and their average separation is about a million light years (ly). The Big Bang theory says that the Universe began expanding about 14 billion years ago. The Sun formed about 5 billion years ago. These numbers and larger numbers can be expressed in standard form and by using prefixes.

2.1 Standard form for large numbers

In standard form, the number is written with one digit in front of the decimal point and multiplied by the appropriate power of 10. For example:

The diameter of the Earth, for example, is 13 000 km.

13 000 km = 1.3 × 10 000 km = 1.3×104 km.

The distance to the Andromeda galaxy is 2 200 000 light years = 2.2 × 1 000 000 ly = 2.2×10

6 ly.

2.2 Prefixes for large numbers

Prefixes are used with SI units (see Topic 1.1) when the value is very large or very small. They can be used instead of writing the number in standard form. For example:

A kilowatt (1 kW) is a thousand watts, that is 1000 W or 103 W.

A megawatt (1 MW) is a million watts, that is 1 000 000 W or 106 W.

A gigawatt (1 GW) is a billion watts, that is 1 000 000 000 W or 109 W.

Prefix Symbol Value Prefix Symbol Value

kilo k 103 giga G 10

9

mega M 106 tera T 10

12



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For example, Gansu Wind Farm in China has an output of 6.8×109 W. This can be written as

6800 MW or 6.8 GW.

Practice questions

1 Give these measurements in standard form:

a 1350 W b 130 000 Pa c 696 × 106

s d 0.176 × 1012

C kg−1

2 The latent heat of vaporisation of water is 2 260 000 J/kg. Write this in:

a J/g b kJ/kg c MJ/kg

2.3 Standard form and prefixes for small numbers

At the other end of the scale, the diameter of an atom is about a tenth of a billionth of a metre. The particles that make up an atomic nucleus are much smaller. These measurements are represented using negative powers of ten and more prefixes. For example:

The charge on an electron = 1.6×10−19

C.

The mass of a neutron = 0.016 75 × 10−25

kg = 1.675×10−27

kg (the decimal point has moved 2 places to the right).

There are a billion nanometres in a metre, that is 1 000 000 000 nm = 1 m.

There are a million micrometres in a metre, that is 1 000 000 μm = 1 m.

Prefix Symbol Value Prefix Symbol Value

centi c 10−2

nano n 10−9

milli m 10−3

pico p 10−12

micro µ 10−6

femto f 10−15

Practice questions

3 Give these measurements in standard form:

a 0.0025 m b 160 × 10−17

m c 0.01 × 10−6

J d 0.005 × 106 m e 0.00062 × 10

3 N

4 Write the measurements for question 3a, c, and d above using suitable prefixes.

5 Write the following measurements using suitable prefixes.

a a microwave wavelength = 0.009 m

b a wavelength of infrared = 1×10−5

m

c a wavelength of blue light = 4.7×10−7

m

2.4 Powers of ten

When multiplying powers of ten, you must add the indices.

So 100 × 1000 = 100 000 is the same as 102 × 10

3 = 10

2 + 3 = 10

5

When dividing powers of ten, you must subtract the indices.

So 1000

100 =

10

1 = 10

−1 is the same as

3

2

10

10 = 10

2 − 3 = 10

−1

But you can only do this when the numbers with the indices are the same.

So 102 × 2

3 = 100 × 8 = 800



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And you can’t do this when adding or subtracting.

102 + 10

3 = 100 + 1000 = 1100

102 − 10

3 = 100 − 1000 = −900

Remember: You can only add and subtract the indices when you are multiplying or dividing the numbers, not adding or subtracting them.

Practice questions

6 Calculate the following values – read the questions very carefully!

a 206 + 10

−3

b 102 − 10

−2

c 23 × 10

2

d 105 ÷ 10

2

7 The speed of light is 3.0×108 m s

−1. Use the equation v = f

λ (where λ is wavelength) to

calculate the frequency of:

a ultraviolet, wavelength 3.0×10−7

m

b radio waves, wavelength 1000 m

c X-rays, wavelength 1.0×10−10

m.

3 Resolving vectors

3.1 Vectors and scalars

Vectors have a magnitude (size) and a direction. Directions can be given as points of the compass, angles or words such as forwards, left or right. For example, 30 mph east and 50 km/h north-west are velocities.

Scalars have a magnitude, but no direction. For example, 10 m/s is a speed.

Practice questions

1 State whether each of these terms is a vector quantity or a scalar quantity: density, temperature, electrical resistance, energy, field strength, force, friction, frequency, mass, momentum, power, voltage, volume, weight, work done.

2 For the following data, state whether each is a vector or a scalar: 3 ms−1

, +20 ms−1

, 100 m NE, 50 km, −5 cm, 10 km S 30° W, 3 × 10

8 ms

−1 upwards, 273 °C, 50 kg, 3 A.

3.2 Drawing vectors

Vectors are shown on drawings by a straight arrow. The arrow starts from the point where the vector is acting and shows its direction. The length of the vector represents the magnitude.

When you add vectors, for example two velocities or three forces, you must take the direction into account.

The combined effect of the vectors is called the resultant.



student worksheet

AQA Physics



This diagram shows that walking 3 m from A to B and then turning through 30° and walking 2 m to C has the same effect as walking directly from A to C. AC is the resultant vector, denoted by the double arrowhead.

A careful drawing of a scale diagram allows us to measure these. Notice that if the vectors are combined by drawing them in the opposite order, AD and DC, these are the other two sides of the parallelogram and give the same resultant.

Practice question

3 Two tractors are pulling a log across a field. Tractor 1 is pulling north with force 1 = 5 kN and tractor 2 is pulling east with force 2 = 12 kN. By scale drawing, determine the resultant force.

3.3 Free body force diagrams

To combine forces, you can draw a similar diagram to the one above, where the lengths of the sides represent the magnitude of the force (e.g., 30 N and 20 N). The third side of the triangle shows us the magnitude and direction of the resultant force.

When solving problems, start by drawing a free body force diagram. The object is a small dot and the forces are shown as arrows that start on the dot and are drawn in the direction of the force. They don’t have to be to scale, but it helps if the larger forces are shown to be larger. Look at this example.

A 16 kg mass is suspended from a hook in the ceiling and pulled to one side with a rope, as shown on the right. Sketch a free body force diagram for the mass and draw a triangle of forces.

Notice that each force starts from where the previous one ended and they join up to form a triangle with no resultant because the mass is in equilibrium (balanced).

Practice questions

4 Sketch a free body force diagram for the lamp (Figure 1, below) and draw a triangle of forces.

5 There are three forces on the jib of a tower crane (Figure 2, below). The tension in the cable T, the weight W, and a third force P acting at X.

The crane is in equilibrium. Sketch the triangle of forces.



student worksheet

AQA Physics



Figure 1 Figure 2

3.4 Calculating resultants

When two forces are acting at right angles, the resultant can be calculated using Pythagoras’s theorem and the trig functions: sine, cosine, and tangent.

For a right-angled triangle as shown:

h2 = o

2 + a

2

sin θ = h

o

cos θ = h

a

tan θ = a

o

(soh-cah-toa).

Practice questions

6 Figure 3 shows three forces in equilibrium.

Draw a triangle of forces to find T and α.

7 Find the resultant force for the following pairs of forces at right angles to each other:

a 3.0 N and 4.0 N b 5.0 N and 12.0 N

Figure 3

4 Rearranging equations

Sometimes you will need to rearrange an equation to calculate the answer to a question. For example, if you want to calculate the resistance R, the equation:

potential difference (V) = current (A) × resistance (Ω) or V = I R

must be rearranged to make R the subject of the equation:

I

VR

When you are solving a problem:

Write down the values you know and the ones you want to calculate.

you can rearrange the equation first, and then substitute the values

or

substitute the values and then rearrange the equation



student worksheet

AQA Physics



4.1 Substitute and rearrange

A student throws a ball vertically upwards at 5 m s−1

. When it comes down, she catches it at the same point. Calculate how high it goes.

step 1: Known values are:

initial velocity u = 5.0 m s−1

final velocity v = 0 (you know this because as it rises it will slow down, until it comes to a stop, and then it will start falling downwards)

acceleration a = g = −9.81 m s−2

distance s = ?

Step 2: Equation:

(final velocity)2 − (initial velocity)

2 = 2 × acceleration × distance

or v2 – u

2 = 2 × g × s

Substituting: (0)2 − (5.0 m s

−1)2 = 2 × −9.81 m s

−2 × s

0 − 25 = 2 × −9.81 × s

Step 3: Rearranging:

−19.62 s = −25

s = 6219

25

.

= 1.27 m = 1.3 m (2 s.f.)

Practice questions

1 The potential difference across a resistor is 12 V and the current through it is 0.25 A. Calculate its resistance.

2 Red light has a wavelength of 650 nm. Calculate its frequency. Write your answer in standard form.

(Speed of light = 3.0 × 108 m s

−1)

4.2 Rearrange and substitute

A 57 kg block falls from a height of 68 m. By considering the energy transferred, calculate its speed when it reaches the ground.

(Gravitational field strength = 10 N kg−1

)

Step 1: m = 57 kg h = 68 m g = 10 N kg−1

v = ?

Step 2: There are three equations:

PE = m g h KE gained = PE lost KE = 0.5 m v2

Step 3: Rearrange the equations before substituting into it.

As KE gained = PE lost, m g h = 0.5 m v2

You want to find v. Divide both sides of the equation by 0.5 m:

m

mgh

50. =

m

mv

50

50 2

.

.

2 g h = v2



student worksheet

AQA Physics



To get v, take the square root of both sides: v = gh2

Step 4: Substitute into the equation:

v = 86102

v = 1360 = 37 m s−1

Practice question

3 Calculate the specific latent heat of fusion for water from this data:

4.03×104 J of energy melted 120 g of ice.

Use the equation:

thermal energy for a change in state (J) = mass (kg) × specific latent heat (J kg−1

)

Give your answer in J kg−1

in standard form.

5 Work done, power, and efficiency

5.1 Work done

Work is done when energy is transferred. Work is done when a force makes something move. If work is done by an object its energy decreases and if work is done on an object its energy increases.

work done = energy transferred = force × distance

Work and energy are measured in joules (J) and are scalar quantities (see Topic 3.1).

Practice question

1 Calculate the work done when the resultant force on a car is 22 kN and it travels 2.0 km.

2 Calculate the distance travelled when 62.5 kJ of work is done applying a force of 500 N to an object.

5.2 Power

Power is the rate of work done.

It is measured in watts (W) where 1 watt = 1 joule per second.

power = taken time

dtransferreenergy or power =

taken time

done work

P = ΔW/Δt Δ is the symbol ‘delta’ and is used to mean a ‘change in’

Look at this worked example, which uses the equation for potential energy gained.

A motor lifts a mass m of 12 kg through a height Δh of 25 m in 6.0 s.

Gravitational potential energy gained:

ΔPE = mgΔh = (12 kg) × (9.81 m s−2

) × (25 m) = 2943 J



student worksheet

AQA Physics



Power =s 6.0

J 2943 = 490 W (2 s.f.)

Practice questions

3 Calculate the power of a crane motor that lifts a weight of 260 000 N through 25 m in 48 s.

4 A motor rated at 8.0 kW lifts a 2500 N load 15 m in 5.0 s. Calculate the output power.

5.3 Efficiency

Whenever work is done, energy is transferred and some energy is transferred to other forms, for example, heat or sound. The efficiency is a measure of how much of the energy is transferred usefully.

Efficiency is a ratio and is given as a decimal fraction between 0 (all the energy is wasted) and 1 (all the energy is usefully transferred) or as a percentage between 0 and 100%. It is not possible for anything to be 100% efficient: some energy is always lost to the surroundings.

Efficiency = inputenergy total

outputenergy useful or Efficiency =

input power total

output power useful

(multiply by 100% for a percentage)

Look at this worked example.

A thermal power station uses 11 600 kWh of energy from fuel to generate electricity. A total of 4500 kWh of energy is output as electricity. Calculate the percentage of energy ‘wasted’ (dissipated in heating the surroundings).

You must calculate the energy wasted using the value for useful energy output:

percentage energy wasted = 100inputenergy total

y)electricit as outputenergy - inputenergy (total

percentage energy wasted = 10011600

4500) - (11600 = 61.2% = 61% (2 s.f.)

Practice questions

5 Calculate the percentage efficiency of a motor that does 8400 J of work to lift a load.

The electrical energy supplied is 11 200 J.

6 An 850 W microwave oven has a power consumption of 1.2 kW.

Calculate the efficiency, as a percentage.

7 Use your answer to question 4 above to calculate the percentage efficiency of the motor. (The motor, rated at 8.0 kW, lifts a 2500 N load 15 m in 5.0 s.)

8 Determine the time it takes for a 92% efficient 55 W electric motor take to lift a 15 N weight 2.5 m.


BRIDGING UNIT: A-Level PSYCHOLOGY


What is psychology? Psychology is a human science. The word ‘science’ is from the Latin word ‘scientia’ which means knowledge. Psychology when broken down is ‘psyche’ which means ‘mind’ and ‘ology’ which means ‘study of’. Put it all together. Psychology is concerned with gaining knowledge about human behaviour with a focus on the

human mind. So it goes beyond just the biological element of humans.

Watch this short video

Approaches in Psychology https://youtu.be/wqzlKoleXn

Or read the following:

https://www.simplypsychology.org/perspective.html TASK:

What are the different approaches/theories (lenses) to studying human behaviour Write very detailed, CONCISE summaries on all EXCEPT Humanistic approach Rank the approaches from favourite to least favourite- Justify your ‘hierarchy’! Apply your knowledge of the learning approaches/theories (behaviourist and social learning theory) to

the scenario on the right

BRIDGING UNIT: A-level RELIGIOUS STUDIES


Read the article on God and the ‘problem of evil’ (Link is below)

https://cct.biola.edu/the-existential-problem-of-evil-and-the-brothers-karamazov/

Make notes on:

1) What the scholars think about God and the problem of evil

2) What you think about God and the problem of evil

3) How Dostoeysky portrays the argument of God and the problem of evil

‘If Evil exists, then God cannot exist’

Then, create a piece that argues for both sides of the argument above

The grid below may assist you when planning

For Against

BRIDGING UNIT: A-level SOCIOLOGY


Follow these instructions and complete any underlined tasks on a separate google doc or sheet of paper.

1. Introducing sociology: Read chapter one of this online textbook. Stop when you get to ‘MAKING CONNECTIONS: SOCIOLOGY IN THE REAL WORLD: https://opentextbc.ca/introductiontosociology/chapter/chapter1-an-introduction-to-sociology/

a. Answer this question in no more than 300 words: What is sociology and why is in important?

2. The sociological imagination: Read this: https://sites.middlebury.edu/utopias/files/2013/02/The-Promise.pdf

a. What are the three most important aspects of the sociological imagination?

3. Sociological perspectives: Watch these videos: a. Functionalism: https://www.youtube.com/watch?v=-83vVeSC2_g b. Interactionism: https://www.youtube.com/watch?v=Ux2E6uhEVk0 c. Marxism: https://www.youtube.com/watch?v=W0GFSUu5UzA d. Feminism: https://www.youtube.com/watch?v=xorqPUYu_SE

What are the key similarities and differences between these perspectives? Which perspective do you prefer and why?

Here’s a bonus video about gender: https://www.youtube.com/watch?v=CquRz_cceH8

Optional extra: How does this video change your perception of gender?

Extra reading which will give you a better foundation of understanding:

Continue the rest of this: https://opentextbc.ca/introductiontosociology/chapter/chapter1-an-introduction-to-sociology/

Here is the foundational text of Marxism:

https://www.marxists.org/archive/marx/works/1848/communist-manifesto/ch01.htm#007

BRIDGING UNIT: A-level SPANISH Contact Email: [email protected]/[email protected]

Google Classroom Code : 2c2cpmk

In Y12 Spanish you will study the following topics :

3.1 Social issues and trends

3.1.1 Aspects of Hispanic society

Students may study all sub-themes in relation to any Spanish-speaking country or countries.

Modern and traditional values (Los valores tradicionales y modernos)

Cyberspace (El ciberespacio)

Equal rights (La igualdad de los sexos)

3.1.2 Multiculturalism in Hispanic society

Students may study all sub-themes in relation to any Spanish-speaking country or countries.

Immigration (La inmigración)

Racism (El racismo)

Integration (La convivencia)

3.3 Works

3.3.2 Films

El laberinto del fauno Guillermo del Toro (2006)

Find below and introduction to the film and some technical information. The film is available on Netflix

and Youtube.

El laberinto del fauno fue una de las grandes revelaciones del cine mundial en 2006. En este año fue la

película de habla hispana con mayor recaudación de la historia. Dirigida por el especialista en cine fantástico

Guillermo del Toro, la película afronta de manera visualmente poderosa y muy original una historia

ambientada en la posguerra española. Así, el mundo de la joven protagonista se traduce en dos universos

paralelos: uno, el fantástico, habitado por terroríficos personajes de leyenda, seres oscuros y mágicos; el otro,

más real, pero no menos terrorífico, el de la ira y las miserias de una guerra con vencedores y vencidos. Entre

las muchas virtudes de esta película ganadora de tres Oscar destacan la fotografía, la banda sonora y las

potentes interpretaciones del trío protagonista, Sergi López y Maribel Verdú e Ivana Baquero.

FICHA TÉCNICA

TÍTULO ORIGINAL El laberinto del fauno

DIRECCIÓN Guillermo del Toro

PAÍS Estados Unidos, España, México

AÑO 2006

DURACIÓN 118 min.

GÉNERO Drama, bélico, fantástico

CALIFICACIÓN No recomendada para menores de 18 años

REPARTO Ivana Baquero, Sergi López, Maribel Verdú, Doug Jones, Ariadna Gil, Álex Angulo y Manolo

Solo.

Join the Google Classroom to have access to resources to keep practicing your Spanish !

Google Classroom Code : 2c2cpmk

BRIDGING UNIT: BTec APPLIED SCIENCE


On the following pages there is some of the content you will be required to know for your Unit 1 Exam, which contributes towards your final grade.

You will need to spend time learning and memorising the content on the pages you have been given. You need to hand in answers to all of the Quick Questions on a sheet of paper.


Revision Guide for BTEC National Applied Science – Principles and Applications 1 Page 2 of 112 © ZigZag Education, 2018

Chapter A: Periodicity and properties of elements

A1 – Structure and bonding in applications in science

In this section you will revisit the structure of atoms, then see how that structure can result in different types of bonding. You will also learn about different types of interaction between molecules which determine the properties of materials. At the end of this section you will learn about important chemical quantities and how to calculate them from a properly balanced equation.

The Electronic Structure of Atoms

The basic model of the atom has two main components – the positively charged nucleus, which contains positive protons and neutral neutrons, and the cloud of negatively charged electrons which encircle it. In all atoms, the number of protons equals the number of electrons. Protons and neutrons are far larger and heavier than electrons.

In the Bohr model of the atom, electrons are arranged in shells around the nucleus, which have higher energy levels as you move further away from the centre. These shells of electrons orbit around the nucleus at fixed distances, rather like planets orbiting the sun. Each shell has a maximum number of electrons that can fit into it – once it is full, extra electrons must go into the next shell.

O816

8 electrons 8 protons

8 neutrons 8p 8n

nucleus electron shells

electron

Na 23

11

Mass number

Atomic number

The number of electrons is equal to the number of protons. The number of protons is given by the atomic number. The number of protons plus neutrons in an atom is given by the mass number.

11 electrons 11 protons

12 neutrons

Definitions Proton a positively charged subatomic particle found in the nucleus of all atoms Neutron an uncharged subatomic particle found in the nucleus of all atoms except hydrogen Electron a negatively charged subatomic particle found orbiting the nucleus in all atoms Bohr model a simple atomic model with a small positively charged nucleus surrounded by

defined rings of negatively charged electrons Shell an energy level occupied by electrons which surrounds the nucleus Orbital a subshell with a certain energy level, which can hold up to two electrons Aufbau principle electrons fill the lowest available energy shells first, and singly occupy orbitals in the

same energy level before pairing up


Shells and orbitals Within a shell, not all electrons have the exact same energy – they are split into subshells which have different energies. Subshells are further divided into orbitals. Each orbital can hold two electrons with different spins.

Electron shell Maximum number of electrons Orbitals in shell 1 2 1 × s 2 8 1 × s, 3 × p 3 18 1 × s, 3 × p, 5 × d 4 32 1 × s, 3 × p, 5 × d, 7 × f 5 50 1 × s, 3 × p, 5 × d, 7 × f, 9 × g

Orbitals are filled according to the Aufbau principle – the lowest available energy level (the one closest to the nucleus) is filled first, before filling ones at higher levels.

Orbitals are labelled s, p, d or f, depending on which subshell they are in – p orbitals are higher in energy than the s orbitals in the same shells, d orbitals are higher than p orbitals, and so on. By arranging electrons according to these rules, the most stable electron configuration is obtained.

Ener

gy

1s

2s

2p

First shell

Second shell

3s

3p

3d 4s

4p

4d

4f

Third shell

Fourth shell

Rules for filling orbitals Each box represents an orbital which can hold two electrons. An electron is represented as an arrow with half a head. The lowest energy levels are filled first. The 4s subshell is lower in energy than the 3d subshell, so is filled first (important for

transition metals). All orbitals in a subshell are filled singly before electrons start to pair up (pairing up requires

more energy due to repulsions between electrons in the same orbital).


a) Use the box method and s/p/d notation to write out the electron configuration of fluorine.

Fluorine has nine electrons in total.

b) Use the box method and s/p/d notation to write out the electron configuration of oxygen.

Oxygen has eight electrons, and in s/p/d notation this is written as 1s22s22p4.

c) Use the box method to write out the electron configuration of iron.

Example

1s

2s

2p

In s/p/d notation, this would be written as:

1s22s22p5

Number of electrons in the

subshell

Number of the electron shell

Type of orbital in

the subshell

1s

2s

2p The 2p subshell can be filled with four electrons in two

ways: with one doubly occupied orbital and two singly

occupied orbitals, or two doubly occupied orbitals and

one empty orbital. However, pairing up electrons is

unfavourable, so the arrangement with the largest

number of unpaired electrons is always used.

1s

2s

2p

3s

3p

4s

3d

Again, as many orbitals as

possible are singly occupied

before electrons start to

pair up.

Iron has 26 electrons, and in

s/p/d notation this is written

as 1s22s22p63s23p64s23d6.

Note that the 4s orbital is

lower in energy than the 3d

orbital, so it gets filled first.

Quick Questions 1 1. How many electrons are there in:

a) one atom of sulfur? b) a full p orbital? c) a full third electron shell?

2. Draw the electron configuration of the following atoms using the box method and s/p/d notation: a) a nitrogen atom b) a vanadium atom

3. Draw a Bohr-model diagram of the atomic structure of magnesium and label the components. 4. What shell is the outer electron of a phosphorus atom in? Explain your answer.

?


Chapter B: Structure and functions of cells and tissues All living organisms consist of cells. This is a fundamental principle of biology, as will be discussed in the following topics. Life cannot exist without the presence of cells, and while some non-cellular substances, such as viruses, may have profound effects on life, they are not considered living, as they are incapable of ‘lifelike’ processes without the presence of other cells. Each cell is specially adapted for a specific function, achieved through a process known as specialisation. Once specialised, cells with similar functions are normally grouped together, with multiple cells that perform a similar function known as tissues. The animal body is composed of only a relatively small number of different tissues, though their arrangement allows great diversity to occur across life. Being able to appreciate the specialisation of a specific cell, as well as the arrangement of cells into tissues, is vital to the understanding of health and of disease. Scientists and medics globally rely on their understanding of tissue structure in order to recognise the processes that are associated with disease and, when fully understood, to target drugs and therapies to restore the normal function of cells.

B1 – Cell structure and function

At the most basic level, cells share astonishing similarities. Cells can be grouped according to the structures that they possess, and can be observed using microscopes.

Cell theory Before 1665, the basic structure of life was not understood. Before this time, scientists believed that life emerged spontaneously, through a process known as spontaneous generation. While we may think that this idea is absurd today, during this era, without microscopes, knowledge would have relied upon what could be seen – the spontaneous growth of grass from bare soil, mould appearing on rotting food, and animals being born from seemingly spontaneously pregnant adults. The invention of the microscope revolutionised our theory of life, and allowed the development of cell theory. Cell theory states that the cell is the most basic level of life possible. There are three key parts of cell theory:

1. All living organisms are composed of cells – either a single cell, or an arrangement of multiple cells.

2. The cell is the most basic unit of structure, and cells organise and form organisms. 3. Cells cannot form spontaneously, and all cells arise from another cell.

Definitions Prokaryote a cell that is lacking membrane-bound organelles, with loose, looped DNA Eukaryote a cell with membrane-bound organelles and structured, organised DNA Nucleoid site where DNA is stored in prokaryotes Tonoplast partially-permeable membrane that surrounds the vacuole of plant cells Centriole tubes of microfilaments that form spindle tubules during mitosis


At first, cell theory was not universally accepted and, therefore, it developed slowly. The discovery of the cell in 1665 by Robert Hooke was one of the key moments. In 1860 Louis Pasteur further built on the theory, showing that bacteria could not generate spontaneously. Now, we are able to show in astonishing detail how organisms are structured, how they develop, and where cells originate.

Ultrastructure Ultrastructure is the term used to describe the fine structure of cells. While observation of a single cell can reveal a lot about an organism, the greatest information is learned from knowledge of the cell’s ultrastructure.

Prokaryotes Prokaryotes, from the Greek meaning ‘before the nucleus’, are sometimes regarded as simple cells. These cell types include bacteria and are, in large part, characterised by the lack of membrane-bound organelles. Prokaryotes lack a distinct nucleus.

Instead, their genetic material is found in a nucleoid, which is like a nucleus, but lacks an enclosed membrane, so DNA is loose in the cell in a single loop.

Prokaryotes also contain plasmids. Plasmids are small, circular loops of DNA. They contain genes that provide a survival advantage, such as those responsible for antibiotic resistance.

Ribosomes are the machinery used by cells to make proteins. The ribosomes in prokaryotes are described as 70S ribosomes, based upon their size (the ‘S’ stands for ‘Svedberg’, a unit of size used for large molecules). At 70S, these are smaller than the ribosomes found in other cell types (eukaryotes have 80S ribosomes).

Many types of prokaryote (but not all) contain a capsule. The capsule is a layer of what is sometimes referred to as ‘slime’ that protects the cell and ensures it does not dry out.

Finally, ribosomes have a cell wall. Prokaryotic cell walls are made of peptidoglycan. This structure protects the cell and ensures it is not adversely affected by environmental conditions.

Plasmid

70S Ribosome

Peptidoglycan cell wall

Nucleoid Capsule

Quick Questions 1 1. Why do viruses not fit the rules of cell theory? 2. Give three features of prokaryotes. 3. How do the ribosomes in prokaryotes differ from the ribosomes in eukaryotes? 4. Which part of a prokaryote is responsible for antibiotic resistance? 5. What is the function of the nucleoid?

?


Eukaryotes Unlike prokaryotes, eukaryotes contain membrane-bound organelles, including an organised nucleus containing the genetic material of the cell. All animal cells and plant cells are eukaryotes. You might be used to seeing 2D representations of cells such as those shown to the right, but at this level, cells appear far more complex. Several organelles are common to all eukaryotes, regardless of their plant or animal origin. An animal cell is shown below with all of the adaptations that are common to all eukaryotes. As well as the organelles described above, plant cells have several additional adaptations.

Nucleus, the genetic material store, with a

nucleolus, which makes RNA and ribosomes.

The cytoplasm, containing chemicals

required by the cell to function correctly.

Mitochondria are the site of cellular respiration

where energy is released for cell processes.

Smooth endoplasmic reticulum (sER) is the site of lipid and carbohydrate synthesis and transport.

Rough endoplasmic reticulum (rER) is the site of protein synthesis and

transport to the Golgi apparatus.

Golgi apparatus receives, modifies and packages

proteins for export from the cell.

Plasma membrane that acts as a barrier, allowing some substances to cross into the cell while blocking others.

80S ribosomes, studded on the rER and loose in the

cytoplasm, are the site of protein synthesis.

Lysosomes are sacs filled with hydrolytic enzymes that break down cellular waste.

Vesicles are sacs filled with substances

requiring transport around or out of cells.

Centrioles produce spindle fibres, which are

essential during cell division.

Cell wall, like the cell wall of the prokaryotes, protects and

supports the cell, though is made from cellulose in plant

cells.

Chloroplasts are the site of photosynthesis, where light energy is used to

produce sugars.

Amyloplast is a membrane-bound sac that

synthesises and stores starch granules.

Vacuole is a membrane-bound sac that maintains pressure within a cell and

stores cell sap.

Tonoplast is the membrane that surrounds

the vacuole that allows some substances to cross into or out of the vacuole.

Plasmodesmata are strands of membrane that pass between

cells allowing transport and communication.

Pits are pores between plant cells that allow water

to enter and leave cells.

Simple cell drawings you

might be familiar with

Animal cell

Plant cell


Chapter C: Waves in communication In the modern world, all communication depends on waves – from long-range radio and television broadcasts, to transmitting data through underground fibre optic cables for sharing information over the Internet. Waves are also used in scientific research and medicine, with spectroscopy, satellite imaging and endoscopy all giving information that wouldn’t ordinarily be available without these technologies. It’s not just technology that uses waves – our own eyes and ears use light waves and sound waves to process and understand the world. All of these applications depend on the various features of waves – their speed, range of frequencies and ability to reflect, refract and diffract.

C1 – Working with waves

In this chapter you will learn about the different features of a wave and the different types of wave. You’ll also learn about some of the applications of waves in industry, and how the features of a wave are used in these applications.

Features of waves There are many different types of wave. Some waves, such as sound waves or the waves you’ll see on the surface of water, move the particles that make up the material the wave is travelling through, called the medium. Other waves, such as light waves, occur in fields. The repeated backwards and forwards motion of particles or field in a wave is called an oscillation. All waves have similar features and can be described in similar ways. Waves are generally described in terms of their displacement of the medium or field as a function of time. The graph on the following page shows the displacement of a single particle as a wave passes through the medium the particle is a part of.

Definitions Wave a regular backwards and forwards motion of a medium which transfers energy from

one location to another Medium the material a wave moves through Oscillation the backwards and forwards motion of a wave Frequency how often an oscillation of a wave occurs


This graph shows how the particle moves with constant motion, up and down in a regular oscillation, always returning to a central point before moving in the opposite direction. This central point is the position the particle would rest in if there were no wave. The diagram above shows a buoy on the ocean. The buoy moves up and down as each wave passes it, but otherwise stays in the same position. It is only the waves themselves that move. The amplitude of the wave is the height of the wave above the surface of the water if it were completely still. The distance between one peak of the wave and another is the wavelength. The time taken for a single wavelength is the periodic time of the wave. The frequency of the wave is the number of wavelengths that pass through a point in a second. Frequency and periodic time are linked by the equation:

𝑓 =1

𝑇

The maximum displacement of the particle from its central equilibrium point is the amplitude of the wave. The speed of the wave is how fast it appears to move and how fast it transfers energy from one place to another.

𝑓 = frequency, in Hz 𝑇 = periodic time, in s

Direction of waves

Exam tip This is how a wave appears on an oscilloscope, or CRO (cathode-ray oscilloscope).

Oscilloscopes display the displacement of a wave (vertically) against time (horizontally).

The display settings of the oscilloscope determine what scale the displacement and time will be seen on.

Wavelength

Amplitude Time D

ispl

acem

ent


Types of wave Waves can be grouped into two main types: transverse and longitudinal. Looking at a wave on the sea surface, it can seem like it’s the water that’s moving towards the shore. In fact, the water is only moving up and down, while it is the wave and the energy it transfers that are moving towards the shore. The oscillations in a transverse wave are at right angles to the direction the wave is travelling in and

transferring energy in.

Examples of transverse waves include light waves and water waves. Not all waves have particles moving in different directions to the wave. Sound waves are made of periodic pressure increases and decreases, compressing particles along the direction of propagation. The oscillations in a longitudinal wave are in the same direction as the direction the wave is travelling in and

transferring energy.

An example of a longitudinal wave is a sound wave.

Quick Questions 1 1. Sketch a graph of displacement against time for a wave. 2. What is meant by each of the following terms: wavelength, frequency, amplitude, periodic time? 3. A wave has a frequency of 12 kHz. Calculate the periodic time of the wave. 4. Describe the main difference between transverse waves and longitudinal waves. 5. Give one example each of a transverse wave and a longitudinal wave.

Particles at top of the wave move down

Direction wave travels in

Central position

Particles at bottom of the wave move up

Particles move up and down, at right angles to the direction of the wave

Direction wave travels in

Most spread out particles move together

Most compressed particles move apart

Particles move left and right, while the wave moves right

?

BRIDGING UNIT: BTec BUSINESS STUDIES (EXT CERT AND DIPLOMA)


BRIDGING UNIT: BTEC HEALTH AND SOCIAL CARE


Unit 1: Human Lifespan Development Level: 3 Preparation for your exam in December 2020

Your research task:

I would like you to explain how an individual GROWS and DEVELOPS during different stages of their lifespan. You must explain the differences between the principles of growth and the principles of development and then relate these 2 principles to the following stages of life:

In infancy (0–2 years) - 1) gross motor skills, fine motor skills, 2) milestones E.G. sitting up, standing, cruising, walking.

In early childhood (3–8 years) - 1) gross and fine motor skills E.G. riding a tricycle, running forwards and backwards, walking on a line, hopping on one foot, hops, skips and jumps confidently, turns pages of a book etc..

In adolescence (9–18 years) 1) puberty: 2) development of primary and secondary sexual characteristics 3) role of hormones in sexual maturity.

In early adulthood (19–45 years) 1) physical strength peaks, pregnancy and lactation occur 2) perimenopause – oestrogen levels decrease Leading to: 3) physical and emotional symptoms,

In middle adulthood (46–65 years) 1) menopause: - causes and effects of female menopause and the role of hormones 2) effects of the ageing process in middle adulthood.

In later adulthood (65+ years) 1) health and intellectual abilities can deteriorate.

You can display this information by creating a timeline like the one below but you must include the key words that are highlighted in RED for each stage and you will need to complete 1 timeline for GROWTH and 1 for DEVELOPMENT:

BRIDGING UNIT: BTec ICT Contact Email: [email protected]

Unit 1 : Information Technology Systems

Equality and Accessibility

Look up https://edu.gcfglobal.org/en/computerbasics/using-accessibility-features/1/

Describe three assistive technology devices which can be used to make input and output more accessible to users with disabilities.

Emerging technologies

Emerging technologies have influenced the daily lives of individuals and the societies in which they live.

(a) Choose one of the following categories:

Communications – e.g. text messaging, email, video calls, instant messaging,

VoIP

Media – HD and 4K video, digital video, video on demand, streaming, digital

music, cloud services

Entertainment – computer games, smart televisions, home cinema systems

Social media – Facebook, Instagram, Snapchat, YouTube etc

(b) Discuss the impact of these technologies on:

The ways in which people interact with each other

Work and employment

Education

Leisure Optional extra:

Improving your digital knowledge can help you find a job, get promoted, or start a whole new

career. Get certified in the Fundamentals of Digital Marketing. Learn the fundamentals of digital

marketing and open up new career opportunities.

There are 26 modules to explore, all created by Google trainers, packed full of practical exercises

and real-world examples to help you turn knowledge into action.

Online course : https://learndigital.withgoogle.com/digitalgarage/course/digital-marketing

BRIDGING UNIT: BTec MEDIA STUDIES


Using the links below, write a definitive 500 word (minimum) argument discussing how much you agree with the following statement. “The internet has changed the way we consume news and television in a positive way”. Use examples from links below or even better, offer your own! Is there a counter argument to consider? If so, why? You MUST discuss both news AND television. https://www.aokmarketing.com/breaking-news-social-media-changed-way-consume-news/ The changing face of online news consumption - Ofcom www.ofcom.org.uk › assets › pdf_file › Scrolling-News

https://www.bbvaopenmind.com/en/articles/first-the-media-then-us-how-the-internet-changed-the-fundamental-nature-of-the-communication-and-its-relationship-with-the-audience/ https://www.theguardian.com/media-network/media-network-blog/2014/jun/10/internet-changing-definition-television http://www.bbc.co.uk/news/av/technology-12141563/how-internet-is-changing-the-way-we-watch-television https://newsroom.cisco.com/feature-content?articleId=1119737

https://www.digitaltrends.com/movies/5-ways-tv-has-gotten-worse-since-online-streaming

BRIDGING UNIT: BTec PERFORMING ARTS


Google classroom code: fwhobja

BTEC Nationals - Tech Level - Extended Certificate in Acting

Total Units : 5

This course is designed for those wishing to pursue a career in acting. You will explore the acting skills and techniques that you will need for an apprenticeship, employment or higher education in the arts.

For more information about this qualification please follow this link:

https://qualifications.pearson.com/content/dam/pdf/BTEC-Nationals/Performing-Arts/2016/specification-and-sample-assessments/9781446938348_BTEC_Nat_Cert_PA_Spec_Iss2C.pdf

Your summer task

“How do you respond as an artist living in an extraordinary time?”

You have been commissioned to devise an engaging piece exploring the impact of Covid-19. This piece will be recorded and should be between 3 -5 minutes.

You need to have a clear idea of your creative intention, the purpose of your piece and who your target audience is. Write your ideas down before you start devising and submit along with your recorded work.

Your work could explore the following:

family community health (personal health or NHS) science (spread, prevention, cure) homelessness faith unemployment arts industry (British theatre industry) mental health

Follow these links for inspiration: https://www.independent.co.uk/life-style/coronavirus-key-workers-black-minority-ethnic-you-clap-for-me-now-video-a9465761.html

https://m.youtube.com/watch?feature=emb_title&time_continue=57&v=Fi6kMjE90Mg

https://m.youtube.com/watch?v=aR_CMTneqtA

This is an opportunity for you to demonstrate just how imaginative you are. There are no limitations! Work should be recorded and uploaded to the classroom. I can’t wait to watch your work!

BRIDGING UNIT: BTec SPORT Contact Email: [email protected]

Unit 1: Anatomy and Physiology. Preparation for exam

Your task:

Use the Google slideshow below to create flashcards and remember the following:

Slides 1-52

1) Major bones to include cranium, clavicle, ribs, sternum, scapula, humerus, radius, ulna, carpals, metacarpals, phalanges, pelvis, vertebral column (cervical, thoracic, lumbar, sacrum, coccyx), femur, patella, tibia, fibula, tarsals, metatarsals. • Type of bone – long, short, flat, sesamoid, irregular.

2) Areas of the skeleton to include axial skeleton, appendicular skeleton, spine, curves of the spine, neutral spine alignment, postural deviations (kyphosis, scoliosis).

3) Process of bone growth – osteoblasts, osteoclasts, epiphyseal plate. A2 Function of skeletal system Understand how the functions of the skeleton and bone types are used in sporting actions and exercise.

Google Slideshow:

https://docs.google.com/presentation/d/1JpGdg9xVAe7v48IeD8vHdQ0zqzj_MAvDHXQ5WZWRjT8/edit?usp=sharing

Documents

YEAR 11 INTO YEAR 12 BRIDGING UNITSfluencycontent2-schoolwebsite.netdna-ssl.com/FileCluster/...1 A burger contains 4 500 000 J of energy. Write this in: a kilojoules b megajoules