maths and further maths a level transition pack

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β€œLearning, in my opinion, is a great treasury” Saint Teresa of Avila, 1587 MATHS and FURTHER MATHS A Level Transition Pack

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Page 1: MATHS and FURTHER MATHS A Level Transition Pack

β€œLearning, in my opinion, is a great treasury” Saint Teresa of Avila, 1587

MATHS and

FURTHER MATHS

A Level

Transition Pack

Page 2: MATHS and FURTHER MATHS A Level Transition Pack

Transition booklet

All of the topics in this booklet are on the Higher GCSE syllabus and are crucial to success in A-level Maths. It is

therefore essential that you revise, practise and develop your skills in these areas. You can use the links to the

website DrFrostMaths.com to help you – you will receive an email inviting you to register and set up a

password (if you haven’t already done so).

You should complete the booklet assessing your work as you go using the answers at the back of the booklet.

We expect to see full working out for each question – naturally any questions where the answer has just

been copied down will not be seen as completed.

The questions with * will be marked during your first lesson and will indicate to you which areas of Maths you

need to ask for help with before the baseline assessment, which will take place in the week commencing

27th September2021. Please complete these on a separate piece of paper with your name clearly written so

that you can hand this work to your teacher on the first lesson.

Extra help is available at www.pearsonactivelearn.com where you can log on using

Username: [email protected]

Password: sbsjAlevel20

This will give you access to the two textbooks that form the basis of the Y12 course and you are encouraged to

familiarise yourself with them.

Next to each exercise below (where applicable) there is a link to a video tutorial on DR Frost.

Page 3: MATHS and FURTHER MATHS A Level Transition Pack

1. Simplify as far as possible a. 3π‘₯ βˆ’ 2𝑦 + 4𝑦

b. 7 – 3π‘₯ + 2 + 4π‘₯ *

c. 5π‘₯ + 2𝑦 – 4𝑦 – π‘₯2

2. Factorise as fully as possible a. π‘₯2 + 5π‘₯

b. 2𝑦2 + 6𝑦

c. π‘Žπ‘π‘ – π‘Žπ‘3 *

d. π‘Žπ‘₯ + π‘Žπ‘¦ + 𝑏π‘₯ + 𝑏𝑦

3. Multiply out the brackets and simplify a. 3π‘₯ + 2(π‘₯ + 1)

b. 5π‘₯ βˆ’ 2π‘₯(π‘₯ βˆ’ 1)

c. 3π‘₯(π‘₯ – 1) – 7π‘₯2 *

d. (π‘₯ βˆ’ 3) (π‘₯ – 5)

e. (2π‘₯ + 𝑦) (π‘₯ – 3𝑦)

4. Simplify the following

a. 6𝑦

3𝑦 b.

4π‘Žπ‘2𝑐3

3π‘Ž3𝑏2𝑐 *

5. Simplify the following

a. 3π‘₯

2 Γ—

2π‘Ž

3π‘₯ * b.

5π‘Žπ‘2

2Γ—

3

2π‘Ž2𝑏

6. Simplify the following

a. 2

π‘Ž Γ·

π‘Ž

2 b.

3(π‘₯βˆ’1)

π‘₯Γ·

6

π‘₯𝑦 *

7. Write as a single fraction

a. π‘₯

2+

π‘₯ + 1

3 * b.

3(π‘₯βˆ’1)

π‘₯ βˆ’

6

π‘₯𝑦

Expanding & Removing Brackets

https://youtu.be/QwAwH5qiA1Y

Simplifying Algebraic Fractions

https://youtu.be/PMXJGwzW1AM

Adding and subtracting Algebraic Fractions

https://youtu.be/PQWQQRRpUlQ

Section 1 – Algebraic Expressions

Page 4: MATHS and FURTHER MATHS A Level Transition Pack

Solve the following equations, give your answers as fractions in mixed number form where necessary.

1. 2π‘₯ – 5 = 11

2. 12 = 2π‘₯ βˆ’ 8

3. 3π‘₯

2= 5

4. 4 =π‘₯

2+ 5

5. π‘₯ + 3( π‘₯ + 1) = 2π‘₯

6. 1

3(π‘₯ + 2) =

1

5(3π‘₯ = 2)

7. 2π‘₯βˆ’1

3=

π‘₯

2 *

8. π‘₯+1

2+

π‘₯βˆ’1

3=

1

6

9.

Make x the subject of each of the following formulae

1. 2π‘₯ = 𝑆

2. 9π‘₯ = 𝑇 + 𝑁

3. V2π‘₯ + 𝑏 = 𝑐

4. 3(π‘₯ – 1) = π‘Ž

5. π΅βˆ’π΄π‘₯

𝐷= 𝐸

6. 𝑄

π‘₯βˆ’πΆ= 𝑇 *

7. π‘š2

π‘₯βˆ’ 𝑛 = βˆ’π‘

8. 𝑔 = βˆšπ‘ βˆ’ π‘₯

9. 𝑑 = √(π‘₯2 βˆ’ π‘˜) βˆ’ 𝑏 *

10. 𝑧π‘₯ = 𝑆(π‘₯ + 𝑑)

11. π‘₯(π‘Ž βˆ’ 𝑐) = 𝑏π‘₯ + 𝑑

12. √(π‘₯2 + π‘Ž2 ) = 𝑏2

Rearranging Formulae https://youtu.be/y-cGpOFCB3Q

Linear Equations in one variable https://youtu.be/l-za234E9F4 https://youtu.be/IHgy9iAk-wU

Section 2 – Linear Equations

Section 3 - Formulae

Page 5: MATHS and FURTHER MATHS A Level Transition Pack

Solve the following pairs of simultaneous equations:

1. 2π‘₯ + 5𝑦 = 24

4π‘₯ + 3𝑦 = 20

2. 3π‘₯ βˆ’ 𝑦 = 9 4π‘₯ βˆ’ 𝑦 = βˆ’14 *

3. 5π‘₯ βˆ’ 7𝑦 = 27 3π‘₯ βˆ’ 4𝑦 = 16

Factorise each of the following quadratics

1. π‘₯2 + 7π‘₯ + 10

2. π‘₯2 + 8π‘₯ + 15

3. π‘₯2 βˆ’ 8π‘₯ + 16

4. π‘₯2 βˆ’ π‘₯ βˆ’ 12

5. π‘₯2 βˆ’ 5π‘₯ βˆ’ 24 *

6. π‘₯2 βˆ’ 49

7. π‘₯2 + 5π‘₯

Factorise each of the following quadratics

1. 2π‘₯2 + 5π‘₯ + 3

2. 3π‘₯2 + 7π‘₯ + 2

3. 3π‘₯2 βˆ’ 11π‘₯ + 6

4. 3π‘₯2 βˆ’ 5π‘₯ βˆ’ 2

5. 2π‘₯2 + π‘₯ βˆ’ 21

6. 12π‘₯2 + 4π‘₯ βˆ’ 5

7. 6π‘₯2 βˆ’ 27π‘₯ + 30 *

8. 4π‘₯2 βˆ’ 25

Write each of the following in the form (π‘₯ + π‘Ž)2 + 𝑏 where π‘Ž and 𝑏 are constants to be determined:

1. π‘₯2 + 4π‘₯ + 7 *

2. π‘₯2 βˆ’ 26π‘₯ βˆ’ 1

3. π‘₯2 + 5π‘₯ βˆ’ 5

4. π‘₯2 βˆ’ π‘₯ + 1

Write each of the following in the form a(π‘₯ + 𝑏)2 + 𝑐 where π‘Ž, 𝑏 and 𝑐 are constants to be determined:

5. 2π‘₯2 βˆ’ 4π‘₯ + 5 *

6. 3π‘₯2 βˆ’ 12π‘₯ + 10 7. 5π‘₯2 + 25π‘₯ βˆ’ 5

https://youtu.be /5p79yczPWQc

https://youtu.be/cbbEFYpeWV8

https://youtu.be/XWNropJ5pn0

https://youtu.be/sQooemMboCw

Section 4 – Simultaneous Equations (both linear)

Section 5 – Factorising Quadratics, x2 (multiplied by 1)

Section 6 – Factorising Quadratics, x2 (not multiplied by 1)

Section 7 – Completing the Square

Page 6: MATHS and FURTHER MATHS A Level Transition Pack

Solve each of the following

1. π‘₯2 + 7π‘₯ + 12 = 0

2. π‘₯2 βˆ’ 8π‘₯ + 12 = 0

3. π‘₯2 + 2π‘₯ βˆ’ 15 = 0 *

4. 6π‘₯2 βˆ’ 13π‘₯ + 6 = 0

5. 10π‘₯2 βˆ’ π‘₯ βˆ’ 3 = 0

6. 6π‘₯2 + 17π‘₯ βˆ’ 3 = 0 7. 9π‘₯2 + 6π‘₯ = 0

8. 4π‘₯2 βˆ’ 1 = 0

Solve the following quadratics, giving your solutions correct to 3 significant figures:

1. π‘₯2 + 7π‘₯ + 5 = 0

2. 10π‘₯2 + 3π‘₯ βˆ’ 2 = 0

3. 5π‘₯2 βˆ’ 9π‘₯ + 2 = 0 *

4. 6π‘₯2 βˆ’ 5π‘₯ = 8

Solve each of the following equations

1. π‘₯ + 5 =14

π‘₯

2. 3

π‘₯βˆ’1+

3

π‘₯+1= 4 *

3. 2

π‘₯βˆ’1βˆ’

1

π‘₯+2=

βˆ’4

π‘₯+3

Quadratic Equations https://youtu.be/Idk4R_baULA

Hint: Form and solve a quadratic Equation

Use the formula:

𝒙 = βˆ’π’ƒ Β± βˆšπ’ƒπŸ βˆ’ πŸ’π’‚π’„

πŸπ’‚

https://youtu.be/q0vb8ybymKI

Section 8 – Solving Quadratic Equations – By Factorising

Section 9 – Solving Quadratic Equations - Formula

Section 10 – Equations with Quadratic Solutions

Page 7: MATHS and FURTHER MATHS A Level Transition Pack

Solve the following pairs of simultaneous equations:𝑦 = π‘₯2 βˆ’ 4π‘₯ + 6 𝑦 = 2π‘₯2 βˆ’ 3π‘₯ + 4

1. 𝑦 = π‘₯ βˆ’ 3 5. 𝑦 + 2π‘₯ = 3 𝑦2 + π‘₯𝑦 + 4π‘₯ = 7 * 𝑦2 + π‘₯𝑦 = 13 βˆ’ 16π‘₯

2. π‘₯ + 𝑦 = 5 6. 3𝑦 βˆ’ 2π‘₯ = 11 3π‘₯𝑦 = 18 π‘₯𝑦 = 2

)

In each diagram, find the value of:

1. * 2.

48π‘œ 23π‘π‘š

7π‘π‘š π‘₯ π‘₯

17π‘π‘š

Finding sides https://youtu.be/r6xuUjwyKek

Finding angles https://youtu.be/5brkGmejvqU

Example/ Advice.

- If both equations have the same letter as their subject (i.e. Q1) then put them equal to each

other, and then solve.

- Otherwise, substitute the linear equation into the non-linear equation, and then solve.

https://youtu.be/2_n1KKPzmm8

Section 11 – Simultaneous Equations (when one is non-linear)

Section 12 – Trigonometry (calculator allowed)

Page 8: MATHS and FURTHER MATHS A Level Transition Pack

1. Simplify:

a) 2𝑝2 Γ— 4𝑝5 b) π‘Ž2 Γ— π‘Ž3 Γ— π‘Ž5 c) 12𝑛7 Γ· 2𝑛2 d) (𝑦3)4 e) f) 𝑀4𝑦3 Γ· 𝑀𝑦2

2. Simplify

a) π‘Ž8 Γ— π‘Žβˆ’6 b) (π‘€π‘˜βˆ’4)3 * c) 2𝑏2

3 Γ— 4𝑏1

4 d) 𝑦

12×𝑦

13

𝑦 e)

4π‘₯23 Γ— 3π‘₯

βˆ’16

6π‘₯34

*

1. Simplify

a) √50 b) √98 c) √27 d) √363

e) √18 + √50 f) 2√8 + √72 * g) √3(2 + √3) h) (3√3 + 1)(2 - 5√3)

2. Express as simply as possible with a rational denominator

a) 1

√8 b)

√5

√15 c)

4√20

3√18 * d)

6βˆ’βˆš8

√2 e)

2

√3βˆ’

√6

√12 *

Laws of indices https://youtu.be/Z_WCKeO-liY

Fractional powers https://youtu.be/E-gq8eq105I

Negative powers https://youtu.be/RBPbRrHSCVs

Negative powers

Simplify https://youtu.be/LDH_VS1fwYE

Add/subtract https://youtu.be/O0Hm8oAgrPI

Multiply/divide https://youtu.be/XiuLvyXyRxY

Rationalise https://youtu.be/wiBovf97ihc

Section 13 - Indices

Section 14 - Surds

Page 9: MATHS and FURTHER MATHS A Level Transition Pack

1. At a coffee shop Americanos cost Β£1.90. Cappuccinos cost Β£2.40. Maddie bought two Americanos

and some cappuccinos. She paid with a Β£20 note and received Β£9.00 change. How many

cappuccinos did she buy? *

2. Jamila has taken five out of six tests. The maximum mark in each test is 20. So far her average (mean) mark

is 12. Is it possible for her to increase her average mark to above 15 after she has taken the final test? *

3. Two work colleagues arrive on the ground floor of a building at 8.01 am for a meeting on the twentieth floor.

The twentieth floor is 60 m above ground level. One decides to take the stairs, the other the lift.

They both arrive on the twentieth floor at 8.07 am. In the meeting room they sketch graphs showing their height

above the ground in terms of time between 8.01 am and 8.07 am.

Which graph do you think shows the person who took the lift? Why ?

What assumptions might have been made in drawing these graphs? *

Problem Solving *

Page 10: MATHS and FURTHER MATHS A Level Transition Pack

4. Calculate the shaded area in the diagram below: *

5. The diagram shows a square and a circle. The corners of the square are on the circle.

The perimeter of the square is 20 cm.

Find the exact value of the area of the circle. *

6. Find the value of x in the diagram below. *

Page 11: MATHS and FURTHER MATHS A Level Transition Pack

Any students hoping to study Further Maths alongside Maths should also complete the following

questions, showing all steps of your method clearly.

1. Simplify these expressions as far as possible.

a. π‘₯2 –π‘₯βˆ’6

π‘₯2+π‘₯βˆ’12 b.

π‘₯2βˆ’49

π‘₯2 –π‘₯βˆ’30Γ·

π‘₯2+5π‘₯βˆ’14

π‘₯2βˆ’36

2. a. Show that the line x + y = 6 is a tangent to the circle x2 + y2 = 18.

a. Show the line and the circle on a diagram.

b. The tangent meets the x axis at P and the y axis at Q. Find the area of the triangle OPQ.

3. Expand and simplify (√a - 3√c)(5√a - √c)

4. a Write 4x2 + 24x + 11 in the form a(x + b)2 + c

b. Hence, or otherwise, write down the coordinates of the turning point of the graph of

y = 4x2 + 24x + 11.

5. Prove algebraically that the sum of the squares of three consecutive numbers less 2 is

always 3 times a square number.

6. The functions f and g are defined as g(x) = 3x and f(x) = 4x + 1 .

2-x

Given that x β‰  2, find the value(s) of x such that g(x) = f(x), giving your answer(s) to 2 decimal

places.

Further Maths

Page 12: MATHS and FURTHER MATHS A Level Transition Pack

7.

8. A triangle has side lengths AB = 18 cm, BC = 15 cm and AC = 8 cm.

a. Find the size of the largest angle, giving your answer to 2 decimal places.

b. Find the area of the triangle, giving your answer to 2 decimal places.

9. a. Sketch the graph of y = sin x for - 180o ≀ x ≀ 360o , showing the points where the graph cuts the axes.

b. Hence find the exact values of x in the interval - 180o ≀ x ≀ 360o for which

sin π‘₯ = βˆ’1

2

Page 13: MATHS and FURTHER MATHS A Level Transition Pack

10. Problem Solving challenge

Can you find the solution ?

Page 14: MATHS and FURTHER MATHS A Level Transition Pack

Section 1

1a) 3π‘₯ + 2𝑦 c) βˆ’π‘₯2 + 5π‘₯ βˆ’ 2𝑦

2a) π‘₯(π‘₯ + 5) b) 2𝑦(𝑦 + 3) d) π‘Ž(π‘₯ + 𝑦) + 𝑏(π‘₯ + 𝑦) = (π‘Ž + 𝑏)(π‘₯ + 𝑦)

3a) 5π‘₯ + 2 b) 7π‘₯ βˆ’ 2π‘₯2 d) π‘₯2 βˆ’ 8π‘₯ + 15 e) 2π‘₯2 + 3𝑦2 βˆ’ 5π‘₯𝑦

4a) 2

5a) b) 15𝑏

4π‘Ž

6a) 4

π‘Ž2

7a) b) 3π‘₯𝑦 βˆ’ 3𝑦 βˆ’ 6

π‘₯𝑦 or

3(π‘₯𝑦 βˆ’ 𝑦 βˆ’ 2)

π‘₯𝑦

Section 2

1. π‘₯ = 8 2. π‘₯ = 10 3. π‘₯ =10

3= 3

1

3 4. π‘₯ = βˆ’2

5. π‘₯ = βˆ’3

2= βˆ’1

1

2 6. π‘₯ = 4 8. π‘₯ = 0

Section 3

1. π‘₯ = 𝑆

2 2. π‘₯ =

𝑇 + 𝑁

9 3. π‘₯ =

π‘βˆ’π‘

𝑉2 4. π‘₯ =

π‘Ž

3+ 1 or

π‘Ž+3

3

5. π‘₯ =π΅βˆ’πΈπ·

𝐴 or

πΈπ·βˆ’π΅

βˆ’π΄ 7. π‘₯ =

π‘š2

π‘›βˆ’π‘ 8. π‘₯ = 𝑐 βˆ’ 𝑔2

10. π‘₯ = 𝑆𝑑

π‘§βˆ’π‘† 11.

𝑑

π‘Žβˆ’π‘βˆ’π‘ 12. π‘₯ = βˆšπ‘4 βˆ’ π‘Ž2

Section 4

1. π‘₯ = 2 , 𝑦 = 4 3. π‘₯ = 4 , 𝑦 = βˆ’1

Section 5

1. (π‘₯ + 2)(π‘₯ + 5) 2. (π‘₯ + 3)(π‘₯ + 8) 3. (π‘₯ βˆ’ 4)2 4. (π‘₯ βˆ’ 4)(π‘₯ + 3)

6. (π‘₯ βˆ’ 7)(π‘₯ + 7) 7. π‘₯(π‘₯ + 5)

Section 6

1. (2π‘₯ + 3)(π‘₯ + 1) 2. (3π‘₯ + 1)(π‘₯ + 2) 3. (3π‘₯ βˆ’ 2)(π‘₯ βˆ’ 3) 4. 3(2π‘₯ βˆ’ 5)(π‘₯ βˆ’ 2)

5. (3π‘₯ + 1)(π‘₯ βˆ’ 2) 6. (2π‘₯ + 7)(π‘₯ βˆ’ 3) 8. (2π‘₯ βˆ’ 5)(2π‘₯ + 5)

Answers

Page 15: MATHS and FURTHER MATHS A Level Transition Pack

Section 7

2. (π‘₯ βˆ’ 13)2 βˆ’ 170 3. (π‘₯ + 21

2)2 βˆ’ 11

1

4 4. (π‘₯ βˆ’

1

2)2 +

3

4

6. 3(π‘₯ βˆ’ 2)2 βˆ’ 2 7. 5(π‘₯ + 21

2)

2βˆ’ 36

1

4

Section 8

1. π‘₯ = βˆ’ 4 π‘œπ‘Ÿ βˆ’ 3 2. π‘₯ = 2 π‘œπ‘Ÿ 6 4. π‘₯ = 2

3 π‘œπ‘Ÿ

3

2

5. π‘₯ = 3

5 π‘œπ‘Ÿ βˆ’

1

2 6. π‘₯ =

1

6 π‘œπ‘Ÿ βˆ’ 3 7. π‘₯ = 0 π‘œπ‘Ÿ βˆ’

2

3 8. π‘₯ =

1

2 π‘œπ‘Ÿ βˆ’

1

2

Section 9

1. π‘₯ = βˆ’ 6.19 π‘œπ‘Ÿ βˆ’ 0.807 2. π‘₯ = βˆ’ 0.622 π‘œπ‘Ÿ 0.322 4. π‘₯ = βˆ’ 0.811 π‘œπ‘Ÿ 1.64

Section 10

1. π‘₯ = βˆ’ 7 π‘œπ‘Ÿ 2 3. π‘₯ = βˆ’7

5 π‘œπ‘Ÿ βˆ’ 1

Section 11 – (with workings out for the questions 1-3)

1. π‘₯2 βˆ’ 4π‘₯ + 6 = 3π‘₯ βˆ’ 4 2. π‘₯ = 𝑦 + 3 β‡’ 𝑦2 + 𝑦(𝑦 + 3) + 4(𝑦 + 3) = 7

π‘₯2 βˆ’ 7π‘₯ + 10 = 0 𝑦2 + 𝑦2 + 3𝑦 + 4𝑦 + 12 = 7 (π‘₯ βˆ’ 2)(π‘₯ βˆ’ 5) = 0 2𝑦2 + 7𝑦 + 5 = 0 π‘₯ = 2 π‘œπ‘Ÿ 5 (2𝑦 + 5)(𝑦 + 1) = 0

∴ 𝑦 = 2 π‘œπ‘Ÿ 11 𝑦 = βˆ’5

2 π‘œπ‘Ÿ βˆ’ 1

∴ π‘₯ = 1

2 π‘œπ‘Ÿ 2

β†’ π‘₯ = 2, 𝑦 = 2 π‘œπ‘Ÿ π‘₯ = 5, 𝑦 = 11 β†’ π‘₯ =1

2, 𝑦 = βˆ’

5

2 π‘œπ‘Ÿ π‘₯ = 2, 𝑦 = βˆ’1

3. π‘₯ = 5 βˆ’ 𝑦 β‡’ 3(5 βˆ’ 𝑦)𝑦 = 18

15𝑦 βˆ’ 3𝑦2 = 18 3𝑦2 βˆ’ 15𝑦 + 18 = 0 𝑦2 βˆ’ 5𝑦 + 6 = 0 (𝑦 βˆ’ 3)(𝑦 βˆ’ 2) = 0 𝑦 = 3 π‘œπ‘Ÿ 2

∴ π‘₯ = 2 π‘œπ‘Ÿ 3

β†’ π‘₯ = 2, 𝑦 = 3 π‘œπ‘Ÿ π‘₯ = 3, 𝑦 = 2

4. π‘₯ =1

2, 𝑦 = 3 π‘œπ‘Ÿ π‘₯ = 3, 𝑦 = 13 5. π‘₯ =

1

2, 𝑦 = 2 π‘œπ‘Ÿ π‘₯ = βˆ’4, 𝑦 = 11

6. π‘₯ =1

2, 𝑦 = 4 π‘œπ‘Ÿ π‘₯ = βˆ’6, 𝑦 = βˆ’

1

3

Page 16: MATHS and FURTHER MATHS A Level Transition Pack

Section 12

2. π‘₯ = 73.07248694 = 73. 10 (3 𝑠. 𝑓. )

Section 13

1a 8𝑝7 1b) π‘Ž10 1c) 6𝑛5 1d) 𝑦12 1e) 𝑀3𝑦

2a) π‘Ž2 2b) 𝑀3π‘˜βˆ’12 2c) 8𝑏11

12 2d) π‘¦βˆ’1

6 2e) 2π‘₯βˆ’1

4

Section 14

1a) 5√2 1 b) 7√2 1 c) 3√3 1 d) 3√11

1 e) 8√2 1 f) 10√2 1g) 2√3 + 3 1 h) -43+√3