Sickmaths a level maths core 1 notes questions with cover

Page 1

Contents p1. Quadratic equations p3. Straight line graphs p5. Indices (Powers) p7. Surds p9. Sketching graphs p11. Transforming graphs p13. Simultaneous equations p15. Arithmetic series p17. Inequalities p19. Discriminant p21. Differentiation p23. Integration p25. The Answers p27. Quotes from Buddha, Jesus, Muhammad, Einstein and Steve Jobs

Any suggestions for improvements? Please contact me at khalidsafir@gmail.com

Factorising, completing the square and the quadratic formula.

1. Quadratic Equations

Factorising is opposite of expanding. 3đ?‘Ľ is HCF of 6đ?‘Ľ and −9đ?‘Ľ 2

Simple (highest common factor) factorising Solve 6đ?‘Ľ = 9đ?‘Ľ 2. 6đ?‘Ľ = 9đ?‘Ľ 2 ⇒ 6đ?‘Ľ − 9đ?‘Ľ 2 = 0 ⇒ 3đ?‘Ľ(2 − 3đ?‘Ľ) = 0 Vid 1 3đ?‘Ľ = 0 2 − 3đ?‘Ľ = 0 2 đ?‘Ľ=0 đ?‘Ľ=

If đ?‘Žđ?‘? = 0 so đ?‘Ž = 0 and/or đ?‘? = 0 therefore 3đ?‘Ľ(2 − 3đ?‘Ľ) = 0 ⇒ 3đ?‘Ľ = 0 and 2 − 3đ?‘Ľ = 0

3

Difference of two squares i.e. đ?’‚đ?&#x;? − đ?’ƒđ?&#x;? = (đ?’‚ − đ?’ƒ)(đ?’‚ + đ?’ƒ) Factorise 4đ?‘Ą 2 − 9. 4đ?‘Ą 2 − 9 = (2đ?‘Ą)2 − 32 = (2đ?‘Ą − 3)(2đ?‘Ą + 3)

Know this pattern. Vid 2

Trinomials have 3 bits (đ?‘Žđ?‘Ľ 2 , đ?‘?đ?‘Ľ and đ?‘?).

đ?&#x;?

Factorising trinomials i.e. đ?’‚đ?’™ + đ?’ƒđ?’™ + đ?’„ = (đ?’‘đ?’™ + đ?’’)(đ?’Žđ?’™ + đ?’?) Factorise 2đ?‘Ľ 2 − 3đ?‘Ľ − 2. Vid 3.1 2đ?‘Ľ 2 − 3đ?‘Ľ − 2 = (2đ?‘Ľ − 1)(đ?‘Ľ + 2) −đ?‘Ľ

Each educated guess at factorising should multiply to get the first and last bits of the original expression, in this case 2đ?‘Ľ 2 and 2. The “double chinâ€? (the curvy lines look like a double chin) multiplies the bits that make the middle bit of the expanded bracket.

+ 4đ?‘Ľ = 3đ?‘Ľ ďƒť

= (2đ?‘Ľ + 1)(đ?‘Ľ − 2)

−3đ?‘Ľ is the correct middle bit so (2đ?‘Ľ + 1)(đ?‘Ľ − 2) is the answer.

đ?‘Ľ − 4đ?‘Ľ = −3đ?‘Ľ ďƒź Solve 2đ?‘Ľ 2 − 11đ?‘Ľ + 12 = 7 by factorising. 2đ?‘Ľ 2 − 11đ?‘Ľ + 12 = 7 2đ?‘Ľ 2 − 11đ?‘Ľ + 5 = 0 (2đ?‘Ľ − 1)(đ?‘Ľ − 5) = 0 2đ?‘Ľ − 1 = 0 đ?‘Ľ − 5 = 0 1 đ?‘Ľ= đ?‘Ľ=5

Rearrange equation to equal 0 to prepare for trick in next step.

Vid 3.2

Factorise using any method. (2đ?‘Ľ − 1)(đ?‘Ľ − 5) = 0 so 2đ?‘Ľ − 1 = 0 and đ?‘Ľ − 5 = 0. 2đ?‘Ľ − 1 = 0 and đ?‘Ľ − 5 = 0 BOTH true for different values of đ?‘Ľ.

2

Turning đ?‘Ľ 2 + đ?‘?đ?‘Ľ + đ?‘? into (đ?‘Ľ + đ?‘?)2 + đ?‘ž or đ?‘Žđ?‘Ľ 2 + đ?‘?đ?‘Ľ + đ?‘? into đ?‘Ž(đ?‘Ľ + đ?‘?)2 + đ?‘ž

Completing the square Write đ?‘Ľ 2 − 8đ?‘Ľ + 7 in the form (đ?‘Ľ + đ?‘?)2 + đ?‘ž Vid 4.2 đ?‘Ľ 2 − 8đ?‘Ľ = (đ?‘Ľ − 4)2 − 16 2 đ?‘Ľ − 8đ?‘Ľ + 7 = (đ?‘Ľ − 4)2 − 16 + 7 = (đ?‘Ľ − 4)2 − 9

Vid 4.1

Write −2đ?‘Ľ 2 − 20đ?‘Ľ + 47 in the form đ?‘Ž(đ?‘Ľ + đ?‘?)2 + đ?‘ž Vid 4.3 −2đ?‘Ľ 2 − 20đ?‘Ľ + 47 = −2(đ?‘Ľ 2 + 10đ?‘Ľ) + 47 = −2[(đ?‘Ľ + 5)2 − 25] + 47 = −2(đ?‘Ľ + 5)2 + 50 + 47 = −2(đ?‘Ľ + 5)2 + 97 Solve đ?‘Ľ 2 − 8đ?‘Ľ + 7 = 16 by completing the square. đ?‘Ľ 2 − 8đ?‘Ľ + 7 = 16 (đ?‘Ľ − 4)2 − 9 = 16 (đ?‘Ľ − 4)2 = 25 đ?‘Ľâˆ’4 = Âą5 đ?‘Ľ = Âą 5 + 4 = 9 and −1 Quadratic formula

đ?‘Ľ 2 + 2đ?‘?đ?‘Ľ = (đ?‘Ľ + đ?‘?)2 − đ?‘? 2 which comes from: (đ?‘Ľ + đ?‘?)2 = đ?‘Ľ 2 + 2đ?‘?đ?‘Ľ + đ?‘? 2 ⇒ (đ?‘Ľ + đ?‘?)2 − đ?‘? 2 = đ?‘Ľ 2 + 2đ?‘?đ?‘Ľ Complete the square for đ?‘Ľ 2 + 10đ?‘Ľ

Vid 4.4

Complete the square. Âą5 means 5 and −5 The quadratic formula. It finds the solution to any quadratic equation.

Vid 5 −đ?‘? Âąâˆšđ?‘?2 −4đ?‘Žđ?‘?

Quadratic formula: if đ?‘Žđ?‘Ľ 2 + đ?‘?đ?‘Ľ + đ?‘? = 0 ⇒ đ?‘Ľ =

Quadratic must be arranged in the form đ?‘Žđ?‘Ľ 2 + đ?‘?đ?‘Ľ + đ?‘? = 0 before using the formula.

2đ?‘Ž

Solve 2đ?‘Ľ 2 + 5 = 11đ?‘Ľ 2đ?‘Ľ 2 + 5 = 11đ?‘Ľ ⇒ 2đ?‘Ľ 2 − 11đ?‘Ľ + 5 = 0 2đ?‘Ľ 2 − 11đ?‘Ľ + 5 = 0 ⇒ substitute đ?‘Ž = 2, đ?‘? = −11, đ?‘? = 5 into đ?‘Ľ = đ?‘Ľ=

−(−11) Âąâˆš(−11)2 −4Ă—2Ă—5 2Ă—2

=

11 Âąâˆš121 − 40 4

=

=

11 Âą 9

11 − 9 2

4

= 4 = 0.5

−đ?‘? Âąâˆšđ?‘?2 −4đ?‘Žđ?‘?

4

=

2đ?‘Ž

11 Âą 9 4

=

11+ 9 4

and

11− 9 4

11 + 9 4 20

= 4 =5

Videos at sickmaths.com > A level > Core 1 > Quadratic Equations

1

Exercise 1a Factorise:: 1. 3. 5.

8đ?‘Ľ 2 − 20đ?‘Ľ 4 − 16đ?‘Ľ 45đ?‘šđ?‘Ľđ?‘§ − 27đ?‘Ľ 2 + 18đ?‘Ľ

Exercise 1đ?‘” Solve:: 2. 4. 6.

3đ?‘Ľđ?‘Ś − 6đ?‘Ś 8xy – 12y đ?‘Ľ 2 đ?‘Ś + 4đ?‘Ľ 2 + đ?‘Ľ 2 đ?‘Ś 2

Exercise 1b Solve by factorising: 1. 3. 5. 7. 9.

đ?‘Ľ 2 + 10đ?‘Ľ = 0 5 − 2đ?‘Ľ = 0 8đ?‘Ľ 2 + 20đ?‘Ľ = 0 12đ?‘Ľ = 6đ?‘Ľ 2 10đ?‘Ľ 2 = 15đ?‘Ľ

đ?‘Ľ 2 − 25 16 − đ?‘Ľ 2 1 − 9đ?‘Ś 2 502 − 492 25đ?‘š2 − 64đ?‘Ą 2 36đ?‘’ 2 − 4đ?‘&#x; 2 đ?‘Ś 4 − 4đ?‘’ 2 12đ?‘? 4 − 75đ?‘’ 2

2. 4. 6. 8. 10.

đ?‘Ľ − 5đ?‘Ľ 2 = 0 21đ?‘Ľ 2 − 7đ?‘Ľ = 0 4đ?‘Ľ 2 − 2đ?‘Ľ = 0 8đ?‘Ľ 2 = −20đ?‘Ľ 4đ?‘Ś 2 − 4đ?‘Ś = 0

2đ?‘Ľ 2 + 7đ?‘Ľ + 3 3đ?‘Ľ 2 + 5đ?‘Ľ + 2 đ?‘Ľ 2 + 4đ?‘Ľ + 3 5đ?‘Ľ 2 − 14đ?‘Ľ − 3 2đ?‘Ľ 2 + 9đ?‘Ľ + 4 4đ?‘Ľ 2 − 4đ?‘Ľ − 3 đ?‘Ľ 2 − 6đ?‘Ľ + 5 6đ?‘Ľ 2 + 23đ?‘Ľ − 4

2. 4. 6. 8. 10. 12. 14. 16.

9đ?‘Ľ 2 − 4 đ?‘Ś 2 − 100 49 − 16đ?‘Ľ 2 5002 − 4992 100đ?‘‘2 − 49đ?‘Ą 2 16đ?‘“ 2 − 9đ?‘Ľ 2 9 − 16đ?‘Ľ 4 2đ?‘š4 − 8đ?‘˘2

2đ?‘Ľ 2 + 12đ?‘Ľ + 10 6đ?‘Ľ 2 + 15đ?‘Ľ + 9 4đ?‘Ľ 2 + 14đ?‘Ľ + 6 20đ?‘Ľ 2 − 50đ?‘Ľ − 30 10đ?‘Ľ 2 − 26đ?‘Ľ − 12

1. 3. 5. 7. 9.

1. 3. 5. 7. 9. 11. 13.

2. 4. 6. 8. 10. 12. 14. 16.

2đ?‘Ľ 2 + 3đ?‘Ľ + 1 7đ?‘Ľ 2 − 8đ?‘Ľ + 1 đ?‘Ľ 2 + 8đ?‘Ľ + 7 2đ?‘Ľ 2 − 11đ?‘Ľ + 5 đ?‘Ľ 2 − 3đ?‘Ľ + 2 2đ?‘Ľ 2 − 7đ?‘Ľ − 5 3đ?‘Ľ 2 + 13đ?‘Ľ + 4 8đ?‘Ľ 2 + 49đ?‘Ľ + 6

2. 4. 6. 8. 10.

2đ?‘Ľ 2 + 6đ?‘Ľ + 4 10đ?‘Ľ 2 + 15đ?‘Ľ + 5 12đ?‘Ľ 2 − 18đ?‘Ľ + 6 20đ?‘Ľ 2 − 110đ?‘Ľ − 50 6đ?‘Ľ 2 + 18đ?‘Ľ + 8

1. 3. 5. 7. 9. 11. 13. 15.

2đ?‘Ľ 2 − đ?‘Ľ − 3 = 0 3đ?‘Ľ 2 + 5đ?‘Ľ + 2 = 0 5đ?‘Ľ 2 − 16đ?‘Ľ + 3 = 0 2đ?‘Ľ 2 − 5đ?‘Ľ = 3 2đ?‘Ľ 2 + 9đ?‘Ľ = −4

đ?‘Ľ 2 + 4đ?‘Ľ + 3 đ?‘Ľ 2 + 6đ?‘Ľ + 5 đ?‘Ľ 2 − 6đ?‘Ľ + 20 4đ?‘Ľ 2 − 16đ?‘Ľ + 18 2đ?‘Ľ 2 − 8đ?‘Ľ + 4

2. 4. 6. 8. 10.

đ?‘Ľ 2 + 12đ?‘Ľ + 40 đ?‘Ľ 2 + 2đ?‘Ľ + 1 đ?‘Ľ 2 − 8đ?‘Ľ + 17 5đ?‘Ľ 2 − 20đ?‘Ľ + 70 3đ?‘Ľ 2 + 18đ?‘Ľ − 9

đ?‘Ľ 2 + 4đ?‘Ľ + 3 = 0 đ?‘Ľ 2 − 6đ?‘Ľ + 5 = 0 đ?‘Ľ 2 − 6đ?‘Ľ = 0 đ?‘Ľ 2 − 6đ?‘Ľ = 7 đ?‘Ľ 2 = 9 − 8đ?‘Ľ 4đ?‘Ľ 2 − 16đ?‘Ľ − 8 = 0 2đ?‘Ľ 2 − 8đ?‘Ľ = 4

2. 4. 6. 8. 10. 12. 14.

đ?‘Ľ 2 + 12đ?‘Ľ + 27 = 0 đ?‘Ľ 2 + 2đ?‘Ľ − 5 = 0 đ?‘Ľ 2 + 8đ?‘Ľ = 0 đ?‘Ľ 2 − 12đ?‘Ľ = 13 đ?‘Ľ 2 = 10đ?‘Ľ − 9 5đ?‘Ľ 2 = 20đ?‘Ľ + 70 3đ?‘Ľ 2 + 18đ?‘Ľ − 9 = 0

2. 4. 6. 8. 10.

2đ?‘Ľ 2 − 9đ?‘Ľ − 5 = 0 đ?‘Ľ 2 − 6đ?‘Ľ + 5 = 0 3đ?‘Ľ 2 + 8đ?‘Ľ + 4 = 0 3đ?‘Ľ 2 = 8đ?‘Ľ − 4 −3đ?‘Ľ 2 − đ?‘Ľ − 4 = 0

2đ?‘Ľ 2 + 7đ?‘Ľ + 3 = 0 3đ?‘Ľ 2 − 11đ?‘Ľ − 4 = 0 đ?‘Ľ 2 + 4đ?‘Ľ + 3 = 0 2đ?‘Ľ 2 = đ?‘Ľ + 3 3đ?‘Ľ 2 = −11đ?‘Ľ + 4 3đ?‘Ľ 2 = 11đ?‘Ľ + 4 9đ?‘Ľ 2 − 4 = 0 đ?‘Ś 2 = 100

2. 4. 6. 8. 10. 12. 14. 16.

3đ?‘Ľ 2 − 8đ?‘Ľ + 4 = 0 2đ?‘Ľ 2 − 11đ?‘Ľ + 5 = 0 2đ?‘Ľ 2 − 7đ?‘Ľ − 5 = 0 đ?‘Ľ 2 + 8đ?‘Ľ + 7 = 0 đ?‘Ľ 2 = −3đ?‘Ľ − 2 −7đ?‘Ľ − 5 = −2đ?‘Ľ 2 16 − đ?‘Ľ 2 = 0 đ?‘š2 − 25 = 0

Exercise 1đ?‘˜ Find length of each shape:

1.

A rectangle has a length 15 đ?‘?đ?‘š less than its width and area 850 đ?‘?đ?‘š2 .

2.

A triangle has length 50 đ?‘?đ?‘š shorter than its perpendicular height and area 300 đ?‘?đ?‘š2 .

3.

A cylinder has length twice its radius and surface area 400 đ?‘?đ?‘š2 .

4.

The surface area of a sphere is 30 times its length (diameter).

5.

A cuboid has a square cross section and its length is twice its height. Its surface area is cuboid is 800 đ?‘š2 . Find its width, accurate to 1 decimal place.

6.

A cuboid has a square cross section and its length is half its height. Its surface area is cuboid is 600 đ?‘š2 . Find its volume, accurate to 1 decimal place.

Exercise 1đ?‘“ Solve by factorising: 1. 3. 5. 7. 9.

(3đ?‘Ľ + 4)(đ?‘Ľ − 5) = 0 (đ?‘š − 4)(đ?‘š + 8) = 0 đ?‘Ľ(4đ?‘Ľ − 1) = 0

Exercise 1đ?‘— Solve using quadratic formula:

Exercise 1đ?‘’ Factorise into the form (đ?‘Žđ?‘Ľ + đ?‘?)(đ?‘?đ?‘Ľ + đ?‘‘) and đ?‘Ž(đ?‘?đ?‘Ľ + đ?‘?)(đ?‘‘đ?‘Ľ + đ?‘’) 1. 3. 5. 7. 9.

2. 4. 6.

Exercise 1đ?‘– Write in the form (đ?‘Ľ + đ?‘?)2 + đ?‘? or đ?‘Ž(đ?‘Ľ + đ?‘?)2 + đ?‘?, then solve.

Exercise 1đ?‘‘ Factorise: 1. 3. 5. 7. 9. 11. 13. 15.

(đ?‘Ľ − 2)(đ?‘Ľ + 3) = 0 (2đ?‘Ś + 1)(đ?‘Ś − 5) = 0 đ?‘?(đ?‘? + 7) = 0

Exercise 1â„Ž Write in the form (đ?‘Ľ + đ?‘?)2 + đ?‘? đ?‘œđ?‘&#x; đ?‘Ž(đ?‘Ľ + đ?‘?)2 + đ?‘?:

Exercise 1đ?‘? Factorise: 1. 3. 5. 7. 9. 11. 13. 15.

1. 3. 5. .

2

Facts about straight line graphs.

2. Straight line graphs

đ?‘Ś=2

Simple straight line graphs (horizontal and vertical ones) đ?‘Ś = đ?‘? graphs are horizontal and pass through (0, đ?‘?) đ?‘Ľ = đ?‘˜ graphs are vertical and pass through (đ?‘˜, 0)

đ?‘Ś = đ?‘šđ?‘Ľ + đ?‘? {

2

−2 đ?‘Ś

Vid 2.1

2

đ?‘Ś

đ?‘Ś = 3đ?‘Ľ − 1

4

horizontal distance between two points

5

đ?‘? = đ?‘Ś intercept (where graph crosses y axis)

5

Vid 3.1

Gradients between two points (coordinates) For (đ?‘Ľ1 , đ?‘Ś1 ) and (đ?‘Ľ2 , đ?‘Ś2 ): đ?‘˘đ?‘? đ?‘‘đ?‘–đ?‘ đ?‘Ąđ?‘Žđ?‘›đ?‘?đ?‘’

across distance

=

−2

3 Vid 2.2

Find gradient and đ?‘Ś intercept for graph 5đ?‘Ś + 2đ?‘Ľ = 20 2 2 5đ?‘Ś + 2đ?‘Ľ = 20 ⇒ đ?‘Ś = − đ?‘Ľ + 4 means gradient = − , đ?‘Ś intercept = 6 5

6

đ?‘Ľ=4

2

vertical distance between two points

Find gradient and đ?‘Ś intercept for graph đ?‘Ś = 3đ?‘Ľ − 1 gradient = 3, đ?‘Ś intercept = −1

Gradient =

đ?‘Ľ

0

−2

Gradient and đ?’š intercept from graph with equation đ?’š = đ?’Žđ?’™ + đ?’„ đ?‘š = gradient =

đ?‘Ś

Vid 1

đ?‘Ś2 −đ?‘Ś1

đ?‘Ľ2 −đ?‘Ľ1

Distance between two points (coordinates) For (đ?‘Ľ2 , đ?‘Ś2 ) and (đ?‘Ľ2 , đ?‘Ś2 ): Pythagoras ⇒ distance = √(đ?‘Ľ2 − đ?‘Ľ1 )2 + (đ?‘Ś2 − đ?‘Ś1 )2 Midpoints between two points (coordinates) For (đ?‘Ľ2 , đ?‘Ś2 ) and (đ?‘Ľ2 , đ?‘Ś2 ): đ?‘Ľ2 + đ?‘Ľ1 đ?‘Ś2 + đ?‘Ś1 2

,

2

đ?‘Ľ 2

đ?‘Ś=− đ?‘Ľ+4

1

5

Find gradient đ?‘” between (3,6) and (7,9) đ?‘Ś 9 3 6 4 đ?‘Ľ 3 7 9−6 3 đ?‘”= = 7−3

Vid 3.2

Midpoint = (middle đ?‘Ľ, middle đ?‘Ś) = (average đ?‘Ľ, average đ?‘Ś) = (

–1

4

Find distance đ?‘‘ between (3,6) and (7,9) đ?‘Ś 9

)

đ?‘‘

3

6

Vid 4

Parallel and perpendicular graphs đ?‘Ž đ?‘Ś đ?‘Ś= đ?‘Ľ+đ?‘?

4 đ?‘Ľ

đ?‘?

đ?‘?

3 7 đ?‘‘2 = (7– 3)2 + (9– 6)2 đ?‘‘ = √(7– 3)2 + (9– 6)2 = 5

đ?‘Ś=− đ?‘Ľ+đ?‘’ đ?‘Ž

đ?‘Ž

đ?‘Ś = đ?‘Ľ+đ?‘‘ đ?‘?

Vid 3.3

đ?‘Ľ Parallel lines have the same gradient đ?‘Ž đ?‘? Perpendicular lines have gradients and − (i.e. negative reciprocals) đ?‘?

Find midpoint đ?‘š between (3,6) and (7,9) đ?‘Ś 9

đ?‘š

7.5 6

đ?‘Ž

đ?‘Ľ

Example 1. đ?‘Ś = 2đ?‘Ľ + 654 and đ?‘Ś = 2đ?‘Ľ − 5 are parallel. 2

3

3

2 1

3

Example 2. đ?‘Ś = đ?‘Ľ + 7 and đ?‘Ś = − đ?‘Ľ + 400 are perpendicular. Example 3. đ?‘Ś = −5đ?‘Ľ + 1 and đ?‘Ś =

5

đ?‘Ľ − 4 are perpendicular.

Finding equation of line from coordinate and gradient Find equation of line that passes through (3,4) and gradient 5. Vid 5.1 Substitute gradient 5 into đ?‘Ś = đ?‘šđ?‘Ľ + đ?‘?: đ?‘Ś = 5đ?‘Ľ + đ?‘? Substitute (3,4) into đ?‘Ś = 5đ?‘Ľ + đ?‘?: 4 = 5 Ă— 3 + đ?‘? so đ?‘? = −11 đ?‘Ś = 5đ?‘Ľ − 11 Find equation of line that passes through (3,4) and gradient 5, give answer in the form đ?‘Žđ?‘Ľ + đ?‘?đ?‘Ś + đ?‘? = 0. Vid 5.2 Substitute gradient 5 into đ?‘Ś − đ?‘Ś1 = đ?‘š(đ?‘Ľ − đ?‘Ľ1 ): đ?‘Ś − đ?‘Ś1 = 5(đ?‘Ľ − đ?‘Ľ1 ) Substitute (3,4) into đ?‘Ś − đ?‘Ś1 = đ?‘š(đ?‘Ľ − đ?‘Ľ1 ): đ?‘Ś − 4 = 5(đ?‘Ľ − 3) đ?‘Ś − 5đ?‘Ľ + 11 = 0 Videos at sickmaths.com > A level > Core 1 > Straight line graphs

7 5

đ?‘š=(

3+7 6+9 2

,

2

) = (5, 7.5)

đ?‘Ś intercepts in example 1 to 3 don’t matter, they could be anything. Example 1: lines have gradient 2 so parallel. 2 3 Example 2: gradients and − are negative 3 2 reciprocals so lines are perpendicular. Example 3: 5 1 1 −5 = − ⇒ negative reciprocal = − − = 1

gradient đ?‘š =

5

đ?‘Śâˆ’đ?‘Ś1 đ?‘Ľâˆ’đ?‘Ľ1

⇒ đ?‘Ś − đ?‘Ś1 = đ?‘š(đ?‘Ľ − đ?‘Ľ1 )

−đ?‘Ś + 5đ?‘Ľ − 11 = 0 would also be correct. 3

5

Exercise 2đ?‘Ž Find gradient and đ?‘Ś intercept for these equations of graphs then draw them: 1

1.

đ?‘Ś =

3.

đ?‘Ś = 3đ?‘Ľ + 2

2

đ?‘Ľâˆ’3

2

2.

đ?‘Ś =

4.

đ?‘Ś = −3đ?‘Ľ + 5

3

đ?‘Ľâˆ’5

5.

đ?‘Ś = đ?‘Ľ+2

6.

đ?‘Ś = −đ?‘Ľ − 3

7.

3đ?‘Ś = 9đ?‘Ľ − 15

8.

3đ?‘Ś = −2đ?‘Ľ − 6

9.

2đ?‘Ľ + đ?‘Ś = 1

10.

đ?‘Ľ+đ?‘Ś=2

11.

8đ?‘Ľ + 4đ?‘Ś = 12

12.

10đ?‘Ľ − 5đ?‘Ś − 15 = 0

13.

5đ?‘Ś =

5

14.

3� = – � + 6

3

đ?‘Ľ − 10

Exercise 2đ?‘‘ Find equations for graphs that pass through the following coordinates and have the following gradients: 1.

Passes through (2, 1) with gradient 5

2.

Passes through (1, 1) with gradient 5.

3.

Passes through (4, 1)

4.

Passes through (1, 4)

5 with gradient − 3

5.

5 (2, −3), gradient 2

6.

5 2

Passes through (−2, 3) with gradient

5 2

7.

Passes through (0, 3) with gradient −2

8.

Passes through (0, 0) with gradient −2

9.

Passes through

10.

Passes through (−2, 3)

9

2

Passes through

with gradient

5 (−2, 3), gradient 2

with gradient −

2 3

Exercise 2đ?‘? Between the coordinate pairs below, find the (đ?‘Ž) gradient (đ?‘?) distance (đ?‘?) midpoint :

Exercise 2đ?‘’ Find equations, in the form đ?‘Ś = đ?‘šđ?‘Ľ + đ?‘? , for graphs that pass through the following pairs of coordinates:

1.

(3,6) and (7,9)

2.

(1,2) and (10,14)

1.

(3,6) and (7,9)

2.

(1,2) and (10,14)

3.

(1,2) and (5,5)

4.

(0,4) and (12,9)

3.

(1,2) and (5,5)

4.

(0,4) and (12,9)

5.

(−1,0) and (4,12)

6.

(−3, −4) and (0,0)

5.

(−1,0) and (4,12)

6.

(−3, −4) and (0,0)

7.

(−3, −4) and (−6, −8)

8.

(−1, −2) and (−10, −14)

7.

(−3, −4) and (−6, −8)

8.

(−1, −2) and (−1, −17)

9.

(3, 4) and (−6, −8)

10.

(−1, 2) and (11, −7)

9.

(3, 4) and (−6, −8)

10.

(−1,2) and (11, −7)

11.

(3, đ?‘Ś) and (đ?‘Ľ, 6) which both lie on 4đ?‘Ś = 3(đ?‘Ľ + 1)

12.

(0, đ?‘Ś) and (đ?‘Ľ, 8) which both lie on đ?‘Ś = đ?‘Ľ 2 − 5đ?‘Ľ + 2

11.

(3, đ?‘Ś) and (đ?‘Ľ, 6) which both lie on 4đ?‘Ś = 3(đ?‘Ľ + 1)

12.

(0, đ?‘Ś) and (đ?‘Ľ, 8) which both lie on đ?‘Ś = đ?‘Ľ 2 − 5đ?‘Ľ + 2

Exercise 2đ?‘? For the graphs below, find equations of graphs that are (đ?‘Ž) parallel with đ?‘Ś intercept 3 (đ?‘?) perpendicular with đ?‘Ś intercept −1: 1.

đ?‘Ś =

3.

3

3

2.

đ?‘Ś =

đ?‘Ś = − đ?‘Ľ+3

4.

đ?‘Ś = − đ?‘Ľâˆ’3

5.

đ?‘Ś = đ?‘Ľ+2

6.

đ?‘Ś = −đ?‘Ľ − 3

7.

2đ?‘Ś = −7đ?‘Ľ − 1

8.

3đ?‘Ś = −4đ?‘Ľ − 35

9.

đ?‘Ś+đ?‘Ľâˆ’9=0

10.

đ?‘Ś+đ?‘Ľ+7=0

5

đ?‘Ľâˆ’2 4 5

5

đ?‘Ľ+2 1

Exercise 2đ?‘“ Find equations, in the form đ?‘Žđ?‘Ľ + đ?‘?đ?‘Ś + đ?‘? = 0, for graphs that pass through the following pairs of coordinates:

1.

(3,6) and (7,9)

2.

(1,2) and (10,14)

3.

(1,2) and (5,5)

4.

(0,4) and (12,9)

5.

(−1,0) and (4,12)

6.

(−3, −4) and (0,0)

7.

(−3, −4) and (−6, −8)

8.

(−1, −2) and (−10, −14)

(3, 4) and (−6, −8)

10.

(−1, 2) and (11, −7)

5

9.

4

3. Indices (Powers)

Vid 2.1

Why 𝑎 𝑥 × 𝑎 𝑦 = 𝑎 𝑥+𝑦 : 𝑎 𝑥 × 𝑎 𝑦 = 𝑎 × 𝑎. .× 𝑎 × 𝑎 × 𝑎. .× 𝑎 = 𝑎 𝑥+𝑦 𝑥 𝑦 𝑥 + 𝑦

Index rules × 𝑎 𝑦 = 𝑎 𝑥+𝑦 Example 1. 72 × 73 = 72+3 = 75

Vid 2.2

Why 𝑎 𝑥 ÷ 𝑎 𝑦 = 𝑎 𝑥−𝑦 : 𝑥

𝑎𝑥

Vid 1.1

Example 2. 8𝑥 2 𝑦 9 × 2𝑥 6 𝑦 4 = 8 × 2 × 𝑥 2 × 𝑥 6 × 𝑦 9 × 𝑦 4 = 16𝑥 8 𝑦13 𝑎 𝑥 ÷ 𝑎 𝑦 = 𝑎 𝑥−𝑦 Example 1. 512 ÷ 53 = 512−3 = 59 Example 2.

8𝑥 2 𝑦 9

÷

2𝑥 6 𝑦 4

=

𝑎×𝑎×𝑎…×𝑎×𝑎×𝑎…×𝑎 𝑎×𝑎×𝑎…×𝑎

𝑦

Vid 1.2

8𝑥 2 𝑦 9 2𝑥 6 𝑦 4

8

𝑥2

2

𝑥6

= ×

×

𝑦9 𝑦4

Vid 2.3

=

4𝑥 −4 𝑦 5

= 𝑎 𝑥−𝑦

𝑥– 𝑦

Why (𝑎 𝑥 )𝑦 = 𝑎 𝑥𝑦 : 𝑦

𝑦

(𝑎 𝑥 )𝑦 = 𝑎 𝑥 × 𝑎 𝑥 … × 𝑎 𝑥 = 𝑎 𝑥+𝑥…+𝑥 = 𝑎 𝑥𝑦

(𝑎 𝑥 )𝑦 = 𝑎 𝑥𝑦

Example 1. (102 )3 = 102×3 = 106 = 1000000

Vid 1.3 Vid 2.4

Example 2. (3𝑥𝑦 5 )2 = 32 𝑥 2 (𝑦 5 )2 = 9𝑥 2 𝑦10 𝑎0 = 1

Example 1. 50 = 1

Vid 1.4

Example 2. 130 = 1 𝑎 −𝑥

𝑏 𝑥

3 −2

𝑏𝑥

( ) = ( ) = 𝑥 Example 1. ( ) 𝑏 𝑎 𝑎 5

5 2

52

3

32

=( ) = 4 −3

Example 2. 4−3 = ( ) 1

𝑥

𝑎𝑥 ÷ 𝑎𝑦 =

𝑦

𝑎𝑦 = ( √𝑎)

2

𝑥

2

Vid 1.5

9

1 3

1

4

43

=

1 64

𝑎

Example 1. 83 = (√8) = 4 1

25

=( ) =

2

3

=

Explaining negative powers: 𝑎3 = 𝑎 × 𝑎 × 𝑎 ÷𝑎 𝑎2 = 𝑎 × 𝑎 ÷𝑎 𝑎1 = 𝑎 ÷𝑎 𝑎0 = 1 ÷𝑎 1 𝑎−1 = 𝑎 ÷𝑎 1 𝑎−2 = 2 𝑎 ÷𝑎 1 𝑎−3 = 3

Vid 1.6

1

Example 2. 162 = (√16) = √16 = ±4

𝑎−𝑛 =

1 𝑎𝑛

𝑥

, if 𝑎 =

𝑦

𝑥 −𝑛

⇒( ) 𝑦

=

1 𝑥 𝑛 ( ) 𝑦

=

𝑥𝑛 𝑦𝑛

Combining index rules 2

2

Solve 8− 3 . 8− 3 =

1 2 83

=

1 3

2

( √8)

=

1 4

Vid 2

Re-writing in index form Write

16

+

3𝑥 2 −2𝑥

𝑥2 √𝑥 16 3𝑥 2 −2𝑥 𝑥2

=

+

16𝑥 −2

√𝑥

+

3𝑥 2 1

𝑥2

=

16𝑥 −2

in the form 𝐴𝑥 𝑚 + 𝐵𝑥 𝑛 + 𝐶𝑥 𝑝

Vid 3

16𝑥 −2 = 16 × −

3 2

2𝑥

3𝑥 2 −2𝑥

1

𝑥2

+ 3𝑥 − 3𝑥

1 2

An advance trick Solve 8 𝑥+2 = 4 𝑥 Vid 4 8 𝑥+2 = 4 𝑥 3 ⇒ (2 )𝑥+2 = (22 )𝑥 ⇒ 23𝑥+6 = 22𝑥 ⇒ 3𝑥 + 6 = 2𝑥 𝑥 = −6

Videos at sickmaths.com > A level > Core 1 > Indices (Powers)

√𝑥 3𝑥 2 1 𝑥2

−

=

2𝑥 1 𝑥2

3𝑥 2 √𝑥

−

1 𝑥2

=

2𝑥 √𝑥

=

1

16 𝑥2 3𝑥 2 1

−

𝑥2

2𝑥 1

𝑥2 1

3

1

= 3𝑥 2−2 − 2𝑥1−2 = 3𝑥 2 − 2𝑥 2

Convert everything to same base i.e. 2 Using (𝑎 𝑥 )𝑦 = 𝑎 𝑥𝑦 𝑎𝑥 = 𝑎𝑦 ⇒ 𝑥 = 𝑦

5

Exercise 3a Find the value of following (as an integer or fraction):

1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21.

34 26 50 110 04 121 134 400 521 2−1 5−2

23.

2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22.

63 43 54 180 01 20 13 07 51 2−3 1−3

( )

24.

( )

25.

( )

26.

( )

27.

( )

28.

( )

5 3 3 5 −2 3 3 −1 4

7 2

𝑎3 × 𝑎2 𝑏 3 × 𝑏 −2 4𝑎3 × 2𝑎2 6𝑥 × 𝑥 −5 2𝑎−3 × 𝑎2 × 2𝑎9 3𝑎3 𝑏 2 × 2𝑎𝑏 −2 𝑎5 ÷ 𝑎2 𝑏 3 ÷ 𝑏 −2 8𝑎3 ÷ −2𝑎2 6𝑥 ÷ 𝑥 −5 14𝑎−3 𝑥 2 ÷ 2𝑎9

2 3 0

3.

83

5.

252

7.

( )

9.

( )

11.

25− 2

13.

8− 3

15.

83

2. 4. 6. 8. 10. 12. 14. 16. 18. 20. 22.

𝑏6 × 𝑏7 𝑛−9 × 𝑛−2 10𝑣 3 × 2.5𝑣 66 7𝑗 9 × −𝑗 −4 8𝑐 7 × 2𝑐 −2 × 𝑐 −1 3𝑥 4 𝑧 × 𝑧 5 𝑥 4 −𝑛3 ÷ 𝑛8 𝑥 −5 ÷ 𝑥 −2 𝑑3 ÷ 4𝑑2 9𝑥 8 ÷ −𝑥 −5 12𝑎−3 𝑥 2 ÷ 3𝑎9

24.

18𝑎3 𝑏 2 6𝑎𝑏 −3

25.

9𝑎3 × 3𝑏 2 3𝑎 × 𝑏 −3

26.

8𝑚8 × 2𝑛 4𝑚 × 𝑛−5

37.

39.

40.

83

4.

325

6.

1253

8.

( )

10.

( )

12.

16− 2

14.

8− 3

16.

8− 3

2.

5 𝑥 = 125

4.

3𝑥 =

6.

( ) =

8.

( ) =

10.

3𝑥 =

12. 14. 16.

25 𝑥 = 5 16 𝑥 = 1 3.1 𝑥 = 3.1

18.

2𝑥 =

16 𝑥−2 = 4 𝑥 162𝑥−1 = 82𝑥 5 𝑦+1 = 25 𝑦+4

1

1

1

1

36

1 2

49

1

64

1 2

25

1

36 − 2 49

1

81 − 2 4

1

1

3

28. 30. 32. 34. 36.

−(𝑥 20 )−2 (2𝑒)5 (3𝑒 7 𝑓 8 )4 3(𝑚8 𝑥 −7 )7 (30𝑥 4 )−2 ÷ 5𝑥

1.

7 𝑥 = 49

3.

7𝑥 =

5.

( ) =

7.

( )

9.

2𝑥 =

11. 13. 15.

2 𝑥 = 512 5𝑥 = 5 8𝑥 = 1

17.

8𝑥 =

19. 21. 23.

4 𝑥+2 = 8 𝑥 27 𝑥−5 = 92𝑥 2 𝑥+2 = 16 𝑥+3

20. 22. 24.

6

26.

25. (𝑟 2 )3 × 𝑟 2 𝑟4

2.

5

2

Exercise 3 d Find x in the following:

27𝑎3 𝑏 2 3𝑎𝑏 −2

(𝑥 2 )3 (4𝑥 2 )3 (7𝑚5 𝑥 3 )2 2(7𝑚5 𝑥 3 )2 (4𝑟 2 )3 ÷ −4𝑟 2

5

83

4

23.

27. 29. 31. 33. 35.

2

1.

2 3 −3

Exercise 3b Simplify: 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21.

Exercise 3 c Find equations for graphs with that pass through the following points with the following gradients:

38.

(2𝑟 2 )3 × 𝑟 2 4𝑟 −4

40.

(2𝑥 2 )4 × (3𝑥 2 )3 2𝑥 −10 × (3𝑥 2 )3

42.

(3𝑟)3 × 𝑟 9 −𝑟 −3 (3𝑥 3 )2 × 𝑥 2 × 𝑥 5 9𝑥 −1 −(𝑥 2 )5 × −(6𝑥 4 )3 3𝑥 −1 × (2𝑥)3

1 49

3 𝑥

9

2

4

2 −𝑥

=

5

25 4

1 8

1 8

= 𝐴𝑒 𝑥

𝑒2

27.

3𝑒−2𝑒 3

29.

2√𝑛

31.

𝑦 3 −2𝑦 2

√𝑒

𝑛2

= 3𝑒 𝑦 + 𝐶𝑒 𝑥

= 𝐴𝑛 𝑥

𝑦3

= 𝑚 − 𝑛𝑦 𝑥

28. 30. 32.

1 27

4 𝑥

64

3

27

3 𝑥

16

2

81

1 √𝑡

1 81

1 16

= 𝑡𝑥

4𝑚+𝑚3 √𝑚 7√𝑛 𝑛−3 2𝑦 2 3√𝑦

= 𝐴𝑚𝑦 + 𝑚 𝑥

= 𝐴𝑛 𝑥 = 𝐴𝑦 𝑒

6

Vid 1.1

4. Surds

In all these examples, we are simplifying surd expressions, which just means writing them in their shortest form.

Surd rules 2

2

√đ?‘Ž Ă— √đ?‘Ž = (√đ?‘Ž) = đ?‘Ž

Example 1. (√9) = 9

Vid 1.2

Square-rooting and squaring are inverse functions so they undo each other.

2

Example 2. √5 Ă— √5 = (√5) = 5

You can split surds into products of the original. One of the surds should have a square number so that it can be simplified.

√đ?‘Ž2 đ?‘? = √đ?‘Ž2 Ă— √đ?‘? = đ?‘Žâˆšđ?‘? Example 1. √18 = √9 Ă— √2 = 3 Ă— √2 = 3√2 Vid 1.3

Example 2. √75 = √25 Ă— √3 = 5√3 đ?‘Ž

√đ?‘? =

√đ?‘Ž

9

3

4

2

Example 1. √ =

√đ?‘?

Prime factor trees can help find factors that are square numbers e.g. Using prime factor tree: 135 = 3 Ă— 3 Ă— 3 Ă— 5 ⇒ 9 is square factor of 135 so √135 = √9 Ă— √15 = 3√15

Vid 1.4

Example 2. √

12 25

=

√4Ă—âˆš3 5

=

2√3 5

Just like 3đ?‘Ľ + 9đ?‘Ľ = 12đ?‘Ľ.

đ?‘Žâˆšđ?‘Ľ + đ?‘?√đ?‘Ľ = (đ?‘Ž + đ?‘?)√đ?‘Ľ Example 1. 3√5 + 9√5 = 12√5 Vid 2.1

Example 2. √147 − √27 = 7√3 − 3√3 = 4√3 đ?‘Žâˆšđ?‘Ľ Ă— đ?‘?√đ?‘Ľ = đ?‘Žđ?‘?đ?‘Ľ

Example 1. 3√5 Ă— 9√5 = 3 Ă— 9 Ă— √5 Ă— √5 = 27 Ă— 5 = 135

Vid 2.2

Example 2.

đ?‘Žâˆšđ?‘Ľ đ?‘?√đ?‘Ľ

đ?‘Ž

=

Example 1.

đ?‘?

(7 + √3)(1 − √3) = 7 − 7√3 + √3 − 3 = 4 − 6√3 3√2 5√2

=

3

4√2 √8

=

4√2 2√2

=2

Rationalise denominators: Basic method: đ?‘˜

=

√đ?‘Ž

đ?‘˜ √đ?‘Ž

Ă—

√đ?‘Ž √đ?‘Ž

=

đ?‘˜âˆšđ?‘Ž

Example 1.

đ?‘Ž

Example 2. Advance method: đ?‘˜ √đ?‘ŽÂąâˆšđ?‘?

=

đ?‘˜ √đ?‘ŽÂąâˆšđ?‘?

Ă—

√đ?‘Žâˆ“√đ?‘?

Ă—

√7−√2

√đ?‘Žâˆ“√đ?‘?

=

đ?‘˜(√đ?‘Žâˆ“√đ?‘?)

5 √3

=

4 5√2

5 √3

=

Ă—

4 5√2

√3 √3

Ă—

=

√2 √2

5√3 Vid 3.2 3

=

4√2 3

Vid 3.3

√7+√2

=

5 √7+√2

√7−√2

=

5(√7−√2) 7−2

=

5(√7−√2) 5

Irrational numbers are numbers like Ď€ that go on for an infinite number of decimal places and do not follow a pattern. More precisely, irrational numbers can’t be put into fractions. Since square roots of numbers (except that of square numbers) are irrational, rationalising the denominator just means getting rid of the square root from the bottom of the fraction. Using √đ?‘Ž Ă— √đ?‘Ž = (√đ?‘Ž)2 = đ?‘Ž. Simplifying with surds means rationalising the denominator and simplifying the fraction.

∓ is the opposite of Âą i.e. đ?‘Ž Âą đ?‘? ∓ đ?‘? means đ?‘Ž + đ?‘? − đ?‘? or đ?‘Ž − đ?‘? + đ?‘?

đ?‘Žâˆ’đ?‘?

Example 1. 5

√147 = √49 Ă— √3 = 7√3 and √27 = √9 Ă— √3 = 3√3. Simplify surds into multiples of a surd, in this case √3 , so that you can add or subtract them.

Cancel by √2. Vid 3.1

5

Vid 2.3

Example 2.

Surds are things with square roots in them e.g. √8. This topic is just about doing tricks with square roots.

Using the difference of two squares. = √7 − √2

Vid 3.4

Example 2. 1 5−√3

=

1 5−√3

Ă—

5+√3 5+√3

=

5+√3 25−3

=

5+√3 22

Videos at sickmaths.com > A level > Core 1 > Surds

7

Exercise 4𝑎 Find the value of the following (as a fraction or decimal): 1.

(√16)2

3. 5. 7. 9. 11. 13. 15. 17.

(√9) (√5)2 (√7)2 √7 × √7 √5 × √5 √3 × √3 (2√7)2 (0√2)2

2

2. 4. 6. 8. 10. 12. 14. 16. 18.

(√36)2 (√4)2 (√17)2 (√6)2 √12 × √12 √8 × √8 √9 × √9 (3√4)2 (5√8)2

Exercise 4𝑑 Simplify: 1. 3. 5. 7. 9. 11. 13. 15.

Exercise 4𝑏 Find 𝑎: 1. 3. 5. 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35.

16 25

36

√

9

= =

𝑎 5 6

2. 4. 6. 8. 10. 12. 14. 16.

𝑎

3√5 + 9√5 = 𝑎√5 3√3 − 17√3 = 𝑎√3 √75 − √27 = 𝑎√3 3√2 + 6√32 − √18 = 𝑎√2 7√2 = √𝑎 3√7 = √𝑎 2√2 = √𝑎 (√2 + 3)(√2 − 5) (√3 + 5)(√3 − 5) (√3 + √5)(√3 + √5)

18. 20. 22. 24. 26. 28. 30. 32. 34. 36.

√72 = 𝑎√2 √800 = 𝑎√2 √28 = 𝑎√7 √12 = 𝑎√3 √32 = 𝑎√2 √200 = 𝑎√2 36

√

49

81

√

4

= =

𝑎 7

5√3 + 2√48 √18 + √200 √60 − √135 √180 − √20 + √25

1. 3. 5. 7. 9.

9 𝑎

6√2 + √2 = 𝑎√2 6√3 − √3 = 𝑎√3 √32 − √8 = 𝑎√2 3√12 + 6√3 − √27 = 𝑎√3 2√5 = √𝑎 2√7 = √𝑎 10√3 = √𝑎 (√2 + 3)(√2 − 5) (1 + √5)(4 − √5) (√3 + 5)(√3 + 5)

Exercise 4𝑐 Simplify: 1. 3. 5. 7.

2.

2√3

8.

5√3 6√5 4√5 √12 √3 √18 √50 10√6 √300

4. 6.

10. 12. 14. 16.

2√3 × 7√3 √5 × √45 √20 × 2√5 8√7 6√7 3√2 9√2 √32 √2 √8 √50 20√8 √400

Exercise 4𝑒 Rationalise the denominator:

√18 = 𝑎√2 √500 = 𝑎√5 √45 = 𝑎√5 √27 = 𝑎√3 √98 = 𝑎√2 √800 = 𝑎√2 √

5√2 × 2√2 √3 × √45 √50 × 3√2

11. 13.

2

2.

√5 7 √2 √5 √6

6. 8.

2√5 3√6 2√5 3√6 2√18 3−√6 √7 7 3−√6

17.

3−√6

21.

√5

√3

15.

19.

4.

√7 √8 √7+√2 −6

1 √3 1

10. 12. 14.

√2 √4 √3 8√5 5√7 7√5 √2 5√18 √6 – 3 √7 7

16.

√6 – 3

18.

3 + √6

20.

√7−√2

20.

√6+2

√6 + 3

√7 √3 √8

Exercise 4𝑓 Simplify: 2. 4. 6. 8.

4√5 + 2√45 √162 + √8 √12 − √75 √18 − √50 + √8

1. 3. 5.

10 √5 √8 √7−√5 √2 √8+3

1. 3. 5.

6 √3 √8 3+√5 √12 3−√5

8

5. Simultaneous equations Simultaneous equations by elimination Solve the simultaneous equations 2đ?‘Ś + đ?‘Ľ = 9 đ?‘Ś+đ?‘Ľ =4 2đ?‘Ś + đ?‘Ľ = 7 1 − đ?‘Ś+đ?‘Ľ =4 2 đ?‘Ś =3 Substitute đ?‘Ś = 3 into 2 : (3) + đ?‘Ľ = 4 đ?‘Ľ=1

Vid 2.1

Solve the simultaneous equations đ?‘“ − 2đ?‘” = 1 2đ?‘“ + 2đ?‘” = 14 đ?‘“ − 2đ?‘” = 1 1 + 2đ?‘“ + 2đ?‘” = 14 2 3đ?‘“ = 15 đ?‘“ = 5 Substitute đ?‘“ = 5 into 1 : (5) − 2đ?‘” = 1 đ?‘”=2 Solve the simultaneous equations 3đ?‘Ž − đ?‘? = 6 đ?‘Ž + 2đ?‘? = 2 3đ?‘Ž − đ?‘? = 6 1 đ?‘Ž + 2đ?‘? = 2 2 1 Ă— 2 ⇒ 6đ?‘Ž − 2đ?‘? = 12 3 + đ?‘Ž + 2đ?‘? = 2 2 7đ?‘Ž = 14 đ?‘Ž = 2 Substitute đ?‘Ž = 2 into 2 : (2) + 2đ?‘? = 2 đ?‘?=0

Vid 1

Vid 2.2

Vid 2.3

Simultaneous equations by substitution Solve the simultaneous equations đ?‘Ś − đ?‘Ľ = 1 Vid 3 đ?‘Ś 2 + đ?‘Ľ 2 = 25 đ?‘Śâˆ’đ?‘Ľ = 1 1 ⇒đ?‘Ś = 1+đ?‘Ľ 3 đ?‘Ś 2 + đ?‘Ľ 2 = 25 2 Substitute 3 into 2 : (1 + đ?‘Ľ)2 + đ?‘Ľ 2 = 25 1 + 2đ?‘Ľ + đ?‘Ľ 2 + đ?‘Ľ 2 = 25 2đ?‘Ľ 2 + 2đ?‘Ľ − 24 = 0 đ?‘Ľ 2 + đ?‘Ľ − 12 = 0 (đ?‘Ľ + 4)(đ?‘Ľ − 3) = 0 đ?‘Ľ = −4 đ?‘Ľ = 3 Substitute đ?‘Ľ = −4 into 3 ⇒ đ?‘Ś = −3 Substitute đ?‘Ľ = 3 into 3 ⇒ đ?‘Ś = 4 Solving simultaneous equations for graphs Find coordinates where graphs đ?‘Ś = 4đ?‘Ľ + 5 and đ?‘Ś = đ?‘Ľ 2 intersect. Vid 4.1 đ?‘Ś = 4đ?‘Ľ + 5 1 and đ?‘Ś = đ?‘Ľ 2 2 Substitute 2 into 1 : đ?‘Ľ 2 = 4đ?‘Ľ + 5 đ?‘Ľ 2 − 4đ?‘Ľ − 5 = 0 Vid 4.2 (đ?‘Ľ + 1)(đ?‘Ľ − 5) = 0 đ?‘Ľ = −1 đ?‘Ľ = 5 Substitute đ?‘Ľ = −1 into 2 ⇒ đ?‘Ś = 1 Substitute đ?‘Ľ = 5 into 2 ⇒ đ?‘Ś = 25 ⇒ (−1,1) ⇒ (5,25)

Using more than one equation together to solve more than one variable (letter). For every variable you want to solve, you will need an equation, so to solve 3 variables you will need 3 equations solved simultaneously. To understand this, if you roll two die and get numbers đ?‘Ľ and đ?‘Ś and then you’re told đ?‘Ľ + đ?‘Ś = 8, that wouldn’t be enough to know đ?‘Ľ and đ?‘Ś. If you’re then told đ?‘Ľđ?‘Ś = 15, then you would know đ?‘Ľ = 3 and đ?‘Ś = 5. Hence, you needed two equations to find đ?‘Ľ and đ?‘Ś. Subtract equations when you have the same thing in both, in this case đ?‘Ľ . This works because subtracting the left sides of two equations should always be same as subtracting right sides of those equations. Add equations when you have the opposite thing in both, in this case −2đ?‘” and +2đ?‘”. This works because the sum of the right sides of two equations should equal the sum of the left sides of those equations. 3đ?‘Ž − đ?‘? = 6 ⇒ 2(3đ?‘Ž − đ?‘?) = 2 Ă— 6 ⇒ 6đ?‘Ž − 2đ?‘? = 12 Double equation 1 to get equation 3 . Now we can add the equations since −2đ?‘? and 2đ?‘? are opposites. Solve linear and quadratic simultaneous equations mostly by substitution method. Rearrange the linear equation to make something the subject, in this case đ?‘Ś, which means đ?‘Ś can be replaced with 1 + đ?‘Ľ in the quadratic equation. Use (đ?‘Ž + đ?‘?)2 = đ?‘Ž2 + 2đ?‘Žđ?‘? + đ?‘? 2 to expand (1 + đ?‘Ľ)2 Simultaneous equations with a linear and a quadratic will produce two pairs of answers, in this case đ?‘Ľ = −4, đ?‘Ś = −3 and đ?‘Ľ = 3, đ?‘Ś = 4. You can use either elimination or substitution method, in this case we’ve use substitution. Solving the same equations graphically simply means reading coordinates of intersection: đ?‘Ś đ?‘Ś = đ?‘Ľ2 đ?‘Ś = 4đ?‘Ľ + 5

(5,25) (– 1,1) đ?‘Ľ i.e. intersection (5, 25) ⇒ đ?‘Ľ = 5 when đ?‘Ś = 25 and intersection (−1,1) ⇒ đ?‘Ľ = −1 when đ?‘Ś = 1 Videos at sickmaths.com > A level > Core 1 > Simultaneous equations

9

Exercise 5𝑎 Solve these simultaneous equations using elimination method to 1d.p:

Exercise 5𝑐 Find coordinates of where the following graphs intersect:

1.

2𝑦 + 𝑥 = 10 𝑦+𝑥 =6

2.

2𝑦 − 𝑥 = 10 𝑦−𝑥 =6

1.

𝑦 = 2𝑥 + 3 𝑦 = −𝑥 + 9

2.

𝑦 = 2𝑥 + 3 𝑦 = −𝑥 + 9

3.

−2𝑡 + 2𝑘 = 90 2𝑡 + 𝑘 = 30

4.

−2𝑦 + 6𝑥 = 8 −2𝑦 + 𝑥 = 13

3.

2𝑦 + 𝑥 = 9 𝑦+𝑥 =4

4.

𝑦−𝑥 =8 2𝑦 − 𝑥 = 18

5.

𝑐 + 3𝑑 = 33 𝑐 + 𝑑 = 27

6.

𝑘 + 3𝑚 = 10 −3𝑚 + 2𝑘 = 20

5.

3𝑚 − 𝑥 = 12 𝑚−𝑥 =4

6.

4𝑓 + 𝑔 = 45 𝑓+𝑔 =9

7.

2𝑎 − 𝑏 = 8 2𝑎 + 2𝑏 = 2

8.

2𝑎 − 2𝑏 = 12 𝑎+𝑏 =4

7.

2𝑘 − 2𝑡 = 9 𝑘 − 2𝑡 = 4

8.

2𝑎 − 3𝑏 = 7 2𝑎 − 2𝑏 = 8

9.

2𝑎 − 𝑏 = 0.5 𝑎 − 𝑏 = 0.4

10.

3𝑎 + 2𝑏 = 1.7 𝑎 − 2𝑏 = 0.3

9.

−𝑏 + 2𝑎 = 7 𝑎−𝑏 = 6

10.

3𝑐 − 𝑚 = 10 𝑐−𝑚 = 4

11.

9𝑐 + 3𝑑 = 33 𝑐+ 𝑑=7

12.

5𝑎 − 𝑏 = 17 2𝑎 + 3𝑏 = 0

11.

−2𝑑 + 4𝑒 = 28 2𝑑 + 3𝑒 = 14

12.

3ℎ + 2𝑎 = 8 −3ℎ + 𝑎 = 4

13.

𝑐−𝑚 = 6 𝑚 + 3𝑐 = 10

14.

3𝑑 + 𝑐 = 2 2𝑐 + 3𝑑 = 7

13.

9𝑐 + 3𝑑 = 33 𝑐+ 𝑑 = 7

14.

5𝑏 + 6𝑐 = 34 𝑏 − 2𝑐 = −6

15.

𝑎−𝑏 =8 2𝑎 + 2𝑏 = 12

16.

3𝑥 + 𝑦 = 10 2𝑥 + 3𝑦 = 9

15.

3𝑥 + 2𝑦 = 13 2𝑥 + 3𝑦 = 7

16.

5𝑝 + 4𝑞 = 34 2𝑝 + 2𝑞 = 14

17.

5𝑑 + 4𝑒 = 20 3𝑑 + 3𝑒 = 12

18.

5𝑑 + 6𝑒 = −16 3𝑑 + 8𝑒 = −14

17.

6𝑥 − 3𝑦 = 3 5𝑥 − 2𝑦 = 4

18.

5𝑥 − 3𝑦 = 1 4𝑥 + 2𝑦 = 14

19.

5𝑎 − 3𝑏 − 14 = 0 2𝑎 + 3𝑏 = 0

20.

5𝑎 − 4𝑏 − 14 = 16 2𝑎 + 3𝑏 = 35

19.

ℎ = 13 − 4𝑖 3𝑖 + 2ℎ = 16

20.

2𝑚 + 3𝑛 = −8 3𝑚 + 2𝑛 = −12

21.

𝑦=7+𝑥 4𝑦 = 25 + 𝑥

22.

𝑥 = 11 − 2𝑦 3𝑦 = 16 − 𝑥

21.

11𝑥 − 3𝑦 = 8 9𝑥 + 4𝑦 = 13

22.

𝑦 + 1 = 2𝑥 𝑦 = 𝑥+2

23.

2𝑦 + 𝑥 2 = 10 𝑦 + 𝑥2 = 6

24.

2𝑦 2 + 6𝑥 = 8 2𝑦 2 + 𝑥 = 13

23.

𝑚 =𝑛+7 𝑚 + 1 = 2𝑛

24.

5𝑎 − 4 = 4𝑏 2𝑏 + 2 = 𝑎

25.

9𝑐 + 3𝑑2 = 33 𝑐 + 𝑑2 = 7

26.

𝑘 2 + 3𝑚 = 10 −3𝑚 + 2𝑘 2 = 20

Exercise 5𝑏 Solve these simultaneous equations using substitution method:

Exercise 5𝑑 Find coordinates of where the following graphs intersect:

1.

𝑦−𝑥 =7 𝑦 2 + 𝑥 2 = 169

2.

𝑚 =𝑛+2 𝑚2 + 𝑛2 = 73

1.

𝑦 = 3𝑥 − 2 𝑦 = 𝑥2

2.

𝑦 = 11𝑥 + 5 𝑦 = 2𝑥 2

3.

2𝑥 + 𝑦 = 4 𝑦2 + 𝑥 2 = 5

4.

𝑚+𝑛 =5 𝑚2 + 𝑛2 = 13

3.

𝑦 = −4𝑥 − 3 𝑦 = 𝑥2

4.

𝑦 = 6𝑥 − 5 𝑦 = 𝑥2

5.

𝑚 =𝑛−3 𝑚2 + 𝑛2 = 17

6.

𝑦 + 𝑥 = 10 𝑦𝑥 = 24

5.

𝑦 = 𝑥+3 𝑦 = 2𝑥 2

6.

𝑦 = 7𝑥 + 5 𝑦 = 2𝑥 2

7.

2𝑦 − 𝑥 = 4 𝑦 2 + 2𝑥 2 = 17

8.

𝑦 = 2𝑥 + 3 𝑦 2 + 𝑥 2 = 90

7.

2𝑥 + 𝑦 = 4 𝑦2 + 𝑥 2 = 5

8.

𝑦 = 2𝑥 + 3 𝑦 2 + 𝑥 2 = 90

10

6. Sketching graphs

Vid 1.1

Shapes of important graphs Quadratic graphs: Vid 2.1 đ?‘Ś Quadratic graphs have:

đ?‘Ľ

Cubic graphs: đ?‘Ś

   

đ?‘Žđ?‘Ľ 2

Equation đ?‘Ś = + đ?‘?đ?‘Ľ + đ?‘?, đ?‘Ž ≠0 One vertex Vertical line of symmetry through vertex đ?‘Ś = đ?‘Ľ 2 (see dotted graph) is most used quadratic graph with vertex at (0,0).

Vid 2.2

đ?‘Ľ

Reciprocal graphs: đ?‘Ś đ?‘Ś=đ?‘Ľ

Cubic graphs have:  Equation đ?‘Ś = đ?‘Žđ?‘Ľ 3 + đ?‘?đ?‘Ľ 2 + đ?‘?đ?‘Ľ + đ?‘‘, đ?‘Ž ≠0  Two vertices except for graphs like đ?‘Ś = đ?‘Žđ?‘Ľ 3 , đ?‘Ž ≠0 (see dotted graph) which flatten out at (0,0).

e.g. đ?‘Ś = 3đ?‘Ľ 2 + 0đ?‘Ľ − 9 A vertex is a point on a graph where its gradient changes from positive to negative (or vice versa).

e.g. đ?‘Ś = 1đ?‘Ľ 3 − 2đ?‘Ľ 2 + 0đ?‘Ľ + 7

Vid 2.3

Reciprocal graphs have: đ?‘˜  Equation đ?‘Ś = , đ?‘˜ ≠0

e.g. đ?‘Ś =

1 đ?‘Ľ

đ?‘Ľ

 �

Sketching graphs means drawing graphs roughly, taking into account their vertex, asymptotes and where they cross the axis. You also have to be aware how graphs transform as their equations change.

Asymptotes (lines which it gets closer to but never reaches) at the đ?‘Ľ and đ?‘Ś axes. Symmetrical about đ?‘Ś = đ?‘Ľ and đ?‘Ś = −đ?‘Ľ.



đ?‘Ś = −đ?‘Ľ

What are functions? Vid 3 Functions are algebraic expressions with a name e.g.  �(�) is a function of x i.e. an expression in �. The function is called �  �(�) is a function of t i.e. an expression in �. The function is called �

Functions usually have one letter names. đ?‘“ is the most popular name for a function. (a) Replace đ?‘Ľ with 5. (b) Replace đ?‘Ľ with đ?‘Ľ + 1 then multiply entire function by 3.

đ?‘“(đ?‘Ľ) = đ?‘Ľ + đ?‘Ľ 3 . Find in terms of đ?‘Ľ: (a) đ?‘“(5) (b) 3đ?‘“(đ?‘Ľ + 1) (a) đ?‘“(5) = 5 + 53 = 130 (b) 3đ?‘“(đ?‘Ľ + 1) = 3[(đ?‘Ľ + 1) + (đ?‘Ľ + 1)3 ] đ?‘“(đ?‘Ľ) = đ?‘Ľ 2 . Find in terms of đ?‘“(đ?‘Ľ): (a) 5(đ?‘Ľ + 1)2 (b) đ?‘Ľ 2 − 10đ?‘Ľ + 26 (a) 5(đ?‘Ľ + 1)2 = 5đ?‘“(đ?‘Ľ + 1) (b) đ?‘Ľ 2 − 10đ?‘Ľ + 26 = (đ?‘Ľ − 5)2 + 1 = đ?‘“(đ?‘Ľ − 5) + 1 Finding where graphs cross or touch the coordinate axes Vid 4.1 đ?‘Ś đ?‘Ś = đ?‘“(đ?‘Ľ) For graphs of đ?‘Ś = đ?‘“(đ?‘Ľ):  đ?‘Ś = đ?‘“(đ?‘Ľ) crosses y axis when đ?‘Ś = đ?‘“(0) đ?‘Ľ  đ?‘Ś = đ?‘“(đ?‘Ľ) crosses x axis when 0 = đ?‘“(đ?‘Ľ) 

Graph touches x axis at duplicate solutions for 0 = đ?‘“(đ?‘Ľ)

Sketch graph of đ?‘Ś = (đ?‘Ľ − 2)(đ?‘Ľ + 3)2 . Vid 4.1 Substitute đ?‘Ś = 0 into đ?‘Ś = (đ?‘Ľ − 2)(đ?‘Ľ + 3)2 ⇒ (đ?‘Ľ − 2)(đ?‘Ľ + 3)2 = 0 ⇒ đ?‘Ľ = 2, −3, −3 i.e. crosses đ?‘Ľ axis at 2, touches đ?‘Ľ axis at −3. Substitute đ?‘Ľ = 0 into đ?‘Ś = (đ?‘Ľ − 2)(đ?‘Ľ + 3)2 ⇒ (0 − 2)(0 + 3)2 = −18 i.e. graph crosses đ?‘Ś axis at −18: đ?‘Ś = (đ?‘Ľ − 2)(đ?‘Ľ + 3)2

đ?‘Ś −3

đ?‘Ľ 2

−18 Videos at sickmaths.com > A level > Core 1 > Sketching graphs

“đ?‘Ś = đ?‘“(đ?‘Ľ)â€? is often used to say “some graph with an equation in terms of đ?‘Ľ.â€?

Duplicate solutions technically means graph crosses đ?‘Ľ axis twice at the same point which practically means graph is touching x axis at that point. In module Core 1 graphs only touch the đ?‘Ľ axis, not đ?‘Ś axis. Solve by factorising: (đ?‘Ľ − 2)(đ?‘Ľ + 3)2 = 0 ⇒ đ?‘Ľ − 2 = 0 and đ?‘Ľ + 3 = 0 twice. As −3 is a duplicate solution, so it touches, rather than crosses, the x axis. Duplicate solutions to đ?‘“(đ?‘Ľ) = 0 doesn’t tell you if graph touches đ?‘Ľ axis from above or below but when you attempt to sketch the graph you will find only one way to do it.

11

Exercise 6đ?‘Ž Copy and complete following tables for given equations then plot and draw the graphs accordingly (ideally, use paper with 5mm squares). 1.

2.

3.

� = �2 � –4 –3 –2 –1 0 �

1

đ?‘Ś = đ?‘Ľ2 − đ?‘Ľ + 2 đ?‘Ľ –4 –3 –2 –1 0 đ?‘Ś

1

� = �3 � –4 –3 –2 –1 0 �

1

2 3

2 3

2 3

Exercise 6đ?‘? 1 For functions (a) đ?‘“(đ?‘Ľ) = đ?‘Ľ 2 (b) đ?‘“(đ?‘Ľ) = đ?‘Ľ 2 (c) đ?‘“(đ?‘Ľ) = (3 − đ?‘Ľ)2 , find the following in terms of đ?‘“(đ?‘Ľ): 1. 2. (đ?‘Ľ + 8)2 (8 + đ?‘Ľ)2

4

4

4

3.

5(đ?‘Ľ − 8)5

4.

(đ?‘Ľ − 4)3

5.

(đ?‘Ľ + 3)2 + 5

6.

−(đ?‘Ľ − 7)2

7.

đ?‘Ľ 2 + 6đ?‘Ľ + 11

8.

đ?‘Ľ 2 + 10đ?‘Ľ + 21

9.

đ?‘Ľ 2 + 4đ?‘Ľ + 10

12.

đ?‘Ľ 2 − 8đ?‘Ľ + 60

11.

đ?‘Ľ

14.

2đ?‘Ľ + 1

3

When plotting above graph, use 4 squares per unit on the đ?‘Ś axis, 1 square per unit on the đ?‘Ľ axis. 4.

đ?‘Ś = đ?‘Ľ 3 − 2đ?‘Ľ + 3 đ?‘Ľ −4−3−2−1 0 đ?‘Ś

1

2 3

Exercise 6đ?‘? 1 For functions (a) đ?‘“(đ?‘Ľ) = đ?‘Ľ 2 (b) đ?‘“(đ?‘Ľ) = đ?‘Ľ 2 2 (c) đ?‘“(đ?‘Ľ) = (3 − đ?‘Ľ) , find the following in terms of đ?‘Ľ:

4

When plotting this graph, use 4 squares per unit on the đ?‘Ľ axis, 4 squares per unit on the đ?‘Ś axis. 5.

đ?‘Ś=

1 đ?‘Ľ

1

1

1

đ?‘Ľ −3−2−1− − − 0 2 3 4 đ?‘Ś

1

1

1

4

3

2

1

2

3

4

When plotting this graph, use 4 squares per unit on the đ?‘Ľ axis, 4 squares per unit on the đ?‘Ś axis. 6.

đ?‘Ś=

2 đ?‘Ľ

1

1

1

đ?‘Ľ −3−2−1− − − 0 2 3 4 đ?‘Ś

1

1

1

4

3

2

1

2

3

đ?‘Ś= −

1 đ?‘Ľ

1

1

1

đ?‘Ľ −3−2−1− − − 0 2 3 4 đ?‘Ś

1

1

1

4

3

2

1

2

3

2đ?‘“(đ?‘Ľ)

2.

2

3.

đ?‘“(3đ?‘Ľ)

4.

đ?‘“(−5đ?‘Ľ)

5.

đ?‘“(đ?‘Ľ − 5)

6.

đ?‘“(4 − đ?‘Ľ)

7.

đ?‘“(−đ?‘Ľ)

8.

−đ?‘“(đ?‘Ľ)

9.

đ?‘“(3đ?‘Ľ) + 1

12.

3 − đ?‘“(5đ?‘Ľ)

11.

đ?‘“(−đ?‘Ľ) + 4

14.

đ?‘“(−√đ?‘Ľ)

4

When plotting this graph, use 4 squares per unit on the đ?‘Ľ axis, 4 squares per unit on the đ?‘Ś axis. 7.

1.

4

When plotting this graph, use 4 squares per unit on the đ?‘Ľ axis, 4 squares per unit on the đ?‘Ś axis.

3

đ?‘“(đ?‘Ľ)

Exercise 6đ?‘‘ Find where the following equations of graphs cross (a) the y axis (b) the x axis (c) if and where it touches the x axis (d) sketch the graphs showing point of intersection with axis. 1. 2. 3. 4. 5. 6. 7. 8.

đ?‘Ś = 2đ?‘Ľ − 5 đ?‘Ś = (đ?‘Ľ − 5)(đ?‘Ľ + 7) đ?‘Ś = (đ?‘Ľ − 1)(đ?‘Ľ + 2)(đ?‘Ľ − 4) đ?‘Ś = (2đ?‘Ľ − 5)(đ?‘Ľ + 7) đ?‘Ś = (đ?‘Ľ − 1)(5đ?‘Ľ + 2)(4 − đ?‘Ľ) đ?‘Ś = đ?‘Ľ(đ?‘Ľ + 2)(đ?‘Ľ − 4) đ?‘Ś = 2đ?‘Ľ 2 + 3đ?‘Ľ + 1 đ?‘Ś = đ?‘Ľ2 − 5

9.

đ?‘Ś=

10.

đ?‘Ś= −5

11.

đ?‘Ś=

12. 13. 14. 15. 16. 17. 18.

đ?‘Ś đ?‘Ś đ?‘Ś đ?‘Ś đ?‘Ś đ?‘Ś đ?‘Ś

7 �–2 2

−5

đ?‘Ľ

= = = = = = =

7 �–3

(đ?‘Ľ − 8)(đ?‘Ľ + 7)2 (đ?‘Ľ + 4)2 (đ?‘Ľ − 1) đ?‘Ľ 2 (đ?‘Ľ − 5) (đ?‘Ľ − 2)đ?‘Ľ 2 đ?‘Ľ(đ?‘Ľ + 3)2 (đ?‘Ľ + 12)3 (đ?‘Ľ − 1)3 12

Graphs can be transformed by sliding and stretching/squeezing.

7. Transforming graphs

Vid 2

Summary of transformations table: Transforming graphs Vid 1.1 Here are the different ways to transform graphs: Vid 1.2 Vid 1.3

move đ?‘Ś direction đ?‘Ś

slide

Vid 1.4

move đ?‘Ľ direction đ?‘Ś −đ?‘Ž +đ?‘Ž

đ?‘Ś = đ?‘“(đ?‘Ľ) + đ?‘Ž đ?‘Ľ đ?‘Ś = đ?‘“(đ?‘Ľ) đ?‘Ś

stretch/ squeeze

Vid 1.5

reflect

Vid 1.6

đ?‘Ľ đ?‘Ś = đ?‘“(đ?‘Ľ)

đ?‘Ś = đ?‘“(đ?‘Ľ + đ?‘Ž) áđ?‘Ž đ?‘Ś Ă—đ?‘Ž

đ?‘Ś = đ?‘“(đ?‘Ľ) đ?‘Ś = đ?‘Žđ?‘“(đ?‘Ľ)

đ?‘Ľ

đ?‘Ľ đ?‘Ś = đ?‘“(đ?‘Ľ) đ?‘Ś = đ?‘“(đ?‘Žđ?‘Ľ)

đ?‘Ś đ?‘Ś = −đ?‘“(đ?‘Ľ)

đ?‘Ś đ?‘Ľ

đ?‘Ś = đ?‘“(đ?‘Ľ)

đ?‘Ľ đ?‘Ś = đ?‘“(đ?‘Ľ)

Summary: change outside function moves graph vertically in the same way i.e. �(�) to �(�) + � moves graph vertically +�. Change inside function moves graph horizontally in the opposite way i.e. �(�) to �(� + �) moves graph horizontally – �.

đ?‘Ś = đ?‘“(−đ?‘Ľ) Vid 1.7

Combining transformations of graphs Vid 3 Find transformations of graph from of đ?‘Ś = đ?‘“(đ?‘Ľ) to đ?‘Ś = 2đ?‘“(đ?‘Ľ) + 1. 0 Enlarges graph by scale factor 2 from đ?‘Ľ axis then translates graph ( ). 1 Find transformations of graph from of đ?‘Ś = đ?‘Ľ 3 to đ?‘Ś = −(2đ?‘Ľ)3 . đ?‘Ś = đ?‘Ľ 3 to đ?‘Ś = −(2đ?‘Ľ)3 is same as đ?‘Ś = đ?‘“(đ?‘Ľ) to đ?‘Ś = −đ?‘“(2đ?‘Ľ) 1 i.e. enlarges graph by scale factor from đ?‘Ś axis then reflects graph about đ?‘Ľ 2 axis. Completing the square to sketch quadratic graphs Vid 4 Sketch graph of đ?‘Ś = đ?‘Ľ 2 + 4đ?‘Ľ + 10 showing coordinate of vertex. đ?‘Ś = đ?‘Ľ 2 + 4đ?‘Ľ + 10 = (đ?‘Ľ + 2)2 + 6 −2 translates graph ( ) 6 For axes intercepts: đ?‘Ś = 0 ⇒ no solution, đ?‘Ľ = 0 ⇒ đ?‘Ś = 10: đ?‘Ś = đ?‘Ľ 2 + 4đ?‘Ľ + 10

đ?‘Ś

For any graph, these changes in function lead to these change in coordinates: Change in function Change in coordinates from đ?‘Ś = đ?‘“(đ?‘Ľ) to: from (đ?‘Ľ, đ?‘Ś) to: đ?‘Ś = đ?‘“(đ?‘Ľ) + đ?‘Ž (đ?‘Ľ, đ?‘Ś + đ?‘Ž) đ?‘Ś = đ?‘“(đ?‘Ľ + đ?‘Ž) (đ?‘Ľ − đ?‘Ž, đ?‘Ś) đ?‘Ś = đ?‘Žđ?‘“(đ?‘Ľ) (đ?‘Ľ, đ?‘Žđ?‘Ś) đ?‘Ľ đ?‘Ś = đ?‘“(đ?‘Žđ?‘Ľ) ( , đ?‘Ś) đ?‘Ž đ?‘Ś = −đ?‘“(đ?‘Ľ) (đ?‘Ľ, −đ?‘Ś) đ?‘Ś = đ?‘“(−đ?‘Ľ) (−đ?‘Ľ, đ?‘Ś)

10 (– 2,6) �

đ?‘Ś = đ?‘“(đ?‘Ľ) to đ?‘Ś = 2đ?‘“(đ?‘Ľ) enlarges graph by scale factor 2 from đ?‘Ľ axis. đ?‘Ś = 2đ?‘“(đ?‘Ľ) to đ?‘Ś = 2đ?‘“(đ?‘Ľ) + 1 translates graph 0 ( ). 1 đ?‘Ś = đ?‘“(đ?‘Ľ) to đ?‘Ś = đ?‘“(2đ?‘Ľ) enlarges graph by scale factor 2 from đ?‘Ś axis. đ?‘Ś = đ?‘“(2đ?‘Ľ) to đ?‘Ś = −đ?‘“(2đ?‘Ľ) reflects graph about đ?‘Ľ axis.

Complete square to compare graph to � = � 2 so that you can find vertex. � = � 2 to � = (� + 2)2 + 6 same as � = �(�) to � = �(� + 2) + 6 � = � 2 to � = (� + 2)2 adds 2 inside the function i.e. equivalent to �(�) to �(� + 2). � = (� + 2)2 to � = (� + 2)2 + 6 adds 6 outside the function i.e. equivalent to �(� + 2) to �(� + 2) + 6. No solution for � = 0 means graph doesn’t cross � axis.

Videos at sickmaths.com > A level > Core 1 > Transforming graphs

13

Exercise 7đ?‘Ž Using đ?‘“(đ?‘Ľ) = đ?‘Ľ 2 , for each of the following expressions (a) sketch its graph on the same axis (b) find coordinate of its vertex (c) write the corresponding equation in terms of đ?‘Ľ

Exercise 7đ?‘‘ đ?‘Ś

đ?‘Ś = đ?‘“(đ?‘Ľ)

5 (3, 2) đ?‘Ľ

1. 3. 5. 7. 9. 11. 13. 15. 17.

đ?‘Ś = −đ?‘“(đ?‘Ľ) đ?‘Ś = đ?‘“(−đ?‘Ľ) đ?‘Ś = đ?‘“(đ?‘Ľ + 5) đ?‘Ś = đ?‘“(2đ?‘Ľ) đ?‘Ś = 2đ?‘“(đ?‘Ľ) đ?‘Ś = 3đ?‘“(4đ?‘Ľ) đ?‘Ś = đ?‘“(−đ?‘Ľ) + 1 đ?‘Ś = 2đ?‘“(đ?‘Ľ − 3) 1 đ?‘Ś = đ?‘“( đ?‘Ľ)

19.

đ?‘Ś = đ?‘“(đ?‘Ľ)

1

2

4

2. 4. 6. 8. 10. 12. 14. 16. 18.

đ?‘Ś = −đ?‘“(đ?‘Ľ) đ?‘Ś = đ?‘“(−đ?‘Ľ) đ?‘Ś = đ?‘“(đ?‘Ľ + 5) đ?‘Ś = đ?‘“(2đ?‘Ľ) đ?‘Ś = 2đ?‘“(đ?‘Ľ) đ?‘Ś = 3đ?‘“(4đ?‘Ľ) đ?‘Ś = đ?‘“(−đ?‘Ľ) + 1 đ?‘Ś = 2đ?‘“(đ?‘Ľ − 3) 1 đ?‘Ś = đ?‘“( đ?‘Ľ)

20.

đ?‘Ś = đ?‘“(đ?‘Ľ)

1

1. 3. 5. 7.

đ?‘Ś = đ?‘“(đ?‘Ľ) + 3 đ?‘Ś = 2đ?‘“(đ?‘Ľ) đ?‘Ś = đ?‘“(3đ?‘Ľ) 3 đ?‘Ś = đ?‘“( đ?‘Ľ)

9. 11.

đ?‘Ś = đ?‘“(đ?‘Ľ + 3) 2 đ?‘Ś = −đ?‘“(đ?‘Ľ + 3) − 2

2

10. 12.

1

đ?‘Ś = −đ?‘“( đ?‘Ľ) + 2 3 đ?‘Ś = đ?‘“(2đ?‘Ľ)

4

Exercise 7đ?‘’ đ?‘Ś 2 đ?‘Ľ −4 1.

đ?‘Ś = −đ?‘“(đ?‘Ľ) đ?‘Ś = đ?‘“(−đ?‘Ľ) đ?‘Ś = đ?‘“(đ?‘Ľ + 5) đ?‘Ś = đ?‘“(2đ?‘Ľ) đ?‘Ś = 2đ?‘“(đ?‘Ľ) đ?‘Ś = 3đ?‘“(4đ?‘Ľ) đ?‘Ś = 2đ?‘“(đ?‘Ľ − 3) 1 đ?‘Ś = đ?‘“( đ?‘Ľ)

3. 5. 7. 9. 11.

đ?‘Ś = đ?‘”(2đ?‘Ľ) 1 đ?‘Ś = đ?‘”( đ?‘Ľ) 2 đ?‘Ś = đ?‘”(4đ?‘Ľ) 1 đ?‘Ś = đ?‘”(− đ?‘Ľ) 2 đ?‘Ś = −2đ?‘”(2 − đ?‘Ľ) 4 đ?‘Ś = đ?‘”(−đ?‘Ľ) 5

Graph of đ?‘Ś = đ?‘”(đ?‘Ľ) crosses axis at (−4,0) and (0,2). For equations below, give coordinates of intersections with the axes. 2. 4. 6. 8. 10. 12.

đ?‘Ś = đ?‘”(đ?‘Ľ − 4) đ?‘Ś = đ?‘”(đ?‘Ľ) − 1 đ?‘Ś = −đ?‘”(đ?‘Ľ − 4) 2 đ?‘Ś = đ?‘”( đ?‘Ľ) 3 đ?‘Ś = −đ?‘”(4đ?‘Ľ + 1) 1 đ?‘Ś = đ?‘”(đ?‘Ľ − 2) 3

2

Exercise 7đ?‘? Describe transformations from graph đ?‘Ś = đ?‘“(đ?‘Ľ) to following graphs: 1. 3. 5. 7. 9. 11. 13.

đ?‘Ś = đ?‘“(đ?‘Ľ) + 1 đ?‘Ś = đ?‘“(đ?‘Ľ + 1) đ?‘Ś = 2đ?‘“(đ?‘Ľ) đ?‘Ś = đ?‘“(2đ?‘Ľ) đ?‘Ś = đ?‘“(−đ?‘Ľ) đ?‘Ś = −đ?‘“(đ?‘Ľ) 1 đ?‘Ś = đ?‘“(đ?‘Ľ)

15.

đ?‘Ś = đ?‘“( đ?‘Ľ)

17.

đ?‘Ś = 2đ?‘“( đ?‘Ľ) 2 đ?‘Ś = −đ?‘“(đ?‘Ľ + 1) đ?‘Ś = −đ?‘“(−đ?‘Ľ) đ?‘Ś = 5 − đ?‘“(đ?‘Ľ) đ?‘Ś = đ?‘“(3 + đ?‘Ľ)

19. 21. 23. 25.

2

2

Exercise 7đ?‘? Using đ?‘“(đ?‘Ľ) = đ?‘Ľ 3 , for each of the following transformations (a) sketch its graph (b) find coordinate to which (0,0) is mapped (c) write the corresponding equation for graph in terms of đ?‘Ľ (d) write the corresponding equation for graph in terms of đ?‘“(đ?‘Ľ) 1. 3. 5. 7. 9. 11. 13. 15.

1

Graph of đ?‘Ś = đ?‘“(đ?‘Ľ) has vertex at (2,3) and crosses axis at (0,5). For equations below, sketch graphs with coordinates of any intersections with the axes or vertices. 2. đ?‘Ś = đ?‘“(đ?‘Ľ + 3) 4. đ?‘Ś = đ?‘“(−đ?‘Ľ) 6. đ?‘Ś = đ?‘“(đ?‘Ľ + 3) − 2 1 đ?‘Ś = đ?‘“( đ?‘Ľ) 8.

3

1

2

3

2. 4. 6. 8. 10. 12. 14.

đ?‘Ś = đ?‘“(đ?‘Ľ) − 4 đ?‘Ś = đ?‘“(đ?‘Ľ − 2) đ?‘Ś = 5đ?‘“(đ?‘Ľ) đ?‘Ś = đ?‘“(31đ?‘Ľ) đ?‘Ś = đ?‘“(−8đ?‘Ľ) đ?‘Ś = −2đ?‘“(đ?‘Ľ) 3 đ?‘Ś = đ?‘“(đ?‘Ľ) − 2

16.

đ?‘Ś = đ?‘“( đ?‘Ľ)

18.

đ?‘Ś = 4đ?‘“( đ?‘Ľ) 5 đ?‘Ś = −đ?‘“(2đ?‘Ľ) + 3 đ?‘Ś = −2đ?‘“(−đ?‘Ľ) đ?‘Ś = 3 − đ?‘“(đ?‘Ľ) đ?‘Ś = đ?‘“(6 − 2đ?‘Ľ)

20. 22. 24. 26.

3

Exercise 7đ?‘“ By writing the following equations in the form đ?‘Ś = đ?‘Ž(đ?‘Ľ − đ?‘?)2 + đ?‘? (a) sketch their graphs (b) find coordinate of its vertex: 1. 3. 5. 7. 9.

đ?‘Ś = đ?‘Ľ 2 + 4đ?‘Ľ + 3 đ?‘Ś = đ?‘Ľ 2 + 6đ?‘Ľ + 5 đ?‘Ś = đ?‘Ľ 2 − 6đ?‘Ľ + 5 đ?‘Ś = 4đ?‘Ľ 2 − 16đ?‘Ľ + 8 đ?‘Ś = 2đ?‘Ľ 2 − 8đ?‘Ľ + 4

1

5

2

14

8. Arithmetic Series

Vid 1.1

Notation used for arithmetic series Vid 1.2 First term = đ?‘Ž Common difference = đ?‘‘ đ?‘› terms of an arithmetic series = đ?‘Ž, đ?‘Ž + đ?‘‘, đ?‘Ž + 2đ?‘‘, đ?‘Ž + 3đ?‘‘, ‌ đ?‘Ž + (đ?‘›â€“ 1)đ?‘‘ đ?‘›đ?‘Ąâ„Ž term = đ?‘ˆđ?‘› = đ?‘Ž + (đ?‘› − 1)đ?‘‘ đ?‘› Last term = đ??ż Sum of first đ?‘› terms = đ?‘†đ?‘› = ∑(expression in đ?‘&#x;)

E.g. 7, 5, 3, 1 has common difference −3 as the difference between each term is −3. đ?‘› is the position of a term. A rule to find any term in the series.

1

Standard trick to find facts about arithmetic series Vid 1.3 6đ?‘Ąâ„Ž term is 14, 4đ?‘Ąâ„Ž term is 10. Find đ?‘›đ?‘Ąâ„Ž term. 6đ?‘Ąâ„Ž term = 14 ⇒ đ?‘Ž + 5đ?‘‘ = 14 _ 4đ?‘Ąâ„Ž term = 10 ⇒ đ?‘Ž + 3đ?‘‘ = 10 2đ?‘‘ = 4 ⇒ đ?‘‘ = 2 ⇒ đ?‘Ž = 4 ⇒ đ?‘›đ?‘Ąâ„Ž term = đ?‘Ž + (đ?‘› − 1)đ?‘‘ = 2đ?‘› + 2 Sum of first đ?’? terms of an arithmetic series: Vid 2.1 đ?’? đ?’? đ?‘şđ?’? = [đ?&#x;?đ?’‚ + (đ?’? − đ?&#x;?)đ?’…] and đ?‘şđ?’? = (đ?’‚ + đ?‘ł) đ?&#x;?

đ?&#x;?

Find sum of the first 60 terms of arithmetic series with đ?‘›th term 3đ?‘› + 2 đ?‘›đ?‘Ąâ„Ž term 3đ?‘› + 2 ⇒ 5, 8, 11 ‌ ⇒ đ?‘Ž = 5, đ?‘‘ = 3. 60 terms ⇒ đ?‘› = 60 đ?‘› 60 đ?‘†đ?‘› = [2đ?‘Ž + (đ?‘› − 1)đ?‘‘] ⇒ đ?‘†60 = [2 Ă— 5 + (60 − 1) Ă— 3] = 5610 2

Number patterns that have a common difference (see below).

Vid 2.2

Series are endless but đ??ż means last term used in a question. If nth term is last term then đ??ż = đ?‘Ž + (đ?‘› − 1)đ?‘‘. đ?‘›đ?‘Ąâ„Ž term = đ?‘Ž + (đ?‘› − 1)đ?‘‘ ⇒ 6đ?‘Ąâ„Ž term = đ?‘Ž + (6 − 1)đ?‘‘ Solve simultaneously (by elimination) đ?‘Ž = 4 means first term is 4, đ?‘‘ = 2 means common difference is 2 đ?‘›

đ?‘†đ?‘› = [2đ?‘Ž + (đ?‘› − 1)đ?‘‘] 2 đ?‘›

= [đ?‘Ž + đ?‘Ž + (đ?‘› − 1)đ?‘‘] 2 Substitute đ??ż = đ?‘Ž + (đ?‘› − 1)đ?‘‘ đ?‘› ⇒ đ?‘†đ?‘› = (đ?‘Ž + đ??ż)

2

2

Find sum of terms 21 to 30 of an arithmetic series, first term 3, last term 101. 20 Vid 2.3 đ?‘› = 20, đ?‘Ž = 3, đ??ż = 101 ⇒ đ?‘†20 = (3 + 101) = 1040 đ?‘› = 30, đ?‘Ž = 3, đ??ż = 101 ⇒ đ?‘†30 =

2 30 2

Substitute đ?‘Ž = 5, đ?‘‘ = 3, đ?‘› = 60 into đ?‘› đ?‘†đ?‘› = [2đ?‘Ž + (đ?‘› − 1)đ?‘‘] 2

(3 + 101) = 1560. đ?‘†30 − đ?‘†20 = 420

đ?‘›

Using đ?‘†đ?‘› = (đ?‘Ž + đ??ż) 2

Proof of formula for the sum of any arithmetic series. Vid 2.5 đ?‘†đ?‘› = đ?‘Ž + đ?‘Ž + đ?‘‘ + đ?‘Ž + 2đ?‘‘ + ‌ + đ?‘Ž + (đ?‘› − 1)đ?‘‘ + đ?‘†đ?‘› = đ?‘Ž + (đ?‘› − 1)đ?‘‘ + ‌..‌‌‌‌‌‌..‌‌‌+ đ?‘Ž 2đ?‘†đ?‘› = 2đ?‘Ž + (đ?‘›â€“ 1)đ?‘‘ + ‌..‌‌‌‌‌‌‌‌.‌ + 2đ?‘Ž + (đ?‘› − 1)đ?‘‘ 2đ?‘†đ?‘› = đ?‘›[2đ?‘Ž + (đ?‘›â€“ 1)đ?‘‘] đ?‘› đ?‘†đ?‘› = [2đ?‘Ž + (đ?‘›â€“ 1)đ?‘‘]

Vid 2.4

2

Using ∑ (sigma) 10

10

Find ∑ 2đ?‘&#x; ∑ 2đ?‘&#x; = 2 Ă— 7 + 2 Ă— 8 + 2 Ă— 9 + 2 Ă— 10 7 100

100

Find ∑ 3đ?‘&#x; + 1 ∑ 3đ?‘&#x; + 1 = 4 + 7 + 10 + ‌ + 301 1

Vid 3.1

7

1

Vid 3.2

The following is a warm up to this proof: Find sum of integers from 1 to 100. đ?‘†100 = 1 + 2 + 3 + ‌ + 100 + đ?‘†100 = 100 + 99 + 98 + ‌ + 1 2đ?‘†100 = 101 + 101 + 101 + ‌ + 101 2đ?‘†100 = 101 Ă— 100 2đ?‘†100 = 10100 10100 đ?‘†100 = 2 đ?‘†100 = 5050 To prove the đ?‘†đ?‘› = đ?‘›(đ?‘Ž + đ??ż) replace đ?‘Ž + (đ?‘› − 1)đ?‘‘ with đ??ż. Sum terms 7 to 10 i.e. substitute đ?‘&#x; for whole numbers 7 to 10 in to 2đ?‘&#x; and sum answers.

⇒ đ?‘Ž = 4, đ?‘‘ = 3, đ?‘› = 100, đ??ż = 301 đ?‘†đ?‘› =

đ?‘› 100 [đ?‘Ž + đ??ż] ⇒ đ?‘†100 = [4 + 301] = 15250 2 2

Recurrence relationships Vid 4.1 đ?‘ˆđ?‘›+1 = đ?‘ˆđ?‘› + 3. đ?‘ˆ1 = 7. Find next two terms. đ?‘ˆ2 = đ?‘ˆ1 + 3 ⇒ đ?‘ˆ2 = 7 + 3 = 10 đ?‘ˆ3 = đ?‘ˆ2 + 3 ⇒ đ?‘ˆ3 = 10 + 3 = 13

Circular rules that make number patterns. Vid 4.2

đ?‘ˆđ?‘›+2 = 2đ?‘ˆđ?‘›+1 − đ?‘ˆđ?‘› . đ?‘ˆ1 = 6, đ?‘ˆ0 = 8. Find next two terms. đ?‘ˆ2 = 2đ?‘ˆ1 − đ?‘ˆ0 ⇒ đ?‘ˆ2 = 2 Ă— 6 − 8 = 4 đ?‘ˆ3 = 2đ?‘ˆ2 − đ?‘ˆ1 ⇒ đ?‘ˆ3 = 2 Ă— 4 − 6 = 2

Vid 4.3

đ?‘ˆ1 , đ?‘ˆ2 etc. means term 1, term 2 etc. unless the pattern starts with đ?‘ˆ0 in which case đ?‘ˆ0 , đ?‘ˆ1 etc. means term 1, term 2 etc.

đ?‘ˆđ?‘›+1 = đ?‘ˆđ?‘› + đ?‘˜. đ?‘ˆ0 = 2, đ?‘ˆ1 = 5. Find đ?‘˜. Vid 4.4 đ?‘ˆ1 = đ?‘ˆ0 + đ?‘˜ ⇒ 5 = 2 + đ?‘˜ ⇒ đ?‘˜ = 3 đ?‘ˆđ?‘› = 10 − đ?‘ˆđ?‘›âˆ’1 . đ?‘ˆ1 = 8. Find đ?‘ˆ100 . Vid 4.5 đ?‘ˆ2 = 2, đ?‘ˆ3 = 8, đ?‘ˆ4 = 2 ⇒ đ?‘ˆđ?‘’đ?‘Łđ?‘’đ?‘› = 2, đ?‘ˆđ?‘œđ?‘‘đ?‘‘ = 8 ⇒ đ?‘ˆ100 = 2 Videos at sickmaths.com > A level > Core 1 > Arithmetic Series

This is a periodic pattern i.e. even terms equal 2 and odd terms equal 8. 15

Exercise 8� Here are the first 4 terms of some arithmetic series. For each, find the common difference, ��ℎ term and 100�ℎ term:

Exercise 8đ?‘‘ For the following saving schemes, find the amount Rudeboi has to pay in the 40đ?‘Ąâ„Ž year, the sum of his payments up to and including the 40đ?‘Ąâ„Ž year and how many years before his savings reach ÂŁ100 000:

1.

2, 4, 6, 8

2.

1, 2, 3, 4

1.

3.

4, 7, 10, 13

4.

3, 8, 11, 14

In the first year Rudeboi pays ÂŁ300. His payments then increase by ÂŁ60 each year, so that he pays ÂŁ360 in the second year.

5.

5, 7, 9, 11

6.

102, 105, 107, 109

2.

7.

7, 8, 9, 10

8.

−2, 3, 8, 13

In the first year Rudeboi pays ÂŁ400. His payments then increase by ÂŁ20 each year, so that he pays ÂŁ420 in the second year.

9.

−2, −4, −6, −8

10.

18, 15, 12, 9

3.

11.

5, 3, 1, −1

12.

−5, −4, −3, −2

In the first year Rudeboi pays ÂŁ1200. His payments then increase by ÂŁ80 each year, so that he pays ÂŁ1280 in the second year.

13.

12, 22, 32, 42

14.

12, 22, 32, 42

4.

15.

−12, −10, −8, −6

16.

−7, −5, −3, −1

In the first year Rudeboi pays ÂŁ6800. His payments then decrease by ÂŁ200 each year, so that he pays ÂŁ6780 in the second year.

Exercise 8đ?‘? Find the sum of the following arithmetic series:

Exercise 8đ?‘’ Find the value of the following:

1.

1.

3.

5.

7.

All the odd numbers from 1 to 299.

2.

With second term 4, third term 7 and 100 terms.

4.

With second term 3, fourth term 7 and last term 101.

6.

With common difference −3, third term 10 and 100 terms.

8.

All the even numbers from 2 to 400. With second term 8, third term 6 and 100 terms.

With second term 30, fourth term 40 and last term 110. With common difference 6, third term 10 and 100 terms.

3.

5.

7.

10

2.

4

∑ 3r

∑ 3r

r=7

r=1

100

4.

200

∑ 2r + 1

∑ 3r

r=0

r=0

50

6.

310

∑4 + r

∑ 4r − 1

r=5

r=7

8.

100

90

∑ 2r − 5

∑ 3r

r=2

r=3

Exercise 8đ?‘? Find the first 4 terms of the following recurrence relationships, state if any have a repetitive pattern and find đ?‘˜ if necessary:

Exercise 8đ?‘“ For an arithmetic series with first term đ?‘Ž, common difference đ?‘‘, last term đ??ż and đ?‘› terms:

1.

đ?‘ˆđ?‘›+1 = đ?‘ˆđ?‘› + 3, đ?‘ˆ1 = 1

2.

đ?‘ˆđ?‘›+1 = đ?‘ˆđ?‘› − 1, đ?‘ˆ1 = 1

1.

3.

đ?‘ˆđ?‘›+1 = 8 − đ?‘ˆđ?‘› , đ?‘ˆ0 = 1

4.

đ?‘ˆđ?‘›+1 = 2đ?‘ˆđ?‘› , đ?‘ˆ1 = 1

3.

5.

đ?‘ˆđ?‘›+1 = đ?‘˜đ?‘ˆđ?‘› , đ?‘ˆ1 = 1, đ?‘ˆ2 = 3

6.

đ?‘ˆđ?‘›+1 = đ?‘ˆđ?‘› + 2đ?‘˜, đ?‘ˆ1 = 7, đ?‘ˆ2 = −2

7.

đ?‘ˆđ?‘›+1 = 3đ?‘ˆđ?‘› − 5 đ?‘ˆ0 = đ?‘˜, đ?‘ˆ1 = đ?‘˜ + 3, đ?‘ˆ2 = 16

8.

đ?‘ˆđ?‘›+1 = đ?‘ˆđ?‘› + 3 đ?‘ˆ1 = 16, đ?‘ˆ2 = đ?‘˜ + 3, đ?‘ˆ3 = 2đ?‘˜

đ?‘ˆđ?‘›+2 = đ?‘ˆđ?‘›+1 + đ?‘ˆđ?‘› + 3 đ?‘ˆ1 = 1, đ?‘ˆ2 = 5

10.

9.

đ?‘ˆđ?‘›+2 = −1 + đ?‘ˆđ?‘›+1 + đ?‘ˆđ?‘› đ?‘ˆ0 = −2, đ?‘ˆ1 = 8

Show �3 = 3� + 3�

5.

7.

đ?‘ˆđ?‘›+2 = đ?‘ˆđ?‘› + 3đ?‘ˆđ?‘›+1 đ?‘ˆ0 = 1

12.

đ?‘ˆđ?‘›+2 = 2đ?‘˜ + đ?‘ˆđ?‘› − đ?‘ˆđ?‘›+1 đ?‘ˆ1 = 6, đ?‘ˆ2 = 8, đ?‘ˆ3 = 1

Show đ?‘†5 − đ?‘†1 = 4đ?‘‘

4. Show that the sum of the last 3 terms is 3đ??ż − 3đ?‘‘ 6. Show the sum of all the even numbers up to the đ?‘Ś đ?‘Ąâ„Ž term.

8.

Show the sum of the last 2 terms is 2đ?‘Ž + 2đ?‘›đ?‘‘ − 3đ?‘‘ Show the sum of all the odd numbers up to the last term đ?‘‹.

Show đ?‘›

11.

2.

đ?‘†đ?‘› = [2đ?‘Ž + (đ?‘› − 1)đ?‘‘] 2

Show đ?‘›

đ?‘†đ?‘› = (đ?‘Ž + đ??ż) 2

16

9. Inequalities

Inequalities are like equations but the solutions have a range of values rather than fixed values.

Simple inequalities Vid 1 Solve 2đ?‘Ľ − 5 1. 2đ?‘Ľ − 5 1 ⇒ 2đ?‘Ľ

Inequalities are mostly solved in the same way as equations. The inequality symbol used in the question should be used throughout the working out.

6⇒�

3

Solve −5đ?‘Ś < 30. −5đ?‘Ś < 30 ⇒ −đ?‘Ś < 6 ⇒ đ?‘Ś < −6 Quadratic inequalities Solve đ?‘Ľ 2 − 10 −3đ?‘Ľ. Vid 2 đ?‘Ľ 2 − 10 −3đ?‘Ľ ⇒ đ?‘Ľ 2 + 3đ?‘Ľ − 10 0 ⇒ (đ?‘Ľ − 2)(đ?‘Ľ + 5) 0 Consider graph of đ?‘Ś = (đ?‘Ľ − 2)(đ?‘Ľ + 5): đ?‘Ś

đ?‘Ś = (đ?‘Ľ − 2)(đ?‘Ľ + 5)

Rearrange inequality with 0 on one side. You should know how to sketch quadratic graphs (see Sketching Graphs chapter). (đ?‘Ľ − 2)(đ?‘Ľ + 5) 0 ⇒ đ?‘Ś 0 đ?‘Ś 0 when đ?‘Ľ − 5 and đ?‘Ľ 2

đ?‘Ľ −5

If you multiply both sides of an inequality by −1 then flip the inequality symbol around.

2

Hence (đ?‘Ľ − 2)(đ?‘Ľ + 5)

0 when đ?‘Ľ

− 5, đ?‘Ľ 2

As đ?‘Ľ 2 − 10 −3đ?‘Ľ ⇒ (đ?‘Ľ − 2)(đ?‘Ľ + 5) 0 đ?‘Ľ − 5, đ?‘Ľ 2 is solution to đ?‘Ľ 2 − 10 −3đ?‘Ľ

Solve (đ?‘Ľ − 2)(3 − đ?‘Ľ) > 0 Vid 3 Consider graph of đ?‘Ś = (đ?‘Ľ − 2)(3 − đ?‘Ľ): đ?‘Ś

đ?‘Ľ 2

3 đ?‘Ś = (đ?‘Ľ − 2)(3 − đ?‘Ľ)

Hence (đ?‘Ľ − 2)(3 − đ?‘Ľ) > 0 when 2 < đ?‘Ľ < 3 Combining inequalities Solve 1 < đ?‘Ľ < 6 and đ?‘Ľ 4. Vid 4 1 4

2 < đ?‘Ľ < 3 means between 2 and 3. You could write this like 2 < đ?‘Ľ, đ?‘Ľ < 3 or 3 > đ?‘Ľ > 2 etc. but 2 < đ?‘Ľ < 3 is the expected way to write it. means “less/greater than but not equal toâ€? means “less/greater than and equal toâ€?

6 Means 1 < đ?‘Ľ < 6 Means đ?‘Ľ

Solution: 4 đ?‘Ľ < 6

6

Solution is where inequalities overlap

Making inequalities from facts Vid 5 The height of a rectangle is đ?‘Ľ, where đ?‘Ľ > 0. The width of the rectangle is 30đ?‘š more than the height. (a) Given that perimeter is less than 200đ?‘š, form a linear inequality in đ?‘Ľ (b) Given area must be greater than 1000đ?‘š2 , form a quadratic inequality in đ?‘Ľ (c) By solving these inequalities, find set of possible values of đ?‘Ľ. đ?‘Ľ + 30 đ?‘Ľ

Area > 1000

Perimeter = 4đ?‘Ľ + 60, Perimeter < 200

(a) Perimeter < 200 and perimeter = 4đ?‘Ľ + 60 ⇒ 4đ?‘Ľ + 60 < 200 ⇒ đ?‘Ľ < 35đ?‘š (b) Area > 1000 and area = đ?‘Ľ(đ?‘Ľ + 30) ⇒ đ?‘Ľ 2 + 30đ?‘Ľ > 1000 ⇒ (đ?‘Ľ + 50)(đ?‘Ľ − 20) > 0 ⇒ đ?‘Ľ < −50, đ?‘Ľ > 20 đ?‘Ľ < −50 not possible. đ?‘Ľ > 20đ?‘š (c) đ?‘Ľ > 20 and đ?‘Ľ < 35 ⇒ 20 < đ?‘Ľ < 35 Videos at sickmaths.com > A level > Core 1 > Inequalities

đ?‘Ľ < −50 not possible as đ?‘Ľ is a length so đ?‘Ľ > 20 is only solution.

17

Exercise 9đ?‘Ž Solve the following inequalities.

Exercise 9đ?‘‘ Solve the following inequalities.

1.

2đ?‘Ľ + 5 > 1

2.

2đ?‘Ľ + 5 > 1

3.

đ?‘Ľ

4.

2đ?‘“

3

−5 1

21

+3<2

5.

5đ?‘? − 1 > đ?‘? + 1

6.

−đ?‘Ľ − 7

7.

−4đ?‘Ľ > 8

8.

−đ?‘Ś < 9

9.

−đ?‘Ľ + 7 > 1

10.

−2đ?‘Ľ + 7 < 1

11.

2(đ?‘’ + 6) < 3đ?‘’ + 1

12.

5đ?‘Ľ − 6

(đ?‘Ľ − 2)(đ?‘Ľ + 5) 0 (1 − đ?‘Ľ)(đ?‘Ľ + 5) > 0 (3 − đ?‘Ľ)(4 − đ?‘Ľ) < 0 đ?‘Ľ 2 + 3đ?‘Ľ − 10 < 0 đ?‘Ľ 2 − 3đ?‘Ľ + 2 > 0 2đ?‘Ľ 2 − 11đ?‘Ľ + 5 0 2đ?‘Ľ 2 + 9đ?‘Ľ < −4 4đ?‘Ľ 2 − 3 4đ?‘Ľ đ?‘Ľ 2 − 6đ?‘Ľ + 7 2 3đ?‘Ľ 2 + 18đ?‘Ľ − 9 < 0

2. 4. 6. 8. 10. 12. 14. 16. 18. 20.

đ?‘Ľ < 10 and đ?‘Ľ

3.

11

(a)

Given that perimeter is less than 18đ?‘š, form an inequality in đ?‘Ś.

(b)

Given area must be greater than 10đ?‘š2 , form a quadratic inequality in đ?‘Ś.

(c)

By solving these inequalities, find set of possible values of đ?‘Ś.

2.

The height of a rectangle is đ?‘Ľ. The width of the rectangle is 30đ?‘š more than the height.

(a)

Given that perimeter is less than 200đ?‘š, form a linear inequality in đ?‘Ľ.

(b)

Given area must be greater than 1000đ?‘š2 , form a quadratic inequality in đ?‘Ľ.

(c)

By solving these inequalities, find set of possible values of đ?‘Ľ.

3.

The width of a rectangle is �. The height of the rectangle is 2� less than the width.

(a)

Given that perimeter is less than 100�, form a linear inequality in �.

(b)

Given area must be less than 63�2 , form a quadratic inequality in �.

(c)

By solving these inequalities, find set of possible values of �.

4.

The height of a rectangle is đ?‘“. The width of the rectangle is 3đ?‘š more than the height.

(a)

Given that perimeter is less than 300đ?‘š, form a linear inequality in đ?‘“.

3đ?‘Ľ + 2

(đ?‘Ľ − 9)(đ?‘Ľ − 7) > 0 (2 − đ?‘Ľ)(đ?‘Ľ + 3) > 0 (1 + đ?‘Ľ)(8 + đ?‘Ľ) < 0 2đ?‘Ľ 2 + 3đ?‘Ľ + 1 0 7đ?‘Ľ 2 − 8đ?‘Ľ + 1 0 đ?‘Ľ 2 + 8đ?‘Ľ + 7 > 0 đ?‘Ľ 2 + 8đ?‘Ľ 0 đ?‘Ľ 2 − 12đ?‘Ľ 13 đ?‘Ľ 2 > 10đ?‘Ľ − 9 5đ?‘Ľ 2 20đ?‘Ľ + 70

Exercise 9đ?‘? Solve the following composite inequalities. 1.

The height of a rectangular field is đ?‘Ś. The width of the field is 3đ?‘š more than its height.

6đ?‘Ľ − 8

Exercise 9đ?‘? Solve the following inequalities. 1. 3. 5. 7. 9. 11. 13. 15. 17. 19.

1.

2.

đ?‘Ą < 7 and đ?‘Ą > 55

đ?‘Ľ > 5 and đ?‘Ľ > 3

4.

đ?‘Ś > 6 and đ?‘Ś > 8

5.

2<đ?‘Ľ

6.

3 < đ?‘š < 8 and đ?‘š < 4

7.

2 < đ?‘Ľ < 5 and đ?‘Ľ − 5 < −2

8.

12 < đ?‘Ś < 19 and đ?‘Ś + 9 > −3 2

(b)

Given area must be greater than 10đ?‘š2 , form a quadratic inequality in đ?‘“.

(đ?‘’ − 1)(9 + đ?‘’) 0 and đ?‘’ < 5

(c)

By solving these inequalities, find set of possible values of đ?‘“.

(5 − đ?‘›)(đ?‘› − 8) > 0 and đ?‘› +5>7

5.

There are 1000 pupils at school, no classes more than 30. Write an inequality to describe the number of classrooms in the school.

5 and đ?‘Ľ 3

3

9.

11.

(3 − đ?‘Ľ)(4 − đ?‘Ľ) and đ?‘Ľ > −7

0

10.

(1 − đ?‘Ľ)(đ?‘Ľ + 8) > 0 and đ?‘Ľ −5>1

12.

2đ?‘Ľ 2 + 7đ?‘Ľ + 3 > 0 and 2đ?‘Ľ + 5 5

14.

3đ?‘Ľ 2 + 5đ?‘Ľ + 2 > 0 and đ?‘Ľ −5>1

16.

3

13.

15.

3

2

2đ?‘Ľ 2 + 3đ?‘Ľ + 1 > 0 and 2 đ?‘Ľ − 6 > −9 3

7đ?‘Ľ 2 − 8đ?‘Ľ + 1 > 0 and 2đ?‘Ľ − + 5 −3 4

18

10. Discriminant The discriminant Vid 0 For equations of the form đ?‘Žđ?‘Ľ 2 + đ?‘?đ?‘Ľ + đ?‘? = 0 the discriminant is đ?‘? 2 − 4đ?‘Žđ?‘?. It is part of the quadratic formula: đ?‘Ľ = So, for đ?‘Žđ?‘Ľ 2 + đ?‘?đ?‘Ľ + đ?‘? = 0: đ?‘? 2 − 4đ?‘Žđ?‘? > 0 ⇒ 2 roots đ?‘? 2 − 4đ?‘Žđ?‘? = 0 ⇒ 1 root đ?‘? 2 − 4đ?‘Žđ?‘? < 0 ⇒ 0 roots

−đ?‘?Âąâˆšđ?‘?2 −4đ?‘Žđ?‘? 2đ?‘Ž

Find set of values of đ?‘˜ for which 2đ?‘Ľ 2 + đ?‘˜đ?‘Ľ + 8 = 0 has no real roots. Vid 1 Comparing đ?‘Žđ?‘Ľ 2 + đ?‘?đ?‘Ľ + đ?‘? = 0 to 2đ?‘Ľ 2 + đ?‘˜đ?‘Ľ + 8 = 0 ⇒ đ?‘Ž = 2, đ?‘? = đ?‘˜, đ?‘? = 8 ⇒ đ?‘? 2 − 4đ?‘Žđ?‘? = đ?‘˜ 2 − 64 No real roots means đ?‘? 2 – 4đ?‘Žđ?‘? < 0 ⇒ đ?‘˜ 2 − 64 < 0 đ?‘˜ 2 − 64 < 0 (đ?‘˜ − 8)(đ?‘˜ + 8) < 0 đ?‘Ś đ?‘Ś = (đ?‘˜ − 8)(đ?‘˜ + 8)

The discriminant decides how many solutions there are to a given quadratic equation. 2 roots means two real solutions (not using imaginary numbers) to a quadratic equation. When the discriminant is positive, it’s square root in the quadratic formula gives two equal and opposite answers and so the rest of the formula gives two answers. When the discriminant is 0 the quadratic formula gives one solution – try it out. If discriminant is negative, you cannot square root it (unless you use imaginary numbers), so the formula can’t be used and so gives no answers. If a function has no roots then its graph doesn’t cross the � axis.

đ?‘˜ –8

8

⇒ −8 < đ?‘˜ < 8 Find values of đ?‘˜ for which 2đ?‘Ľ 2 + đ?‘˜đ?‘Ľ + 8 = 0 has one root. Vid 2 Comparing đ?‘Žđ?‘Ľ 2 + đ?‘?đ?‘Ľ + đ?‘? = 0 to 2đ?‘Ľ 2 + đ?‘˜đ?‘Ľ + 8 = 0 ⇒ đ?‘Ž = 2, đ?‘? = đ?‘˜, đ?‘? = 8 ⇒ đ?‘? 2 − 4đ?‘Žđ?‘? = đ?‘˜ 2 − 64 One root means đ?‘? 2 – 4đ?‘Žđ?‘? = 0 ⇒ đ?‘˜ 2 − 64 = 0 ⇒ đ?‘˜ = Âą8 Show that, for all values of đ?‘˜, the roots of 2đ?‘Ľ 2 + đ?‘˜đ?‘Ľ + 5 = 6 are real and different. Vid 3 2đ?‘Ľ 2 + đ?‘˜đ?‘Ľ + 1 = 6 ⇒ 2đ?‘Ľ 2 + đ?‘˜đ?‘Ľ − 1 = 0 ⇒ đ?‘? 2 − 4đ?‘Žđ?‘? = đ?‘˜ 2 + 1 đ?‘˜ 2 > 0 ⇒ đ?‘˜ 2 + 1 > 0 ⇒ roots are real and different.

Videos at sickmaths.com > A level > Core 1 > Discriminant

đ?‘˜ 2 0 as squaring any number makes it 0 or positive, so đ?‘˜ 2 + 1 1 ⇒ đ?‘˜ 2 + 1 > 0. As đ?‘˜ 2 + 1 is the discriminant so đ?‘˜ 2 + 1 > 0 ⇒ roots are real and different.

19

Exercise 10đ?‘Ž Find the value(s) of đ?‘˜ for which:

Exercise 10đ?‘?

1.

(đ?‘˜ + 1)đ?‘Ľ 2 − 8đ?‘Ľ + 2 = 0 has a repeated root.

2.

đ?‘Ľ 2 + 4đ?‘Ľ + đ?‘˜ + 1 = 0 has no real roots.

3.

2đ?‘Ľ 2 + (đ?‘˜ − 3)đ?‘Ľ + 50 = 0 has repeated roots.

4.

2đ?‘Ľ 2 + (đ?‘˜ + 1)đ?‘Ľ + 2 = 0 has two distinct and real roots.

3.

4.

6.

9đ?‘Ľ 2 + (đ?‘˜ − 3)đ?‘Ľ + 4 = 0 has two real and distinct roots.

7.

đ?‘Ľ 2 + 7đ?‘Ľ − 2đ?‘˜đ?‘Ľ = −9 has imaginary roots only.

8.

đ?‘Ľ 2 + đ?‘˜đ?‘Ľ = −1 has two roots.

1.

By completing the square, find in terms of đ?‘˜, the roots of the equation đ?‘Ľ 2 + 2đ?‘˜đ?‘Ľ = 5. Give answer as surd.

2. Find in terms of đ?‘˜, the roots of the equation đ?‘Ľ 2 − 8đ?‘˜đ?‘Ľ + 8 = 5. Give answer as surd. Show that, for all values of đ?‘˜, the roots of đ?‘Ľ 2 + 3đ?‘˜đ?‘Ľ − 7 = 0 are real and different.

5.

Show that đ?‘˜đ?‘Ľ 2 − 2đ?‘˜đ?‘Ľ + đ?‘˜ = 0 has only one root. 5.

−4(đ?‘Ľ 2

9.

đ?‘˜đ?‘Ľ =

10.

2đ?‘Ľ 2 + (3 − đ?‘˜)đ?‘Ľ = −3 − đ?‘˜ has equal roots.

11.

đ?‘Ľ 2 + 2 + đ?‘˜ = −2đ?‘˜đ?‘Ľ has equal roots.

12.

đ?‘˜đ?‘Ľ 2 + 4đ?‘Ľ + 3 + đ?‘˜ = 0 has two roots that are real.

13.

đ?‘˜đ?‘Ľ 2 + 3đ?‘Ľ = 4 − đ?‘˜ has imaginary roots only.

14.

đ?‘˜(đ?‘Ľ 2 − 1) + 2đ?‘Ľ + 1 = 0 has two real and distinct roots.

Show that, for all values of đ?‘˜, the roots of 2đ?‘˜đ?‘Ľ 2 + 5đ?‘˜ = 0 has no real solutions. 6.

+ 1) has two different roots. 7.

The equation đ?‘˜đ?‘Ľ 2 + 3 = đ?‘˜đ?‘Ľ has real roots. Show that đ?‘˜(đ?‘˜ − 12) > 0 The equation −4đ?‘˜đ?‘Ľ 2 + đ?‘Ľ − đ?‘˜ = 0 has no roots. 1 1 Show that − < đ?‘˜ < 4

4

8. Write down the values of đ?‘˜ for which the equation đ?‘˜đ?‘Ľ 2 + 5đ?‘˜đ?‘Ľ + 2 = 0 has equal roots. 9. đ?‘“(đ?‘Ľ) = đ?‘Ľ 2 + 4đ?‘Ľ − 8đ?‘? has equal roots, find đ?‘? 10. đ?‘“(đ?‘Ľ) = đ?‘˜đ?‘Ľ 2 − 2đ?‘Ľ + 3 has equal roots, find đ?‘˜

15. 3đ?‘Ľ(1 − đ?‘˜) + 3 = −đ?‘˜đ?‘Ľ 2 has no real roots. 16. 17. 18.

đ?‘˜đ?‘Ľ 2 + đ?‘˜ = 8đ?‘Ľ − 2đ?‘Ľđ?‘˜ has no real roots. đ?‘˜đ?‘Ľ 2 + 1 = đ?‘˜(4đ?‘Ľ − 3) has two different roots. r

đ?‘Ľ 2 + đ?‘˜đ?‘Ľ + 8 = đ?‘˜ has no real solutions for đ?‘Ľ. đ?‘Ľ 2 + đ?‘˜đ?‘Ľ + (đ?‘˜ + 3) = 0 has different real roots. root find the possible values of b

20

Vid 0

11. Differentiation Notation đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ

If � = �(�) is the equation of a graph, �′(�) is the formula for the gradient on any point � on that graph. The meaning of � ′ ′(�) will be explained in module C2.

Vid 1

Differentiating � = �(�) ⇒

đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ

đ?‘‘đ?‘Ś

= �′(�), differentiating

đ?‘‘đ?‘Ľ

⇒

đ?‘‘2đ?‘Ś đ?‘‘đ?‘Ľ 2

= �′′(�)

means gradient of a tangent to the curve đ?‘Ś = đ?‘“(đ?‘Ľ): đ?‘Ś inaccurate tangent

Tangents are straight lines that touch a curve. Practically, it’s impossible to draw such a line without intersecting the curve in two places. You have to use the technique demonstrated here.

tangent đ?‘‘đ?‘Ś

P đ?‘Ś = đ?‘“(đ?‘Ľ)

Differentiation is a way of finding the gradient of curvy graphs. Differentiation is part of something called calculus.

đ?‘‘đ?‘Ľ and đ?‘‘đ?‘Ś are infinitely small.

đ?‘‘đ?‘Ľ đ?‘Ľ

A tangent at a point đ?‘ƒ is got by connecting it to the closest point to đ?‘ƒ on the same curve, horizontally đ?‘‘đ?‘Ľ away and vertically distance đ?‘‘đ?‘Ś away. Hence đ?‘‘đ?‘Ś gradient of tangent is . đ?‘‘đ?‘Ľ

How to differentiate Vid 2.1 đ?‘‘đ?‘Ś đ?‘Ś = đ?‘Žđ?‘Ľ đ?‘› + â‹Ż + đ?‘?đ?‘Ľ + đ?‘? ⇒ = đ?‘›đ?‘Žđ?‘Ľ đ?‘›âˆ’1 + â‹Ż + đ?‘? + 0

đ?‘Žđ?‘Ľ đ?‘› differentiates to đ?‘›đ?‘Žđ?‘Ľ đ?‘›âˆ’1 . đ?‘?đ?‘Ľ differentiates to đ?‘?. Constants i.e. numbers, differentiate to 0.

Find gradient on graph đ?‘Ś = 5đ?‘Ľ 2 – 2đ?‘Ľ + 1000 at (3, 1039). Vid 2.2 đ?‘‘đ?‘Ś đ?‘Ś = 5đ?‘Ľ 2 − 2đ?‘Ľ + 1000 ⇒ = 2 Ă— 5đ?‘Ľ 2−1 − 2 + 0 = 10đ?‘Ľ − 2.

5đ?‘Ľ 2 differentiates to 10đ?‘Ľ. −2đ?‘Ľ differentiates to −2. 1000 differentiates to 0.

đ?‘‘đ?‘Ľ

�=3⇒

đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ

đ?‘‘đ?‘Ľ

= 30 − 2 = 28

Find coordinate on graph đ?‘Ś = 4đ?‘Ľ −2 – 3đ?‘Ľ where gradient is – 4. đ?‘‘đ?‘Ś đ?‘Ś = 4đ?‘Ľ −2 – 3đ?‘Ľ ⇒ = −2 Ă— 4đ?‘Ľ −2−1 − 3 = −8đ?‘Ľ −3 − 3

Substitute đ?‘Ľ = 3 into 10đ?‘Ľ − 2 ⇒ 10(3) − 2 = 28. So gradient is 28.

Vid 2.3

Substitute đ?‘Ľ = 2 into 4đ?‘Ľ −2 – 3đ?‘Ľ ⇒ đ?‘Ś = 5 Hence đ?‘Ś = 5 when đ?‘Ľ = 2 ⇒ (2, −5)

đ?‘‘đ?‘Ľ đ?‘‘đ?‘Ś

gradient = −4 ⇒ = −8đ?‘Ľ −3 − 3 = −4 ⇒ đ?‘Ľ = 2 đ?‘‘đ?‘Ľ ⇒ đ?‘Ś = 4(2)−2 − 3(2) = −5 ⇒ (2, −5)

đ?‘Ś= đ?‘Ś = đ?‘“(đ?‘Ľ) , where đ?‘“(đ?‘Ľ) = đ?‘“(đ?‘Ľ) =

16 đ?‘Ľ2

+

3đ?‘Ľ 2 −2đ?‘Ľ √đ?‘Ľ

=

16 đ?‘Ľ2

+

16đ?‘Ľ −2

3đ?‘Ľ 2 −2đ?‘Ľ √đ?‘Ľ

, find �’(�)

3 2

+ 3đ?‘Ľ − 2đ?‘Ľ ⇒ đ?‘“’(đ?‘Ľ) =

đ?‘‘đ??´ đ?‘‘đ?‘&#x;

đ?‘‘đ??´ đ?‘‘đ?‘&#x;

9

−32đ?‘Ľ −3

1 2

tangent đ?‘Ľ

� = �2 ⇒

đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ

+ đ?‘Ľ −đ?‘Ľ 2

đ?‘Ś = đ?‘Ľ2 tangent đ?‘Ľ

= 2�, � = 3 ⇒

đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ

= 16đ?‘Ľ −2 +

đ?‘‘đ?‘Ś

đ?‘‘đ?‘Ľ

1

−

3 2

2đ?‘Ľ √đ?‘Ľ 2đ?‘Ľ 1

đ?‘Ľ2

1

đ?‘‘đ?‘Ś đ?‘‘đ?‘Ľ đ?‘‘đ??´

= �′(�) so

Tangents are straight lines and so have equations of the form đ?‘Ś = đ?‘šđ?‘Ľ + đ?‘?. đ?‘Ś = 6đ?‘Ľ + đ?‘? is the equation, at this stage, of the tangent at đ?‘Ľ = 3.

Vid 4

=6 1

normal gradient = − ⇒ đ?‘Ś = − đ?‘Ľ + đ?‘? 6

1 đ?‘Ľ2

−

đ?‘‘đ?‘&#x;

đ?‘‘đ?‘Ś

= 2�, � = 3 ⇒

√đ?‘Ľ 3đ?‘Ľ 2

đ??´ = đ?‘“(đ?‘&#x;) ⇒ = đ?‘“′(đ?‘&#x;) đ?‘‘đ?‘&#x; Hence differentiating đ??´ = đ?œ‹đ?‘&#x; 2 đ?‘‘đ??´ ⇒ = 2 Ă— đ?œ‹đ?‘&#x; 2−1 = 2đ?œ‹đ?‘&#x;

= 2(3) = 6

đ?‘‘đ?‘Ľ

3đ?‘Ľ 2

= 16đ?‘Ľ −2 + 3đ?‘Ľ − 2đ?‘Ľ 2 đ?‘Ś = đ?‘“(đ?‘Ľ) ⇒

tangent gradient = 6 ⇒ đ?‘Ś = 6đ?‘Ľ + đ?‘? substitute (3, 9) into đ?‘Ś = 6đ?‘Ľ + đ?‘? ⇒ (9) = 6(3) + đ?‘? ⇒ đ?‘? = −9 ⇒ đ?‘Ś = 6đ?‘Ľ − 9

� = �2 ⇒

√đ?‘Ľ

1 2

Vid 2.5

Find equation of normal at (3, 9) on curve with equation đ?‘Ś = đ?‘Ľ 2 . đ?‘Ś

3đ?‘Ľ 2 −2đ?‘Ľ

−

Finding equations of tangents and normal Vid 3 Find equation of tangent at (3, 9) on curve with equation y = x2. đ?‘Ś = đ?‘Ľ2

+

= 16đ?‘Ľ −2 +

= 2đ?œ‹đ?‘&#x;

đ?‘Ś

đ?‘Ľ2

Vid 2.4

1 2

đ??´ = đ?œ‹đ?‘&#x; 2 , where đ??´ and đ?‘&#x; are variables. Find

16

6

Normals are straight lines perpendicular to their tangents, so a normal’s gradient is the negative reciprocal of its tangent.

1

substitute (3, 9) into đ?‘Ś = − đ?‘Ľ + đ?‘? normal

1

⇒ (9) = − (3) + đ?‘? ⇒ đ?‘? = 6 1

19

6

2

⇒ đ?‘Ś= − đ?‘Ľ +

6 19 2

Videos at sickmaths.com > A level > Core 1 > Differentiation

21

Exercise 11đ?‘Ž đ?‘‘đ?‘Ś Find for each of the following:

Exercise 11đ?‘‘ Find the equation of the (a) tangent and (b) normal to each of the following graphs at the given coordinates:

đ?‘‘đ?‘Ľ

1.

đ?‘Ś = 2đ?‘Ľ + 5

3.

đ?‘Ś = 6đ?‘Ľ 5 + đ?‘Ľ −2

5. 7. 9. 11.

đ?‘Ś=

−3đ?‘Ľ 5

−

2 3

đ?‘Ľ4

2

2.

đ?‘Ś = đ?‘Ľ − 5đ?‘Ľ 2

1.

4.

đ?‘Ś = −2đ?‘Ľ −4 + 1

3.

6.

9

1

8.

2

5.

1

đ?‘Ś = đ?‘Ľ + đ?‘Ľ −1 − √đ?‘Ľ 3

7. 10.

4

đ?‘Ś=

3

đ?‘Ś = đ?‘Ľ −5 + 5đ?‘Ľ −2 − 6đ?‘Ľ 4

đ?‘Ľ(đ?‘Ľ −1

1 4

1

2

5

đ?‘Ś = đ?‘Ľ2 +

14.

đ?‘Ś = (đ?‘Ľ + 1)3

3

−

2 5

− đ?‘Ľ )

12.

đ?‘Ľ2

9.

−

6

11.

đ?‘Ľ −1

1

13.

đ?‘Ś = −4đ?‘Ľ −2 − 5đ?‘Ľ −2 + 6

đ?‘Ś = đ?‘Ľ −2 − 2 at 1 (−1, − )

1

2

2

2

1

đ?‘Ś = đ?‘Ľ −1 + đ?‘Ľ −5

4.

1

(−1, − )

3

7

đ?‘Ś = đ?‘Ľ 3 + đ?‘Ľ −3 − 4

đ?‘Ś = −12 at (−1, −12)

đ?‘Ś = đ?‘Ľ 4 − đ?‘Ľ at 2

đ?‘Ś = đ?‘Ľ −3 + đ?‘Ľ −4

2.

đ?‘Ś = 2đ?‘Ľ − 6 at (6,2)

6.

đ?‘Ś = đ?‘Ľ 3 + đ?‘Ľ at đ?‘Ľ = 2 đ?‘Ś=

1 √đ?‘Ľ

đ?‘Ś = đ?‘Ľ 3 + đ?‘Ľ −2 at đ?‘Ľ=2

8.

at đ?‘Ľ = 4

đ?‘Ś=

10.

đ?‘Ś = đ?‘Ľ 2 − đ?‘Ľ at đ?‘Ś = 2

√đ?‘Ľ đ?‘Ľ

at đ?‘Ľ = 9

đ?‘Ś = đ?‘Ľâˆšđ?‘Ľ at đ?‘Ś = 2

12.

đ?‘Ś = đ?‘Ľ(đ?‘Ľ − 5) on đ?‘Ľ axis

đ?‘Ś = đ?‘Ľ 2 on đ?‘Ľ axis

Exercise 11đ?‘? Given đ?‘Ś = đ?‘“(đ?‘Ľ), find (a) đ?‘“′(đ?‘Ľ) (b) 1.

đ?‘“(đ?‘Ľ) = √đ?‘Ľ +

3.

đ?‘“(đ?‘Ľ) =

5.

6đ?‘Ľ 2 3

√đ?‘Ľ

5 đ?‘Ľ

+ 2đ?‘Ľ −2

3−đ?‘Ľ

đ?‘“(đ?‘Ľ) =

đ?‘‘ 2đ?‘Ś

6√đ?‘Ľ

7.

đ?‘“(đ?‘Ľ) = 2đ?‘Ľ 2 −

9.

đ?‘“(đ?‘Ľ) = (đ?‘Ľ − 2)(đ?‘Ľ + 3)2

11.

đ?‘“(đ?‘Ľ) =

13.

đ?‘“(đ?‘Ľ) = −

3

2.

đ?‘“(đ?‘Ľ) = 1 − √đ?‘Ľ

4.

đ?‘“(đ?‘Ľ) =

6.

đ?‘Ľ2

Exercise 11đ?‘‘

đ?‘‘đ?‘Ľ 2

đ?‘“(đ?‘Ľ) =

đ?‘Ľ3 6√đ?‘Ľ

−

đ?‘Ľ2

−

đ??´ = đ?œ‹đ?‘&#x; 2 . Find đ?‘&#x; at

3.

Find coordinates of points on graph đ?‘Ś = đ?‘Ľ 3 + 9đ?‘Ľ with gradient 0.

4.

Find gradients on graph đ?‘Ś = (đ?‘Ľ − 1)(đ?‘Ľ + 2)(đ?‘Ľ − 5) where it intercepts the coordinate axes.

5.

Find another coordinate that has the same gradient as (1,2) on graph đ?‘Ś = 3đ?‘Ľ 3 − đ?‘Ľ.

6.

Curve đ??ś has equation đ?‘Ś = đ?‘Ľ 2 . The normal to

6đ?‘Ľ 2

đ?‘“(đ?‘Ľ) =

10.

đ?‘“(đ?‘Ľ) = (đ?‘Ľ − 3)3

12.

đ?‘“(đ?‘Ľ) =

14.

đ?‘“(đ?‘Ľ) =

√đ?‘Ľ

2.

1

8.

đ?‘Ľ

The gradient on đ?‘Ś = đ?‘Ľ 2 − 6 at (đ?‘›, 2) is 7. 2 Find đ?‘›.

đ?‘Ľ2

đ?‘Ľ 2 −đ?‘Ľ+1

−đ?‘Ľ 2

√đ?‘Ľ 5

1

đ?‘Ľ

4đ?‘Ľ 2

+

2 3√đ?‘Ľ

đ?‘Ś = 2đ?‘Ľ − 6 at (6,2)

√đ?‘Ľ(2đ?‘Ľ+5)2 2

5đ?‘Ľâˆ’1 đ?‘Ľ −2

−1 2

đ?‘Ľ √đ?‘Ľ

3.

đ?‘Ś = đ?‘Ľ 4 − đ?‘Ľ at (−1, − )

4.

đ?‘Ś = đ?‘Ľ −2 − 2 at (−1, − )

5.

đ?‘Ś = đ?‘Ľ 3 + đ?‘Ľ at đ?‘Ľ = 2

6.

đ?‘Ś = đ?‘Ľ 3 + 2đ?‘Ľ −2 at đ?‘Ľ = 2

7.

đ?‘Ś=

8.

đ?‘Ś=

9.

đ?‘Ś = đ?‘Ľ 2 − đ?‘Ľ at đ?‘Ś = 2

10.

đ?‘Ś = đ?‘Ľ(√đ?‘Ľ − đ?‘Ľ) at đ?‘Ś = 2

11.

đ?‘Ś=

12.

đ?‘Ś=

13.

đ?‘Ś = (đ?‘Ľ − 3)2 on đ?‘Ľ axis

15.

3)2

1

1

2

2

at đ?‘Ľ = 4

at đ?‘Ś = −17

đ?‘Ś = (đ?‘Ľ −

on đ?‘Ś axis

14. 16.

2

7.

Curve đ??ś with equation đ?‘Ś = đ?‘Ľ 2 . The normals to đ??ś at đ?‘Ľ coordinates −1 and 3 intercept at đ??ľ. Find coordinates of đ??ľ.

8.

The tangent to curve đ??ś at point đ??´ with coordinate (1,1) interests đ??ś again at point đ??ľ. Curve đ??ś has equation đ?‘Ś = đ?‘Ľ 3 . Find distance from đ??´ to đ??ľ.

9.

Curve đ??ś has equation đ?‘Ś = đ?‘“(đ?‘Ľ). = 2đ?‘Ľ . đ?‘‘đ?‘Ľ Find the equation of the tangent on curve đ??ś at point (1,1).

10.

đ?‘ = đ?‘˘đ?‘Ą + đ?‘Ą 2 where đ?‘ and đ?‘Ą are variables;

1 2

√đ?‘Ľ đ?‘Ľ

1

1

đ?‘Ś = −12 at (−1, −12)

3

= 0.

đ??ś at (1, ) intercepts it again at point đ??ľ. Find 2 coordinates of đ??ľ.

2.

đ?‘Ľâˆ’3

đ?‘‘đ?‘&#x;

4

(đ?‘Ľâˆ’1)2

1.

1

đ?‘‘đ??´

√đ?‘Ľ

Exercise 11đ?‘? Find the gradient(s) on each graph at given points:

√đ?‘Ľ

1

1.

at đ?‘Ľ = 9

2−đ?‘Ľ+√đ?‘Ľ √đ?‘Ľ

at đ?‘Ś = −17

đ?‘Ś = đ?‘Ľ(đ?‘Ľ − 5) on đ?‘Ľ axis

đ?‘‘đ?‘Ś

1 2

and � is a constant. Find � when

đ?‘‘đ?‘ đ?‘‘đ?‘Ą

= 2.

đ?‘Ś = đ?‘Ľ(đ?‘Ľ − 5) on đ?‘Ś axis 22

Vid 0

12. Integration Notation âˆŤ đ?‘“(đ?‘Ľ) đ?‘‘đ?‘Ľ means integrate đ?‘“(đ?‘Ľ)

đ?‘? is a constant of integration, which can be any number. To explain this: đ?‘‘đ?‘Ś đ?‘Ś = đ?‘“(đ?‘Ľ) + đ?‘? ⇒ = đ?‘“‘(đ?‘Ľ) đ?‘‘đ?‘Ľ As integration is the reverse of differentiation: âˆŤ đ?‘“‘(đ?‘Ľ)đ?‘‘đ?‘Ľ = đ?‘“(đ?‘Ľ) + đ?‘?

Vid 1

How to integrate đ?’‚ âˆŤ đ?’‚đ?’™đ?’? đ?’…đ?’™ = đ?’?+đ?&#x;? đ?’™đ?’?+đ?&#x;? + đ?’‘ âˆŤ đ?’ƒ đ?’…đ?’™ = đ?’ƒđ?’™ + đ?’’ âˆŤ đ?’‚đ?’™đ?’? + â‹Ż + đ?’ƒ đ?’…đ?’™ = âˆŤ đ?’‚đ?’™đ?’? đ?’…đ?’™ + â‹Ż + âˆŤ đ?’ƒ đ?’…đ?’™ đ?’‚ = đ?’™đ?’?+đ?&#x;? + â‹Ż + đ?’ƒđ?’™ + đ?’„

đ?‘? is sum of constants of integration from each part of the expression. 5

đ?’?+đ?&#x;?

Find âˆŤ 5đ?‘Ľ 2 − 2đ?‘Ľ + 1000 đ?‘‘đ?‘Ľ. Vid 2 5 đ?‘Ľ 2+1 − âˆŤ 5đ?‘Ľ 2 − 2đ?‘Ľ + 1000 đ?‘‘đ?‘Ľ = = Find âˆŤ 16

âˆŤ đ?‘Ľ2 +

16

+

3đ?‘Ľ 2 −2đ?‘Ľ

đ?‘Ľ2 √đ?‘Ľ 3đ?‘Ľ 2 −2đ?‘Ľ √đ?‘Ľ

đ?‘‘đ?‘Ľ

2+1 5 3 đ?‘Ľ − 3

2 1+1

đ?‘Ľ1+1 + 1000đ?‘Ľ + đ?‘?

3

1

1

âˆŤ 16đ?‘Ľ −2 + 3đ?‘Ľ 2 − 2đ?‘Ľ 2 đ?‘‘đ?‘Ľ =

Vid 3 3

−1

âˆŤ 5đ?‘Ľ 2 đ?‘‘đ?‘Ľ = đ?‘Ľ 3 + đ?‘˜ 3 âˆŤ −2đ?‘Ľ đ?‘‘đ?‘Ľ = − đ?‘Ľ 2 + đ?‘™ âˆŤ 1000 đ?‘‘đ?‘Ľ = 1000đ?‘Ľ + đ?‘š đ?‘˜ + đ?‘™ + đ?‘š = đ?‘?

đ?‘Ľ 2 + 1000đ?‘Ľ + đ?‘?

1

đ?‘‘đ?‘Ľ = âˆŤ 16đ?‘Ľ −2 + 3đ?‘Ľ 2 − 2đ?‘Ľ 2 đ?‘‘đ?‘Ľ =

Integration is the inverse of differentiation. Integration and differentiation, together, are called calculus.

5

2

2

= 3

1 −1 1 −1

Ă— 16đ?‘Ľ −1 +

1 5 2

2

5

Ă— 3đ?‘Ľ 2 − 5

1 3 2

3

Ă— 2đ?‘Ľ 2 + đ?‘?

2

3

Ă— 16đ?‘Ľ −1 + Ă— 3đ?‘Ľ 2 − Ă— 2đ?‘Ľ 2 + đ?‘? 5

3

Ă— 16đ?‘Ľ −1 + Ă— 3đ?‘Ľ 2 − Ă— 2đ?‘Ľ 2 + đ?‘? 6

5

5

4

3

3

= −16đ?‘Ľ −1 + đ?‘Ľ 2 − đ?‘Ľ 2 + đ?‘? 5

3

How to find the constant of integration Vid 4 A curve has equation đ?‘Ś = đ?‘“(đ?‘Ľ) and passes through (2, 7). Given đ?‘“’(đ?‘Ľ) = 2đ?‘Ľ, find đ?‘“(đ?‘Ľ). đ?‘“(đ?‘Ľ) = âˆŤ 2đ?‘Ľ đ?‘‘đ?‘Ľ = đ?‘Ľ 2 + đ?‘? substitute (2, 7) into đ?‘“(đ?‘Ľ) ⇒ (7) = (2)2 + đ?‘? ⇒ đ?‘? = 3 đ?‘“(đ?‘Ľ) = đ?‘Ľ 2 + 3

As integration is the reverse of differentiation, đ?‘“(đ?‘Ľ) = âˆŤ đ?‘“‘(đ?‘Ľ) đ?‘‘đ?‘Ľ (2,7) is on the line đ?‘Ś = đ?‘“(đ?‘Ľ) means when đ?‘Ľ = 2, đ?‘Ś = 7. We know đ?‘Ś = đ?‘Ľ 2 + đ?‘?. Substituting đ?‘Ľ = 2 and đ?‘Ś = 7 into đ?‘Ś = đ?‘Ľ 2 + đ?‘? ⇒ đ?‘? = 3 ⇒ đ?‘Ś = đ?‘Ľ 2 + 3.

Videos at sickmaths.com > A level > Core 1 > Integration

23

Exercise 12𝑎 Integrate the following:

Exercise 12𝑐 Find 𝑓(𝑥) in each case:

1.

∫ 7𝑥 2 + 6𝑥 − 5 𝑑𝑥

2.

∫ 3𝑥 4 + 6 − 5𝑥 2 𝑑𝑥

3.

∫ −4𝑥 − 3𝑥 5 + 6 𝑑𝑥

4.

∫ −𝑥 − 6𝑥 5 + 6 𝑑𝑥

5.

∫ 3𝑥 −4 − 8 𝑑𝑥

6.

∫ 3𝑥 + 2𝑥(𝑥 − 9) 𝑑𝑥

7.

∫ 𝑥 2 + 5 𝑥 7 + 9𝑥 − 1 𝑑𝑥

2

8.

∫

∫ −3𝑥 −2 𝑑𝑥

10.

9. 11.

∫ 6 𝑑𝑥

13.

∫ − 5 𝑥 7 + 7.5 𝑑𝑥

15.

∫2𝑥

1

3

−

3 4

− 1.8𝑥 𝑑𝑥

1 2

𝑥 2 +9𝑥 2

− 1 𝑑𝑥

A curve has equation 𝑦 = 𝑓(𝑥) and passes through (2, 7). 𝑓′(𝑥) = 2𝑥.

2.

A graph has equation 𝑦 = 𝑓(𝑥) and passes through (1, 0). 𝑓′(𝑥) = 3𝑥 2 + 6𝑥 − 5.

3.

A curve has equation 𝑦 = 𝑓(𝑥) and passes 𝑑𝑦 through (2, 11). = 3𝑥 2 . 𝑑𝑥

1

∫ 𝑥 −2 𝑑𝑥

4.

12.

∫ −8 𝑑𝑥

14.

∫ 0.8𝑥 7 𝑑𝑥

16.

1.

A graph has equation 𝑦 = 𝑓(𝑥) and passes through (4,40). 𝑓 ′ (𝑥) = 6√𝑥.

∫(𝑥 − 5)3 𝑑𝑥

Exercise 12𝑏 Integrate the following: 7𝑥 2

1.

∫

+ 6√𝑥 𝑑𝑥

3.

∫ 𝑥 −2 − 2√𝑥 𝑑𝑥

5.

∫

7.

∫

9.

∫ 𝑥3 −

11.

∫

∫

1

4.

∫ 𝑥 −2 − 2√𝑥 𝑑𝑥

𝑑𝑥

6.

∫

𝑑𝑥

8.

∫

𝑑𝑥

10.

∫ 𝑥3 −

√𝑥+2𝑥 −4 −2𝑥 3 𝑑𝑥 3√𝑥

12.

∫

14.

∫

6

5+9.6𝑥 2 3𝑥−2𝑥 4 √𝑥 6

3𝑥−2𝑥 4 √𝑥

∫(𝑥 − 4)3 𝑑𝑥

𝑥3

+ 6√𝑥 − 5 𝑑𝑥

6

1

5+9.6𝑥 2 3𝑥−2𝑥 4 √𝑥 6

1

13.

7𝑥 2

2.

𝑥3

𝑑𝑥 𝑑𝑥

3𝑥−2𝑥 4 √𝑥

𝑑𝑥

1

√𝑥+2𝑥 −4 −2𝑥 3 𝑑𝑥 3√𝑥 (2𝑥+1)2 𝑥 −1

𝑑𝑥

24

13. The Answers These are clearly not the answers to the questions in the book but they are the answers to our problems, our dreams and what the world is searching for. This is ancient spiritual wisdom that is logical and simple. You might be wondering what wisdom is doing at the back of a maths book. Well, the first universities the world saw came from Morocco and Spain a thousand years ago, where they taught spiritual ideas alongside everything else. They were in a society hundreds of years ahead of the world. Without this spiritual wisdom we will make material progress but we will also destroy the world, as we are.

The secret of happiness You will not find paradise (find peace of mind, live a beautiful life, make paradise on Earth and find it in the next life) until you believe. You will not believe (have hope, realise the true reality of the world, become aware of God) until you love each other. You will not love each other until you offer each other peace (to wish people well from your heart, words and actions) when you meet each other.

How to change the world Your leaders come from among you, meaning, if your leaders are good or corrupt, they are a reflection of your society. They are not aliens from space but people from among you. If they are corrupt that means your society is corrupt. The corruption in your leaders is only more obvious then in society because we focus on them instead of ourselves. Of course, if people have a habit of being dishonest to themselves and reflecting over their behaviour then when they become leaders, they will continue. When a person’s powers increase, they can hurt people with greater ease. My brothers and sisters, if you truly want to stop the suffering of your brothers, sisters, the animals and the environment, you must change. Change starts with being honest to yourself. Honesty comes from humility – the ability the accept truth from wherever it comes. Humility comes from destroying your ego.

How to fix the economy Interest, the money paid for borrowing money, destroys the economy. Firstly, it is unnatural, as it assumes the money has the right to make more money (if you put money in the bank, you get paid interest, at no risk – unlike other business transactions). Interest sucks up money from the poor to the rich, as it is the rich that lend and the poor that borrow. It also creates an unnatural need for the economy to keep growing to survive – the economy has to raising more and more money to pay back interest it took to fund its businesses. Stock piling money also harms the economy. Money is to the economy what blood is to the body. If large amounts of money is kept in bank accounts unused (as it is by rich people to amass interest) then the economy suffers. Such unused money should be taxed to encourage it to be reinvested in the economy. These principles were used in an Austrian town called Worgel during the last world recession that preceded world war one and made that tiny town very rich. This experiment was shut down by the big banks of Austria.

How to save the environment It’s important to remind you that we damage the environment from our actions, it’s not just the fault of our governments. We have known that we have been damaging the environment long before we knew about global warming. Here in the UK we have long been building tall chimneys so the fumes land in Scandinavia and destroy their forests. One of the most obscene ways we are damaging the environment is the way we treat animals to get meat. We feed them drugs to keep them eating to get them fat to kill them. We make enough space for the huge number of animals we kill by destroying thousands of miles of forest in countries in which people are poor and it’s easy to take their land. The methane from the millions of animals we kill (while they’re still alive) is one the greatest contributors to global warming. The question is, why do we continue? The answer is we do it for the same reason we take drugs. We take drugs, knowing they’re bad for us, because of the eternal spiritual principle that if we forget God we will

forget our own souls. Forgetting God means to ignore God, not to be grateful to God. When we forget God we forget our spiritual needs, our connection to nature and the other living things on this planet. We can only forget God if we are dishonest to ourselves and that’s where the journey of self-destruction begins. When we are dishonest to ourselves, we spend our lives crazily chasing unnecessary desires at the cost of our health and happiness. To change we must stop blaming governments and big companies (even though they are also to blame, but as I explained earlier, they have come into existence because of our corrupt behaviour) and to look at yourself. The reason why we are in this mess is because of our selfishness which has led to us ignoring spiritual values which get in the way of our selfish desires. The simple solution is to believe in God again because this will awaken truths within us that helps us unite as brothers and sisters to make a beautiful world everyone.

Proving God exists I have already explained why we need God - because when we forget God we remove ourselves from our essential spirit and so end up becoming zombies that destroy ourselves and the world. Proving God is easy – God is proof of God. God responds to anyone who sincerely wishes to know God and listen to God. But to want to even want to talk to God you must be convinced that God should exist. That again is easy. If you compare the probability of God existing versus God not existing, the honest conclusion must be God should exist. There are millions of incredibly complicated creatures, each made of amazing organs and bits, for them to be made by a series of accidents is very unlikely. You may convince yourself that it is still possible for these amazing creatures to have been made by a series of accidents but you have to admit it is more likely that they weren’t. God makes it simple to prove God’s existence, as long as we’re honest to ourselves, so that we can benefit from knowing God. Some of the causes for not believing in God is the belief that believing in God is not logical or against scientific evidence. In fact, that opposite is true, it is entirely scientific (if science is defined as following evidence). For example, people say evolution disproves the existence of God, not true. Evolution simply describes the way living things were made in stages over millions of years. This is exactly in line with the description of God in the Quran (Muslim Holy Scripture), which claims to also support all other previous religious scriptures such as the Torah (Jewish Holy Scripture) and the New Testament.

How to be successful in life To become successful means to do great things with your life and be happy with it. You can do supposedly great things like make tonnes of money and be extremely miserable, so you have to be careful. Generally, the key to being successful at anything is consistency. The path to success is full ups and downs because you have to learn a lot of stuff that other people didn’t want to learn. They, the people who weren’t successful, didn’t want to learn the lessons you will have to in order to be successful, because they didn’t want to learn the necessary tough lessons. So, by being consistent, you learn, you have to be patient, have a long term view and so many other things that come with being consistent. But to also be happy as well be successful, whatever it is that you do, it should have an overall benefit to humanity. For example, when people trade cocoa beans on the stock market, they only do it mainly to make money. Trading like this is very dull and you will find traders tend not to be very clever (which is different to what you see in films), nobody else will really do these kinds of jobs. Also, even though they have lots of money, traders tend to go into debt. This is because they haven’t realised what really makes people happy and so they keep spending more and more money, never finding happiness. Alternatively, you could do something that makes the world a better place. By doing so you automatically begin to make your business viable because people want stuff that makes their lives better. At the same time, you will feel energised and happy doing your job as being good is the essential nature and need of human beings.

Ask me your questions I most probably have the answers to your questions. I know what I know because I keep trying to align my spirit with nature and God. When you do that, everything becomes simple and obvious. I’m definitely not perfect and my answers will only go so far. I am like the first rung of the ladder, use me to go even higher and I will learn from you. I want to answer your questions because I love you. So feel free to ask me questions by commenting on my videos which can be found on the home page at sickmaths.com

14. Quotes from Buddha, Jesus, Muhammad, Einstein and Steve Jobs Here’s to the crazy ones – the misfits, the rebels, the troublemakers, the round pegs in the square holes. The ones who see things differently – they’re not fond of rules. You can quote them, disagree with them, glorify or vilify them, but the only thing you can’t do is ignore them because they change things. They push the human race forward, and while some may see them as the crazy ones, we see genius, because the ones who are crazy enough to think that they can change the world, are the ones who do. Do not consider any act of kindness insignificant, even meeting your brother with a cheerful face. Imagination is more important than knowledge. Look for me among the weak and infirm, for you are succoured and provisioned because of those among you who are weak and helpless. If you want to be perfect, go, sell your possessions and give to the poor, and you will have treasure in heaven. Thousands of candles can be lighted from a single candle, and the life of the candle will not be shortened. Happiness never decreases by being shared. Any intelligent fool can make things bigger, more complex, and more violent. It takes a touch of genius — and a lot of courage — to move in the opposite direction. The two things that made early generations of this nation happy are abstinence and certitude, and the two things that would destroy the ending generations of this nation are stinginess and having high (materialistic) aspirations. But I say to you, Love your enemies and pray for those who persecute you, so that you may be sons of your Father who is in heaven; for he makes his sun rise on the evil and on the good, and sends rain on the just and on the unjust. Do not dwell in the past, do not dream of the future, concentrate the mind on the present moment.

Turn static files into dynamic content formats.

Create a flipbook
Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.