C2

Ch1 Algebra and

functions

Ch 2Sine and Cosine

rule

Ch 3Exponentials

and Logarithms

Ch 4Coordinate geometry

Ch 5The binomial expansion

Ch 6Radians

Ch 7Trigonometric

functions

Ch 8Differentiation

Ch 9Trigonometric

Identities

Ch 10Integration

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Chapter 1 – Algebra and functions egwgwgw

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Chapter 2 – Cosine and Sine rule egwgwgw

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Chapter 3 – Exponentials and logarithms egwgwgw

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Chapter 4 – Coordinate geometry in the xy plane egwgwgw

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Chapter 5 - The Binomial Theorem The binomial theorem tells us how to expand brackets like (3 - 2x)11 quickly.

If we have some brackets to a power then there are three ways we can expand them.

If n (the power) is quite small then we can just expand them as normal. (a + b)2 = (a + b)(a + b)

= 1a2 + 2a1b1 + 1b2

(a + b)3 = (a + b)(a + b)(a + b)

= 1a3 + 3a2b1 + 3a1b2 + b3

(a + b)4 = (a + b)(a + b)(a + b)(a + b)

= 1a4 + 4a3b1 + 6a2b2 + 4a1b3 + b4

After this it gets a little time consuming so we start to use…

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Pascal’s triangle If we look at the first few expansions then we start to notice a pattern in the coefficients (the numbers in front of the x, x2, x3, etc). (a + b)2 = 1a2 + 2a1b1 + 1b2 (a + b)3 = 1a3 + 3a2b1 + 3a1b2 + 1b3 (a + b)4 = 1a4 + 4a3b1 + 6a2b2 + 4a1b3 + 1b4 The coefficients when n (the power) is four are the same as the numbers on the fifth row of Pascal’s triangle.

Pascal’s triangle

1 1 1

1 2 1 1 3 3 1

1 4 6 4 1

Each number is equal to the two above it added together.

This predicts that (a + b)5 = 1a5 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + 1b5

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Expand (2x + 3y)4 ----------------------------------------------------------------------- Pascal’s triangle says (a + b)4 = 1a4 + 4a3b1 + 6a2b2 + 4a1b3 + 1b4 (2x + 3y)4 = 1.(2x)4 + 4.(2x)3.(3y)1 + 6.(2x)2(3y)2 + 4.(2x)1(3y)3 + 1.(3y)4

= 16x4 + 4.8x3.3y + 6.4x2.9y2 + 4.2x1.27y3 + 1.81y4 = 16x4 + 96x3y + 216x2y2 + 216xy3 + 81y4

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The Binomial theorem For brackets with powers much bigger than this like (3 + 2x)17 we start to use the Binomial theorem.

(1 + x)7 = 1 + 7x + 7 X 61 X 2

x2 + 7 X 6 X 51 X 2 X 3

x3 + .

.

= 1 + 7x + 21x2 + 35x3 + .

.

(1 + x)n = 1 + nx + n(n – 1)1 X 2

x2 + n(n – 1)(n – 2)1 X 2 X 3

x3 + . .

.

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(1 + x)n = 1 + nx + n(n – 1)1 X 2

x2 + n(n – 1)(n – 2)1 X 2 X 3

x3 + . . .

(1 + 2x)9 = 1 + 9 X (2x) + 9 X 81 X 2

(2x)2 + 9 X 8 X 71 X 2 X 3

(2x)3 + .

= 1 + 18x + 36 X 4x2 + 84 X 8x3 + .

= 1 + 18x + 144x2 + 672x3 + .

(1 + x2)10 = 1 + 10 X (x

2) + 10 X 9

1 X 2 (x

2)2 + 10 X 9 X 8

1 X 2 X 3 (x

2)3 +

. . .

= 1 + 5x + 45 x x2

4 + 120 x (x3

8) + .

. . = 1 + 5x + 11.25x2 + 15x3 + .

. .

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This formula only works if the number is 1. If the number isn’t 1 then we need to make it 1 by bringing it out of the bracket.

(2 + x)6 = [2(1 + x2)]6

= 26 x (1 + x2)6

= 64 x (1 + 6 X (x2) + 6 X 5

1 X 2 (x

2)2 + 6 X 5 X 4

1 X 2 X 3 (x

2)3 + .

= 64 x (1 + 3x + 15 x x2

4 + 20 x (x3

8) + . .

. = 64 x (1 + 3x + 3.75x2 + 2.5x3 + . .

. = 64 + 192x + 3.75x2 + 240x3 + . .

.

(1 + x)n = 1 + nx + n(n – 1)1 X 2

x2 + n(n – 1)(n – 2)1 X 2 X 3

x3 + . . .

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We can use the binomial theorem to calculate high powers of numbers, like (0.98)10, quickly. (1 − 2x)10 = 1 − 20x + 180x2 − 960x3 +..

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Compare 1 – 2x and 0.98

1 – 2x = 0.98 x = 0.01 So if we substitute x = 0.01 into our expansion for (1 – 2x)10 we get an approximation for (0.98)10.

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(1 − 2x)10 = 1 − 20x + 180x2 − 960x3 + . . .

= 1 − 20(0.01) + 180(0.01)2 −

960(0.01)3

= 1 − 0.20 + 180 x 0.0001 − 960 X

0.000001

= 1 − 0.20 + 0.018 − 0.00096

= 0.81704

and the real value of (0.98)10 is 0.81707 so we get a very

good approximation just using the first four terms of the

expansion. We don’t need to go any further than x3

because (0.01)4 is 0.00000001 which is so small it is

irrelevant.

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We can also write the formula using combination and factorial notation. 9! (9 factorial) = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

Bob has seven friends but only three spaces in his car. How many different ways can he choose three people to give a lift to out of a group of seven?

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Arrangements where order doesn’t matter are called combinations.

We can calculate the numbers of ways to choose 3 people from a group of 7 people by using factorial and

7C3 = 7!

4! 3! = �73�

these are all just different ways of writing the same thing.

nCr = n!

(n – r)! r! = �nr�

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(1 + x)n = nC0 + nC1 x1 + nC2 x2 + . . . + nCr xr +. . . +

nCnxn

(1 + x)n = 1 + nx + n(n – 1)1 X 2

x2 + n(n – 1)(n – 2)1 X 2 X 3

x3 + . . .

(1 + x)n = �n0� + �n1�x1 + �73�x

2 + . . . + �73� xr +. . . +

nCn�73�xn

These are just three different ways of writing the same thing. Notice that nC1 (the number of ways of choosing 1 person from n) is n and nCn (the number of ways of choosing n people from n people) is 1. If you have 8 friends, how many different ways can you choose 1 of them? 8. If you have 8 friends, how many different ways can you choose 8 of them? 1.

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Chapter 6 – Radian measure and its applications ufuful

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Chapter 7 – Geometric sequences and series Arithmetic sequences are adding (or taking away) number patterns. We always call the first number a and the number we add on each time d. Take the sequence with a = 4 and d = 3 1st 2nd 3rd 4th 4, 7, 10, 13, = 4, 4 + 1X3, 4 + 2X3, 4 + 3x3… = a, a + d, a + 2d, a + 3d… Notice that the number of ‘d’s is always one less than the term number. In general the nth term is a + (n-1)Xd so the 10th term here would be 4 + 9X3 = 31.

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A Geometric sequence is a multiplying number pattern (though we could be multiplying by a fraction so the numbers are going down). We always call the first number a and the number we multiply by each time r. Take the sequence with a = 3 and r = 2 1st 2nd 3rd 4th 3, 6, 12, 24, = 3, 3 X 21, 3 X 22, 3 X 23,… = a, a X r1, a X r2, a X r3,… Notice that the number of ‘r’s is always one less than the term number. In general the nth term is a X rn-1 so the 10th term here would be 4 X 39 = 78,732.

1st 2nd 3rd 4th … nth a, ar, ar2, ar3, … arn-1

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Notice that when we divide one term by another the ‘a’s cancel. This can be very useful when answering questions. 3rd term 2nd term

= ar2

ar1 = r

The second term of a geometric sequence is 2 and the fifth term is 54. Find the common ratio and the 10th term. ------------------------------------------------------------- 5th term = ar4 = 54 2nd term = ar1 = 2 ----------------------------------------------------------------

ar4

ar1 = 542

r3 = 27 r = 3

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If we end up with r2 = something the answer could be positive or negative.

The second term of a geometric sequence is 5 and the fourth term is 20.

Find the common ratio and the 10th term. ----------------------------------------------------------- ar3 = 20 ar1 = 5 --------------------------------------------------------------

ar3

ar1 = 205

r2 = 4 r = +2 or -2

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We can use geometric sequences to solve interest questions.

Ann invests an amount of money at a rate of interest of 4% per year. After 5 years it was worth £10,000. How much did she invest at the start? ------------------------------------------------------------- This is a geometric sequence with a = amount and r = 1.04. ---------------------------------------------------------------- After 5 years = 6th term = ar5 = 10,000 a X (1.04)5 = 10,000 a X 1.21665 = 10,000

a = 10,0001.21665

a = £8219.29

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What is the first term in the geometric progression

3, 6, 12, 24, . . . to exceed 1 million?

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This is a geometric sequence with a = 3 and r = 2

nth term = arn-1 = 3 X 2n-1

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3 X 2n-1 > 1,000,000

log(3X2n-1 ) > log 1,000,000

log 3 + log 2n-1 > log 1,000,000

log 3 + (n-1)log 2 > log 1,000,000

(n-1)log 2 > log 1,000,000 – log 3

n – 1 > log 1,000,000 – log 3log 2

n > 19.35

n = 20 as n must be an

integer.

We can use the laws of logarithms to solve geometric sequence questions.

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Up till now we have been looking at geometric sequences which involves each term by itself. We are now going to look at geometric series which are the sum of the sequences.

Sum of the first n terms of a geometric series = Sn Sn = a + ar + ar2 + . . . + arn-3 + arn-2 + arn-1

Sometimes it is more convenient to write it as

Sn = a(rn – 1)r – 1

by multiplying top and bottom by – 1.

Sn = a(1 - rn)1 - r

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Proof of formula We are going to - write out the sum of the first n terms - multiply both sides by r - take the bottom line away from the top line - put both sides into brackets - divide through ------------------------------------------------------------------

Sn = a + ar + ar2 + . . . + arn-3 + arn-2 + arn-1

rSn = ar + ar2 + ar3 + . . . + arn-2 + arn-1 +

arn

Sn – rSn = a −

arn

Sn – rSn = a - arn

(1 – r)Sn = a(1 – rn)

Sn = a(1 - rn)

1 - r

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Find the sum of the 1st 10 terms of the following series

2 + 6 + 18 + 54 + . . .

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This is a Geometric Series with a = 2 , r = 3 and n = 10.

Sn = a(1 - rn)

1 - r

Sum of the first 10 terms = S10

S10 = 2(1 - 310)

1 - 3 = 59, 048

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Find the how many terms are in the following series

1024 − 512 + 256 − 128 + . . . – 2 + 1

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This is a Geometric Series with a = 1024 , r = − 12

but what is n?

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nth term = arn-1

nth term = 1

arn-1 = 1

1024 X (− 12)n-1 = 1

(− 12)n-1 = 1

1024

Log(− 12)n-1 = 1

1024

(n-1)Log(− 12) = Log 11024

(n-1) = Log 1

1024

Log − 12

n = 1

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What is the least value of n such that the sum

1 + 2 + 4 + 8 + . . .

exceeds 2,000,000?

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This is a geometric series with a = 1 and r = 2

Sn = a(rn – 1)

r - 1

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a(rn – 1)

r - 1 > 2,000,000

1(2n – 1)

2 - 1 > 2,000,000

2n - 1 > 2,000,001

2n > 2,000,001

log 2n > log 2,000,001

n log 2 > log 2,000,001

n > log 2,000,001log 2

n > 20.9

n = 21 as n must be an integer.

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Find ∑ (3 X 2r)

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= 3 x 21 + 3 X 22 + 3 X 23 + . . . + 3 X 210

= 3 X (21 + 22 + 23 + . . . + 210)

This is 3 X a geometric series with a = 2, r = 2 and

n = 10

--------------------------------------------------------------

Sn = a(rn – 1)

r - 1

= 2(210 – 1)

2 - 1

= 6138

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If we look at the series with a = 3 and r = 0.5 3 + 1.5 + 0.75 + 0.375 + . . . If we look at the series or sum at certain points Sum of 1st 5 terms = S5 = 5.8125 Sum of 1st 10 terms = S10 = 5.9994 Sum of 1st 20 terms = S20 = 5.999994 As the number of terms we are adding goes up the sum goes up as well but not by a lot. In fact it will never actually get to 6, it will just get closer and closer and closer for ever. We say that ‘it tends to 6 ‘ or ‘it has a limit of 6’ or ‘the sum to infinity is 6’ This series is convergent. Not all series converge, some diverge. If the series is going to converge then r (the number we multiply by) must be between – 1 and + 1. If it isn’t then the series will be divergent.

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Sum to infinity of a geometric series

S∞ = a1 - r

for lrl < 1

Proof of formula

Take a number less than 1, say 12

(½)2 = 0.25 (½)3= 0.125 (½)4 = 0.06125

The bigger n gets the smaller (½)n gets. It gets nearer and

nearer (but never actually gets to) zero. We say as n tends

to infinity (½)n tends to zero.

As n » ∞, (½)n » 0

And this is true for any r between – 1 and + 1.

So in the formula rn disappears, (1 – 0) = 1 and

Sn = a(1 - rn)

1 - r becomes

S∞ = a

1 - r

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Find the sum to infinity of the series

120 + 40 + 10 + 2.5 + . . .

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This is an infinite geometric series with

a = 120 and r = 0.25

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S∞ = a

1 - r

= 120

1 – 0.25

= 160

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Find the sum to infinity of the series

1 + 1p +

1p2 +

1p3 + . . .

-------------------------------------------------------------

This is an infinite geometric series with

a = 1 and r = 1p

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S∞ = a

1 - r

= 1

1 - 1p

(multiply top and bottom by p)

= p

p - 1

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The sum of the first 4 terms of a geometric series is 15

and the sum to infinity is 16. Find the possible values of r.

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Sum of 1st n terms Sum to infinity

Sn = a(1 - rn)

1 - r S∞ =

a1 - r

Sum of the first 4 terms Sum to infinity

a(1 – r4)

1 - r = 15

a1 - r

= 16

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If we divide the 1st equation by the 2nd we get rid of the a

and the 1 – r (remember to divide we can flip the 2nd

fraction and multiply).

1 – r4 = 1516

r4 = 116

r = + 12 or r = − 1

2

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Chapter 8 – Graphs of trigonometric functions ufufuli

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Chapter 9 – Differentiation ufuful

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Chapter 10 – Trigonometric identities and simple equations fuful

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Chapter 11 - Integration ufufu

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