# Pure Maths How-Tos

Pure Maths with some Statistics (as specified)

Teacher recommended

- Mathematics
- Statistics, averages and distributionsAlgebra and functionsGraphs and transformationsLogarithms and exponentialsProbabilitySequences and seriesVectors
- AS
- AQA

- Created by: Betsy_2018
- Created on: 12-12-16 21:07

## How To Find Polynomials

(1) You are given an expression stating that a term *X *is part of this expression

(2) This *X* is therfore equal to the original expression when multiplied with p*(X*)

**Example**

x^3 + 12x^2 + 34 - 12 = (x + 6) x p(x)

- = (x+6) x (Ax^2 + Bx + C)
- = Ax^3 + Bx^2 + Cx + 6x^2 + 6Bx + 6C
- = x^3(A) + x^2(B + 6) + x(6B + C) +6C

Therefore

- A = 1
- B + 6 = 12 | B = 6
- 6B + C = C + 36 = 34 | C = -2 | (can also be applied to '-12')

P(x) = x^2 + 6x - 2

## Vertex And The Line Of Symmetry

To do this, you must complete the square on an equation.

Example

{y = x^2 + 4x + 12} = {(x + 2)^2 - 2^2 + 12} = {(x + 2)^2 + 8}

**Vertex**

Of (x + 2)^2 + 8, the *x* value of the **vertex **is *'-1x'* what is inside the bracket. They *y* value is what is outside the bracket

The vertex is (-2, 8)

**Line of symmetry**

The line of symmetry is sumply what the *x *equals when on the graph to split the graph in two identical lines. In this case, x = -2

## Intersections In Quadratic Graphs

There will be 1, 2 or 0 intersections.

(1) You will be given 2 expressions, which you will need to turn into equations

(2) You will need to find a solution to these simultaneous equations. It may give you co-ordinates: one set, two sets or no sets.

**Example**

{x + 2y = 3} = {x = 3 - 2y}

{x^2 + 3yx = 10}

Therefore, via substitution, (2y + 1)(y+1) = 0

- When (y= -1/2), ( x= 4)
- When (y= -1), (x= 5)

## Quadratic Inequalities

If the graph's equation >0, you are looking for values above (y= 0)

If the graph's equation <0, you are looking for values below (y= 0)

*You are advised to always draw a graph for these questions, as it is very easy to make mistakes.*

(1) Factorise to give 2 values of x | *use the quadratic formula | use discriminant (for algebraic)*

(2) plot these on a graph, highlight the area of the line above or below the x axis, depending on whether the equation >0 or <0

**Example**

x^2 + 3x - 10 ≥ 0

(x + 5)(x - 2) ≥ 0

x= -5, 2

Equation is *above the line* and is a *positive* graph so the inequality is (x≤5), (x≥2)

## Dividing a Polynomial

Some can easily be separated into their constituents and cancelled out.

Others will require more complex methods of division. 2 ways of dividing polynomials are:

- Long division
- The Box Method

Both will give you a quotient, divisor and a remainder

**Example**

{x^2 + 2x + 1} ÷ {x + 1)

____ *x + 1*

x+1 | x^2 + 2x + 1

_____- (x^2 + x)

______________0

## Factor Theorem

**Factor Theorem**

**P(x) = []x^(n) + []x^(n-1) +... []x + []**

(1) substitute the divisor in (flip the symbol in the brackets) into p(x)

(2) if it equals 0, it is a factor

(3) find out the rest of the divisors (quotient) to find the rest of the factors for this equation.

**Remainder Theorem**

**P(x) = (x - a)Q(x) + Remainder**

This is the same concept as the factor theorm. (x-a) will have its symbol flipped in order for it to go into P(x). This makes the answer {P(a) = 0} and the {remainder = 0}

## Laws of Indices

- Multiplying: ____________________(x^n)(x^m) = x^(n+m)

- Dividing: ______________________ x^n/x^m =_(x^n-m)

- To the Power of 0: _______________x^0 = 1

- Multiplying by another indice:______(x^n)^m = x^nm

- Indices as a fraction: x^n/m = ______(m√x)^n

- Negative indices: ________________x^-n = 1/x^n

- Base > x
**ª** - _______ ^ index

## Exponentials and Expressions as Logarithms

**Exponentials**

- graph
- where a > 0
- always passes thrpugh (0, 1)
- steepness increases with greater base
- never touches x axis

**Expressions as Logarithms**

Log*a*N = x

*a^*x = n

e.g Log*10*1000 = 3

10^3 = 1000

## Laws of Logarithms

- Log
*a*(XY) = Log*a*X + Log*a*Y

- Log
*a*(X/Y) = Log*a*X - Log*a*Y

- Log
*a*(X)^n = N(Log*a*X)

- Log
*a*(1/X) = -Log*a*X

- Log
*a*A = 1

## Changing the Base (Logarithms) and Taking Logs

**Log aX = (LogbX) / Logba **

**bLog aX = (b) / Logxa**

- always use base 10
- state that you are working to base ten (unless specified otherwise)

**Taking Logs**

take logs using the index, multiplictaion, division and negative laws

e.g 7^(x+1) = 3^(x+2)

(x+1)Log7 = (x+2)Log3

xLog7 + Log7 = xLog3 + 2Log3

x = (2Log2 - Log7) / (Log7 - Log3) [This can be put into a calculator]

## Sketching Cubic Graphs

{y = ax^3 + bx^2 + cx + d} = (x+a)(x+b)(x+c)

- From here you can find all of the points where the graph crosses the x/y axis
- Can be called a 'repeated root' graph [e.g (-1, 0) (0, 1) (1,0)
- The graph is a 'squiggle' shape
- goes up for positive, and down for negative

**Example**

y = x^3 - 2x^2 - x + 2

_ = (x+1)(x-1)(x-2)

0 = (-1,0) (1,0) (2, 0)

= 2 [when x=0]

TOTAL CO-ORDINATES: = (-1,0) (1,0) (2, 0) (0, 2)

Plot on a graph

## Translations

**Move in the x axis**

| a | = [when a>0 = move right] = [when a<0 = move left]

| b |

**Move in the y axis**

| a | = [ when b>0 = move up] = [when b<0 = move down]

| b |

**Formula**

y - b = f(x - a)^n

**Example: y = (x-5)^2 by -4 | 0**

y - 0 = (x - 5 - (-4))^2

y = (x-1)^2

## Factorials

n! = n x (n-1) x (n-2)...**(n-n+1)**

= n factorial

- if you divide a factorial by another factorial, many terms can be cancelled out
- 0! = 1 =
**(0 - 0 + 1)**

**N choose R**

*n*= number of objects*r*= number of different objects that can be chosen from*n*- represented by:

(*n*!) */ r*!(*n - r*)!

**Example**

| 4 | = 4! / (3!)(4-3)! = 24 / 6(1) = 4

| 3 |

## Pascal's Triangle

(a + b)^n

- use the relevant row of pascal's traingle (n=x)

**Example**

The co-efficent of x^2 in the expansion of (2-cx)^3 is 294. Find the value of c.

[a = 2] [b = cx] [n = 3] look up n = 3

3ab^2

3(2)(cx)^2 = 6(c^2)(x^2)

6c^2 = co-efficient of x^2

6c^2 = 294

c =7

## Binomial Expansion

(a + b)^2 = a^n + | n | a^(n-1)b + | n | a^(n-2)b^2... + b^n

___________________| 1 |_____________| 2 |

- in the absence of a calculator, you must do the formula on the previous card ('choosing')

**Example**

(2x + y)^3

[a= 2x] [b= y] [n= 3]

(2x)^3 + (3C1)(2x^2)(y) + (3C2)(2x)(y^2) + (y^3)

= 8x^3 + 6x^2(3y) + (12x)(6y^2) + (y^3)

## Suffix Notation

- This is to do with terms of a sequence
- terms go: t1, t2, t3...

**Example**

*Un = n(n + 2) - find U3*

U3 = (3)(5) = 15

**Example**

*Un-1 = (n-1)(n-1+2) - find an expression for Un - Un - 1*

- imagine the expression without '-1'; it would be (n)(n+2) = Un

= n(n + 2) - (n - 1)(n-1+2)

## Inductive Sequences

- Inductive sequences are ones that have terms that include the value of the previous term

Un = Un-1 ... U1 = U0...

Un+1 = Un+1-1... U2 = U1...

**Example**

U1 = 3 | Un = Un-1 + 2n+1

Find U2

U2 = (3) + (2(2)+1) = 8

## Limit of a Sequence

U∞ = L [this is a lot like taking Logs]

**Example**

*Un+1 = 2 - 1/3Un [U1 = 3]*

Therefore, U2 = 2 - 1/3U1

____________U2 = 2 - 1/3(3) = 1

U∞ = L

L = 2 - 1/3L

4/3L = 2

L = 2/4/3 = 3/2

Un => 1.5 as n => ∞

## Arithmetic Sequences/Series [nth term]

- Specific term Un = Un-1 + d [inductive]
- Nth term = a + (n-1)d

**Example**

*Find the first number after 1000 in the sequence with [a = 6], [d = 2.5]*

Un = a + (n-1)d

___= 6 + 5/2(n - 1)

5/2(n - 1) > 994

n - 1 > 397.6

n > 398.6

n = 399, U399 = 1001

## Sum of an Arithmetic Series

**Sum of the first n terms**

- 1/2 (n)(a +
*L*)

**Sum of the first (IN) natural numbers**

- 1/2(n)(n+1)

**Sum of the sequence**

- common difference or first term can be found

- use simultaneous equations

- (1/2)n [2a + (n-1)d]

## Sigma Notation

_x

∑ n^z

n = y

[x = final value of n] [y = first value of n] [z = nth term]

Look at the sequence, find a common term for 'z'. This could be n^x, or dividing each term to make n^x, n^x+1...

**Examples**

2^2 + 3^2 + 4^2 + 5^2...10^2 [x = 10, y= 2, z= n^2]

100

∑n^2

n = 2

1 + 3 + 9 + 27 = 3^0 + 3^1 + 3^2 + 3^3

x = 27, y = 1, nth = n-1

## Geometric Series

**Nth term**

- Un = ar^(n-1)

**Sum**

- Sn = a(1-r^n) / (1 - r)

**Example**

*Find the least number of terms of the geometric series 16 + 20 + 25... required to give a sum greater than 25,000 [a + ar + ar^2...]*

20/16 = 5/4

Sn = a(1-r^n)/1-r = (16(1 - 5/4^n) / 1 - 5/4) > 25000

Cancel out, take logs and solve to find n

n = 27

## Convergent Geometric Series

- a geometric series with common ratio r, converges when | r | < 1
- conergent geometric series has a sum to infinity
- S∞ = a / (1 - r)

**Example**

Find the least number of terms of the geometric series 16 + 20 + 25... required to give a sum greater than 25,000

a + ar + ar^2...

20/16 =

S∞ = a/1-r = 16 / 1 -

## Writing Equations in the Form ax + by + c = 0

- The equation may require cancelling down or scaling
- Rearrange the equation so that all terms are equalling 0
- a= the multiplier for x, and b= the multiplier for y

**Example**

y = -1/2x - 3

2y = -x - 3

0 = -x - 2y - 3

x + 2y + 3 = 0

## Straight Line Geometry

- Line Length (Distance beteen 2 points): √(x2- x1) + (y2- y1)

- Midpoint: (x1 +x2)/2, (y1 + y2)/2

- Gradient: Change in y / change in x

- Gradient of [OP: a/b] is perpendicular to [OP': -a/b]

- Equation of lime passing through (x1, y1), gradient m: y - y1 = m(x - x1)

## Circle Geometry

**Equation of a circle with centre (0, 0), radius r**

x^2 + y^2 = r2

The x and y are any set of co-ordinates that lay on the circle

**Equation of a circle with centre (a, b), radius r**

(x - a)^2 + (y - b)^2 = r^2

For when the circle does have the centre (0, 0) (has 'moved away')

**Finding a point outside of a circle circumference**

Use the (x-a)^2 + (x-b)^2 = r^2

The outer-circle point is a new 'radius', so just apply its co-ordinates to the formula

## Applying Translations on Circles

*Translation = moving a curve without altering its shape by vector | a |*

*______________________________________________________>.>.| b |*

Apply *| a |* to *x^2 + y^2 = r^2*

*.>.>.>.| b |*

*This will give you the equation (x - a)^2 + (y - b)^2 = r^2*

In general, if a circle is already translated, you just apply the translation further.

(x-p)^2 + (y - q)^2 = r^2 by a translation a b

= (x - p - a)^2 + (y - q - b)^2 = r^2

*r^2 does not change*

## Finding the Equation of a Circle Using its Propert

*This may involve finding the midpoint (should it be the centre), length of the diameter etc*

**Example**

Circle with traingle P(2, 12), Q(-6, 2) and R(12, -10)

(1) Find midpoint of diameter co-ordinates

(2) Find the distance between midpoint and circumference co-ordinate (radius)

(3) use the equation (x-a)^2 + (y-b)^2 = r^2

- midpoint = [(2+12)/2, (14+-10)/2] = (7, 2)
- distance = √(2-14)^2 + (12--10)^2 = √26
- equation = (x-7)^2 + (x-2)^2 = √26^2

## Length of Tangents

- P = point of tangents meeting
- A = 1 point on the circumference
- B = 1 point on the circumference
- C = centre of the circle

*B and A are interchangable*

*You will be given the co-ordinates *

PA^2 = CP^2 - CA^2

## Differentiation To Find Tangent and Normal Equatio

(1) Given the equation of a circle and the co-ordinate of contact

(2) differentiate the equation and apply the co-ordinates to it

(3) this is the gradient

(4) apply the orignal co-ordinates and the new gradient to *y-y1 = m(x-x1)*

*The normal is the same as the tangent, except it is perpendicular, meaning its gradient is -1/m*

**Example**

Curve y=x^2 + 1 on point (3, 10)

Differentiated = 2x [2(3) = 6]

Tangent equation = y = 6x - 8

Normal equation = y = -1/6x + 10.5

## Triangles (simple)

**Finding the line of symmetry of a triangle**

- Find midpoint of line of equilateral traingle
- find gradient of line between midpoint and point opposite the line
- find equation using gradient and a point on that line

**Showing that a triangle has a right angle**

- find the gradients of 2 lines
- multiply together
- if it = -1, they are perpendicular, and so must have a right angle between them

## Trigonometry

Remember: SOH CAH TOA

Sine Rule (length) = [a/SinA = b/SinB]

Sine Rule (angle) = [SinA/a = SinB/b]

Cosine Rule (length) = a^2 = b^2 + c^2 - 2bcCosA

Cosine Rule (angle) = CosA = b^2 + c^2 - a^2 / 2bc

- Sin30 = 1/2
- Sin45 = 1/(√2)
- Sin60 = (√3)/2
- Cos30 = (√3)/2
- Cos45 = 1/√2
- Cos60 = 1/2
- Tan30 = (√3)/3
- Tan45 = 1
- Tan60 = √3

## Related discussions on The Student Room

- I don't understand statistics »
- Engineering Mathematics »
- The central limit theorem and its importance to statistical estimation »
- How does A level maths work? »
- A level maths OCR pure maths and statistics help »
- Normal Distribution »
- Need help finding past paper for maths »
- Edexcel AS Further Maths (New Spec) - Statistics Discussion »
- Notnek learns stats »
- New draft specs for 2017 A-Level published »

## Comments

Report

Report