# Maths c1 revision

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• Created by: Anna
• Created on: 12-01-13 10:17

## Surds and Indices

a^m x a^n = a^m+n

a^m/a^n = a^m-n

(a^m)^n = a^mn

a^1/m = m√a

a^-m = 1/a^m

a^m/n = n√a^m = (n√a)^m

a^0 = 1

(√a)^2 = √a √a = a

√ab = √a √b

√a/b = √a/√b

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## Algebra

Brackets/ Common factors

a (b + c + d) = ab + ac + ad

(a + b)(c + d) = ac + ad + bc + bd

(a + b)^2 = (a + b)(a + b) = a^2 + 2ab + b^2

(x + y + z)(a + b + c + d) = x(a + b + c + d) + y(a + b + c + d) + z(a + b + c + d)

(y + a)^2 (x - a)^3 + (x - a)^2 = (x - a)^2((y + a)^2 (x - a) + 1)

Algebraic fractions

a/x + b/x + c/x = a + b + c/x

ax + ay/az = a(x + y)/az = x + y/z

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## Proof

Exhaustian

Prove "for any integer of x, the value of f(x) = x^3 + x + 1 is an odd integer"

• even integer, 2n, substitued into equation: (2n)^3 + 2n + 1 = 4(2n^3 + n) + 1. the sum and product of any integers are also integers, multiplied by 2 is even, +1 makes it odd.
• odd integer, 2m + 1, substituted into function: (2m + 1)^3 + 2m + 1 + 1 = 2 (4m^3 + 6m^2 + 4m) +3 is and odd integer too.

Prove "if x^2 is even, then x must be even"

• if x is odd, 2k + 1, (2k + 1)^2 = 4k^2 + 4k + 1 if x is odd, so is x^2. so if x^2 is even so must x.

Counter- example

Disprove "for any pair of real numbers x and y, if x>y then x^2 + x > y^2 + y

• ONE CASE IS NEEDED. eg. when x = 2 and y = -4
• 2 > -4 but 6 < 12 , so the statement is untrue
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Quadatic formula: x = -b (+/-) √b^2 - 4ac / 2a

Discriminant

• > 0. two distinct real roots
• = 0. one repeated root
• < 0. no real roots

Graphs

coefficent of x^2 = positive or negative graph. u-shaped positive, n-shaped negative.

when y = 0, the x values determine where the graph crosses that axis, vice versa

the minimum/maximum co-ordinates are found from Completing the square

• in the form f (x) = (x - a)^2 + b
• the min/max point is (a,b)

Almost quadratics: x^4 + 6x^2 + 9 = 0 can be written as (x^2)^2 - 6(x^2) + 9 or even y^2 - 6y +9 if y=x^2

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## Completing the square

QUADRATIC. (2x^2 + 8x - 5)

TAKE OUT THE COEFFIENT OF X^2. 2(x^2 + 4x) - 5

HALF THE COEFFIENT OF X AND WRITE IN SQUARE FORM.  2(x + 2)^2 - 5

EXPAND BRACKETS. (x + 2)^2 = (x^2 + 4x + 4) - 5

..

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## Inequalities

Solved like equations, but need to keep inequalities correct way round

Quadratic inequalities can be shown on a graph.
-x^3 + 2x + 3 >/= (x + 1)(x - 3) so x = -1 and 3

> or = to 0 means that the quadratic is satisfied when -1 </= x </= 3

• if dividing an inequality by a negative, the sign changes direction
• also do not divide by a variable, eg x or y, as it may be negative

Simultaneous equations

• first try and get the eqautions to match, e.g. to have a shared coefficient of x or y, you can then add or take away these varibles to get 0.
• this will allow you to find the value of the other variable, and therefore the first too after substituting it into the eqaution.
• you can also substitute equations into one another or make them equal each other in order to solve an equation, especially when finding points of intersection e.g. x^2 - 4x + 5 = 2x - 3
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## Co-ordinate geometry

the equation of a line between two points A (x1,y1) B (x2,y2)

m = change in y/change in x = y1-y2 / x1-x2
parallel lines have equal gradients
perpendicular lines have negative reciprocal gradients e.g. their sum is -1

Midpoint co-ordinates

(x1+x2)/2 , (y1+y2)/2

Distance between the points uses Pythagorus

√ (x2 - x1)^2 + (y2 - y1)^2

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## Circles

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

radius r and centre (a,b) or x^2 + y^2 = r^2 if centre is (0,0)

the form: x^2 + y^2 - 6x - 4y + 4 = 0 can be changed to the familiar form using Completing the square. this simplifies to (x - 3)^2 + (y + 2)^2 = 9. i.e. the centre = (3,-2) and r = 3

properties

• angles in a semicircle are right angles
• perpendicular from the centre to a chord bisects the chord
• a radius and tangent to the same point will meet at right angles
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## Graph sketching and transformations

n^2 - u shaped graph
-n^2 - n shaped graph

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