# FMSQ Maths

##### by Marcus Burgess 04/30/2010

# FMSQ Maths

by Marcus Burgess# 1. Calculus

## 1.1. Integration

### 1.1.1. Integration is the reverse of differentiation.

### 1.1.2. Be able to integrate kxⁿ where n is a positive integer or 0, and the sum of such functions

### 1.1.3. Be able to find a constant of integration.

### 1.1.4. Be able to find the equation of a curve, given its gradient function and one point.

## 1.2. Differentiation

### 1.2.1. Be able to differentiate kxⁿ where n is a positive integer or 0, and the sum of such functions

### 1.2.2. The gradient function dy/dx gives the gradient of the curve and measures the rate of change of y with x

### 1.2.3. The gradient of the fuction is the gradient of the tangent at that point.

### 1.2.4. Be able to find the equation of a tangent and normal at any point of a curve.

### 1.2.5. Be able to use differentiation to find stationary points on a curve.

### 1.2.6. Be able to determine the nature of a stationary point.

### 1.2.7. Be able to sketch a curve with known stationary points.

## 1.3. Kinematic applications

### 1.3.1. Be able to use differentiation and integration with repect to time to solve simple problems involving variable acceleration.

### 1.3.2. Be able to recognise the special case where the use of constant acceleration of formulae is appropriate.

### 1.3.3. Be able to solve problems using the formulae.

## 1.4. Definite Integrals

### 1.4.1. Definition of definite integrals

### 1.4.2. Definition of indefinite integrals

### 1.4.3. Be able to evaluate definite integrals

### 1.4.4. Be able to find the area between a curve, two ordinates and the x axis.

### 1.4.5. Be able to find the area between two curves.

# 2. Algebra

## 2.1. Manipulations of equations

### 2.1.1. Simplifying expressions including: algebraic fractions, square roots and polynomials

2.1.1.1. Algebraic Fractions

2.1.1.1.1. x/n=y

2.1.1.1.2. x=ny

2.1.1.2. Square roots.

2.1.1.2.1. √54

2.1.1.2.2. √(9x6)

2.1.1.2.3. 3√6

2.1.1.3. Polynomial

2.1.1.3.1. 5x²+25x+15

2.1.1.3.2. 5(x²+5x+3)

## 2.2. Remanider Theorem

### 2.2.1. Remainders of a polynomial up to order 3 when divided by a linear factor

## 2.3. Factor Theorem

### 2.3.1. Linear factors of a polynomial up to order 3

## 2.4. Solution of equations

### 2.4.1. Use of brackets

### 2.4.2. Solving linear equations with one unknown

2.4.2.1. x+4=6 x=2

### 2.4.3. Solving quadratic equations by factorisation, the use of formula and by completing the square

2.4.3.1. Factorisation

2.4.3.1.1. x²+6x+8

2.4.3.1.2. (x+4) (x+2)

2.4.3.2. Quadratic Formula

2.4.3.2.1. x=-b±√(b²-4ac) ÷ 2a

2.4.3.3. Completing the square.

2.4.3.3.1. (x-(b/2))²-(b/2)²+c

### 2.4.4. Solving a cubic equation by factorisation

2.4.4.1. Use factor theorum to find one factor, and then factorise the quadratic you are left with.

### 2.4.5. Solving two linear simultaneous equations in 2 unknowns

2.4.5.1. 6y+4x=-22 y+3x=8

2.4.5.2. y=8-3x

2.4.5.3. Subsitute the rearranged equation into the first one.

2.4.5.4. 6(8-3x)+4x=-22

2.4.5.5. 48-18x+4x=-22

2.4.5.6. 48-14x=-22

2.4.5.7. -14x=-70

2.4.5.8. 14x=70

2.4.5.9. x=5

### 2.4.6. Solving two simultaneous equations where one equation is linear and the other is quadratic.

2.4.6.1. x²+3x+2=0 x+y=5

2.4.6.2. x=5-y

2.4.6.3. Subsitute the rearranged equation into the first one.

2.4.6.4. (5-y)²+3(5-y)+2=0

2.4.6.5. 15-3y+2+y²-10y+25=0

2.4.6.6. 42-13y+y²=0

2.4.6.7. (y-6)(y-7)=0

2.4.6.8. y=6 or y=7

2.4.6.9. ஃ x=-1 or x=-2

### 2.4.7. Set up and solve problems leading to linear, quadratic and cubic equations in one unknown, and to simultaneious linear equations in two unknowns.

## 2.5. Inequalities

### 2.5.1. Ability to manipulate inequalities

2.5.1.1. x+4>2x+7

2.5.1.2. -3>x

### 2.5.2. Solving linear and quadratic inequalities algebraically and graphically

2.5.2.1. Step 1 Write down your inequality.

2.5.2.2. Step 2 Solve your inequality for x or y. You can solve for either as long as you are consistent. Normally, you solve for x.

2.5.2.3. Step 3 Create a table of values for y. Normally, you will choose three values. Three points are all that is needed for straight lines. You can choose any three values you wish. If your inequality solved with fractions, you may want to choose values divisible by the denominator or anything to turn the fraction into a whole number. Zero is also a good choice for a value.

2.5.2.4. Step 4 Solve your inequality for each value of y you chose. When solving, you can replace the inequality symbol with an =.

2.5.2.5. Step 5 Graph the points on your graph paper and connect the dots using a ruler.

2.5.2.6. Step 6 Determine where to shade the graph. Choose a point above or below your line. Substitute your point in place of the x and y in your original inequality. If the point makes your inequality true, then shade that side of the line. If the point makes your inequality false, then shade the opposite side of the line.