A-Level Maths Revision Notes

PATCHWAY COMMUNITY COLLEGE

Core Mathematics 1, 3, 4 and Statistics 1

Jamie Lake

May 15, 2014

Jamie Lake ([email protected]) 1

CONTENTS

1 Core Mathematics 1 2

1.1 Rules of Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Difference of squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Rules of Surds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Functions of the Quadratic Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.5 Equations and Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.6 Graph Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.7 Cartesian coordinate geometry in the (x, y) plane . . . . . . . . . . . . . . . . . . 61.8 Sequences and Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.9 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.10 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9


2.1 Algebraic fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Exponential and log functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.4 Numerical Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.5 Graph Transformations and Moduli . . . . . . . . . . . . . . . . . . . . . . . . . . 152.6 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.7 Further trigonometric identities and their applications . . . . . . . . . . . . . . . 192.8 Diifferentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20


3.1 Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2 Coordinate geometry in the (x, y) plane . . . . . . . . . . . . . . . . . . . . . . . . 243.3 The binomial expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.4 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.5 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.6 Even more ****** integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.6.1 Integration examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4 Statistics 1 39

4.1 Mathematical models in probability and statistics . . . . . . . . . . . . . . . . . . 394.2 Representation and summary of data - location . . . . . . . . . . . . . . . . . . . 404.3 Representation and summary of data - Measures of dispersion . . . . . . . . . . 414.4 Representation of data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.5 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

5 Graphs - the ones you'll need to remember 45

6 Notation 45

7 Special sets 46


This book may contain notation which you are unfamiliar with. There is a reference in theback of the book, section 6 called notation - familiarise yourself with this notation, it’ll makeyour work more readable. There are a list of graphs in the back of the book too. Section C1may need some additional content added. The worked solutions in C4 need to be reformat-ted. The section S1 needs completing and C2 needs to be added.

This document was produced using LATEX. a document preperation system designed for high-quality typesetting. It includes features designed for the production of technical and scien-tific documentation. LATEX is freeware and is the de facto standard for the communication andpublication of scientific documents.


1 CORE MATHEMATICS 1

1.1 RULES OF INDICES

am ×an = am+n

am ÷an = am−n

a−m = 1

am (1)

a1m ≡ am−1 = m

pa

anm = m

pan

(am)n = amn

a0 = 1

(1.1)

(1) can be seen as reciprocating or finding the reciprocal of a.

1.2 DIFFERENCE OF SQUARES

x2 − y2 = (x + y)(x − y) (1.2)

1.3 RULES OF SURDS

The square root of a prime number is a surd.

pab =p

a ×p

b√a

b=

papb

(1.3)

You can use these rules to simplify surds:

p405 =p

9×p45

= 3p

45(1.4)

To rationalise a fractions of the following form:

1pa

1

a ±pb

(1.5)

You must use the following method respectively: Mutiply top and bottom by:

pa

a ∓p

b(1.6)


1.4 FUNCTIONS OF THE QUADRATIC FORM

The general form of a quadratic equation is

y = ax2 +bx + c a 6= 0 (1.7)

where a, b, c are constants.

You can solve quadratic equations with three methods:

• factorisation and setting each factorial equal to 0

• completing the square:

x2 +bx = (x + b

2)2 − (

b

2)2 (1.8)

• using the quadratic formula:

x1,2 = −b ±√

(b2 −4ac)

2a(1.9)

A quadratic equation will always have two solutions, which may be equal. When a > 0 thegraph assumes a shape whereby the two branches continue upwards, in the positive y direc-tion to positive infinity. When a < 0 the graph assumes a shape whereby the two branchescontinue downwards, in the negative y direction to negative infinity. The value of c will al-ways provide you with the point of intersection with the y axis. To work out the point ofintersection(s) with the x axis, you simply need to find the roots. The red plots are the graphsfor which a is < 0. The green plots have a c value of −4 and therefore intercept y =−4 whenx = 0. Fig. 1, 2 and 3 demonstrate curves (green) of equation, respectively:

x3 +x −4

x2 +x −4

1

x +2+3

(1.10)

You can see that Fig. 3 has an asymptotic curve, that is, hyperbolic. The two red, dashed linesdemonstrate where the equation is undefined for y . For example, the x value of 2, gives anundefined value of y , and conversely the y value of 3 gives an undefined value of x.


−1 1

−9−8−7−6−5−4−3−2−1

12345

y-intercept

root

Fig. 1.

−2 −1 1 2

−8−7−6−5−4−3−2−1

12345678

y-intercept

root root

Fig. 2.

−3 −2 −1 1 2 3 4 5 6

−3

−2

−1

1

2

3

4

5

6

7

8

9

Fig. 3.

A quadratic equation will have:

• equal roots when:

b2 = 4ac ≡ b2 −4ac = 0 (1.11)


• real roots when:

b2 ≥ 4ac ≡ b2 −4ac ≥ 0 (1.12)

• real different roots when:

b2 > 4ac ≡ b2 −4ac > 0 (1.13)

• no real roots when:

b2 < 4ac ≡ b2 −4ac < 0 (1.14)

1.5 EQUATIONS AND INEQUALITIES

You can solve simultaneous equations by elimination or substitution. You can use the sub-stitution method to solve simultaneous equations where one equation is linear and the otheris quadratic. You usually start by finding an expresion for x or y from the linear equation.When you multiply or divide an inequality by a negative number, you need to change theinequality sign to its opposite.To solve a quadratic inequality you:

• solve the corresponding equation, then

• sketch the graph of the quadratic function, then

• use your sketch to find the required set of values

1.6 GRAPH TRANSFORMATIONS

• f (x +a) is a horizontal translation of −a. This can also be seen as a translation left of aunits.

• f (x)+a is a vertical translation of a. This can be seen as moving the graph up a units.

• f (ax) is a horizontal stretch of scale factor of a−1 or 1a . This can be seen as multiplying

all the x coordinates by (a−1) and leaving y unchanged.

• a f (x) is a vertical stretch of scale factor a. This can be seen as multiplying all the ycoordinates by a and leaving x unchanged.

• − f (x) is a reflection in the x axis.

• f (−x) is a reflection in the y axis.


1.7 CARTESIAN COORDINATE GEOMETRY IN THE (X, Y ) PLANE

Equations of straight lines are given in the general form:

y = mx + c (1.15)

where m is the gradient and c is the intercept on the y axis. They can also be given in thegeneral form:

ax +by + c = 0 (1.16)

The gradient of a straight line is given as the rate of change of y with respect to x. So, by takingtwo different points on the line, say, (x1, y1) and (x2, y2), you can get:

m = y2 − y1

x2 −x1(1.17)

Given that you know the gradient, you can find the equation of a line that passes through thepoint with coordinates (x1, y1) by using the formula:

y − y1 = m(x −x1) (1.18)

You can find the equation of the lines that passes through the coordinates (x1, y1) and (x2, y2)by using the formula:

y − y1

y2 − y1= x −x1

x2 −x1(1.19)

If there is a line of gradient m, and a line lies ⊥ to it, that lines must have a gradient of:

− 1

m(1.20)

This is known as the negative reciprocal of m. It is due to this fact that if two lines are ⊥, thenthe product of their gradients must equal -1.

1.8 SEQUENCES AND SERIES

A series of numbers following a set rule is called a sequence. 3, 7, 11, 15, 19,... is an exampleof a sequence. Each number in a sequence is called a term. The nth term of a sequence issometimes called the general term. A sequence can be expressed as a formula for the nthterm. For example, the formula Un = 4n+1 produces the sequence 5, 9, 13, 17,... by replacingn with 1, 2, 3, 4, etc. in 4n +1. A sequence can be expressed by a recurrence relationship. Forexample, the same sequence 5, 9, 13, 17,... can be formed from Un+1 =Un +4, given U1 = 5.Arecurrence relationship of the form:

Uk+1 =Uk +n k ≥ 1 n ∈Z (1.21)


is called an arithmetic sequence.All arithemtic sequences can be put in the form:

a + (a +d)+ (a +2d)+ (a +3d)+ (a +4d)+ (a +5d)+ (a +6d) (1.22)

where a is the 1st term, (a +d) is the 2nd tem, and so on... The nth term of an arithemticseries is:

a + (n −1)d =Un (1.23)

where a is the first term and d is the common difference. The formula for the sum of anarithemtic series is:

Sn = n

2[2a + (n −1)d ]

or Sn = n

2(a +L)

(1.24)

where a is the first term, d is the difference, n is the number of terms and L is the last term inthe series.

To prove that Sn = n2 [2a + (n −1)d ] you can do:

Sn = a + (a +d)+ (a +2d)+ (a +3d)+ ...+a + (n −1)d (1.25)

and if L = the last term, then,

Sn = a + (a +d)+ (a +2d)+ (a +3d)+ ...+ (L−3d)+ (L−2d)+ (L−d)+L (1.26)

if you reverse the order of the last equation, then,

Sn = L+ (L−d)+ (L−2d)+ (L−3d)+ ...+ (a +3d)+ (a +2d)+ (a +d)+a (1.27)

adding the two equations together, gives,

2Sn = a +L+ (a +L)+ (a +L)+ (a +L)+ ...+ (a +L)+ (a +L)+ (a +L)+a +L (1.28)

noticing that (a +L)+ ...+ (a +L) is simply, (a +L) n times:

2Sn = n(a +L)

Sn = n

2(a +L)

Sn = n

2[2a + (n −1)d ]

(1.29)

Σ signifies ’sum of’. You can use Σ to write a series in a more compact and concise way:

10∑r=1

(5+2r ) = 7+9+ ...+25 (1.30)


1.9 DIFFERENTIATION

The gradient of a curve y = f (x) at a specific point is equal to the gradient of the tanget tothe curve at the point (Fig. 4.). In Figure 4 you can see that the curve u0 has a point A onit, and from this point a line extends, tangentially, called I0. I0 is the tangent to the curve atA. Finding the gradient of I0 will give you the gradient of u0 at the point A. In other words,mI0 ≡ mu0 at A .

f (x)

x

u0

u1

I1

I0

B

A

Fig. 4.

The gradient, m can be calculated from the gradient function, f ′(x) [said f dash of x] or d yd x

[said d y by d x]. Differentiation can be seen as reducing the power by 1 and multiplying bythe orignial power:

f (x) = xn , f ′(x) = nxn−1, f or n ∈R (1.31)

It can similarly be shown that if f (x) = axn then f ′(x) = naxn−1. It can also be shown that if

y = f (x)±g (x) then d yd x = f ′(x)±g ′(x). A second order derivative can be written as d 2 y

d x2 [said dtwo y, by d x squared] or f ′′(x) in function notation, [said f double dash of x]. You can find therate of change of a given function, f (x), at a particular point by using f ′(x) and substituting inthe value for x. Figure 5 shows a curve, f (x), in red, in green are the tangents at given points,and in blue, are the normals at given points. The normals are always 90◦ to the tangent at agiven point along f (x).


Fig. 5.

The equation of the tanget to curve y = f (x) at point A(a, f (a)) is:

y − f (a) = f ′(a)(x −a) (1.32)

The equation of the normal to the curve y = f (x) at point A(a, f (a)) is:

y − f (a) =− 1

f ′(a)(x −a) (1.33)

1.10 INTEGRATION

Integration can be used to find the area bound beneath a function of x between two pointsx0 and x1. It can be seen as increasing the power by 1, and dividing by the new power. Everytime you integrate you must leave a +c at the end of the integration because there will alwaysbe a constant of integration. You can find the constant of integration, c, when you are givenany point (x, y) that the curve passes through. You must use algebra to simplify expressionsbefore you integrate them. You can use rules of indices to write the expression in a suitableformat to integrate. For example, you may not know the integral of

px when you’re first

asked, but rewritten as x12 , it is simplified and you can apply the rules beneath to work it out.∫

xnd x = xn+1

n +1+ c (n 6= −1) (1.34)


∫kxnd x = kxn+1

n +1+ c (n 6= −1) (1.35)

∫(kxn)+ (l xm)d x = kxn+1

n +1+ l xm+1

m +1+ c (n 6= −1, m 6= −1) (1.36)


2.1 ALGEBRAIC FRACTIONS

You can cancel down algebraic fractions by finding factors that are common to both the nu-merator and denominator. When an algebraic expression has fractions in the numerator ordenominator, it’s sensible to multiply both by the same number to create an equivelant frac-tion.

12 x +113 x + 2

3

≡ ( 12 x +1)×6

( 13 x + 2

3 )×6

= 3x +6

2x +4

= 3(x +2)

2(x +2)

= 3

2

(2.1)

You multiply fractions together by finding the product of the numerator and dividing by theproduct of the denominator. When there are factors common to both the numerator anddenominator cancel down first.

a

b× c

d= ac

bd(2.2)

x +1

2× 3

x2 −1

= x +1

2× 3

(x +1)(x −1)

= 3

2(x −1)

(2.3)

To divide by a fraction, simply multiply by the reciprocal of that fraction:

a

b÷ a

c≡ a

b× c

a

= c

b

(2.4)


When adding or subtracting algebraic fractions you need to have the same denominator:

a

x+b = a

x+ b

1

= a

x+ bx

x

= a +bx

x

(2.5)

3

x +1− 4x

x2 −1

= 3

x +1− 4x

(x +1)(x −1)

= 3(x −1)

(x +1)(x −1)− 4x

(x +1)(x −1)

= 3(x −1)−4x

(x +1)(x −1)

= −x −3

(x +1)(x −1)

(2.6)

An improper algebraic fraction is one whose numerator has a degree equal to or larger thanthe denominator, e.g.

x2 +5x +8

x −2or

x3 −8x +5

x2 +2x +1(2.7)

You can change them into mixed number fractions by long division or by using the remaindertheorem:

x2 +4x +12

x −3)

x3 +x2 −7−x3 +3x2

4x2

−4x2 +12x

12x −7−12x +36

29

[Long Division]

Therefore,

x3 +x2 −7

x −3≡ x2 +4x +12+ 29

x −3(2.8)

The remainder theorem states that any polynomial F (x) can be put in the form:

F (x) ≡Q(x) × di vi sor + r emai nder (2.9)


where Q(x) is the quotient and is how many times hte divisor divides into the function.For example, if we divide x3 +x2 −7 by x −3 again but this time use the remainder theorem:We first need to set up the identity:

F (x) ≡Q(x) × di vi sor + r emai nder

x3 +x2 −7 ≡ (Ax2 +B x +C )(x −3)+D(2.10)

and now solve to find constants A, B, C and D. As the divisor is a linear expression and F (x) isa cubic polynomial then Q(x) must be a quadratic polynomial and the remainder must be aconstant.Let x = 3:

27+9−7 = (9A+3B +C )×0+D

D = 29(2.11)

Let x = 0:

0+0−7 = (A×0+B ×0+C )× (0−3)+D

−7 =−3C +D

−7 =−3C +29

−3C =−36

C = 12

(2.12)

Compare the coefficients in x3, so you have:

x3 +x2 −7 ≡ Ax3 −3Ax2 +B x2 +12x −3B x −7

A = 1(2.13)

Compare the coefficients in x2, so you have:

1 =−3A+B

=−3+B

B = 4

(2.14)

Therefore:

x3 +x2 −7 ≡ (1x2 +4x +12)(x −3)+29

= x3 +x2 −7

x −3≡ x2 +4x +12+ 29

x −3

(2.15)

2.2 FUNCTIONS

A mapping transforms one set of numbers into a different set of numbers. The mappingcan be described in words or through an algebraic equation. It can also be represented bya Cartesian graph. A function is a special mapping such that every element of set A (the


domain, or x values) is mapped to exactly on element of set B (the range, or y values). Afunction is written as:

f (x) = 2x +1 or f : x 7→ 2x +1 (2.16)

A one-to-one function will have one x value, to one y value. A many-to-one function, hasatleast two x values that make the same y value. If set A has a value that maps to more thanone value in set B, it isn’t a function. You can establish whether a given expression is a func-tion, by seeing if it passes the vertical line test. If you imagine a vertical line passing throughall the x values, if it intersects the graph more than once for a given x value, then it can beproven that it isn’t a function. y = 1

x is not a function because it is undefined for 10 . Not all el-

ements of set A get mapped to elements in set B because x = 0 doesn’t get mapped anywhere.Many mappings can be made into functions by restricting the domain. If we consider thegraph of y =p

x:

−0.5 0 0.5 1 1.5 2

−1

0

1

Fig. 6.If the domain is all of the real numbers x ∈R, then this is not a function because values ofx less than 0 do not get mapped anywhere. If you restrict the domain to x ≥ 0, all of set Agets mapped to exactly one element in set B. You can combine two or more basic functionsto make a new more complex function: Given two functions, f (x) and g (x) the combinationof the two functions would be f g (x) [said f of g of x]. f g (x) means apply g first, followed byf. f g (x) is called a composite function. The inverse function performs the opposite opera-tion to the function. It takes elements of the range and maps them back into elements of thedomain. For this reason, inverse functions only exist for one-to-one functions. The inversefunction of f (x) is written f −1(x).


Find the inverse of the function: f (x) = 3x−1 , [x ∈R, x 6= 1]:

Let y = f (x)

y = 3

x −1y(x −1) = 3

y x − y = 3

y x = 3+ y

x = 3+ y

y

f −1(x) = 3+x

x

(2.17)

2.3 EXPONENTIAL AND LOG FUNCTIONS

Exponential functions are ones of the form y = ax . Graphs of these functions all pass through(0,1) because a0 = 1 for any number a.

Function f (x) Gradient Function f ′(x)1x 0×1x

2x 0.7×2x

3x 1.1×3x

4x 1.4×4x

You can see from the table above that as a increases for the function, so does the gradientfunction. There is a number between 2 and 3 such that the gradient function is the same asthe function. This number is approximately equal to 2.718 and is represented byt the letter ’e’.It’s an irrational number representing a number that exists in the real world. The exponentialfunction y = ex is therefore the function in which the gradient is identical to the function.For this reason it is often referred to as the exponential function. All exponential graphs willfollow a similar pattern. It is worth keeping in mind that as: x →∞,ex →∞, when x = 0,e0 = 1and as x =−∞,ex → 0.The inverse of the exponential function is the natural log. That is, loge (x) and is often writtenas ln. The graph of ln(x) will be a reflection of ex in the line y = x. You can see the graphs ofboth of these functions in the rear of this booklet.

2.4 NUMERICAL METHODS

You can find approximations for the roots of the equation f (x) = 0 graphically. This is some-times the most tedious method, because you have to draw the graph and state which two xvalues the line intersects the x axis between. In general, though, if you find an interval inwhich f (x) changes sign, then the interval must contain a root of the equation f (x) = 0. Theonly exception to this is when f (x) has a discontinuity in the interval, e.g. f (x) = 1

x has a dis-continuity at x = 0. If you have two functions that equal eachother, then there corresponding


roots will be the points of intersection of the two graphed functions. Consider ln x = 1x , it has

only one root:

y = 1x y = ln x ln x = 1

x

Fig. 7.

You can use iteration to find an approximation for the root of the equation f (x) = 0. To solvean equation of the for f (x) = 0 by an iterative method, rearrange f (x) = 0 into a form x = g (x)and use the iterative formula: xn+1 = g (xn). Different rearrangements of the equation f (x) =0 give iteration formulae that may lead to different roots of the equation, for this reason, youwill usually be given the rearrangement in the exam. If you choose a value x0 = a that is closeto a root, the sequence: x0, x1, x2, x3, x4... does not necessarily converge to that root. In fact,it might not converge to a root at all. When this is the case, the sequence x0, x1, x2, x3, x4... issaid to be divergent.

2.5 GRAPH TRANSFORMATIONS AND MODULI

The modulus of a number a, is written as |a|, is its positive numerical value. For example,|5| = 5 and also | − 5| = 5. It’s sometimes known as the absolute value and is shown on thedisplay of some calculators as Abs. A modulus function is, in general, a function of the typey = | f (x)|. When f (x) ≥ 0, | f (x)| = f (x) and when f (x) < 0, | f (x)| = − f (x).Consider drawingthe graph of y = x (ignore the modulus). For the part of the line below the x axis (the negativevalues of y), reflect in the x axis. For example this will change the y value −3 into the y value3.Consider the graph y = |x2 −3x −10|. For the part of the curve below the x axis (the negativevalues of y), reflect in the x axis. For example, this will change the y value −3 into the y value3.

| f (x)| 6= f (|x|) (2.18)

You need to be able to sketch the graph of the function y = f (|x|). Consider the graph ofy = |x| − 2. The graph intersects the y axis at −2. Reflect in the y axis. It can be seen as avertical translation of −2 units.It’s also important to understand how to solve equations involving a modulus. Consider |2x−32 | = 3. Sketch the graph of the LHS and RHS of the equation, the intersection of the graphson the positive arm (right most arm) of the sketch will be the solution of +(2x − 3

2 ) = 3 andthe intersection of the graphs on the negative arm (left most arm) of the sketch will be the


solution of −(2x − 32 ) = 3.

2x − 3

2= 3

2x = 9

2

x = 9

4

(2.19)

and

−(2x − 3

2) = 3

−2x + 3

2= 3

−2x = 3

2

x =−3

4

(2.20)

The same method can be assumed for an equation of the form |a+x| = |b+c|. You must alsobe able to see an equation and recognise the constituent transformations that map it from aknown function of x. For example, y = (x −2)2 +3, you know the graph of y = x2 and you cansee that the two transformations taking place are f (x −2) and f (x)+3. See section 1.6 for alist of graph transformations.

2.6 TRIGONOMETRY

You need to know the functions, secθ, cscθ and cotθ. That is, secant, cosecant and cotangentrespectively. The functions, secθ, cscθ and cotθ are defined as:

secθ = 1

cosθ

cscθ = 1

sinθ

cotθ = 1

tanθ= cosθ

sinθ

(2.21)

• secθ is undefined for values of θ at which cosθ = 0.

• cscθ is undefined for values of θ at which sinθ = 0.

• cotθ is undefined for values of θ at which tanθ = 0.

What you may not be comfortable with from Core Mathematics 2 is dealing with cosn θ butthis can be rewritten as (cosθ)n for n ∈Z+. That is, providing n belongs to the set of positiveintegers (see the back page). For example, cos−1 6= 1

cosθ . You may recall the rules SOH, CAHand TOA. With the inverse functions, these are now flipped:


• secθ = HA .

• cscθ = HO .

• cotθ = AO .

You also need to know the graphs that the three functions form.

−10 −5 5 10

−4

−2

2

4

6

Fig. 8. y = secθ y = cosθThe graph of y = secθ, θ ∈ R, has symmetry in the y axis and repeats itself every 360◦. It hasvertical asymptotoes at all the values of θ for which cosθ = 0, i.e. at θ = 90◦+180n◦,n ∈Z. Itcan be said that the function is asymptotic at θ = 90◦ or undefined.

−10 −5 5 10

−4

−2

2

4

6

Fig. 9. y = cscθ y = sinθThe graph of y = cscθ, θ ∈R, has vertical asymptotes at all the values of θ for which sinθ = 0,i.e. at θ = 180n◦,n ∈Z, and the curve repeats itself every 360◦.


−10 −5 5 10

−10

−5

5

10

Fig. 10. y = cotθ y = tanθThe graph of y = cotθ, θ ∈R is a little more complicated and has vertical asymptotes at all thevalues of θ for which sinθ = 0, i.e. at θ = 180n◦,n ∈Z, and the curve repeats itself every 180◦.

You need to be able to simplify expressions, prove identities and solve equations involvingall the trigonometric functions you’ve learned about up until now. There are a few identitiesthat will provde useful when tackling problems that ask you to prove the LHS is equal to theRHS or vice-versa. The first is one that you have come across from Core Mathematics 2:

cos2θ+ sin2θ = 1(a)

1+ tan2θ = sec2θ(b)

1+cot2θ = csc2θ(c)

tanθ = sinθ

cosθ

(2.22)

The identities b and c are derived from identity a by dividing by, in b cos2θ and in c sin2θ.You need to able to use the inverse trigonmetric functions, arcsin x, arccos x and arctan x andtheir graphs.

x

y

−π2 −1 π

21

−1

π2

1

−π2

0

Fig. 12.The graph of y = arcsin x (Fig. 11.) can be drawn by first drawing y = sin x (Fig. 9.), withthe restricted domain of −π

2 ≤ x ≤ π2 . This is then a one-to-one function. Reflect in the line of

y = x. The domain of arcsin x is −1 ≤ x ≤ 1; the range is −π2 ≤ arcsin x ≤ π

2 . Remember that therange of the inverse function will be the domain of the original function. Also, remember thatthe x and y coordinates of points interchange when reflecting in y = x. For example: (π2 ,0) →


(0, π2 ). The same method of sketching can be assumed for the other two inverse trigonmetricfunctions.

2.7 FURTHER TRIGONOMETRIC IDENTITIES AND THEIR APPLICATIONS

You need to know and be able to use the addition formulae:

sin(A±B) ≡ sin A cosB ±cos A sinB

cos(A±B) ≡ cos A cosB ∓ sin A sinB

tan(A±B) ≡ tan A± tanB

1∓ tan A tanB

(2.23)

Remember that when you’re dealing with trigonometric functions like this, sin(θ+α) 6= sin(θ)+sin(α).With the above trigonmetric functions, it’s possible to derive double angle formulae by seeingthat sin(2A) ≡ sin(A+ A):

sin2A ≡ 2sin A cos A

cos2A ≡ cos2 A− sin2 A ≡ 2cos2 A−1 ≡ 1−2sin2 A

tan2A ≡ 2tan A

1− tan2 A

(2.24)

Expressions of the form a sinθ±b cosθ = c can be rewritten in terms of a sine only or a cosineonly (by using the ’R formulae’ or the general formula). Remember that if c = 0 then theequation reduces to the form tanθ = k. Recall that R =

pa2 +b2.

a sinθ±b cosθ ≡ R sin(θ±α)

a cosθ±b sinθ ≡ R cos(θ∓α)(2.25)

If we look at expressingp

3sinθ−cosθ in the form R sin(θ−α), R > 0 and 0 ≤α< π2 then, first,

we must set the two equal to eachother

p3sinθ−cosθ ≡ R sin(θ−α)

=p3sinθ−cosθ ≡ R sinθcosα−R cosθ sinα

(2.26)

By comparing coefficients of sinθ and cosθ respectively we find that:

R cosα=p3

R sinα= 1(2.27)

Taking the above terms, squarring them, and adding them, gives:

R2 cos2α+R2 sin2α=p3

2 +12

= R2(cos2α+ sin2α) = 12 +p3

2 (2.28)


We know that cos2α+ si n2α= 1:

R2 = 4

R =±2(2.29)

We know that a constraint of the formulae is that R > 0 so we can reject −2. Thus, R = 2. Withthe same two terms that we used to find R, we can also find α. We know that tanα= sinα

cosα . Sodividing these terms will eliminate R and gives the expected result:

tanα= 1p3

α= tan−1 1p3

α= π

6

(2.30)

So, we now know R and α, the two variables:

p3sinθ−cosθ ≡ 2sin(θ− π

6(2.31)

The products of sines and/or cosines can be expressed as the sum or difference of sines orcosines. The following formulae are derived from the relevant addition formulae in 2.27.

sin(A+B)+ sin(A−B) ≡ 2sin A cosB

sin(A+B)− sin(A−B) ≡ 2cos A sinB

cos(A+B)+cos(A−B) ≡ 2cos A cosB

−[cos(A+B)−cos(A−B)] ≡ 2sin A sinB

(2.32)

When dealing with proofs of the form cosS+cosTsinS−sinT ≡ cot S−T

2 , for example, it’s useful to knowa rule you can use to manipulate the addition of two like sines or cosines. These formulaecan be derived from 2.32 and are so called the factor formulae because they reduct sums ordifferences of sines or cosines into a product of sines and/or cosines - they’re a factors of thesums or differences.

sinP + sinQ ≡ 2sinP +Q

2cos

P −Q

2

sinP − sinQ ≡ 2cosP +Q

2sin

P −Q

2

cosP +cosQ ≡ 2cosP +Q

2cos

P −Q

2

cosP −cosQ ≡−2sinP +Q

2sin

P −Q

2

(2.33)

2.8 DIIFFERENTIATION

Often, when differentiation in Core 3, you’ll be given a function of a function to differentiate,this could be y = sin(x2) or y = (4x2 + 3x + 9)3. The chain rule can be used to differentiate


a function of a function. This can also be seen as the rule used to differentiate compositefunctions. It might be helpful to remember the chain rule as the inside-outside rule.

y = [ f (x)]n ⇒ d y

d x= n[ f (x)]n−1 f ′(x)

y = f [g (x)] ⇒ d y

d x= f ′[g (x)]g ′(x)

(2.34)

This says, in layman’s terms, if you have any function of x raised to a power n then the differ-ential of that function to the n will be n lots of the original function to the n−1, multiplied bythe differential of the function. For example:

y = (4x2 +9)3 ⇒ d y

d x= 3(4x2 +9)2 ×8x ⇒ d y

d x= 24x(4x2 +9)2 (2.35)

If you have a function multiplied by a function then the second function evaluated at theinside of the derivative of the first function multiplied by the derivative of the second functionwill leave you with the derivative of the composite function.

y = sin x2 ⇒ d y

d x= cos x2 ×2x ⇒ d y

d x= 2x cos x2 (2.36)

Another form of the chain rule states that:

d y

d x= d y

du× du

d x(2.37)

Where y is seen as a function of u, and u is a function of x. Also, always remember that aparticular case of the chain rule is the result:

d y

d x= 1

d yd x

(2.38)

y = 3sin5 x ≡ 3(sin x)5 ⇒ d y

d x= 3×5× (sin x)4 × (cos x) ≡ 15sin4 x cos x (2.39)

y = 2t an4(5x −π) ≡ 2(tan(5x −π))4 ⇒ d y

d x= 2×4× tan(5x −π)3 × (sec(5x −π))2 ×5

= 40tan3 (5x −π)sec2 (5x −π)(2.40)

The product rule can be employed when two functions are multiplying eachother, instead ofcomposites. It says that if y = uv then:

d y

d x= u

d v

d x+ v

du

d x(2.41)


You may not always have a product of functions, you may have a division of two functions;when a function u(x) is divided by another function v(x), to form a rational function. Whenthis is the case, a third rule is available which is called the quotient rule. If y = u

v then: (noticev is taken to be the denominator in this case)

d y

d x= v du

d x −u d vd x

v2(2.42)

It’s important to know the standard results for differentiating functions as some of these willnot be given to you in your exam booklet. If:

y = e f (x) ⇒ d y

d x= f ′(x)e f (x)

y = l n[ f (x)] ⇒ d y

d x= f ′(x)

f (x)

y = sin f (x) ⇒ d y

d x= f ′(x)cos f (x)

y = cos f (x) ⇒ d y

d x=− f ′(x)sin f (x)

y = tan f (x) ⇒ d y

d x= f ′(x)sec2 f (x)

y = csc f (x) ⇒ d y

d x=− f ′(x)csc( f (x)cot f (x)

y = sec f (x) ⇒ d y

d x= f ′(x)sec f (x) tan f (x)

y = cot f (x) ⇒ d y

d x=− f ′(x)csc2 f (x)

(2.43)

The derivatives available in the formula booklet are those for: tan(kx), sec(x), cot(x), csc(x)and the quotient rule.

If y = (3x2 +5)13 and we let u = (3x2 +5), then y = u

13 . So we can see that du

d x = 6x and d ydu =

13 u− 2

3 . Then using the chain rule,

d y

d x= d y

du× du

d xd y

d x= 1

3u− 2

3 ×6x

d y

d x= 2x(3x3 +5)−

23

(2.44)

If we now consider 5y3 −2y−1 = x: (notice that the subject of the equation is x, not y)

d x

d y= 15y2 + 2

y≡ d x

d y= 15y4 +2

y2

d y

d x= y2

15y4 +2

(2.45)


You can see that f (x) = x5√

(x2 +4) is the product of two function, so to find f ′(x) we mustuse the product rule. If we let u = x5 and now let v = (x2 +4)

12 . You can see that the differen-

tiate v , you’ll need to use the chain rule, so:

du

d x= 5x4

d v

d x= 2x × 1

2(x2 +4)−

12

d y

d x= u

d v

d x+ v

du

d x

= x5 ×x(x2 +4)−−12 +

√(x2 +4)×5x4

= x6 +5x4(x2 +4)√(x2 +4)

= 6x6 +20x4√(x2 +4)

= 2x4(3x2 +10)√(x2 +4)

(2.46)

A slightly harder example may be f (x) = ln(csc x +cot x)

f ′(x) = (−csc x cot x −csc2 x)× 1

(csc x +cot x)

=−csc x(csc x +cot x)× 1

(csc x +cot x)

=−csc x

(2.47)


3.1 PARTIAL FRACTIONS

An algebraic fraction can be written as a sum of two or more simpler fractions. THis tech-nique is called splitting into partial fractions. An expression with two linear terms in thedenominator can be split by converting the fraction into two seperate, linear factors.

11

(x −3)(x +2)≡ A

(x −3)+ B

(x +2)4

(x +1)(x −3)(x +4)≡ A

(x +1)+ B

(x −3)+ C

(x +4)

(3.1)

An expression with three or more linear terms can be split by converting into three seperate,linear factors.An improper fraction is one where the index of the numerator is equal to or higher than theindex of the denominator. An improper fraction must be divided first to obtain a number and


a proper fraction before you can express it in partial fractions.1

x2 +3x +2)

x2 +3x +4−x2 −3x −2

2

[Long Division]

x2 +3x +4

x2 +3x +2= 1+ 2

x2 +3x +2

= 1+ A

(x +1)+ B

(x +2)

(3.2)

An expression with repeated terms in the denominator can be split by converting into theform of the first linear factor, the second first order linear factor and then the second secondorder differential:

6x2 −29x +−29

(x +1)(x −3)2 ≡ A

(x +1)+ B

(x −3)+ C

(x −3)2(3.3)

You can also use methods described in section 2.1 (Core Mathematics 3 - Algebraic Fractions).

3.2 COORDINATE GEOMETRY IN THE (x, y) PLANE

To find the Cartesian equation of a curve given parametrically you eliminate the paramterbetween the parametric equations. To find the area under a cruve given parametrically youuse: ∫

yd x

d td t (3.4)

You may be asked to find an equation for t in terms of x given two equations:

x = 1

t +1

y = 1

t −2

(3.5)

If we rearrange x = 1t+1 to make t the subject of the formula:

x = 1

t +1x × (t +1) = 1

t +1 = 1

x

t = 1

x−1

(3.6)


You may now want to find the cartesian equation of the curve, you can do this by substitutingyour rearranged t equation into a y = t equation.

y = 1

( 1x −1)−2

= 11x −3

= 11x − 3x

x

= 1

( 1−3xx )

= x

1−3x

(3.7)

It won’t be unusual for you to be given parametric equations with trigonmetric functionsinvolved:

x = 2−4cosθ

y = 3sinθ+4

0 ≤ θ < 2π

(3.8)

If we’re asked to find the exact coordinates of the points where the parametric equations’ plotmeets the y-axis then this could not be easier. x will be equal to 0 when the plot intersectsthe y-axis. So you can rearrange the parametric equation with x as the subject in terms of θand use the constraint on θ to find the two values for the intersection, which happen to be, inthis case: θ = π

3 , 5π3 . Substitute the values of θ back into your equations with y as the subject

and work through the equation until a solution for y is found.A slightly more taxing task may be to find a Cartesian equation of the ellipse formed by theplot of the parametric equations. A rearrangement of the first parametric equation gives:x +4cosθ = 2. Further rearranging can reduce the parametric equation to be in terms of thetrigonemtric function only: cosθ = 2−x

4 . The same can be done with the parametric equation

with y as the subject, yielding: sinθ = y−43 . Now you have two parametric equations in terms

of the trigonometric functions: sinθ and cosθ. The two functions are linked by the identityexpressed in 2.22(a). So, by squaring both of the functions and adding them, we can see thatthe parametric equations will equate to 1.(

2−x

4

)2

+(

y −4

3

)2

= 1 (3.9)

Using 3.4 it’s possible to find the area, R bound by two parametric equations. Given twoparametric equations: x = 2t

12 , y = t (9− t ), it’s possible to work out y d x

d t and hence intergrate


this to find the area bound by the Cartesion plot of the parametric equations.

yd x

d t= t (9− t )

d

d t(2t

12 )

= t (9− t )× t−12

= t 1− 12 (9− t )

= t12 (9− t )

(3.10)

If we define a domain for t , say, 0 ≤ t ≤ 9, then it’s possible to implicitly integrate our resultfrom 3.10 to find the are bound by the curve.∫

yd x

d td t =

∫ 9

0t

12 (9− t )d t

=∫ 9

09t

12 − t

32 d t

=[

9(32

) t32 − t

52(5

2

)]9

0

=[

6t32 − 2

5t

52

]9

0

= (6(9)32 − 2

5(9)

52 )− (6(0)

32 − 2

5(0)

52 )

= (6(27)− 2

5(243))− (0−0)

= 644

5

(3.11)

The area of R in this case, is 64 45 .

3.3 THE BINOMIAL EXPANSION

The binomial expansion can be used to express an exact expression if n is a positive integer,or an approximate expression for any other rational number. The below expansion is true forwhen n is negative or a fraction and is only valid if |x| < 1.

(1+x)n = 1+nx + n(n −1)x2

2!+ n(n −1)(n −2)x3

3!+ ... (3.12)

(1+2x)3 = 1+3× (2x)+3×2× (2x)2

2!+3×2×1× (2x)3

3!+3×2×1×0× (2x)4

4!= 1+6x +12x2 +8x3

(3.13)

Here the expansion is finite and exact.√(1−x) = (1−x)

12 = 1+ 1

2×−x + 1

2×−1

2× (−x)2

2!+ 1

2×−1

2×−3

2× (−x)3

3!+ ...

= 1− 1

2x − 1

8x2 − 1

16x3 − ...

(3.14)


Here the expansion is infinite and approximate.It’s possible to adapt the binomial expansion to include expressions of the form (a +bx)n bysimply taking out a common factor of a: e.g.

1

3+4x= (3+4x)−1 =

[3

(1+ 4x

3

)]−1

= 3−1(1+ 4x

3

)−1 (3.15)

You can use previously learnt knowledge of partial fractions to expand more difficult expres-sions, e.g.

7+x

(3−x)(2+x)= 2

(3−x)+ 1

(2+x)

= 2(3−x)−1 + (2+x)−1

= 2

3

(1− x

3

)−1+ 1

2

(1+ x

2

)−1

(3.16)

You may be asked to state the range of values for x for which the expansion you have found isvalid. To do this, using 3.12 we can see that |x| < 1. If we take our expansion and oberve thecoefficient of x. For example, if we have the expansion (1+2x)

12 , then the expansion is valid

if |2x| < 1 ⇒|x| < 12 .

3.4 DIFFERENTIATION

When a relation is described by parametric equations:

• You differentiate x and y with respect to the parameter t .

• Then you use the chain rule rearranged into the form:

d y

d x= d y

d t÷ d x

d t(3.17)

When a relation is described by an implicit equation:

• Differentiate each term in turn, using the chain rule and product and quotient rules asappropriate.

• Using the chain rule:

d

d x(yn) = nyn −1

d y

d x(3.18)

• By the product rule:

d

d x(x y) = x

d y

d x+ y (3.19)


In an implicit equation:

• Note that when f (y) is differentiated with respect to x it becomes f ′(y) d yd x .

• A product term such as f (x)g (y) is differentiated by the product rule and becomes

f (x)g ′(y) d yd x + g (y) f ′(x).

You can differentiate the function f (x) = ax , where a is a constant:

d

d x(ax ) = ax ln a (3.20)

You can use the chain rule once, or several times to connect the rates of change in a ques-tion involving more than two variables. You can set up simple differential equations frominformation given in context. This may involve using connected rates of change, or ideas ofproportion.

The easiest way to learn this is through examples:

1) Find the gradient of the curve with parametric equations: x = t 2

2t+1 , y = 12t+1 , at the point(1

8 , 12

).

Differentiate both equations: d xd t = (2t+1)2t−t 2(2)

(2t+1)2 , d yd t = −2

(2t+1)2

∴ d yd x = −2

(2t 2+2t ) , using 3.17. So, at(1

8 , 12

), t = 1

2 and thus d yd x =−4

3 .

2) Use implicit differentiation to find d yd x in terms of x and y where x4 −6x2 y +4y3 = 2:

So, 4x3 −12x y −6x2 d yd x +12y2 d y

d x = 0.

It’s important to know where the terms −12x y − 6x2 d yd x come from. If we differentiate the

quantity −6x2 y with respect to x, we must use the product rule. If we set u =−6x2 and v = y

then we know that the derivatives of both of these terms is: u′ =−12x and v ′ = d yd x (differen-

tiating any y term with respect to x yields a d yd x ). If we apply the product rule, d y

d x = u′v + v ′u,

then we can get: −6x2 d yd x −12x y . Taking our result from above, it’s now necessary to collect

d yd x terms, factorise, and make d y

d x the subject of the equation.

So, we now have (12y2 −6x2) d yd x = 12x y −4x3, and dividing both sides by (12y2 −6x2), gives:

d yd x = 12x y−4x3

12y2−6x2 = 6x y−2x3

6y2−3x2

3) Given that the volume V cm3 of a hemisphere is related to its radius r cm by the formulaV = 2

3πr 3 and the hemisphere expands so that the rate of increase of its radius in cm s−1 is

given by drd t = 2

r 3 , find the exact value of dVd t when r = 3.

In these questions, it’s important to think about what you have and what you do not have.You know you need to end up with dV

d t . You have drd t , the only other piece of information you

have is an equation with the variables V and r , so the only possible derivative is in terms ofdVdr . Lo and behold, dV

dr × drd t allows dV

d t to fall out nicely. So, you now know you want some-thing along the lines of:


dVd t = dr

d t × dVdr .

So, if we look at differentiating V = 23πr 3:

dVdr = 3× 2

3πr 2 = 2πr 2. So we now have all of the constituent quantities for the derivation of:dVd t = 2πr 2 × 2

r 3 = 4πr and ofcourse when r = 3 the exact value of dV

d t = 4π3 .

4) The curve C is given by the equations y = 2t , x = t 2 + t 3 where t is a parameter. Find theequation to the normal to C at the point P on C where t =−2.So, first, differentiate both parametric equations with respect to t :So we have, d y

d t = 2 and d xd t = 2t +3t 2. Using 3.17 we can divide the two equations, as in the

chain rule, to find an equation for d yd x :

d yd x = 2

2t+3t 2

The question tells us that it wants the equation of the normal to C at the point P . At P , the tcoordinate is equal to −2. If we substitute −2 into our gradient function, we’ll get a result ofthe gradient of the tanget at that point. The negative reciprocal of that output will leave uswith the gradient of the normal at that point, because the two lines are ⊥. So,d yd x = 2

2(−2)+3(−2)2 = 14

Taking the negative reciprocal:The gradient of the normal is −4.We now have the gradient, we now need to account for the two other variables in the straight-line equation y = mx + c to find c. We have two parametric equations in terms of t that haveour unkown variables as the subject, so, we substitute t =−2 into them to obtain a value forboth x and y :When t =−2, x =−4 and y =−4. Now, using the equation: y − y1 = m(x −x1):y − (−4) =−4(x − (−4)) ≡ y +4x +20 = 0.

5) At time t seconds the surface area of a cube A cm2 and its volume is V cm3. The volumeof the cube is expanding at a uniform rate of 2 cm3s−1. Show that a d A

d t = k A− 12 , where k is a

constant to be determined.Questions like this might require a little forethought. If the side of the cube is x cm, thenV = x3, that is, base × height × depth. We can also see that A = 6x2, because a cube has sixfaces and one face is x2, that is, base × height.If we think about what the following statement means: “The volume of the cube is expandingat a uniform rate of 2 cm3s−1.” In other words, the rate of change of volume with respect totime, is 2 cm3s−1. So, re-written: dV

d t = 2.

The only two things we can use at the moment, are the dVd t term and by differentiating V = x3,

dVd x = 3x2. So, by the chain rule: d x

d t = 23x2 .

If A = 6x2, then d Ad x = 12x and so d A

d t = 12x × 23x2 = 8

x ∵ d Ad t = d A

d x × d xd t .

Rearranging A = 6x2 to find x we find: x =√

A6 . Substituting x back into our d A

d t gives 8p

6pA

.

The question originally asks for d Ad t in the form k A− 1

2 , where k is a constant. We can now dothis:d Ad t = 8

p6A− 1

2 . So, k = 8p

6.


6) Given that y2(x2 +x y) = 12, evaluate d yd x when y = 1.

When y = 1, x2 +x = 12 ∴ x = 3 or −4.Now, to differentiate our equation, first, expand the brackets:

dd x (x2 y2 +x y3) = d

d x 12 ∴ dd x (x2 y2 +x y3) = 0

To differentiate our L.H.S. we must first apply the rules of the product rule and then the rulesof implicit differentiation:Let’s look at each individual term:

dd x (y2x2): Here, we’ll first apply the product rule by letting u = y2 and v = x2, giving us:

u′ = 2y d yd x and v ′ = 2x.

Applying the product rule to this: u′v + v ′u:

2x y2 +x22y d yd x

Now, dd x (x y3): Very much the same thing, we’ll first apply the product rule by letting u = x

and v = y3, giving us: u′ = 1 and v ′ = 3y2 d yd x .

Applying the product rule to this: u′v + v ′u:

x2y3 d yd x +x y2 d y

d xAdding the constituent parts back together, we get the nasty looking:

x22y d yd x +2x y2 +x3y d y

d x + y3 = 2x y2 + y3 + (x2y +x3y2) d yd x

This leaves us with: −2x y2−y3

x22y+x3y2 = d yd x .

When x = 3, y = 1, ∴ d yd x = −7

27 .

When x =−4, y = 1, ∴ d yd x = 7

20 .

3.5 VECTORS

(The information and things I can do here is diagrammatically limited by the constraints ofboth time and my expertise with LATEX- the document markup language used to prepare andtypeset this PDF).

• A vector is a quantity that has both magnitude and direction.

• Vectors that are equal have both the same magnitude and the same direction.

• Two vectors are added using the ’triangle law’.

b

a

a +b

• Adding the vectors ~PQ and ~QP gives the zero vector 0 ( ~PQ + ~QP = 0).

• The vector −a has the same magnitude as the vector a but is in the opposite direction.


• Any vector ∥ to the vector a may be written as λa, where λ is a non-zero scalar.

• a −b is defined to be a + (−b).

• A unit vector is a vector which has magnitude (or modulus) 1 unit.

• If λa +µb =αa +βb, and the non-zero vectors a and b are not parallel, then λ=α andµ=β.

• The position vector of a point A is the vector ~O A, where O is the origin. ~O A is usuallywritten as the vector a.

• ~AB = b −a, where a and b are the position vectors of A and B respectively.

• The vectors i , j and k are unit vectors parallel to the x-axis, the y-axis and the z-axisand in the direction of x increasing, y increasing and z increasing, respectively.

• The modulus (or magnitude) of xi + y j is√

x2 + y2.

• The vector xi + y j + zk may be written as the column matrix

xyz

• The distance from the origin to the point (x, y, z) is

√x2 + y2 + z2.

• The distance between the points (x1, y1, z1) and (x2, y2, z2) is√

(x1 −x2)2 + (y1 − y2)2 + (z1 − z2)2.

• The modulus (or magnitude) of xi + y j + zk is√

x2 + y2 + z2.

• The scalar product of two vector a and b is written as a.b, and defined by a.b = |a||b|cosθ,where θ is the angle between a and b.

• If a and b are the position vectors of the points A and B , then:

cos(AOB) = a.b

|a||b| (3.21)

• The non-zero vectors a and b are ⊥ if and only if a.b = 0.

• If a and b are ∥, a.b = |a||b|. In particular, a.a = |a|2.

• If a = a1i +a2 j +a3k and b = b1i +b2 j +b3k, then:

a.b =a1

a2

a3

.

b1

b2

b3

= a1b1 +a2b2 +a3b3 (3.22)

• A vector equation of a straight line passing through the point A with position vector a,and ∥ to the vector b, is r = a + tb, where t is a scalar parameter.


• A vector equation of a straight line passing through the points C and D , with positionvectors c and d respectively, is r = c + t (d − c), where t is a scalar parameter.

• The acute angle θ between two straight lines is given by:

cos(θ) =∣∣∣∣ a.b

|a||b|∣∣∣∣ (3.23)

1) The vector a is directed due north and |a| = 6.The vector b is directed due east and |b| = 4.(a) Find the value of |a +2b|.Vector 2b is ∥ to b (with magnitude 8).

|a +2b|2 = |a|2 +|2b|2= 62 +82 = 100

|a +2b| = 10

(3.24)

(b) Find the direction of a +2b, as a bearing to the nearest degree.Using simple trigonometry and drawing or imagining the triangle, we can see that tanθ =86 ⇒ θ = 53.1◦.So, as a bearing, the direction of a +2b is 053ci r c , to the nearest degree.

2) Given that 5λa + (1−λ)b = µa + (µ−3)b, where a and b are non-∥, non-zero vectors, findthe value of λ and the value of µ.By comparing coefficients we can see that: 5λ=µ and (1−λ) = (µ−3).So,

1−λ= 5λ−3

6λ= 4

λ= 2

3

µ= 10

3

(3.25)

3) Given that a = 3i +2 j , b = 10i +12 j and c =−6i +10 j , find |a+b+c|, and hence find a unitvector in the same direction as a +b + c.

a +b + c = 3i +2 j +10i +12 j −6i +10 j

= 7i +24 j

|a +b + c| =√

72 +242 = 25

(3.26)

Unit vector = a+b+c|a+b+c| = 1

25 (7i +24 j )


4) Calculate the distance between the points (3,6,−4) and (7,1,3), giving your answer in itssimplest form.Distance:

=√

(3−7)2 + (6−1)2 + (−4−3)2

=√

(16+25+49) =p90

=p9p

10 = 3p

10

(3.27)

5) The points A and B have position vectors (5i −2 j +3k) and (8i +8 j + tk) respectively, and| ~AB | = 5

p5.

Find the possible values of t .

~AB =8

8t

− 5−23

= 3

10t −3

| ~AB | =

√32 +102 + (t −3)2

=√

(9+100+ t 2 −6t +9)

=√

(t 2 −6t +118) = 5p

5

∴ t 2 −6t +118 = 125

= t 2 −6t −7 = 0

= (t −7)(t +1) = 0

(3.28)

6) Given that a = 7i −4 j +2k and b = 2i +8 j + tk:(a) find the value of t for which a and b are ⊥,

a.b = 7−42

.

28t

= 14−32+2t

= 2t −18

(3.29)

If a and b are perpendicular, 2t −18 = 0 and thus t = 9(b) find, to the nearest degree, the angle between a and b when t = 11.When t = 11:

a.b = 2t −18 = 4

|a| =√

72 + (−4)2 +22 =p69

|b| =√

22 +82 +112 =p189

cosθ = a.b

|a||b| =4p

69p

189

θ = 88◦

(3.30)


7) The points A and B have position vectors (5i +8 j −4k) and (8i +2 j +5k) respectively.(a) Find a vector equation for the line l which passes through A and B .

a = 5

8−4

, b =8

25

b −a =8

25

− 5

8−4

= 3−69

(3.31)

Using the fact that a vector equation of a straight line passing through the points A and B ,with position vectors a and b respectively, is r = a + t (b −a), where t is a scalar parameter:

l : r = 5

8−4

+ t

3−69

(3.32)

(b) Given that the point with coordinates (p,4p, q) lies l , find the value of p and the value ofq . 5

8−4

+ t

3−69

= p

4pq

(3.33)

p = 5t

4p = 8−2t

q =−4+3t

(3.34)

(3.35)

(3.36)

From (3.43), t = p −5Substituting into (3.44):

4p = 8−2(p −5)

4p = 8−2p +10

6p = 18

p = 3

(3.37)

So, looking at (3.43) we know that t =−2.Now, substituting t back into (3.45), it follows that q =−10.

p = 3

t =−2

q =−10

(3.38)

(3.39)

(3.40)


8) The lines l1 and l2 have vector equations:

r =5

04

+ t

3−42

r = 5−19

+ s

2−33

(3.41)

(a) Show that l1 and l2 intersect.

r =5+3t−4t

4+2t

, r = 5+2s−1−3s9+3s

At an intersection,5+3t

−4t4+2t

= 5+2s−1−3s9+3s

(3.42)

Equation 1 Equation 25+3t = 5+2s −4t =−1−3s3t = 2s 4t −1 = 3s32 t = s 8

2 t −1 = 92 t

32 (−2) = s −1 = t

2s =−3 t =−2

If the lines intersect, then 4+2t = 9+3s must be true, i.e. the z-components must be equal:4+2t

?= 9+3s

4+2t = 4−4 = 0

9+3s = 9−9 = 0

(3.43)

(3.44)

(3.45)

The z-components are equal, so the lines l1 and l2 do intersect.

(b) Find the coordinates of their point of intersection.

Intersection point: r =5+3t−4t

4+2t

=−1

80

Coordinates: (−1,8,0)

(3.46)


(c) Find the acute angle between l1 and l2, giving your answer in degrees to 1 decimal place.

Use the direction vectors: a = 3−42

,

2−33

Find the dot product of a and b a.b = 3−42

.

2−33

= 6+12+6 = 24

Find the magnitude of both a and b |a| =√

32 + (−4)2 +22 =p29

|b| =√

22 + (−3)2 +32 =p22

cosθ = a.b

|a||b| =24p

29p

22

Angle between l1 and l2 is 18.2◦ (1 d.p.)

(3.47)

3.6 EVEN MORE ****** INTEGRATION

You should be familiar with the following integrals. The +c has been omitted here but shouldbe used and included in the examination because it can often gain an extra mark.

y = ex ⇒∫

f (x) d x = ex

y = 1

x⇒

∫f (x) d x = ln |x|

y = sin x ⇒∫

f (x) d x =−cos x

y = cos x ⇒∫

f (x) d x = sin x

y = tan x ⇒∫

f (x) d x = ln |csc x|

y = cot x ⇒∫

f (x) d x = ln |sin x|

y = sec x ⇒∫

f (x) d x = ln |sec x + tan x|

y = csc x ⇒∫

f (x) d x =− ln |csc x +cot x|

y = sec2 x ⇒∫

f (x) d x = tan x

y =−csc x cot x ⇒∫

f (x) d x = csc x

y = sec x tan x ⇒∫

f (x) d x = sec x

y =−csc2 x ⇒∫

f (x) d x = cot x

(3.48)


Also, not forgetting that in general an integration with x can be seen as:

y = xn ⇒∫

f (x) d x = xn+1

n +1(3.49)

Using the chain rule in reverse you can obtain generalizations of the above formulae:∫f ′(ax +b) d x = 1

af (ax +b)+ c∫

(ax +b)n d x = 1

a

(ax +b)n+1

n +1+ c∫

eax+b d x = 1

aeax+b + c∫

1

ax +bd x = 1

aln |ax +b|+ c∫

cos(ax +b) d x = 1

asin(ax +b)+ c∫

sec2(ax +b) d x = 1

atan(ax +b)+ c∫

csc(ax +b)cot(ax +b) d x =− 1

acsc(ax +b)+ c∫

csc2(ax +b) d x =− 1

acot(ax +b)+ c∫

sec(ax +b) tan(ax +b) d x = 1

asec(ax +b)+ c

(3.50)

Sometimes trigonometric identities can be useful to help change the expression into one youknow how to integrate. e.g. to integrate sin2 x or cos2 x use formula for cos2x, so:∫

sin2 x d x =∫ (

1

2− 1

2cos2x

)d x (3.51)

You can use partial fractions to integrate expressions of the type: x+8(x−1)(x+2) .

There are two patterns shown below that should be remembered.∫f ′(x)

f (x)d x = ln | f (x)|+ c∫

f ′(x)[ f (x)]n d x = 1

n +1[ f (x)]n+1; n 6= 1

(3.52)

Sometimes you can simplify an integral by changing the variable. This process is called inte-gration by substitution.

Integration by parts is done as follows:∫u

d v

d xd x = uv −

∫v

du

d xd x (3.53)


The area of region enclosed by y = f (x), the x-axis and x = a and x = b is given by:

area =∫ b

ay d x (3.54)

The volume of revolution about the x-axis between x = a and x = b is given by:

volume =π

∫ b

ay2 d x (3.55)

Seperation of variables for first order differential equations:

d y

d x= f (x)g (y) ⇒

∫1

g (y)d y =

∫f (x) d x (3.56)

You may have to use the trapezium rule (as in Core 2 mathematics but applied to more com-plicated functions). The trapezium rule (in the formula booklet)∫ b

ay d x ≈ 1

2h[y0 +2(y1 + y2 + ...+ yn−1)+ yn]

where h = b −a

nand yi = f (a + i h)

(3.57)

3.6.1 INTEGRATION EXAMPLES

Find: (a)∫

sec2(4x +1) d x (b)∫ 3

1+2x d x.

(a) Let I =∫

sec2(4x +1) d x

then I = 1

4tan(4x +1)+ c

(b) Let I =∫

3

1+2xd x

Consider y = ln|1+2x|

thend y

d x=

∫1

1+2x×2

So I = 3

2ln |1+2x|+ c

(3.58)

Find: (a)∫ ( x+1

x2+2+5

)d x and (b)

∫cos(4x −1)sin2(4x2 −1) d x. In part (a) we’ll use the fact that


∫ (f ′(x)f (x)

)d x = ln | f (x)|+ c and adjust the constant. In part (b)

(a) Let I =∫ (

x +1

x2 +2+5

)d x

Let y = ln |x2 +2x +5|d y

d x= 2x +2

x2 +2x +5= 2(x +1)

x2 +2x +5

so I = 1

2ln |x2 +2x +5|+ c

(b) Let I =∫

cos(4x −1)sin2(4x −1) d x

Let y = si n3(4x = 1)

d y

d x= 3si n2(4x −1)cos(4x −1)×4 [Using the chain rule]

so I = 1

12sin3(4x −1)+ c

(3.59)

Find∫

cot2 3x d x

cot2 3x = csc2 3x −1

so I =∫

cot2 3x d x =∫

(csc2 3x −1) d x

so I =−1

3cot(3x −1)+ c

(3.60)

Find, letting u = tan x the integral of sec2 x etan x with respect tp x.

I =∫

sec2 xetan x d x

Let u = tan x

∴du

d x= sec2 x

∴ du = sec2 x d x∫sec2 xetan x d x =

∫etan x du

∴∫

eu du

= eu + c

= etan x + c

(3.61)

4 STATISTICS 1

4.1 MATHEMATICAL MODELS IN PROBABILITY AND STATISTICS

It’s important to understand the importance of developing a respected model. A lot of thetools you’ll learn about in the Statistics unit can be applied to data and datum to analyse and


compare datasets to statistical and scientific models. You’ll need to know the differences andexamples of normal and discrete distributions. A discrete distribution is a statistical distribu-tion whose variables can take on only discrete values. Discrete variables can only take certainvalues from a finite set. This is typically the number of something; the number of cars, starsor hairs it discrete because it is impossible to have 3 1

2 stars, etc. Normal distributions havemany convenient properties, so random variates with unknown distributions are often as-sumed to be normal, especially in physics and astronomy. Although this can be a dangerousassumption, it is often a good approximation due to a surprising result known as the centrallimit theorem. This theorem states that the mean of any set of variates with any distributionhaving a finite mean and variance tends to the normal distribution. Many common attributessuch as test scores, height, etc., follow roughly normal distributions, with few members at thehigh and low ends and many in the middle. Normal distributions have variables that can takeon continuous variables. Continuous variables can take on any value. The time taken to run100m can take any value, such as 41.359238 seconds, or 295.38 seconds.

• A mathemtical model is a simplified mathematical version of a problem devised to de-scribe, or make predictions about, a real-world situation.

• A prediction is a result or outcome predicted by a mathematical model.

• An observed outcome is results of observations of the real-world e.g. experimentaldata.

• A statistical test is a test used to compare a prediction with an observed.

• Refining the model can be seen as steps taken to make predictions closer to he observedoutcomes.

• Steps in modelling -

– Devise a model.

– Use it to make predictions.

– Collect experimental data from the real-world.

– Compare predictions with collected data and test statistically to see how well themodel describes the real world.

– If necessary refine the model.

– When satisfied accept the model.

4.2 REPRESENTATION AND SUMMARY OF DATA - LOCATION

• Data is a series of observations, measurements or facts.

• A variable is that which is measured or observed e.g. height.

• Types of variable


– Qualitative variable - one with measurements given as numbers.

– Quantitative variable - one to which numbers cannot be assigned.

– Discrete variable - one taking only specified values.

– Continuous variable - one taking any value within a given range.

• Frequency distributions show the values of a variable and how often each occurred.

• Cumulative frequency is obtained by adding the frequencies one at a time across a rowof frequencies.

Frequency Cumulative frequency1 14 58 13

• Grouped frequency distribution is a table in which frequencies are associated withgroups or classes rather than single observations.

• Terminology: consider the class 8−10 shown here:

class widthlower class boundary lower class limit class mid-point upper class limit upper class boundary7.5 8 9 10 10.5

• Mode - the value of a variable that occurs most frequently.

• Median - the middle value of an ordered set of data.

• Mean - the sum of all observations divided by the total number of observations.

– Population mean = µ = Σxn or Σ f x

Σ f

– Sample mean = x = Σxn or Σ f x

Σ f

• Coding - transforms a variable into a simpler one and makes for easier arithmetic, e.g.y = x−a

b ⇒ x = by +a.

4.3 REPRESENTATION AND SUMMARY OF DATA - MEASURES OF DISPERSION

• Range - the value obtained when the smallest observation in a data set is subtractedfrom the largest one.

• Quartiles - divides the data into four equal parts:

– 25% of observations are less than or equal to first quartile Q1

– 50% of observations are less than or equal to the second quartile (median) Q2


– 75% of observations are less than or equal to the third quartile Q3

• Interquartile range (IQR) - the value obtained when the lower quartile is subtractedfrom the upper quartile

IQR =Q3 −Q1 (4.1)

• Semi-interquartile range (SIQR) - half the interquartile range

SIQR = 1

2(Q3 −Q1) (4.2)

• Percentiles - divide the data into 100 equal parts

• Population - a collection of individual items or individuals

• Sample - a subset of a population used to represent that population. 20 people on adesert island, 5 of whom suffer from diabetes and 15 do not. 20 is the population. 5 isthe sample.

• Variance - the variance of a population of observations x1, x2, ..., xn is the mean of thesum of the squared deviations from their mean, µ:

∴ σ2 =∑

x2

n−µ2 or

∑f x2∑

f−µ2 (4.3)

• Standard deviation (σ) - the positive square root of the variance (pσ2). Shows how

emuch variation or dispersion from the average or mean or expected value exists.

4.4 REPRESENTATION OF DATA

• Stem and leaf diagram - each row represents a stem and is indicated by the number tothe left of the vertical line. The digits to the right of the vertical line are leaves associatedwith the stem.

> stem(5 | 0 means 50)

5 | 0 1 1 2 4

6 | 0 1 2 6

7 | 1 2 4 8

• Outlier - an extreme value which does not fit into the main body of the data. Any rulesto identify outliers MUST be specified in S1 question papers.

• Histogram - used to represent a continuous variable which has been summarised by agroup frequency distribution. Each group or class is represented by a bar with widthequal to the class width and area proportional to its frequency.


C l ass F r equenc y C l ass W i d th F r equenc y densi t y = F r equenc yC l ass wi d th

7.58−1010.5 12 3 4

The histogram is polotted using class boundaries and frequency densities. The fre-quency density is plotted on the y axis.

• Box plot - used to represent data and allows comparisons. You’ll usually be asked tocompare box plots. The left-most line is the smallest value. Moving right, you have thefirst quartile Q1, right again is the median Q2, right again is the third quartile Q3, rightagain is the largest value.

0 0.5 1 1.5 2 2.5

Index 0

Index 1

Index 2

• Skewness - an indicator of the shape of a distribution.

– Symmetrical (Index 2): mode = median = meanor Q2 −Q1 =Q3 −Q2

– Positive skew (Index 1): mode < median < meanor Q2 −Q1 <Q3 −Q2

– Negative skew (Index 0): mean < median < modeor Q2 −Q1 >Q3 −Q2

– Given also by: 3(mean−medi an)st and ar d devi ati on or x−Q2

σ

4.5 PROBABILITY

• If p is probability then 0 ≤ p ≤ 1

• Sample space - all possible outcomes

• Event - set of possible outcomes


• P (A∩B) - probability of both A and B . (The intersection of A and B)

• P (A∪B) - probability of A or B or both. (The union of A and B)

• P (A|B) - probability of A given B has happened. (Conditional probability)

• P (A′) - probability of not A. (The probability that A does not happen)

• P (A)+P (A′) = 1

• Addition rule: P (A∪B) = P (A)+P (B)−P (A∩B) (In formula book)

• Multiplication rule: P (A∩B) = P (A|B)×P (B) or often seen as P (A|B) = P (A∩B)P (B) or

P (B |A) = P (A∩B)P (A) (In formula book)

• Independent events - if A and B are independent events then P (A∩B) = P (A)×P (B)

• Mutually exclusive events - if A and B are mutually exclusive events then P (A∩B) = 0.

• Do not confuse the two previous points. An event can never be both mutually exclusiveand independent.

• Venn diagrams


5 GRAPHS - THE ONES YOU’LL NEED TO REMEMBER

y = c y = x y = x2 y = x3

y = 1x y =p

x y = ex y = ln x

y = sin x y = cos x y = tg x y = cotg x

6 NOTATION

• ∴ therefore, x2 = 9∴ x =±3

• ∵ because

• ± plus or minus

• ∓ minus or plus

• = is equal to

• ≡ is identical to

• ' is approximately equal to

•?= is used for proofs, “is a = b” can be rewritten as “a

?= b”.

• > is greater than


• ≥ is greater than or equal to

• < is less than

• ≤ is less than or equal to

• ∞ infinitely large

• ⇒ implies

• ⇐ is implied by

• ⇔ implies and is implied by

• ∈ is a member of

• |x| is the modulus of x,

{|x| = x f or x ≥ 0|x| = −x f or x < 0

•(n

r

)the binomial coefficient, n!

r !(n−r )!

• A stroke through a symbol negates it. i.e. 6= means ’is not equal to’.

• : such that

• +ve positive

• −ve negative

• ∥ parallel

• ⊥ perpendicular

• [a,b] the interval x : a ≤ x ≤ b

• (a,b] the interval x : a < x ≤ b

• (a,b) is the interval x : a < x < b

• ∝ proportional to

7 SPECIAL SETS

• P, denoting the set of all primes: P= 2,3,4,5,11,13,17, ...

• N, denoting the set of all natural numbers: N= 1,2,3, ... (sometimes defined containing0).

• Z, denoting the set of all integers (whether positive, negative or zero): Z= ...,−2,−1,0,1,2, ....


• Q, denoting the set of all rational numbers (that is, the set of all proper and improperfractions): Q = a/b : a,b ∈Z,b 6= 0. For example, 1/4 and 11/6. All integers are in thisset since every integer can be expression as the fraction a/1.

• R, denoting the set of all real numbers. This set includes all rational numbers, togetherwith all irrational numbers (that is, numbers that cannot be rewritten as fraction, suchas

p2, as well as transcendental numbers such as π, e and numbers that cannot be

defined).

• C, denoting the set of all complex numbers: C= a +bi : a,b ∈R. For example, 1+2i ∈C.


A-Level Maths Revision Notes

Education

Transcript of A-Level Maths Revision Notes