TJP TOP TIPS FOR OCR ADDITIONAL MATHS · Add Maths: Algebra . T J Price Page 2 12/10/2008 . Quadratics – solve using formula . a b b ac x 2. 2. 4. Should be self-explanatory. Watch

TJP TOP TIPS

FOR

OCRADDITIONAL

MATHS© Dr T J Price, 2012

Add Maths: Algebra

T J Price Page 1 12/10/2008

Surds – simplify, rationalise the denominator, combine

Simplify (find the biggest square number) 3532532575

Rationalise the denominator (type i) 535

515

5

5

5

15

5

15

Rationalise the denominator (type ii) 7

23

)2(3

23

23

23

23

1

23

122

Combine by multiplying or dividing 31

91273;981273

Combine by adding or subtracting 363235333

67527

Indices – positive, zero, negative, fractional; relation to roots and reciprocals

Negative power = reciprocal (‘1 over’, or flip a fraction) 8273

233

32

2

2 ;9

1

3

13

Fractional power = root 452/1

162533/1 ;32727

Zero power = 1 (always) 12730

Multiply by adding indices 743 777

Divide by subtracting indices 143 777

Do powers of powers by multiplying indices 1243 7)7(

Quadratics – number of solutions, discriminant acb 42

The discriminant of the quadratic equation 02 cbxax is defined as acb 42 .

It is inside the ‘rooty bit’ in the quadratic formula, and it determines the number of solutions as follows:

Discriminant > 0 two distinct solutions [ acb 42 give two values]

Discriminant = 0 one repeated solution [ acb 42 is zero, so we get ab 2/ , twice]

Discriminant < 0 no real solutions [ acb 42 does not exist]

Quadratics – complete the square, factorise

‘Completed square form’ looks like qpxa 2)( , or perhaps qpxa 2)( .

The safe way to do these questions is to multiply this out and then equate the coefficients, e.g.

Write 20183 2 xx in the form qpxa 2)( :

qapapxax

xx

22

2

2

20183 7so2033,3so1832,3 2 qqppa .

Add Maths: Algebra


Quadratics – solve using formula a

acbbx

2

42

Should be self-explanatory. Watch out for:

Negative signs; if b is negative, -b is positive and so is b2.

Also if either a or c is negative, be careful with –4ac; it will be positive.

Don’t forget to divide everything by 2a.

Quadratics in disguise – convert and solve

If an equation contains ‘thing squared’, ‘thing’ and numbers, it is a quadratic in disguise.

E.g. 034 tt becomes 0342 xx if we substitute tx .

Then solve 0)1)(3(342 xxxx so 1or3 xx .

Finally, txtx 2so giving 1or9 tt .

Simultaneous linear/quadratic – solve; zero discriminant implies tangent

If it’s y=… and y=… , bolt them together, rearrange to get a quadratic, solve for x, then get y.

xxy

xy

3

24

2 gives 2432 xxx or 022 xx so 0)1)(2( xx

Then 1,2 xx and 6,10 yy respectively.

If not, get y=… or x=… from one equation and then substitute into the other, solve for x and y.

25

1

22 yx

xy gives 25)1(so1 22 xxxy or 25122 2 xx

so 0)3)(4(or012or02422 22 xxxxxx

Then 3,4 xx and 4,3 yy respectively.

N.B. if your equations have a repeated solution (discriminant = 0) then the line is a tangent.

Inequalities – solve using a sketch graph

Rearrange the inequality to get zero on one side, then factorise the other side.

Sketch the graph and see where it is above/below the y-axis.

E.g. solve xx3

03 xx

0)1)(1()1( 2 xxxxx

From the graph, 01or1 xx

-1 0 1

Add Maths: Polynomials & Graphs


Polynomials

A polynomial is a sum of terms that are multiples of non-negative integer powers of the variable, e.g.

6x³ – 4x + 7 [think of 7 as 7x0].

These are polynomials in x And these aren’t 1742

32 5xx

x

23.5

0

x

3

x 7.14x

Order

The order of a polynomial is the highest power present in it; quadratics are order 2 and cubics are order 3.

Factorising Polynomials

Some polynomials can be factorised, e.g. 223 )1(2 xxxxx .

This factorised form makes it easy to sketch the graph of the polynomial, using the following tips:

)( ax corresponds to an x-axis crossing point at ax , 2)( ax corresponds to an x-axis touching point at ax , 3)( ax corresponds to an x-axis cubic (flat) crossing point at ax .

So 32 )4()1)(2( xxxy look like this:

How do we know which way up the graph goes?

Right-way-up (unreflected or positive) functions go up on the right,

Upside-down (reflected or negative) functions go down on the right.

So sketch the graph from right to left, putting in crossing/touching points as required.

You can easily reverse the whole process to ‘suggest an equation for a graph’.

One more thing; to find where the graph cuts the y-axis, simply put in x = 0 and work out y.

So the above graph cuts the y-axis at 128)4()1(2 32y .

-2 1 4 x

y



Using graphs to solve inequalities

To solve 0)4()1)(2( 32 xxx using our graph, we look for where the graph’s height is positive,

i.e. 4,2 xx .

For 0)4()1)(2( 32 xxx , our solution would change to 4,1,2 xxx .

And for 0)4()1)(2( 32 xxx , we get two ‘sandwiches’ 41,12 xx .

Adding and subtracting polynomials

Just add/subtract matching powers – easy!

Multiplying polynomials

Draw up a grid

Put the coefficients along the top and right (insert ‘0’ for any missing powers)

Fill in the boxes by multiplying

Add along the diagonals sloping down to the left

Read off the answer

N.B. Bigger powers are always to the left or higher up.

Example: calculate )54()273( 2 xxx .

3 -7 2

12 -28 8 4

12 -15 35 -10 -5

-43 43 -10

Answer: 10434312 23 xxx

Dividing polynomials

Tip: set up the question as for long division arithmetic (insert ‘0’ for any missing powers).

Example: calculate )2()273( 2 xxx .

Answer: 133x , remainder 28.

3x – 13

x + 2 3x² – 7x + 2 Look at biggest powers; x goes into 3x² 3x times

3x² + 6x Now subtract 3x lots of x + 2

– 13x + 2 Look at biggest powers; x goes into –13x –13 times

– 13x – 26 Now subtract –13 lots of x + 2

28 Remainder is 28



Factor Theorem

If (x – a) is a factor of a polynomial p(x) , then p(a) = 0.

E.g. (x + 1) = (x – (–1)) is a factor of 33)( 23 xxxxp because 03)1()1(3)1()1( 23p .

This is obvious if you think about graphs and x-axis crossing/touching points;

Each bracket (x – a) corresponds to a crossing/touching point at x = a, i.e. the height of the graph is zero.

Remainder theorem

The remainder when we divide a polynomial p(x) by (x – a) is p(a).

E.g. if we divide 33)( 23 xxxxp by (x – 2) the remainder is 1532232)2( 23p .

Using the Factor Theorem to factorise polynomials

To factorise 652)( 23 xxxxp ,

List the ±ve factors of –6 (the constant term). They are ±1, ±2, ±3, ±6.

Work out p(a) where a is each of these factors in turn; if p(a)=0, we have found a factor (x – a).

Handy hint; start with the easiest, smallest factors first; one of them is bound to work.

Once you have found two factors, you can deduce the third one by considering p(0) (see below).

Working through this example, we try:

p(1) = –8

p(–1) = 0 so (x + 1) is a factor

p(2) = 0 so (x – 2) is a factor

To find the third bracket, write ))(2)(1(652)( 23 cxxxxxxxp and then consider p(0).

Now p(0) = c)2(16 so c must be 3.

Therefore )3)(2)(1()( xxxxp .

If the coefficient of 3x is greater than 1, things are not quite this simple. The question will usually give you

a hint to get you started, e.g. ‘show that (2x – 3) is a factor of…’, so you check that p(1.5) = 0.

Finding unknown coefficients using the Factor and Remainder Theorems

When 62)( 23 bxaxxxf is divided by (x – 1) there is no remainder, and when f(x) is divided by

(x + 1) the remainder is 10. Find the values of a and b.

Solution: No need to do long division – just use the factor and remainder theorems!

If (x – 1) is a factor, then f(1) should equal zero.

So 061112 23 ba or 2 + a + b = 0.

If dividing by (x + 1) gives remainder 10, then f(–1) = 10.

So 106)1()1()1(2 23 ba or –2 + a – b + 6 = 10.

Now solve simultaneously: adding, 2a + 6 = 10 so a = 2 and b = –4.

Add Maths: Binomial


Meaning of nCr

r

nC means the number of different ways of choosing r distinct objects from a total of n objects.

E.g. if 2 different cakes are to be chosen from a menu of 5 cakes, there are 2

5C ways of doing so.

r

rnnnn

rnr

nCr

n

...321

)1)...(2)(1(

)!(!

! [the second way is easier to work out for large n]

N.B. nn ...321! and is called ‘n factorial’.

Going back to our example, 102

20

21

452

5C .

[Why? The first cake can be one of 5, the second can be one of the remaining 4, and we divide by 2

because ‘éclair, then millefeuille’ is the same as ‘millefeuille, then éclair’.]

The numbers produced by r

nC occur in Pascal’s Triangle, being the (r+1)th entry in the (n+1)th row.

Expanding (1 + x)n or (a + b)n – positive integer n

n

n

n

nnnnnn

xxnnn

xnn

nx

xCxCxCxCCx

...321

)2)(1(

21

)1(1

...)1(

32

3

3

2

210

nnnnn

n

n

nnnnnnnnnn

bbannn

bann

bnaa

bCbaCbaCbaCaCba

...321

)2)(1(

21

)1(

...)(

33221

33

3

22

2

1

10

(It’s not as bad as it looks!)

[Hint: the powers of a and b add up to n each time.]

For small n, we can use Pascal’s Triangle directly: e.g. 323 331)1( xxxx .

Expanding (2 – 3x)n, etc.

To expand more complicated binomials, substitute to get nba )( and then replace a and b at the end.

E.g. 44 )()32( bax with a = 2, b = –3x 4322344 464)( babbabaaba (using Pascal’s Triangle)

432

432234

812162169616

)3()3(24)3(26)3(242

xxxx

xxxx

N.B. Watch minus signs, and remember to raise everything in the brackets to the required power…

Finding a particular term

Sometimes we do not need to find every single term of a binomial expansion.

E.g. find the term in x3 in the expansion of (3 – x)

7:

33

34

337

3

7

28358135

)(3321

567

)()3(

xx

x

xC

Add Maths: Binomial


Applying Binomials to probability

If there are n independent trials (occurrences) of some event, each having two outcomes (success or

failure) so that

p = prob(success) and q = prob(failure) = 1 – p,

then the probability of getting r successes in n trials is

P(X = r) = )( rnr

r

n qpC .

This is made up of

r

nC ways of getting r successes out of n trials (imagine the tree diagram),

r successes, probability pr and

(n-r) failures, probability q(n-r)

[X stands for ‘the number of successes’ and you may see the strange-looking X ~ B(n,p) which means

‘X has the Binomial distribution for n trials where p is the probability of success in each trial’.

If the events are not independent, then X will not be Binomial; there are many other distributions…]

Worked Example:

A multiple-choice paper has 10 questions, each with five options. Find the probability that a student will

get (by guesswork alone)

(a) no questions right,

(b) at least two questions right.

For this problem, n = 10, p = 0.2, q = 0.8

(a) P(X = 0) = 100

0

10 8.02.0C = 0.107374

(b) P(X ≥ 2) = 1 – P(X = 0) – P(X = 1)

Since P(X = 1) = 91

1

10 8.02.0C = 0.268435

P(X ≥ 2) = 1 – 0.107374 – 0.268435 = 0.624191

Add Maths: Co-ordinate Geometry


Co-ordinates – midpoint, intersection of lines, distance between two points (x1,y1) & (x2,y2)

Midpoint = average of the two end points 2

,2

2121 yyxx.

Intersection of lines: see solving simultaneous linear/quadratics.

Distance between two points: use Pythagoras on the change in x and the change in y.

Distance = 2

12

2

12 )()( yyxx

Gradient of line segment = Rise Run = (change in y) (change in x) =)(

)(

12

12

xx

yy.

Straight lines – gradient for parallel, perpendicular 121mm

Parallel lines have the same gradient.

Perpendicular lines (at right-angles) have gradients that multiply to make –1.

Alternatively, the gradient at 90 to m is –1 m.

If m is a fraction, flip it upside down and make it the negative of what it was before.

Equation of straight line with gradient m through point (x0, y0) is )( 00 xxmyy

If (x0, y0) is any known point on the line and m is its gradient, then )( 00 xxmyy .

This is a very commonly-used equation, e.g. for tangents and normals.

Quadratics – sketch factorised or completed square form

If )3)(2( xxy , this is a U-shaped graph cutting the x-axis at –2 and 3.

If 3)2( 2xy , this is a U-shaped graph with its turning point at (-2,3);

(it is 2xy translated by –2 in the x direction and +3 in the y direction).

N.B. this graph does not cut the x-axis at all.

Factorised graphs – sketch, use to solve inequalities

If 0)3)(2)(4( xxx , sketch the graph )3)(2)(4( xxxy and see where it

rises above the x-axis.

Answer: 3or24 xx .

-4 -2 3

Add Maths: Co-ordinate Geometry


[Powers of x – sketch positive, negative]

2xy 3xy xy /1 2/1 xy

[Transformations of graphs – translate, stretch, reflect]

Expressions in brackets change the graph horizontally, and do the opposite of what you might expect.

f(x) + a translate up by a f(x + a) translate left by a

f(x) – a translate down by a f(x – a) translate right by a

af(x) stretch vertically by a f(ax) squash horizontally by a

-f(x) reflect vertically in y=0 f(-x) reflect horizontally in x=0

Circles – sketch, identify centre and radius

The pictured circle with centre (a, b)

and radius r has equation

222 )()( rbyax .

A circle touches an axis if one coordinate

is equal to r. In this picture, a = r and b < r.

Circles – complete the square to put into brackets

222 )()( rbyax

The centre of the above circle is (a, b) and its radius is r.

E.g. if 07104 22 yyxx

Then 075)5(2)2( 2222 yx

So 362547)5()2( 22 yx which is a circle centre (2, -5), radius 6.

N.B. the number on the right is radius squared, not radius.

Is a point is inside a circle or not? Compare its distance from the circle’s centre with the radius.

(a, b)

radius r

Add Maths: Linear Programming

T J Price Page 10 12/10/2008

In the world of business and industry, it is desirable to maximise profit or the number of customers and to

minimise costs or the time taken for manufacture.

Linear programming is a mathematical technique used to tackle such problems, and these methods were

developed between 1945 and 1955 by American mathematicians.

In the real world there are not only quantities to be maximised and minimised (as above), but also

constraints such as the size of workforce, quantity of raw materials, etc., as well as obvious facts like we

can’t manufacture a negative number of items.

Mathematically, we set up an axis on our graph for each variable (quantity which can be controlled and

changed) and then shade out regions matching our constraints (graphical inequalities). The best solution

to the problem is then the corner of our remaining feasible region that gives the best value of our

objective function (the quantity we are trying to maximise or minimise).

Using Linear Programming to solve a Puzzle Problem:

A carpenter makes tables and chairs. Each table can be sold for a profit of £30 and each chair for a profit

of £10. The carpenter can afford to spend up to 42 hours per week working and takes six hours to make a

table and three hours to make a chair. Customer demand requires that he makes no more than four times

as many chairs as tables. Tables take up four times as much storage space as chairs and there is room for

at most five tables each week.

Variables

t = number of tables made per week

c = number of chairs made per week

Constraints

Total work time: 6t + 3c ≤ 42 Also t ≥ 0, c ≥ 0

Customer demand: c ≤ 4t (non-negativity constraints)

Storage space: ¼c + t ≤ 5

Objective function

Maximise P = 30t + 10c

Set up axes for t (horizontal) and c (vertical) and shade out the unwanted regions given by the

inequalities.

Then draw an iso-profit line (a line of constant profit, e.g. 30t + 10c = 30) and move your ruler parallel to

this to find the corner of the remaining region giving the optimal solution.

See over for the graph…

Handy hints:

To plot a line such as 6t + 3c = 42, set t=0 and find c (=14), then set c=0 and find t (=7).

So we draw a straight line between (0, 14) and (7, 0).

To plot a line such as c = 4t, note that the gradient is 4 and the c-intercept is zero. Make sure the quantity

on the left is on the vertical axis.

Add Maths: Linear Programming

T J Price Page 11 12/10/2008

The optimal solution is at t=3, c=8. (Place a ruler along the iso-profit line and move it parallel and up to

the right until you are just about to leave the feasible region.)

So the carpenter should make 3 tables and 8 chairs per week, giving a profit of

30×3 + 10×8 = £170 per week.

Quirky questions

If the question asks for (or requires) a solution with integer values (e.g. number of tables and chairs) and

the coordinates of the best corner of the feasible region are not integers, we must instead find the best

feasible integer solution near that corner.

If the question has three variables that are related, don’t try to plot a 3-D graph!

Instead, eliminate one variable. E.g. if x + y + z = 10, then replace ‘z’ with ‘10 – x – y’.

0 1 2 3 4 5 6 7 8 9 10 0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

feasible region

iso-profit line 6t + 3c ≤ 42 1/4c + t ≤ 5

c ≤ 4t

t no of tables

optimal solution

c no of chairs

Add Maths: Trigonometry

T J Price Page 12 12/10/2008

Definitions of sin x, cos x, tan x – SOHCAHTOA

In a right-angled triangle:

opposite adjacent oppositesin , cos , tan

hypotenuse hypotenuse adjacent

These are used to solve right-angled triangles as studied in IGCSE. Sine and Cosine Rule and area of triangle

Use these if a triangle does not have a right angle.

Pick the sine rule if you know any opposite pair of angle & side length.

Sine rule: C

c

B

b

A

a

sinsinsin (or flip it upside down to find an unknown angle)

Otherwise use the cosine rule; the first version to find a side and the second version to find an angle.

Cosine rule: bc

acbAAbccba

2cosorcos2

222222

Trig in 3 Dimensions

Find a RAT (Right Angled Triangle) in the 3D diagram and use trigonometry on it.

For a long diagonal, you can use the triple Pythagoras formula: length = 222 zyx

1) If each sloping edge of a square-based pyramid has a length of 80m and makes an angle of 48°

with the base, find (i) the height EG, (ii) the base area ABCD, (iii) the area of a sloping face GBA

and (iv) the angle EFG between a sloping face and the base.

(i) Drop a vertical from G to E to make a RAT (GEA) with hyp=80m, opp=height, θ=48° so

height EG = 80sin48° = 59.45m

(ii) The distance from E to A (using the same RAT) is 80cos48° = 53.53m. Base area = AB² =

AE² + EB² = 53.53² + 53.53² (by Pythag) = 5731m²

(iii) Each sloping face is an isosceles triangle (GBA) with base √5731 = 75.70m and sides 80m.

Chop it down the middle to make a RAT (GFA) and use Pythag to get height GF = √(80² – 37.85²)

= 70.48m. So area = 37.85×70.48 = 2668m²

(iv) GEF is a RAT; GE = 59.45m and GF = 70.48m. So angle EFG = sin-1

(59.45/70.48) = 57.5°

NB we could have used cosine/sine rule for some of these instead…

2) Find the long diagonal of a cuboid with sides 7cm, 9cm and 13cm.

Diagonal = √(7² + 9² + 13²) = √299 = 17.3cm

48°

80m

A B

C D

E

F

G


T J Price Page 13 12/10/2008

Trig graphs – solving trig equations

Learn these three graphs!

1) Solve cos(x) = 0.5 for 3600 x .

A calculator gives x = cos-1

(0.5) = 60°.

The graph indicates another solution at 360 – 60 = 300°.

So x = 60° or 300°

2) Solve tan(2x) = √3 for 3600 x . First substitute φ = 2x;

A calculator gives φ = tan-1

(√3) = 60°.

The graph shows other solutions at 240°, 420°, 600°, etc. (repetition every 180°).

Since x = φ/2, we get x = 30°, 120°, 210°, 300° (next one would be 390°, out of range)

0

1

-1

y = cos(x)

180 360 90 270 450 540 630

Repeats every 360°; C shaped over first 360°

y = tan(x)

180 360 0

1

-1

540 720 90 270

Repeats every 180°; goes to infinity at 90°, 270°, etc.

y = sin(x)

180 360 0

1

-1

540 720

Repeats every 360°; S shaped over first 360°


T J Price Page 14 12/10/2008

Trig Identities – sin² θ + cos² θ ≡1, sin θ ÷ cos θ ≡ tan θ

Consider this triangle with angle and hypotenuse 1:

By the definition of tan,

cos

sintan .

By Pythagoras, 1cossin 22

These trig identities can be used to solve trig equations.

1) Solve 0cos3sin xx for 3600 x .

Divide through by cos x to get 03tan x

Rearrange to get 3tan x

Use inverse function to get )3(tan 1 x = –71.6°

But this is out of range, so use tan repetition every 180° to get x = 108.4° or 288.4°.

2) Solve 03cos3sin2 xx for 3600 x .

Substitute xx 22 cos1sin to get 03cos3cos1 2 xx

Tidy up to get 02cos3cos2 xx

And swap all signs (multiply by –1) 02cos3cos2 xx

Put xy cos (quadratic in disguise) 0232 yy

Factorise 0)2)(1( yy

Solutions are 1cos xy or 2cos xy

But cos x is never –2, so solve cos x = –1 to get x = 180°.

sin

cos

1

Add Maths: Differentiation

T J Price Page 15 12/10/2008

Differentiate powers of x – positive integer or zero powers 1then If nn nx

dx

dyxy

Always rewrite to get ‘x to the power of something’ before proceeding…

E.g. 23 3 , xdx

dyxy 62 ,96)3( 22 x

dx

dyxxxy

14 , 3425

xdx

dyxx

x

xxy

Gradient at a point – limiting case of sequence of chords

To find the gradient at a given point, find dy/dx and then substitute for x.

E.g. gradient of 452 3 xxy at 1x is 151656 22xdx

dy.

The gradient at a point can be considered as the limiting case of a sequence of chords through two points

that get closer and closer to each other.

E.g. if we draw a chord through P and one

other point moving down from R towards

P, the gradient of this chord approaches

the gradient of the curve at P itself.

Equation of tangent/normal

To find the equation of the tangent to a curve at a given point, find

i) the co-ordinates of this point, (x0, y0)

ii) the gradient, m (using dy/dx at this point)

Then substitute into )( 00 xxmyy .

For a normal, do the same apart from using m = –1 curve gradient.

E.g. Find the normal to xxy 42 3 at 2x .

82422 3y so x0 = 2, y0 = 8.

2042646 22xdx

dy so m = -1/20.

101

201

201 8or)2(8 xyxy .

P

Q

R

Add Maths: Differentiation

T J Price Page 16 12/10/2008

Turning points – locate maximum, minimum

Turning points occur only where the gradient is zero, i.e. 0dx

dy.

A turning point can be a maximum (top of a hill) or a minimum (bottom of a valley).

To distinguish them, you can:

i) use common sense (e.g. 753 2 xxy will have a minimum because it is a +ve quadratic)

ii) use a sketch graph

iii) use the second derivative (see below)

E.g. find the turning points of 162 3 xxy

5or3and1so1giving066 22 yxxxdx

dy

So the turning points are (1, -3) and (-1, 5).

Second derivatives – distinguish max/min

At a turning point, find the second derivative to distinguish between a maximum and a minimum.

If 02

2

dx

yd then the turning point is a minimum (think ‘positive’ = = ‘minimum’).

If 02

2

dx

yd then the turning point is a maximum (think ‘negative’ = = ‘maximum’).

If 02

2

dx

yd we cannot tell; use another method (e.g. sketch graph).

E.g. with the above example, xdx

yd12

2

2

so (1, -3) is a minimum and (-1, 5) is a maximum

Applying derivatives – kinematics, max/min problems, rates of change

Learn that if x = displacement, v = velocity and a = acceleration, then adt

dvandv

dt

dx.

So velocity is rate of change of displacement, and acceleration is rate of change of velocity.

E.g. if 624and612,34 223 tattvttx .

You may also be asked to maximise a volume or minimise an area depending on x.

Do this by setting 0,0dx

dA

dx

dV, etc. and solving the equation.

Add Maths: Integration

T J Price Page 17 12/10/2008

Integration is the opposite process to differentiation. I.e. ‘what differentiates to give…?’

Integrate powers of x – positive integer or zero powers cxdxyxy n

n

n 1

11then If

Always rewrite to get ‘x to the power of something’ before proceeding…

E.g. cxdxxxy 4

4133 ,

cxxxdxxxxxxy 9396 ,96)3( 23

31222

cxxdxxxxxx

xxy 2

215

5144

25

,

Finding the constant of integration, c

If you are given the values of x and y, you can substitute these to find c.

E.g. If 43xdx

dy, find y if the curve passes through (2, 3).

cxxdxx 443 2

23

At x=2, y=3 we get c2423 2

23 so c = 5

Answer: 542

23 xxy

Definite integrals – area under a curve

The area under a curve can be found using integration. This time you include limits on the integral.

E.g. Find the area under the curve xxy 49 2 between x=1 and x=3.

223233

1

233

1

2 units94599)1213()3233(]23[49 xxdxxx

N.B. we don’t need ‘+c’ with a definite integral because it cancels out.

Definite integrals – watch out for:

Areas below the x-axis which come out as a negative integral; the area is positive.

The area between two curves; do the integral of (top curve – bottom curve).

Mixture of areas below and above the x-axis; split the integral into separate parts.

Applying integrals – kinematics

Learn that if x = displacement, v = velocity and a = acceleration, then dtavdtvx , .

So displacement is the integral of velocity, and velocity is the integral of acceleration.

Don’t forget ‘+c’! Use information about the initial displacement or velocity to find c.

Add Maths: Formulae

T J Price Page 18 12/10/2008

The following formulae must be learned off by heart!

The remainder theorem

Given the polynomial f(x), the remainder when f(x) is divided by (x – a) is f(a).

The factor theorem

If (x – a) is a factor of a polynomial f(x), then x = a is a root (solution) of the equation f(x) = 0.

Solution of quadratic equations

By formula and completing the square.

Formula: for the quadratic equation a

acbbxcbxax

2

4,0

22 .

Binomial expansion

Know how to derive terms in the expansions of (1 + x)n and (a + b)

n where n is a positive integer.

The coefficients may be expressed in the form !

!( )!

n

r

n nC

r r n r

Be aware that n n

r n r

n nC C

r n r

Coordinate Geometry – Straight line

Know the forms of the straight line equation, including y = mx + c and y – y0 = m(x – x0).

Parallel lines: m1 = m2 , Perpendicular lines: m1 × m2 = –1 or m2 = –1/m1

2 2

1 1 2 2 1 2 1 2

1 2 1 21 1 2 2

Distance between two points ( , ), ( , )

Midpoint of two points ( , ), ( , ) ,2 2

x y x y x x y y

x x y yx y x y

Coordinate Geometry – Circle

The circle 2 2 2 has centre ( , ) and radius .x a y b r a b r

Trigonometry

In a right-angled triangle:

opposite adjacent oppositesin , cos , tan

hypotenuse hypotenuse adjacent

Identities: cos

sintan , 1cossin 22

Sine rule: C

c

B

b

A

a

sinsinsin Cosine rule:

bc

acbAAbccba

2cosorcos2

222222

Add Maths: Formulae

T J Price Page 19 12/10/2008

Calculus

1d

d

dStationary points occur when 0

d

Area under curve between and is d

n n

b

a

yy ax nax

x

y

x

x a x b y x

Kinematics

Constant acceleration formulae:

If s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time, then:

atuv tvus )(21 2

21 atuts 2

21 atvts asuv 222

Calculus (if acceleration is not constant):

If x = displacement, v = velocity, a = acceleration, t = time, then:

dt

dxv

dt

dva dtvx dtav

Documents

TJP TOP TIPS FOR OCR ADDITIONAL MATHS · Add Maths: Algebra . T J Price Page 2 12/10/2008 . Quadratics – solve using formula . a b b ac x 2. 2. 4. Should be self-explanatory. Watch