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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
T J Price Page 2 12/10/2008
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
T J Price Page 3 12/10/2008
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
Add Maths: Polynomials & Graphs
T J Price Page 4 12/10/2008
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
Add Maths: Polynomials & Graphs
T J Price Page 5 12/10/2008
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
T J Price Page 6 12/10/2008
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
T J Price Page 7 12/10/2008
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
T J Price Page 8 12/10/2008
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
T J Price Page 9 12/10/2008
[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
Add Maths: Trigonometry
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°
Add Maths: Trigonometry
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