Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Additional Mathematics Exam 2013
Completing the square:

-It involves writing the equation in the form (x-k)2=m
...

 For example:
Solve the equation x2-4x-7=0
 (x-2)2-4-7=0
 (x-2)2=11
 x-2=±√11
 x=2±√11
-The minimum or maximum value of a quadratic expression can be
found by completing the square
...

-The functions f(x)=(x-a)2+b has a minimum value of b when x=a
...
If you do then you must reverse the
symbol
...

 For example:
3
Simplify
2
3 2

2

Polynomials:
Multiplication of polynomials:
-Multiplying out two polynomials is the same process as multiplying out
two linear expressions; everything in one bracket is multiplied by
everything in the other
...

 (x2-3x+4) x (2x-3)

 x2(2x-3)-3x(2x-3)+4(2x-3)
 2x3-3x2-6x2+9x+8x-12
 2x3-9x2+17x-12
Division of polynomials
-As with numbers, when one polynomial is divided into another it ‘goes’
a number of times, leaving a remainder
...

 For example:
Divide (x3-3x2+4x-1) by (x-1)
...


Remainder Theorem
-If a polynomial f(x) is divided by x - a, then the remainder, R= f(a)
...

 The remainder is f(2)=-8-12+2-3= -5
...

 For example:
Show that (x-2) is a factor of f(x)= x3+2x2-x-2
 f(-2)=-8+8+2-2=0, so (x+2) is a factor
...


Binomial Expansion
-The binomial expansion is the expansion of an expression such as
(a+b)n, where n is a positive integer
...


The Binomial Expansion
-When (a+b)n is expanded, the first term is an, the last is bn and the

Cr a n-r b r
...


The Coefficients
-The coefficients can be found in two ways:
1) The first way, when many or all of the terms are required and n is
not too large, uses Pascal’s triangle:
1
1
1
1
2
1
1
3
3
1
1
4
6
4
1
1
5
10
10
5
1
1
6
15
20
15
6
etc
...

-For a binomial distribution with n events, P (r successes) is given by the
appropriate term in the binomial expansion (p+q)n
...
510

2) A spinner has five side, numbered 1 to 5
...
What is
the probability that the side with number 5 lands on the table
exactly three times in five spins?


5

C3

× 0
...
82

 10 × 0
...
64 = 0
...

3) A normal dice is thrown three times
...
0046 = 0
...
Show that AB and CD are parallel
...

-Perpendicular lines are lines with gradients m1 and m2 such that
m1×m2=-1
...


7-3
4
=
5- 2
3
0-3
3
 Gradient of AC=
= 6-2
4
4
3
 Since
× - =-1, the lines AB and AC are perpendicular
...






2)





(x-a)2 + (y-b)2 = r2 with r=5 and (a,b)=(2,3)
...

Find the centre and radius of the circle x2+y2-2x+4y-4=0
(x2-2x) + (y2+4y) – 4 = 0
(x-1)2 + (y+2)2 – 4 = 5
(x-1)2 + (y+2)2 = 9
Centre (1,-2) and radius 3
...

-It can then be seen which of the vertices will give the maximum
value for P
...

-The objective function is the function that is to be maximised or
minimised
...

-However, there are two angles in the range of 0° £ q £ 360°
...

 Cosine curve= 360 – x
 Sine curve= 180 - x
 Tan curve= 180 + x

Identities and equations
Identities:
-There are two identities which are true for all angles, q
...
Find the exact value of cos q
...


sin q
=2
cosq
 tan q = 2
 q = 63
...
4 = 243
...

3) Solve the equation

3sin 2 q + sin q -1 = 0

 3x + x -1 = 0
 a= 3, b= 1 and c= -1
2

-1± 1+12
6
-1± 13

6
 sin q =

 -0
...
4343
 q = -50
...
7
 The first value gives the solutions 180+50
...
1 and
360-50
...
9
...
7 and 18025
...
3
...
7°, 154
...
1° and 309
...


Area and the Sine and Cosine Rules

Area:
-The formula for the area of a triangle is:
Area = ½absinC
...
Find the area of the triangle
...
5 x sin50 x 6 x 8 = 18
...

Sine Rule:
-In any triangle ABC with sides a, b and c:

a
b
c
=
=
sin A sin B sinC

-This can be rewritten as:

sin A sin B sinC
=
=
a
b
c


For example:
In a triangle ABC, a=5cm, A=56° and B=47°
...


5
b
=
sin 56 sin 47
5
= 4
...

 sin 47 ´
sin 56


Cosine Rule:
-In any triangle ABC with sides a, b and c:

a 2 = b 2 + c 2 - 2bc cos A

-Cosine can be made the subject of the formula:

cos A =


b2 + c2 - a2
2bc

For example:
The three sides of a triangle are 6, 7 and 8cm
...


6 2 + 72 - 82
2´6´7
-1
 cos (0
...
5°
...


-If

y = ax n then the gradient function is


For example:

dy
= nax n-1
...


dy
= 2x +12x 3
...

 For example:
Find the gradient of the curve

dy
dx

y = 4x 2 - x 3 at point (2,8)
...

dx
dy
=16 - 2 = 4
...

-The gradient of the tangent is the gradient of the curve at that
point
...

-Perpendicular lines have gradients m1 and m2 such that m1×m2=-1
Tangents:
-To find the equation of the tangent at a point on a cruve:
1
...

dx

2
...

3
...

 For example:
Find the equation of the tangent to the curve
point (2,7)
...

y = 10x + c
7 = 20 + c
c = -13
y =10x -13
y -10x +13 = 0

Normals:
-To find the equation of the normal at a point on a cruve:
1
...

dx

2
...

3
...

4
...

 For example:
Find the equation of the normal to the curve

y = x 3 - 3x 2 - 6x +11 at point (3,-7)
...

1
 y =- x+c
3
1
 -7 = - ´ 3+ c
3
 c = -6
1
 y =- x-6
3
1
 y+ x+6 = 0
3
Stationary points
-A stationary point on a curve is a point where the gradient is 0
...

-There are the minimum and maximum points:

Sketching a Curve:
-The essential details of a curve to be sketched are:
 The co-ordinates and nature of any turning points
...

-To determine whether a turning point is a maximum or a minimum you
can calculate the value of the gradient on each side of the turning
point and at the turning point
...

dy
= 6x 2 - 6x -12

dx
2
2
 0 = 6x - 6x -12 = x - x - 2 = 0
 ( x - 2 ) ( x +1) = 0
 x= 2 or -1


(2 ´ 23 ) - (3´ 2 2 ) - (12 ´ 2) + 24 = 4

 Turning point is when x=2 and y=4
...


Integration
-The reverse of differentiation:
If

dy
a n+1
= ax n then y =
x +c
dx
n +1

-C is called the constant of integration and if you differentiate a
constant you get 0, so when you integrate you need to write it in
...

 y=

6 4
x + c = 1
...
Find the equation of the curve, given that it
dx

passes through point (1,2)
...

2
3) Find ò (6x - 4x)dx
...

-Given two numbers, called the upper and lower limits, you evaluate
the function for the upper and the lower limit and subtract
...

 For example:

ò

3
2

(6x 2 - 4x)dx

é2x 3 - 2x 2 ù
ë
û2
3
2
3
2
 (2 ´ 3 - 2 ´ 3 ) - (2 ´ 2 - 2 ´ 2 )
 (54 -18) - (16 - 8)
3



 36-8=28
Area:
-The area between a curve y=f(x), the two lines x=a and x=b and the
x-axis is

ò

b
a

f (x)dx
...

 Area =

ò

4
1

(-x 2 + 4x + 3)dx

é 1 3
ù
2
ê- x + 2x + 3x ú
ë 3
û1
1
1
(- ´ 43 + 2 ´ 4 2 + 3´ 4) - (- ´13 + 2 ´12 + 3´1)
3
3
64
1
(- + 32 +12) - (- + 2 + 3)
3
3
68 14
- = 18units 2
3 3
4






-Note that if the curve lies beneath the x-axis then the numerical
value of the area will be negative
...

Area Between Two Curves:

ò

ò

- Area= (top curve)dx- (bottom curve)dx


For example:
1) Find the area enclose by the curve y=5+2x-x2, the x-axis
and the lines x+y=5
...

 5-x=5+2x-x2
...

 Area =


ò

3
0

ò

3
0

(-x 2 + 2x + 5)dx - (5 - x)dx

(-x 2 + 3x)dx

é 1 3 3 2ù
 ê- x + x ú
ë 3
2 û0
1 3 3 2
 (- ´ 3 + ´ 3 ) - 0
3
2
2
 (-9 +13
...
5units
3

2) Find the area between the curve y=13-6x+x2 and the curve
y=3+6x-x2
...

 Area under 'top' curve =

ò

5
1

(-x 2 + 6x + 3)dx

é 1 3
ù
2
 ê- x + 3x + 3x ú
ë 3
û1
æ 1 3
ö æ 1 3
ö
2
2
 ç - ´ 5 + 3´ 5 + 3´ 5÷ - ç - ´1 + 3´1 + 3´1÷
è 3
ø è 3
ø
æ 125
ö æ 1
ö
+ 75 +15÷ - ç - + 3+ 3÷
 çè 3
ø è 3
ø
145 17 128
2
- =
= 42

3
3
3
3
5
2
 Area under 'bottom' curve = ò (x - 6x +13)dx
5

1

é1 3
ù
2
ë3
û1
æ1 3
ö æ1 3
ö
2
2
ç ´ 5 + 3´ 5 +13´ 5÷ - ç ´1 + 3´1 +13´1÷
è3
ø è3
ø
æ 125
ö æ1
ö
- 75 + 65÷ - ç - 3+13÷
ç
è 3
ø è3
ø
95 31 64
1
- =
= 21
3 3
3
3
2
1
1
Total area = 42 - 21 = 21
3
3
3
5

 ê x - 3x +13x ú





-If the curve lies under the x-axis then the area will be negative
...

 Can be negative
...

-Velocity:

Vector quantity
...

If motion along a line remains in the same direction then it can
also be called speed
...

 Cannot be negative
...

 Interchangeable with distance if motion remains in the same
direction
...

-Distance:
 Cannot be negative
...
Find the average speed and velocity
...

 Displacement= 0
...

time taken
2
displacement 0
= = 0mph
...

s = displacement (metres)
...

v = current velocity (metres/second)
...

-Formulae:

v = u + at

æu+vö
s = tç
÷
è 2 ø
1
s = ut + at 2
2

v 2 = u2 + 2as


For example:

1) A car decelerates uniformly from 15ms-1 to rest in 10
seconds
...

 u= 15, v= 0, t=10
 v = u + at




0 =15 +10a
10a = -15
a = -1
...
5m/s2

æ 15 + 0 ö
÷
è 2 ø
 7
...

 s = 10 ç

Acceleration due to gravity:
 For example:
A ball is dropped from a window 30 metres above ground
...
8ms-2
...
8
 s = ut +




1 2
at
2

1
30 = 0t + ´ 9
...
9t
30
= t2
4
...
47 , so it took 2
...






v 2 = u2 + 2as
v 2 = 0 2 + 2 ´ 9
...
2 , so its velocity

Variable Acceleration:



For example:

when it landed was 24
...


A body moves along a straight line
...
The acceleration t seconds passing O is
given by a=2+t
...


1 2
t +c
2
1
4 = 2 ´ 0 + ´ 02 + c
2
c=4
1
v = 2 ´ 5 + ´ 52 + 4
2
v = 26
...
5m/s after 5 seconds
...
8m after 5 seconds

---