Core 3 / Core 4 Workshops
A* Booster Course
Dear Student,
ELITE Tuition is running a series of two day, EdExcel Mathematics C3 / C4 workshops, designed to ensure
students know all of the examination tricks necessary to secure the A*. Students are taught in small groups by
an EdExcel Examiner, paying specific attention to exam technique and how to answer the A* grade questions.
The cost of the workshop, for both days including all workshop material is £250.00. However, students from
West London schools are eligible for a 40% discount, making the cost £150.00. Each day will look, into detail,
60 examination questions from a bank of EdExcel questions not commonly available to students.
The two-day workshops will be running in early June 2011. The structure of the days will be as follows:
Day 1: Core 3
11th June
Day 2: Core 4
18th June
11:00am
Introduction
11:00am
Introduction
11:30am
C3 Workshop Part 1
11:30am
C4 Workshop Part 1
1:30pm
Analysis of Trick Questions
1:30pm
Analysis of Trick Questions
2:00pm
Lunch
2:00pm
Lunch
3:00pm
C3 Workshop Part 2
3:00pm
C4 Workshop Part 2
5:00pm
Top Tips for C3 Success
5:00pm
Top Tips for C4 Success
5:30pm
Finish
5:30pm
Finish
If you are interested in attending, please fill in the section below and enclose a cheque of £150.00 made
payable to ELITE Tuition, and send these to ELITE Tuition, 1 Castle Road, Northolt, Middlesex, UB5 4SD.
For further information, please call 0800 612 9545 or visit www.elitetuition.com.
Best wishes,
Rachael Gill
(School Liaisons Officer)
All details must be completed below.
Surname of Student:
Date of Birth:
First Name(s):
/
/
Email:
Home Address:
Post Code:
Home Telephone Number:
Student Mobile:
Name of School / College:
I confirm that I will be attending the ELITE Tuition EdExcel Core 3 / Core 4 Revision Workshop “A* Booster
Course” on the Saturdays 11th and 18th June at the venue selected: Merchant Taylors School. I enclose a
cheque of sum £150.00 made payable to ELITE Tuition.
Signed:
Print Name:
ELITE Tuition, 1 Castle Road, Northolt, Middlesex, UB5 4SD
Tel: 0800 612 9545
Email: [email protected]
Web: www.elitetuition.com
C4 Binomial Expansion Problem
Question:
The function f (x) is defined as f ( x ) ≡
6x
, x < 1.
x − 4x + 3
a)
Express f (x) as partial fractions.
b)
Show that for small values of x,
8
26 3
f ( x ) ≈ 2x + x 2 +
x
3
9
2
(3 marks)
(5 marks)
Answer:
a)
Writing as partial fractions (assuming not top heavy):
f ( x) ≡
6x
x − 4x + 3
(Factoring the denominator)
2
x 2 − 4x + 3 ≡ ( x − 1) ( x − 3)
⇒ f ( x) ≡
6x
A
B
≡
+
( x − 1) ( x − 3) ( x − 1) ( x − 3)
Multiplying through:
6x ≡ A ( x − 3) + B ( x − 1)
Select values of x to eliminate A and B.
Let x = 1
6 (1) = A (1 − 3) + B (1 − 1)
6 = −2A + B ( 0 )
∴ A = −3
Let x = 3
6 ( 3) = A ( 3 − 3) + B ( 3 − 1)
18 = A ( 0 ) + 2B
⇒ f ( x) ≡
∴B = 9
9
3
−
( x − 3) ( x − 1)
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b)
Re-writing our expression for f (x):
f ( x ) ≡ 9 ( x − 3) − 3( x − 1)
−1
−1
We cannot perform a binomial expansion here unless it is in the form (1 + ax ) . So if we rewrite it a little more:
n
f ( x ) ≡ 9 ( −3 + x ) − 3( −1 + x )
−1
−1
Factoring to put into the form of (1 + ax ) :
n
⎛
f ( x ) ≡ 9 × −3−1 ⎜ 1 −
⎝
1⎛
f ( x) ≡ 9 × − ⎜1 −
3⎝
⎛
f ( x ) ≡ −3 ⎜ 1 −
⎝
x⎞
⎟
3⎠
∴ f ( x ) ≡ 3(1 − x )
−1
−1
x⎞
⎟
3⎠
x⎞
⎟
3⎠
−1
−1
⎛
− 3 × −1−1 ⎜ 1 −
⎝
1⎛
− 3 × − ⎜1 −
1⎝
⎛
+ 3⎜ 1 −
⎝
x⎞
⎟
1⎠
x⎞
⎛
− 3⎜ 1 − ⎟
⎝
3⎠
x⎞
⎟
1⎠
x⎞
⎟
1⎠
−1
−1
−1
−1
Expanding the first term:
2
3
4
5
Use (1 − x ) = 1 + x + x + x + x + x + ...
−1
3(1 − x ) = 3 ⎡⎣1 + x + x 2 + x 3 + ...⎤⎦
−1
3(1 − x ) = 3 + 3x + 3x 2 + 3x 3 + ...
−1
Expanding the second term:
−1
⎡ x x2 x3
⎤
= −3 ⎢1 + + +
+ ...⎥
⎣ 3 9 27
⎦
−1
2
3
x⎞
x
x
⎛
−3 ⎜ 1 − ⎟ = −3 + −x − − + ...
⎝
3⎠
3 9
⎛
−3 ⎜ 1 −
⎝
x⎞
⎟
3⎠
Adding the two results:
f ( x ) ≡ 3 + 3x + 3x 2 + 3x 3 + ... − 3 + −x −
x2 x3
8
26 3
− + ... = 2x + x 2 +
x + ...
3
9
3
9
Yet when x is small:
∴ f ( x ) ≈ 2x +
8 2 26 3
x +
x
3
9
Found this useful? Attend our A* Booster C3 / C4 Workshop.
ELITE Tuition, 1 Castle Road, Northolt, Middlesex, UB5 4SD
Tel: 0800 612 9545
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C4 Parametric Integration Problem
Question:
An ellipse is defined be the following parametric equations
x = 4 cosθ , y = 2 sin θ , 0 ≤ θ ≤ 2π .
which meets the positive coordinate axes at the points A and B
c) Find the values of the parameter θ at the points A and B.
(2 marks)
d)
Show that the area of the ellipse is 8π.
(8 marks)
Answer:
a)
It is always useful to draw a diagram when solving problems.
Hence, A is found when x = 0 and B is found when y = 0.
Thus: When x = 0
0 = 4 cosθ
π
⇒θ =
2
b)
When y = 0
0 = 2 sin θ
⇒θ = 0
Using the equation for parametric equations:
b
b'
a
a'
∫ y dx = ∫ y
y = 2 sin θ
&
dx
= −4 sin θ
dθ
dx
dθ
dθ
0
⇒ Area =
∫ 2 sinθ × −4 sinθ
π
2
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dθ
0
∫ −8 sin
⇒ Area =
π
π
θ dθ
(Remove – sign and flip limits)
2
2
∫ 8 sin
⇒ Area =
2
2
(Use sin 2 θ =
θ dθ
0
π
Area =
2
1
∫ 8 × 2 (1 − cos 2θ )
1
(1 − cos 2θ ) )
2
dθ
0
π
2
Area = 4 ∫ 1 − cos 2θ dθ
(Integrate)
0
π
1
⎡
⎤ 2
Area = 4 ⎢θ − sin 2θ ⎥
2
⎣
⎦0
⎧⎡ π 1
1
⎛π⎞⎤ ⎡
⎤⎫
Area = 4 ⎨ ⎢ − sin 2 ⎜ ⎟ ⎥ − ⎢ 0 − sin 2 ( 0 ) ⎥ ⎬
⎝ 2⎠⎦ ⎣
2
⎦⎭
⎩⎣ 2 2
⎧⎡ π
⎫
⎤
Area = 4 ⎨ ⎢ − 0 ⎥ − [ 0 − 0 ]⎬
⎦
⎩⎣ 2
⎭
Area = 2π
Since the shaded area is a quarter of the area of the ellipse, the total area of the ellipse would be
8π.
∴ Total Area = 8π
Found this useful? Attend our A* Booster C3 / C4 Workshop.
ELITE Tuition, 1 Castle Road, Northolt, Middlesex, UB5 4SD
Tel: 0800 612 9545
Email: [email protected]
Web: www.elitetuition.com
C4 Differential Equation Problem
Question:
Given that y = 0 when x = 0, solve the differential equation:
dy
= e x + y cos x
dx
(10 marks)
Answer:
First one needs to separate the variables:
dy
= e x + y cos x
dx
(Use e x + y = e x e y )
dy
= e x e y cos x
dx
(Separate the variables)
e− y dy = e x cos x dx
(Integrate both sides)
∫e
−y
dy = ∫ e x cos x dx
Integrating the left hand side first:
−e− y = ∫ e x cos x dx
(Ignoring the constant from LHS)
Note: If you ignore the integrating constant from the left hand side, you MUST remember to
include it when integrating the right hand side.
Integrating the right hand side by parts:
Let:
Use I = uv − ∫ v
u = ex
dv
= cos x
dx
du
= ex
dx
v = sin x
du
dx from the C4 Formula Bookmark:
dx
−e− y = e x sin x − ∫ e x sin x dx
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Integrate
∫e
Let:
x
sin x dx by parts again:
u = ex
dv
= sin x
dx
du
= ex
dx
v = − cos x
Substituting back in:
−e− y = e x sin x − ⎡⎣ −e x cos x − ∫ −e x cos x dx ⎤⎦
Simplifying the negative signs.
−e− y = e x sin x − ⎡⎣ −e x cos x + ∫ e x cos x dx ⎤⎦
−e− y = e x sin x + e x cos x − ∫ e x cos x dx
The trick behind this question is to notice that
∫e
x
cos x dx is −e− y :
Therefore:
−e− y = e x sin x + e x cos x − ⎡⎣ −e− y ⎤⎦ + c
−e− y = e x sin x + e x cos x + e− y + c
−2e− y = e x sin x + e x cos x + c
1
e− y = − ⎡⎣ e x sin x + e x cos x ⎤⎦ + c
2
Applying boundary conditions, let x = 0 and y = 0.
1
e− ( 0 ) = − ⎡⎣ e( 0 ) sin ( 0 ) + e( 0 ) cos ( 0 ) ⎤⎦ + c
2
1
1 = − [ 0 + 1] + c
2
⇒c=
3
2
Therefore
1
3
e− y = − ⎡⎣ e x sin x + e x cos x ⎤⎦ +
2
2
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C4 Vectors Problem 01
Question:
The line l1 has vector equation
⎛ 3 ⎞
r= ⎜ 1 ⎟ +λ
⎜
⎟
⎝ 2 ⎠
⎛ 1 ⎞
⎜ −1 ⎟
⎜
⎟
⎝ 4 ⎠
and the line l2 has vector equation
⎛ 0 ⎞
r= ⎜ 4 ⎟ +µ
⎜
⎟
⎝ −2 ⎠
⎛ 1 ⎞
⎜ −1 ⎟ ,
⎜
⎟
⎝ 0 ⎠
where λ and µ are parameters.
The lines l1 and l2 intersect at the point B and the acute angle between l1 and l2 is θ.
(a) Find the coordinates of B.
(4)
(b) Find the value of cos θ , giving your answer as a simplified fraction.
(4)
The point A, which lies on l1, has position vector a = 3i + j + 2k.
The point C, which lies on l2, has position vector c = 5i – j – 2k.
The point D is such that ABCD is a parallelogram.


(c) Show that ⎢ AB ⎢= ⎢ BC ⎢.
(3)
(d) Find the position vector of the point D.
(2)
Answer:
a)
At point B the vector equations for the lines are equal. Thus:
(This leads us to the equations)
[1]
[2]
[3]
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Generally, one solves equations [1] & [2] simultaneously and substitute into equation [3] to find
the values of λ and µ. Fortunately, this question has an easy route.
Taking equation [3]:
Substituting this into the line equation for l1.
b)
To find the angle, we use the equation:
It is important to note that the vectors that are dotted together must be the direction components
of the lines only.
Let a equal the direction component of l1,
Let b equal the direction component of l2,
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Thus
c)
(Substituting the vectors for a and b)
(Substituting the vectors for c and b)
Since both
and
equal
,
.
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d)
This question is remarkably easy to answer and the technique used to answer this question arises
often. But as always, it is a good idea to draw a diagram.
.
.
l2
D
C
.
B
.
l1
A
O
From the diagram above, it is easy to see that the position vector of D or
simply:
if one prefers is
Thus,
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C4 Vectors Problem 02
Question:
Referred to a fixed origin O, the points A and B have position vectors (i + 2j – 3k) and
(5i – 3j) respectively.
(a) Find, in vector form, an equation of the line l1 which passes through A and B.
(2)
The line l2 has equation r = (4i – 4j + 3k) + µ (i – 2j + 2k), where µ is a scalar parameter.
(b) Show that A lies on l2.
(1)
(c) Find, in degrees, the acute angle between the lines l1 and l2.
(4)
The point C with position vector (2i – k) lies on l2.
(d) Find the shortest distance from C to the line l1.
Answer:
a) The line is given by the equation
Thus:
a)
One can write line l2 as:
If point A lies on l2 then:
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(4)
Be equating coefficients we arrive at three equations:
1)
2)
3)
Since
b)
for all three coefficients, point A exists on the line l2.
Using the formula
Using only the direction components of the two lines:
˚
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c)
It is often best to draw a diagram to visualise how best to answer this question.
l1
.
.
A
19.5 ˚
Distance
.
C
l2
O
There are many ways in which one can answer this question, but the best way to answer this
question is to use trigonometry.
Since the shortest line linking point C to the line l1 will meet the line l1 at a right angle, we have a
right angle triangle.
Thus:
Distance =
(Thus determining
)
units
Distance =
Distance = 1.00 units
Found this useful? Attend our A* Booster C3 / C4 Workshop.
ELITE Tuition, 1 Castle Road, Northolt, Middlesex, UB5 4SD
Tel: 0800 612 9545
Email: [email protected]
Web: www.elitetuition.com