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
C3 Trigonometry Problem
Question:
a) Find R and α for:
sin x + 3 cos x = Rsin ( x + α )
b)
for 0 ≤ α ≤
π
2
Find the values of x which satisfy:
π⎞
⎛
sin x + 3 cos x + 2 sin ⎜ 3x − ⎟ = 0
⎝
3⎠
Given 0 ≤ x ≤ 2π .
Answer:
a)
Given sin x + 3 cos x = Rsin ( x + α ) , we expand the RHS.
sin x + 3 cos x = Rsin x cos α − Rsin α cos x
By comparing coefficients:
R cos α = 1
Rsin α = 3
(1)2 + (
⇒R=
3
)
2
∴R = 2
⎛ 3⎞
⇒ α = tan −1 ⎜
⎟
⎝ 1 ⎠
π
∴α =
3
Therefore:
π⎞
⎛
sin x + 3 cos x = 2 sin ⎜ x + ⎟
⎝
3⎠
b)
π⎞
⎛
Given that sin x + 3 cos x + 2 sin ⎜ 3x − ⎟ = 0 , we substitute our answer for part a).
⎝
3⎠
π⎞
⎛
Since sin x + 3 cos x = 2 sin ⎜ x + ⎟ , this implies
⎝
3⎠
π⎞
π⎞
⎛
⎛
2 sin ⎜ x + ⎟ + 2 sin ⎜ 3x − ⎟ = 0
⎝
⎠
⎝
3
3⎠
Please turn over…
ELITE Tuition, 1 Castle Road, Northolt, Middlesex, UB5 4SD
Tel: 0800 612 9545
Email: [email protected]
Web: www.elitetuition.com
π⎞
π⎞
⎛
⎛
Factoring 2 sin ⎜ x + ⎟ + 2 sin ⎜ 3x − ⎟ = 0 , we get:
⎝
⎝
3⎠
3⎠
π⎞
π⎞⎤
⎡ ⎛
⎛
2 ⎢sin ⎜ x + ⎟ + sin ⎜ 3x − ⎟ ⎥ = 0
⎝
3⎠
3⎠ ⎦
⎣ ⎝
On the ELITE Tuition Bookmark, you will see the equation:
⎛ A + B⎞
⎛ A − B⎞
sin A + sin B = 2 sin ⎜
cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎟⎠
π
3
π
π
A + B x + 3 + 3x − 3 4x
=
=
= 2x
2
2
2
π
π
2π
A − B x + 3 − 3x + 3 −2x + 3
π
=
=
= −x +
2
2
2
3
If we let A = x +
Therefore:
And
π
3
and
B = 3x −
π⎞
π⎞⎤
π⎞⎤
⎡ ⎛
⎡
⎛
⎛
2 ⎢sin ⎜ x + ⎟ + sin ⎜ 3x − ⎟ ⎥ = 2 ⎢ 2 sin ( 2x ) cos ⎜ −x + ⎟ ⎥ = 0
⎝
⎝
3⎠
3⎠ ⎦
3⎠ ⎦
⎣ ⎝
⎣
π⎞
⎛
4 sin ( 2x ) cos ⎜ −x + ⎟ = 0
⎝
3⎠
(Divide both sides by 4)
π⎞
⎛
sin ( 2x ) cos ⎜ −x + ⎟ = 0
⎝
3⎠
Therefore:
sin ( 2x ) = 0
⇒ 2x = 0, π , 2π , 3π , 4π
π
3π
∴ x = 0, , π ,
, 2π
2
2
or
π⎞
⎛
cos ⎜ −x + ⎟ = 0
⎝
3⎠
Thus: x = 0,
π
π
π
3π
=
, − , −
3
2
2
2
π 5π 11π
∴x = − ,
,
6 6
6
⇒ −x +
(Ignore −
π
)
6
π 5π
3π 11π
,
, π,
,
, 2π
2 6
2
6
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
C3 Trigonometry Problem 2
Question:
c)
d)
1 − tan 2 θ
Prove that
≡ cos 2θ .
1 + tan 2 θ
(4 marks)
Hence, or otherwise, prove that:
π
tan 2 = 3 − 2 2 .
8
(5 marks)
Answer:
a)
For this question it is useful to use the identity tan 2 θ ≡ sec 2 θ − 1
Thus
=
1 − tan 2 θ
and substituting tan 2 θ ≡ sec 2 θ − 1 gives us:
2
1 + tan θ
(
)
1 + ( sec θ − 1)
1 − sec 2 θ − 1
2
1 − sec 2 θ + 1
=
1 + sec 2 θ − 1
2 − sec 2 θ
=
sec 2 θ
2
sec 2 θ
=
−
= 2 cos 2 θ − 1
2
2
sec θ sec θ
(Use cos 2θ = 2 cos 2 θ − 1 )
∴= cos 2θ
Please turn over…
ELITE Tuition, 1 Castle Road, Northolt, Middlesex, UB5 4SD
Tel: 0800 612 9545
Email: [email protected]
Web: www.elitetuition.com
b)
When asked to solve tan 2
π
= 3 − 2 2 using the result from part a), one must look out for the
8
trick:
Taking our result from part a) :
1 − tan 2 θ
≡ cos 2θ
1 + tan 2 θ
One needs to solve for tan 2 θ :
1 − tan 2 θ
≡ cos 2θ
1 + tan 2 θ
(
1 − tan 2 θ ≡ cos 2θ 1 + tan 2 θ
(Multiply through)
)
1 − tan θ ≡ cos 2θ + cos 2θ tan 2 θ
1 − cos 2θ ≡ cos 2θ tan 2 θ + tan 2 θ
2
(Re-arrange)
(Factorise)
1 − cos 2θ ≡ tan 2 θ ( cos 2θ + 1)
1 − cos 2θ
≡ tan 2 θ
cos 2θ + 1
⇒ tan 2 θ ≡
1 − cos 2θ
1 + cos 2θ
Now we let θ =
π
and substitute into above:
8
π
1 − cos
π
4
tan 2 =
8 1 + cos π
4
2 −1
2 = 2 −1
2 +1
2 +1
2
2 −1
2 −1 2 − 2 − 2 +1
×
=
2 −1
2 +1
2 −1
1
π
2 =
tan 2 =
1
8 1+
2
(Recall cos
π
1
=
)
4
2
1−
tan 2
π
=
8
∴ tan 2
(Rationalise the denominator)
π
= 3− 2 2
8
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
C3 Trigonometry Problem 3
Question:
e) Using the identities
cos ( A + B ) = cos A cos B − sin Asin B
cos ( A − B ) = cos A cos B + sin Asin B
to prove that:
⎛ P + Q⎞
⎛ P − Q⎞
cos P + cosQ = 2 cos ⎜
cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎟⎠
f)
Find, in terms of π, the values of x in the interval 0 ≤ x ≤ 2π for which
cos x + cos 2x + cos 3x = 0 .
Answer:
a)
For this question, we proceed by taking the difference between the two identities:
cos ( A + B ) − cos ( A − B ) = ( cos A cos B − sin Asin B ) − ( cos A cos B + sin Asin B )
Thus,
cos ( A + B ) − cos ( A − B ) = cos A cos B − sin Asin B − cos A cos B + sin Asin B
cos ( A + B ) − cos ( A − B ) = 2 cos A cos B
The trick to this question is making the substitutions:
Let P = A + B and let Q = A − B .
Therefore cos P + cosQ = 2 cos A cos B . Now we need A & B in terms of P & Q.
Taking P = A + B we get B = P − A . If we substitute this into Q = A − B :
Q = A − ( P − A)
Q = 2A − P
⇒A=
P+Q
2
Similarly, taking P = A + B we get A = P − B . If we substitute this into Q = A − B
Q = ( P − B) − B
Q = P − 2B
⇒B=
P−Q
2
⎛ P + Q⎞
⎛ P − Q⎞
Therefore cos P + cosQ = 2 cos ⎜
.
cos ⎜
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
Please turn over…
ELITE Tuition, 1 Castle Road, Northolt, Middlesex, UB5 4SD
Tel: 0800 612 9545
Email: [email protected]
Web: www.elitetuition.com
(4 marks)
(7 marks)
c)
When asked to solve cos x + cos 2x + cos 3x = 0 , there is a very hidden trick here.
If we re-write the equation as cos 2x + cos 3x + cos x = 0 , let P = 3x and Q = x :
cos 2x + cos 3x + cos x = 0
⎛ 3x + x ⎞
⎛ 3x − x ⎞
cos 2x + 2 cos ⎜
cos ⎜
=0
⎟
⎝ 2 ⎠
⎝ 2 ⎟⎠
cos 2x + 2 cos 2x cos x = 0
(Factorise cos 2x out)
cos 2x (1 + 2 cos x ) = 0
Thus either cos 2x = 0 or cos x = −
1
2
When cos 2x = 0 :
2x = 0, π , 2π , 3π , 4π
π
3π
⇒ x = 0, , π ,
, 2π
2
2
1
When cos x = − :
2
⇒x=
2π 4π
,
3
3
Therefore:
x = 0,
π 2π
3π 4
,
, π,
, π , 2π
2 3
2 3
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
C3 Differentiation Problem
Question:
a)
The curve with the equation y =
1 2
x − 3ln x , x > 0 , has a stationary point A. Find the
2
exact x coordinate of A.
(3 marks)
b)
Determine the nature of the stationary point.
(2 marks)
c)
d)
Show that the y coordinate of A is
3
(1 − ln 3) .
2
(2 marks)
Find an equation for the tangent to the curve at the point where x = 1 , giving your answer
in the form ax + by + c = 0 , where a, b and c are integers.
(3 marks)
Answer:
a)
To find the stationary point of A, we need to differentiate y:
dy
3
= x−
dx
x
3
x− =0
x
3
x=
x
2
x =3
⇒ x = 3 taking the positive as x > 0 .
b)
(At stationary point,
dy
= 0)
dx
(Solve for x)
d2 y
To determine the nature of the stationary point, we need 2 :
dx
2
d y
3
dy
3
= x − , this implies 2 = 1 + 2 . Thus substituting in x = 3 :
Since
dx
x
dx
x
d2 y
dx 2
= 1+
x= 3
3
( 3)
2
=2
Which is positive, implying the stationary point is a minima.
Please turn over…
ELITE Tuition, 1 Castle Road, Northolt, Middlesex, UB5 4SD
Tel: 0800 612 9545
Email: [email protected]
Web: www.elitetuition.com
c)
Substitute x = 3 into the equation for y.
1 2
x − 3ln x
2
2
1
y=
3 − 3ln
2
1
3
y = − 3ln 32
2
3 3
y = − ln 3
2 2
(Substitute in x = 3 )
y=
( )
∴y =
d)
( 3)
(Use log power law)
(Factoring
3
out)
2
3
(1 − ln 3)
2
To find the equation of a tangent, we need to use the equation y − y1 = m ( x − x1 ) :
We are given x1 which is x = 1 .
To find y1 , one substitutes x = 1 into y =
y1 =
1 2
(1) − 3ln (1)
2
∴ y1 =
1
2
To find m, we substitute x = 1 into
dy
3
= x−
dx
x
dy
dx
dy
dx
1 2
x − 3ln x
2
= (1) −
x =1
dy
:
dx
(Substitute in x = 1 )
3
(1)
= −2
x =1
Since we are looking at a tangent problem, m = −2 .
y − y1 = m ( x − x1 )
1
y − = −2 ( x − 1)
2
2y − 1 = −4 ( x − 1)
2y − 1 = −4 x + 4
(Multiply through by 2)
(Re-arrange)
∴ 4 x + 2y − 5 = 0
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
C3 Differentiation Problem 02
Question:
a) Use the derivatives of sin x and cos x to prove that:
d
( cot x ) = −cosec 2 x
dx
(4 marks)
b)
Show that the curve with equation
y = e x cot x
has no turning points.
(5 marks)
Answer:
a)
Use the fact that cot x =
u = cos x
u ' = − sin x
Let
⇒
cos x
dy vu '− uv'
=
and that
.
sin x
dx
v2
v = sin x
v' = cos x
d
− sin 2 x − cos 2 x
( cot x ) =
dx
(sin x )2
(
sin 2 x + cos 2 x
d
cot
x
=
−
(
)
dx
(sin x )2
(Factoring the negative)
)
(Use sin 2 x + cos 2 x = 1 )
d
1
( cot x ) = − 2
dx
sin x
∴
b)
d
( cot x ) = −cosec 2 x
dx
Using the product rule:
( )
dy d x
d
=
e × cot x + ( cot x ) × e x
dx dx
dx
dy
= e x × cot x + −cosec 2 x × e x
dx
Please turn over…
ELITE Tuition, 1 Castle Road, Northolt, Middlesex, UB5 4SD
Tel: 0800 612 9545
Email: [email protected]
Web: www.elitetuition.com
dy
= e x cot x − e x cosec 2 x
dx
(
dy
= e x cot x − cosec 2 x
dx
At stationary points,
(Factorising)
)
dy
= 0.
dx
Thus:
(
)
e x cot x − cosec 2 x = 0
Which implies ex = 0 which only occurs when x = - ∞ or that:
cot x − cosec 2 x = 0
Searching for values of x which would satisfy this:
cot x − cosec 2 x = 0
cos x
1
− 2 =0
sin x sin x
(Multiplying both sides by sin2 x )
sin x cos x − 1 = 0
sin x cos x = 1
Use the double angle identity sin 2x = 2 sin x cos x.
1
sin 2x = 1
2
⇒ sin 2x = 2
Since the since function must produce an answer between -1 ≤ x ≤ 1, there are no solutions to
the above equation.
Thus implies there are no stationary points for the function y = ex cot x.
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