ADVANCED GCE UNIT
Additional materials:
Answer booklet (8 pages)
Graph paper
MEI Examination Formulae and Tables (MF2)
4756/01
Morning
Time: 1 hour 30 minutes
INSTRUCTIONS TO CANDIDATES
• Write your name, centre number and candidate number in the spaces provided on the answer booklet.
• Answer all the questions in Section A and one question from Section B.
• You are permitted to use a graphical calculator in this paper.
• Final answers should be given to a degree of accuracy appropriate to the context.
INFORMATION FOR CANDIDATES
• The number of marks is given in brackets [ ] at the end of each question or part question.
• The total number of marks for this paper is 72.
ADVICE TO CANDIDATES
• Read each question carefully and make sure you know what you have to do before starting your answer.
• You are advised that an answer may receive no marks unless you show sufficient detail of the working to indicate that a correct method is being used.
HN/3
This document consists of 4 printed pages.
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2
Section A (54 marks)
Answer all the questions
1 (a) A curve has polar equation r a e k q for 0 q p , where a and k are positive constants.
The points A and B on the curve correspond to q 0 and q p respectively.
(i) Sketch the curve.
[2]
(ii) Find the area of the region enclosed by the curve and the line AB.
(b) Find the exact value of ı
0
2
1
1
3 + 4 x
2
[4]
[5]
(c) (i) Find the Maclaurin series for tan x , up to the term in x
3
.
(ii) Use this Maclaurin series to show that, when h is small,
h
4 h tan x d x x
3 h 7 h 3 .
[4]
[3]
2 (a) You are given the complex numbers w = 3 e
1
12 p j and z = 3 j.
(i) Find the modulus and argument of each of the complex numbers w , z and w
.
z w
(ii) Hence write in the form z a b j, giving the exact values of a and b .
[5]
[2]
(b) In this part of the question, n is a positive integer and q is a real number with 0 q p
.
n
(i) Express that e
1
2 j q
+ e
1
2 j q in simplified trigonometric form, and hence, or otherwise, show
1 + e j q
= 2 e
1
2 j q cos
1 q .
2
Series C and S are defined by
C
Ê n ˆ
Ë 1 ¯ cos q +
Ê n ˆ
Ë 2 ¯ cos 2 q +
Ê
Ë n ˆ
3 ¯ cos 3 q + º +
Ê n ˆ
Ë n ¯ cos n q ,
S =
Ê n ˆ
Ë 1 ¯ sin q +
Ê n ˆ
Ë 2 ¯ sin 2 q +
Ê n ˆ
Ë 3 ¯ sin 3 q + º +
Ê n ˆ
Ë n ¯ sin n q .
[4]
S
(ii) Find C and S , and show that
C tan
1
2 n q .
[7]
© OCR 2007 4756/01 Jan 07

3
3 Let P =
Ê
Á
Á
Ë
4 2 k ˆ
1 1 3
1 0 1
˜
˜
¯
(where k 4 ) and M =
Ê 2 2 6 ˆ
Á 1 3 1 ˜ .
1 2 2
(i) Find P
1 in terms of k , and show that, when k 2, P
1
=
Ê 1 2 4 ˆ
1
Á
2
4 6 10
1 2 2
˜ .
[6]
(ii) Verify that
Ê 4 ˆ Ê 2 ˆ
Á 1 ˜
1
˜
, Á 1 ˜ and
0
˜
Ê
Á
1
2 ˆ
3 ˜ are eigenvectors of M , and find the corresponding eigenvalues.
[4]
(iii) Show that M n
=
Ê 4 6 10 ˆ
Á 2 3 5 ˜
0 0 0
+ 2 n 1
Ê
Á
2 4 4 ˆ
3 6 6 ˜ .
1 2 2
[8]
Section B (18 marks)
Answer one question
Option 1: Hyperbolic functions
4 (i) Show that arcosh x = ln
( x + x
2
1
)
.
[5]
(ii) Find
Û
Ù ı numbers.
1
4 x
2
9 d x , giving your answer in the form a ln b , where a and b are rational
[5]
(iii) There are two points on the curve y cosh x
2 sinh x at which the gradient is
1
9
.
Show that one of these points is
( ln
(
1 + 2 ,
) 1
3 in a similar form.
2
)
, and find the coordinates of the other point,
[8]
4756/01 Jan 07
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© OCR 2007

4
Option 2: Investigation of curves
This question requires the use of a graphical calculator.
5 Cartesian coordinates ( x , y ) and polar coordinates ( r , q ) are set up in the usual way, with the pole at the origin and the initial line along the positive x -axis, so that x r cos q and y r sin q .
A curve has polar equation r k cos q , where k is a constant with k 1.
(i) Use your graphical calculator to obtain sketches of the curve in the three cases k 1, k 1.5 and k 4.
[5]
(ii) Name the special feature which the curve has when k 1.
[1]
(iii) For each of the three cases, state the number of points on the curve at which the tangent is parallel to the y -axis.
[2] d x
(iv) Express x in terms of k and q , and find Hence find the range of values of k for which there d q
.
are just two points on the curve where the tangent is parallel to the y -axis.
[4]
The distance between the point ( r , q ) on the curve and the point ( 1, 0 ) on the x -axis is d .
(v) Use the cosine rule to express d
2 in terms of k and q , and deduce that k
2 d
2 k
2
1.
[4]
(vi) Hence show that, when k is large, the shape of the curve is very nearly circular.
[2]
Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (OCR) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity.
OCR is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate
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© OCR 2007 4756/01 Jan 07

31

4756
1(a)(i)
B1
B1
(ii)
(b)
Area is
=
=
⎡
⎢
⎣
−
1
2 r
2 d θ a
2
4 k
= e − 2 k θ a 2
4 k
( 1 − e
− 2 k π
)
⌠
⎮
0
π
⎥
⎦
⎤
π
0
1
2 a
2
(e
− k θ
)
2 d θ
⌠
⌡
0
2
1
3 +
1
4 x
2 d x =
⎡
⎢
2
1
3 arctan
⎛
⎝
2 x
3
⎞
⎟⎟
⎤
⎥
1
2
0
=
1
2
π
3
=
12 3 arctan
⎛
⎝
1
3
⎞
⎟⎟
(c)(i)
(ii)
OR
3 tan θ
M1
A1
⌠
⎮
1 π
6 is
0
2
=
12
π
1
3
3 d θ A1
M1
A1 f ( x ) = tan x , f ( 0 ) = 0 f ′ ( x ) = f ′′ ( x ) = sec 2 x
2 sec
2
, x f ′ ( 0 ) = 1 tan x , f ′′ ( 0 ) = 0 f ′′′ ( x ) = 2 sec
4 x + 4 sec
2 x tan
2 x , tan x x x 3
3!
= + 1
3 x
3 f ′′′ ( 0 ) = 2
+ ... )
⌡ h
4 h tan x
= x
⎡
⎢ x d x ≈
+ 1
9 x
⌠
⎮ h
4 h
3
( 1
⎤
⎥
4 h h
+ 1
3 x
2
) d x
= ( 4 h
= 3 h +
+ 64
9
7 h
3 h
3
) − ( h + 1
9 h
3
)
B1
M1
A1
B1 ft
M1
A1 ft
A1 ag
M1
A1
M1
A1
M1
A1A1
M1
A1
2
Correct shape for 0 ≤ θ ≤ 1
2
π
Correct shape for
1
2
π ≤ θ ≤ π Requires decreasing r on at least one axis
Ignore other values of θ
4
5
For ∫ ( e
− k θ
)
2 d θ
For a correct integral expression including limits (may be implied by later work) (Condone reversed limits)
Obtaining a multiple of e
− 2 k θ as the integral
For arctan
For
1 and
2 3
2 x
3
Dependent on first M1
For any tan substitution
For ∫
2
1
3 d θ
For changing to limits of θ
Dependent on first M1
4
Obtaining f ′′′ ( x )
For f (0) and f (0) correct ft requires x
3 term and at least one other to be non-zero
Obtaining a polynomial to integrate
For x + 1
9 x
3 ft requires at least two non-zero terms
3
32

4756
2(a)(i)
(ii)
(b)(i) w z w z w z
=
=
3
2
(cos 1
4
π
2
3
2
+
2
3
+
2 j sin 1
4
π ) j e
− 1
2 j θ
= 3 ,
= 2 ,
= 3
2
, arg w arg z
=
= −
− 1
12
π
1
3
π arg w z
= ( − 1
12
π ) − ( − 1
3
π ) =
1 + e j θ = e
1
2 j θ
( e
− 1
2 j θ
= e
2
1 j θ
( 2 cos
+
1
2
θ ) e
1
2 j θ
)
1
4
π
+ e
1
2 j θ
=
=
(cos
2 cos
1
2
θ
1
2
θ
− j sin 1
2
θ ) + (cos 1
2
θ + j sin 1
2
θ )
OR 1 + e j θ = 1 + cos θ + j sin θ
=
=
2 cos 2
2 cos
1
2
θ +
1
2
θ (cos
2 j sin
1
2
θ +
1
2 j
θ cos sin
1
2
θ
1
2
θ )
M1
= 2 e
2
1 j θ cos 1
2
θ A1
M1
A1
M1
A1
M1
A1 ag
B1
B1B1
B1B1 ft
5
Deduct 1 mark if answers given in form r (cos θ + jsin ) but modulus and argument not stated.
Accept degrees and decimal approxs
Accept 1.125
+ 1.125 j
2
4
For either bracketed expression
(ii)
C + j S = 1 +
⎛
⎜⎜ n
1
⎞
⎟⎟ e j θ
= ( 1 + e j θ
) n
+
⎛
⎜⎜ n
2
⎞
⎟⎟ e
2 j θ + ...
+
⎛
⎜⎜ n n
⎞
⎟⎟ e n j θ
S
C
= 2 n e 2
1 n θ j cos n 1
2
θ
C =
S =
2 n cos( 1
2
2 n sin( 1
2 n θ ) cos n θ ) cos n n 1
2
1
2
θ
θ
=
2 n
2 n sin( cos(
1
2
1
2 n θ ) n θ ) cos n cos n
1
2
1
2
θ
θ
= sin( cos(
1
2
1
2 n θ ) n θ )
= tan( 1
2 n θ )
M1
M1A1
M1
A1
A1
B1 ag
7
Using (i) to obtain a form from which the real and imaginary parts can be written down
Allow ft from C + j S = e
1
2 n θ real function of n and θ j
× any
33

4756
3 (i)
(ii) det P
= 1 ( 6 − k )
= 4 − k
− 1 ( 4 −
P
− 1 =
4
1
− k
− 1
⎜
⎝
−
4
1 2
2 )
2
− 4 − k
When k = 2 , P
− 1
M
4
⎜
⎝
1
1
⎟
⎠
=
0
⎜
⎝
0
0
⎟
⎠
=
4
0 ⎜
⎝
1
1
⎟
⎠
OR
=
1
2
− 1
⎜
⎝
−
4
1
6 − k k −
2
12 ⎟
⎠
2
− 6
2
2
M ⎜
⎝
1
0
⎟
⎠
2
= ⎜
⎝
1
0
⎟
⎠
2 4 2
M ⎜
⎝
−
3
1
⎟
⎠
= ⎜
⎝
−
6
2
⎟
⎠
= 2 ⎜
⎝
−
3
1 ⎠
⎟
⎟
Eigenvalues are 0, 1, 2
Eigenvalues are 0, 1, 2
4
−
2
10 ⎟
⎠
2
= 1 ⎜
⎝
1
0
⎟
⎠
(iii)
M n
4
= ⎜
⎝
1
1
=
1
2
⎛
⎜
0
0
0
2
1
0
2
1
0
2 0 0 0
−
3
1
⎟
⎠
⎜
⎝
0
0
1
0
3
2
× n + 1
2 n
⎞
⎟
− 2 n
− 1
⎜
⎝
4
− 1
0
2 n
⎟
⎠
2
− 6
2
− 1
1
2
⎜
⎝
4
− 1
4
−
2
10 ⎟
⎠
2
− 6
2
=
⎛
⎜ 2 −
4
3
−
×
2 n
2 n − 1
2 n − 1
−
− 3
6
+
+
3
2 n + 1
× 2 n
− 2 n
− 10 +
− 5 + 3
2
× n + 1
2 n
⎞
⎟
− 2 n
4
− 10
2
⎟
⎠
4
= ⎜
⎝
2
0
− 6
− 3
0
− 10
− 5
0
⎟
⎠
+
− 2
2 n − 1 ⎜
⎝
−
1
3
4
−
6
2
4
−
6
2
⎟
⎠
B1B1
M1A1
B1 ft
M1
A1
A1 ag
M1
A2
A1
M1
A1
M1
M1
A1 ft
B1 ag
6
Evaluating at least three cofactors
Fully correct method for inverse
Ft from wrong determinant
Correctly obtained
M1
A1A1A1
4
For one evaluation
Obtaining an eigenvalue (e.g. by solving − λ 3 + 3 λ 2 − 2 λ = 0 )
Give A1 for one correct
Verifying given eigenvectors, linking with eigenvalues correctly
8
4 2 2 0 0 0
For ⎜
⎝
1
1
1
0 −
3
1 ⎠
⎟
⎟ and ⎜
⎝
0
0
1
0
0
2 n
⎟
⎠ seen
(for B2, these must be consistent)
For
S D S
− 1 (M1A0 if order wrong) or
1
2
4
⎜
⎝
1
1
2
1
0
2
−
3
1
⎟
⎠
0
⎜
⎝
−
4
2 n
0
2
− 6 n + 1
0
−
2 n
10
+ 1
⎟
⎠
Evaluating product of 3 matrices
Any correct form
34

4756
OR Prove
M n
When
= +
1, A
2 n − 1
B
B
M
by induction
B1
Assuming
M k + 1 =
A M
+
= +
2 k − 1
B M
2 k − 1
B
,
M1A2
2 k − 1 (2 ) A1A1
=
2 k
B
⇒
A1
True for 1 ; hence true for all positive integers n A1 or
M k + 1 =
M A
+ 2 k − 1
M B
Dependent on previous 7 marks
35

4756
4 (i) If
(ii) y = arcosh x , e 2 y − 2 x e y
Since y ≥ 0 , x = cosh y = 1
2
( e y
+ 1 = 0
+ e
− y
) e y =
2 x ±
= x ±
4 x
2
2
− 4 x 2 − 1 e y ≥ 1 , so arcosh x = y = ln( x + e y = x + x
2 − 1 ) x
2 − 1
⌠
⎮
3 .
9
2 .
5
1
4 x
2 − 9 d x =
⎡
⎢
1
2 arcosh
⎛
⎝
= 1
2
( arcosh 2 .
6 − arcosh 5
3
)
2 x
3
⎞
⎠
⎤
⎥
3 .
9
2 .
5
= 1
2 ln( 2 .
6 +
=
=
1
2
1
2
(ln ln 5
3
5 − ln 3 )
2 .
6
2 − 1 ) − ln(
5
3
+ 25
9
− 1 )
OR
⎡
⎢⎣
1
2 ln( 2 x + 4 x
2
M2
− 9 )
⎤
⎥⎦
3 .
9
2 .
5
A1A1
=
= 1
2
1
2 ln 15 − ln 5
3
1
2 ln 9
A1
(iii) d y d x
=
=
(
(
2 + sinh x ) sinh
+ x − sinh
(cosh x )
2
2 sinh
2 + x − 1 sinh x )
2
( 2 x )(cosh x ) d y d x
=
1
9 when 18 sinh x − 9 sinh
2 x − sinh
14 x = sinh
1 ,
= x
( 2
13
+
+
13 sinh
= 0 x )
2
When sinh x = 1 , cosh x = ln( 1 + 2 ) ,
2 ,
3
2 x = ln( 1 +
When sinh x = 13 , cosh x = ln( 13 +
170 , x = ln( 13 +
170 ) ,
170
15
2 )
170 )
M1
A1
M1
M1
M1
A1 ag
A1A1
M1
M1
M1
A1
A1 ag
M1
A1
M1
A1A1
5
5
1
2
and + must be correct
For arcosh (or any cosh substitution)
For
1 and
2 x
2 3
(or 2 x = 3cosh u and ∫ 1
2 d u )
(or limits of u in logarithmic form)
For ln( kx + k 2 x 2 − ...
)
Give M1 for ln( k x + k x − ...)
For 1
2 and ln( 2 x + 4 x
2 − 9 )
(or ln( x + x
2 − 9
4
)
8
Using quotient rule
Any correct form
Quadratic in sinh x (or product of two quadratics in e x )
Solving quadratic to obtain at least one value of sinh x (or e x )
Obtaining x in logarithmic form
(must use a correct formula for arsinh)
SR B1B1 for verifying y = 1
3 and d y x
=
1 d 9 when x =
2 ln(1 + 2)
36

4756
Alternatives for Q4 (i) cosh ln( x + x 2 − = 1
2
(e ln( x + x
2 − 1) + e
− ln( x + x
2 − 1)
= 1
2
( x + x
2
1 x + x 2 − 1
)
)
Since ln( x + x
2
=
=
1
2 x
( x + x
2 x x
2 − 1 )
1) 0, arcosh x = ln( x + x
2 − 1)
M1
M1
M1
A1
A1
If y = arcosh x then ln( x + x 2 y + cosh 2 y − 1)
= ln(cosh y + y
since sinh y > 0
=
= y
M1
M1
A1
M1
A1
5
5
37

4756
5 (i) k = 1 k = 1 .
5 k = 4
B1
B1
B1
B1
B1
5
General shape correct
Cusp at O clearly shown
General shape correct
‘Dimple’ correctly shown
(ii) Cusp B1
1
(iii)
When k = 1 , there are 3 points
When
When k k
=
=
1 .
5
4
, there are 4 points
, there are 2 points B2
2
Give B1 for two cases correct
(iv)
(v)
(vi) x = k d x
= d θ
= cos θ +
− k sin
− sin
θ cos
2
θ ( k
−
+
2
2
θ cos cos
θ
θ sin
)
θ
= 0 when θ = 0 , π , or
For just two points, k ≥ 2 d
2 = r
2 + 1
2 − 2 r cos θ
= ( k + cos θ ) 2 + 1 − 2 ( k + cos θ ) cos θ
= k
2 + 1 − cos
2 θ ( = k
2 + sin
2 θ )
Since 0 ≤ cos 2 θ k
2 ≤
≤ 1 , d
2 ≤ k
2 + 1
When k is large, k 2 + 1 ≈ cos θ = − 1
2 k k , so d ≈ k
Curve is very nearly a circle,
with ( 1 , 0 ) and radius k
B1
B1
M1
A1
M1
A1
M1
A1 ag
M1
A1
4
4
2 or
Allow k > 2
0 ≤ sin
2 θ ≤ 1
38

Report on the units taken in January 2007
4756 - Further Methods for Advanced Mathematics (FP2)
General Comments
There was a wide range of performance on this paper, with about a quarter of the candidates scoring 60 marks or more (out of 72), and about a quarter scoring less than 30 marks. The standard integrals involving inverse trigonometric and hyperbolic functions were handled particularly well, but marks were often lost through carelessness when differentiating trigonometric functions. Many candidates used very long methods to find eigenvalues, and some were unable to complete the paper as a result.
Each of the three questions in Section A had an average mark of about 11 or 12 (out of 18). In
Section B, Q.4 (on hyperbolic functions) was chosen by almost all the candidates, and the average mark was about 10.
Comments on Individual Questions
1 Polar curve and Maclaurin series
In part (a)(i) the curve was usually sketched correctly, although it sometimes spiralled in the wrong direction and often went through the origin. In part (a)(ii) most candidates used
1
2 r 2 d θ with the correct limits, although a substantial minority forgot to square r even though they had written it down. The work was very often completed correctly, but the integration of e − 2 k θ proved to be surprisingly challenging with factors of k going astray and answers such as −
1
2 k θ e − 2 k θ
or even
− 2 k
1
θ + 1 e − 2 k θ + 1 were quite common.
The integral in part (b) was usually found confidently and correctly; the only difficulty was with the factor
1
in front of the integral.
2 3
In part (c), the method for finding the Maclaurin series was well understood, but the triple differentiation of tan x very often went wrong. The first derivative was sometimes written as
1
before proceeding, making the work much more difficult cos 2 x
or even
2
1 cos 2 x
, and obtained very efficiently. When the Maclaurin series had been found correctly, the final part (ii) was usually also completed correctly.
2 Complex numbers
Part (a) was well answered; most candidates were able to work with modulus and argument correctly, although solutions were quite often spoilt by careless errors (such as an incorrect argument for z ).
The identity in part (ii) was usually handled successfully.
In part (iii), almost all candidates realised that they should consider C + j S , and there were very many fully correct solutions. However, a fair proportion of candidates failed to recognise the resulting series as binomial, and were determined to use the formulae for a geometric series, thereby losing most of the marks for this part.
16

Report on the units taken in January 2007
3 Matrices
In part (i) almost all candidates knew a method for finding the inverse matrix, and the process was very often completed accurately.
In part (ii), it was expected that, for each of the given vectors e , the candidates would evaluate Me and see that this is a multiple of e . Many did this, but a large number of candidates found the characteristic equation, then the eigenvalues, and finally the eigenvectors. This did sometimes yield the correct results, but it must have been very time-consuming. Another common method was to write ( M − λ ) = and use one component to find λ ; however, this does not establish that e is an eigenvector unless the other two components are checked, and this was rarely done.
Many candidates knew how to answer part (iii) by forming the product SD S − 1 , although the order of the product was often wrong, and inaccuracies in evaluation frequently prevented the emergence of the given answer.
4 Hyperbolic functions
Most candidates knew how to tackle the standard proof in part (i), although few gave the correct reason (arcosh x ≥ 0 ) for discarding the other root.
The integral in part (ii) was very often found correctly, although the factor 1
2 frequently omitted.
was quite
In part (iii) the differentiation was usually done correctly. Setting the gradient equal to 1
9 gives a quadratic in sinh x which was often solved correctly; then the logarithmic form of arsinh was usually correctly employed to obtain the values of x . Many wrote the gradient in terms of e x and e
− x , and rarely made any further progress; and a common error was to equate the gradient to zero instead of 1
9
. Some made heavy weather of finding the values of y ; from sinh x = 1 and x = ln(1 + 2) they evaluated cosh x as cosh(ln(1 + 2)) instead of using + 2 x .
5 Investigation of curves
There were only a few attempts at this question, and all of these scored less than half marks.
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