General Certificate of Education Advanced Level Examination January 2013

Mathematics

MFP2

Unit Further Pure 2 Wednesday 23 January 2013

9.00 am to 10.30 am

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For this paper you must have: * the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator.

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Time allowed * 1 hour 30 minutes

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Instructions * Use black ink or black ball-point pen. Pencil should only be used for drawing. * Fill in the boxes at the top of this page. * Answer all questions. * Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. * You must answer each question in the space provided for that question. If you require extra space, use an AQA supplementary answer book; do not use the space provided for a different question. * Do not write outside the box around each page. * Show all necessary working; otherwise marks for method may be lost. * Do all rough work in this book. Cross through any work that you do not want to be marked.

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Information * The marks for questions are shown in brackets. * The maximum mark for this paper is 75. Advice * Unless stated otherwise, you may quote formulae, without proof, from the booklet. * You do not necessarily need to use all the space provided.

P59577/Jan13/MFP2 6/6/6/

MFP2

2 1 (a)

Show that 12 cosh x  4 sinh x ¼ 4e x þ 8ex

(b)

(2 marks)

Solve the equation 12 cosh x  4 sinh x ¼ 33 giving your answers in the form k ln 2 .

(5 marks)

Two loci, L1 and L2 , in an Argand diagram are given by pffiffiffi L1 : jz þ 6  5ij ¼ 4 2

2

L2 : argðz þ iÞ ¼

3p 4

The point P represents the complex number 2 þ i . (a)

Verify that the point P is a point of intersection of L1 and L2 .

(2 marks)

(b)

Sketch L1 and L2 on one Argand diagram.

(6 marks)

(c)

The point Q is also a point of intersection of L1 and L2 . Find the complex number that is represented by Q. (2 marks)

3 (a)

(b)

Show that

1 1 A  ¼ , stating the value of the constant A. 5r  2 5r þ 3 ð5r  2Þð5r þ 3Þ (2 marks)

Hence use the method of differences to show that n X

1 n ¼ ð5r  2Þð5r þ 3Þ 3ð5n þ 3Þ r¼1

(c)

Find the value of 1 X

1 ð5r  2Þð5r þ 3Þ r¼1

(02)

(4 marks)

(1 mark)

P59577/Jan13/MFP2

3

The roots of the equation

4

z 3  5z 2 þ kz  4 ¼ 0 are a, b and g . (a) (i)

Write down the value of a þ b þ g and the value of abg .

(ii) Hence find the value of a 2 bg þ ab 2 g þ abg 2 .

(2 marks) (2 marks)

The value of a 2 b 2 þ b 2 g 2 þ g 2 a 2 is 4 .

(b) (i)

Explain why a, b and g cannot all be real.

(ii) By considering ðab þ bg þ gaÞ2 , find the possible values of k.

5 (a)

e y  ey Using the definition tanh y ¼ y , show that, for jxj < 1 , e þ ey   1þx 1 1 tanh x ¼ ln 2 1x d 1 ðtanh1 xÞ ¼ . dx 1  x2

(b)

Hence, or otherwise, show that

(c)

Use integration by parts to show that ð1 2

4 tanh

1

0

(1 mark) (4 marks)

(3 marks)

(3 marks)

 m 3 x dx ¼ ln n 2

where m and n are positive integers.

(5 marks)

A curve is defined parametrically by

6

x ¼ t3 þ 5 ,

y ¼ 6t 2  1

The arc length between the points where t ¼ 0 and t ¼ 3 on the curve is s. (a)

Show that s ¼

ð3 3t

pffiffiffiffiffiffiffiffiffiffiffiffiffi t 2 þ A dt , stating the value of the constant A.

(4 marks)

0

(b)

(4 marks) Turn over

s

(03)

Hence show that s ¼ 61 .

P59577/Jan13/MFP2

4

The polynomial pðnÞ is given by pðnÞ ¼ ðn  1Þ3 þ n3 þ ðn þ 1Þ3 .

7 (a) (i)

Show that pðk þ 1Þ  pðkÞ , where k is a positive integer, is a multiple of 9. (3 marks)

(ii) Prove by induction that pðnÞ is a multiple of 9 for all integers n 5 1 . (b)

(4 marks)

Using the result from part (a)(ii), show that nðn2 þ 2Þ is a multiple of 3 for any positive integer n. (2 marks)

pffiffiffi Express 4 þ 4 3 i in the form reiy , where r > 0 and p < y 4 p . (3 marks) pffiffiffi (b) (i) Solve the equation z 3 ¼ 4 þ 4 3 i , giving your answers in the form reiy , where r > 0 and p < y 4 p . (4 marks) pffiffiffi (ii) The roots of the equation z 3 ¼ 4 þ 4 3 i are represented by the points P, Q and R on an Argand diagram. pffiffiffi Find the area of the triangle PQR, giving your answer in the form k 3 , where k is an integer. (3 marks) pffiffiffi (c) By considering the roots of the equation z 3 ¼ 4 þ 4 3 i , show that

8 (a)

cos

2p 4p 8p þ cos þ cos ¼0 9 9 9

(4 marks)

Copyright Ó 2013 AQA and its licensors. All rights reserved.

(04)

P59577/Jan13/MFP2

Key to mark scheme abbreviations M m or dM A B E or ft or F CAO CSO AWFW AWRT ACF AG SC OE A2,1 –x EE NMS PI SCA c sf dp

mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result correct answer only correct solution only anything which falls within anything which rounds to any correct form answer given special case or equivalent 2 or 1 (or 0) accuracy marks deduct x marks for each error no method shown possibly implied substantially correct approach candidate significant figure(s) decimal place(s)

No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.

MFP2 - AQA GCE Mark Scheme 2013 January series

MFP2 Q 1(a)

Solution 1 x cosh x  (e  e  x ) 2 1 or sinh x  (e x  e  x ) 2 12cosh x  4sinh x =

Marks

Total

Comments or 12cosh x  6(e x  e x ) or 4sinh x  2(e x  e x )

M1

6(e x  e x )  2(e x  e x ) 12cosh x  4sinh x  4e x  8e  x (b)

A1 cso

2

AG

4e x  8e  x  33

 4e  33e  8 2x

x

( 0)

M1

 (e x  8)(4e x  1) ( 0)   e x   8,

attempt to multiply by ex to form quadratic in ex

e  x

1 4

 x   3ln 2  x    2ln 2 Total

m1

factorisation attempt (see below) or correct use of formula

A1

correct roots

A1 A1

5 7

MFP2 - AQA GCE Mark Scheme 2013 January series

MFP2 (cont) Q 2(a)

Solution

Marks

4  4i  16  16  32  4 2

B1

3 4

B1

arg(–2+ 2i) =   tan 1 (1) =

Total

Comments

verification that 2  i  6  5i  4 2 2

verification that arg (z+ i) =

3 4

Im

Re

(b) Circle

M1

freehand circle sketched

Centre at – 6 + 5i

A1

clear from diagram or centre stated

Cutting Re axis but not cutting Im axis

A1

“Straight” line

M1

freehand line

Half line from 0 – i

A1

not horizontal or vertical but end point at 0 – i must be clear from diagram/stated

gradient –1 (approx)

A1

6

making 45 to negative Re axis and positive Im axis

(c) Calculation based on fact that L2 passes through centre of L1

Q represents

A1

–10 + 9i

Total

 4  idea of vector   from centre 4

M1 2

10

must write as a complex number

MFP2 - AQA GCE Mark Scheme 2013 January series

MFP2 (cont) Q 3(a)

Solution

Marks

5r  3   5r  2  1 1   5r  2 5r  3  5r  2  5r  3

M1



(b)

5

 5r  2  5r  3

Attempt to use method of differences

1  1 k    3 5n  3    5n  3   3  k   3(5n  3) 

S 

1 15

Total

Comments

condone omission of brackets for M1 2

A=5

M1

at least 2 terms of correct form seen

A1

correct cancellation leaving correct two fractions

m1

1   5n  3   3  n Sn    5  3(5n  3)  3(5n  3) (c)

A1cso

Total

A1cso

4

B1

1 7

attempt to write with common denominator 1 AG k  used correctly throughout 5

MFP2 - AQA GCE Mark Scheme 2013 January series

MFP2 (cont) Q 4(a)(i)

(ii)

Solution      5    4

Marks

Total

B1 B1

2

 2   2   2    (     )

M1

= 5  4  20 (b)(i)

A1

Comments

2

FT their results from (a)(i)

1

argument must be sound

If  ,  ,  are all real then

 2  2   2 2   2 2  0 Hence  ,  ,  cannot all be real

E1

      k

B1

  k

M1

correct identity for

(ii)

     

PI

2

   2   2( 2   2   2  ) 2

  4  2(20)

A1 A1 cso

k  6

Total

4

9

  

2

substituting their result from (a)(ii) must see k=…

MFP2 - AQA GCE Mark Scheme 2013 January series

MFP2 (cont) Q

Solution e  e y x  tanh y  y e  e y xe y  xe  y  e y  e  y

Marks

Total

Comments

y

5(a)

M1

or

xe 2 y  x  e 2 y  1

 ( x  1)e y  e y (1  x)  ( x  1)  e2 y (1  x) e2 y 

(b)

A1

1 x 1 1 x   y  ln   1 x 2 1 x 

A1cso

1 1 y  ln(1  x)  ln(1  x) 2 2 dy 1 1   d x 2(1  x) 2(1  x) 1 x 1 x 2 1    2 2(1  x)(1  x) 2(1  x ) 1  x 2

3

AG

3

AG

M1 A1 A1cso

Alternative 1 d y 1 (1  x) d  1  x       M1 d x 2 (1  x) d x  1  x 

d y 1 (1  x) (1  x)  (1  x)    A1 d x 2 (1  x) (1  x)2 dy 1   A1 cso d x 1  x2 (c)

 4 tanh

1

4x dx 1  x2 4 x tanh 1 x  2ln(1  x 2 )

x dx  4 x tanh 1 x  

A1

 12 ln 3

B1

must simplify logarithm to ln3

ln 3  2ln 34

A1

any correct form

tanh Value of integral =

1 1 2

M1

3  ln  4  2  3

Total

A1cso

5 11

all working must be correct

MFP2 - AQA GCE Mark Scheme 2013 January series

MFP2 (cont) Q 6(a)

Solution

dx  3t 2 dt

Marks

dy  12t  dt

Comments

B1

both correct

 dx   dy  4 2       9t  144t  dt   dt 

M1

 dx   dy  ‘their’       dt   dt 

9t 4  144t 2  dt 

A1

OE

2

s

2

2

3

s   3t t 2  16 dt

A1cso

0

(b)

Total

3

4

3

A1

(t 2  16) 2 3

A = 16

where k is a constant; ft their A

M1

k (t 2  A) 2

3

m1

25 2  16 2 = 61

A1 cso Total

2

F(3) – F(0) 4 8

AG

MFP2 - AQA GCE Mark Scheme 2013 January series

MPC1 (cont) Q

Solution

Marks

Total

Comments

7(a)(i) p(k  1)  p(k )  k 3  (k  1)3  (k  2)3

 (k  1)3  k 3  (k  1)3

M1

 (k  2)3  (k  1)3  k 3  6k 2  12k  8  (k 2  3k 2  3k  1)  9k 2  9k  9  9(k 2  k  1) which is a multiple of 9 (since k2 + k + 1 is an integer )

A1

A1cso

(ii) p(1) = 1 + 8 = 9  p(1) is a multiple of 9

p(k+1) = p(k) + 9(k 2  k  1) or p(k+1) = p(k) + 9N Assume p(k) is a multiple of 9 so p(k) = 9M, where M is integer  p( k  1) = 9M + 9N = 9(M + N)  p(k  1) is a multiple of 9

multiplied out & correct unsimplified

3

correct algebra plus statement

B1

result true for n = 1

M1

p(k+1) = … and result from part (i) considered and mention of divisible by 9 must have word such as “assume” for A1

A1

Result true for n = 1 therefore true for n = 2, n = 3 etc by induction. ( or p(n) is a multiple of 9 for all integers n 1 )

E1

convincingly shown 4

(b) p(n)  (n  1)3  n3  (n  1)3

must earn previous 3 marks before E1 is scored

need to see this OE as evidence

= 3n  6n 3

or 3n(n 2  2)

B1

p(n) = 3n(n 2  2) & p(n) is a multiple of 9. Therefore n(n 2  2) is a multiple of 3 (for any positive integer n.) Total

both of these required E1

2 9

plus concluding statement

MFP2 - AQA GCE Mark Scheme 2013 January series

MFP2 (cont) Q 8(a)

Solution

Marks

r 8

Total

Comments

B1

tan 1 

4 3  or  seen 4 3  

or

2 3

A1

(b)(i) modulus of each root = 2

3

4  4 3i  8 e

4 2 8 , , 9 9 9

A2

1 2 (ii) Area = 3   PO  OR  sin 2 3 1 2  3   2  2  sin 2 3 =3 3

4

A1 if 3 “correct” values not all in requested interval

()4 2 8   2  cos  cos  cos  9 9 9  

e

cos

, 2e

i

2 9

, 2e

i

8 9

A1

correct values of lengths in formula 3

E1 M1

must be stated explicitly in form r (cos  i sin  )

A1

isolating real terms ; correct and with “2”

4 4  cos explicitly stated to 9 9 earn final A1 mark

or cos

A1cso Total TOTAL

4 9

Correct expression for area of triangle PQR

4 4  cos  isin seen earlier 9 9

2 4 8  cos  cos 0 9 9 9

i

M1

A1cso

(c) Sum of roots (of cubic) = 0 Sum of 3 roots including Im terms

4 9

2 3

use of De Moivre – dividing argument by 3

2e

i

i

B1 M1

  



marked as angle to Im axis with 6 “vector” in second quadrant on Arg diag

M1

4 14 75

AG

Scaled mark unit grade boundaries - January 2013 exams A-level Maximum Scaled Mark

A*

LAW UNIT 3

80

66

MD01

MATHEMATICS UNIT MD01

75

-

63

57

52

47

42

MD02

MATHEMATICS UNIT MD02

75

68

62

55

49

43

37

MFP1

MATHEMATICS UNIT MFP1

75

-

69

61

54

47

40

MFP2

MATHEMATICS UNIT MFP2

75

67

60

53

47

41

35

MFP3

MATHEMATICS UNIT MFP3

75

68

62

55

48

41

34

MFP4

MATHEMATICS UNIT MFP4

75

68

61

53

45

37

30

MM1B

MATHEMATICS UNIT MM1B

75

-

58

52

46

40

35

MM2B

MATHEMATICS UNIT MM2B

75

66

59

52

46

40

34

MPC1

MATHEMATICS UNIT MPC1

75

-

64

58

52

46

40

MPC2

MATHEMATICS UNIT MPC2

75

-

62

55

48

41

35

MPC3

MATHEMATICS UNIT MPC3

75

69

63

56

49

42

36

MPC4

MATHEMATICS UNIT MPC4

75

58

53

48

43

38

34

MS1A

MATHEMATICS UNIT MS1A

100

-

78

69

60

52

44

MS/SS1A/W

MATHEMATICS UNIT S1A - WRITTEN

75

58

34

MS/SS1A/C

MATHEMATICS UNIT S1A - COURSEWORK

25

20

10

MS1B

MATHEMATICS UNIT MS1B

75

-

60

54

48

42

36

MS2B

MATHEMATICS UNIT MS2B

75

70

66

58

50

42

35

MEST1

MEDIA STUDIES UNIT 1

80

-

54

47

40

33

26

MEST2

MEDIA STUDIES UNIT 2

80

-

63

54

45

36

28

MEST3

MEDIA STUDIES UNIT 3

80

68

58

48

38

28

18

MEST4

MEDIA STUDIES UNIT 4

80

74

68

56

45

34

23

Code LAW03

Title

Scaled Mark Grade Boundaries and A* Conversion Points A B C D E 60

54

48

43

38