General Certificate of Education Advanced Level Examination January 2013

Mathematics

MPC3

Unit Pure Core 3 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.

P58909/Jan13/MPC3 6/6/6/

MPC3

2

It is given that yðxÞ satisfies the differential equation

1

where and (a)

dy ¼ f ðx, yÞ dx pffiffiffiffiffiffiffiffiffiffiffiffiffi f ðx, yÞ ¼ 2x þ y yð3Þ ¼ 5

Use the Euler formula yrþ1 ¼ yr þ hf ðxr , yr Þ with h ¼ 0:2 , to obtain an approximation to yð3:2Þ, giving your answer to four decimal places. (3 marks)

(b)

Use the formula yrþ1 ¼ yr1 þ 2hf ðxr , yr Þ with your answer to part (a), to obtain an approximation to yð3:4Þ, giving your answer to three decimal places. (3 marks)

2 (a)

(b)

Write down the expansion of e3x in ascending powers of x up to and including the (1 mark) term in x 2 . Hence, or otherwise, find the term in x 2 in the expansion, in ascending powers of x, of

3

32 3x e ð1 þ 2xÞ .

(4 marks)

It is given that the general solution of the differential equation d2 y dy 2 þy¼0 2 dx dx is y ¼ e x ðAx þ BÞ . Hence find the general solution of the differential equation d2 y dy þ y ¼ 6e x  2 2 dx dx

(02)

(5 marks)

P56804/Jan13/MFP3

3

ð1 Explain why

4 (a)

x 4 ln x dx is an improper integral.

(1 mark)

0

ð1 Evaluate

(b)

x 4 ln x dx , showing the limiting process used.

(6 marks)

0

Show that tan x is an integrating factor for the differential equation

5 (a)

dy sec2 x þ y ¼ tan x dx tan x Hence solve this differential equation, given that y ¼ 3 when x ¼

(b)

(2 marks) p . 4

(6 marks)

It is given that y ¼ lnðe3x cos xÞ .

6 (a) (i)

Show that

(ii) Find

dy ¼ 3  tan x . dx

d4 y . dx 4

(3 marks)

(3 marks)

(b)

Hence use Maclaurin’s theorem to show that the first three non-zero terms in the 1 1 expansion, in ascending powers of x, of lnðe3x cos xÞ are 3x  2 x 2  12 x 4 . (3 marks)

(c)

Write down the expansion of lnð1 þ pxÞ , where p is a constant, in ascending powers of x up to and including the term in x 2 . (1 mark)

(d) (i)

  3x  1 e cos x lim Find the value of p for which x ! 0 2 ln exists. 1 þ px x

  3x  1 e cos x lim (ii) Hence find the value of x ! 0 2 ln when p takes the value found in 1 þ px x part (d)(i). (4 marks)

s

(03)

Turn over

P56804/Jan13/MFP3

4 7 (a)

Find the general solution of the differential equation d2 y dy  6 þ 10y ¼ e2t 2 dt dt giving your answer in the form y ¼ f ðtÞ .

(6 marks)

1

(b)

Given that x ¼ t 2 , x > 0 , t > 0 and y is a function of x, show that d2 y d2 y dy ¼ 4t þ 2 dx 2 dt 2 dt

(5 marks)

1

(c)

Hence show that the substitution x ¼ t 2 transforms the differential equation x

2 d2 y dy  ð12x 2 þ 1Þ þ 40x 3 y ¼ 4x 3 e2x 2 dx dx

into d2 y dy  6 þ 10y ¼ e2t 2 dt dt (d)

Hence write down the general solution of the differential equation x

(04)

(2 marks)

d2 y dy 2 3 3 2x 2 þ 40x  ð12x þ 1Þ y ¼ 4x e dx 2 dx

(1 mark)

P56804/Jan13/MFP3

5

The diagram shows a sketch of a curve.

8

P

O

N

Initial line

The polar equation of the curve is s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 1 p r ¼ sin 2y 2 þ cos y , 0 4 y 4 2 2 The point P is the point of the curve at which y ¼

p . 3

The perpendicular from P to the initial line meets the initial line at the point N . (a) (i)

Find the exact value of r when y ¼

p . 3

(2 marks)

pffiffiffi 3 3 sec y . (ii) Show that the polar equation of the line PN is r ¼ 8

(2 marks)

pffiffiffi k 3 , where k is an integer. (iii) Find the area of triangle ONP in the form 128

(2 marks)

(b) (i)

Using the substitution u ¼ sin y , or otherwise, find

ð

sinn y cos y dy , where n 5 2 . (2 marks)

(ii) Find the area of the shaded region bounded by pffiffithe ffi line OP and the arc OP of the

curve. Give your answer in the form ap þ b 3 þ c , where a, b and c are constants. (8 marks)

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

(05)

P56804/Jan13/MFP3

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.

MFP3 - AQA GCE Mark Scheme 2013 January series

Q Solution 1(a) y(3.2) = y(3) + 0.2 2  3  5

Marks M1

= 5 + 0.2 × 11

A1

= 5.66332… = 5.6633 to 4dp

A1

(b) y(3.4) = y(3) + 2(0.2){f [3.2, y(3.2)]}

M1

Total

3

A1F

…. = 5 + 2(0.2) 2  3.2  5.6633...

Comments

Condone >4dp

Ft on cand’s answer to (a)

(= 5 + (0.4) 12.0633... ) = 6.389 to 3dp

A1

Total

3

6

2 (a) e3x = 1 + 3x + 4.5x2

(b)

B1

1  2 x 3 / 2 = 1  3x  15 x 2

CAO Must be 6.389 Ignore higher powers beyond x2 throughout this question

1

M1

1  2 x 3 / 2 =1±3x+kx2 or 1+kx±7.5x2 OE

A1

1−3x+7.5x2 OE (simplified PI)

M1

Product of c’s two expansions with an attempt to multiply out to find x2 term

2

e3x 1  2 x 

3 / 2

= (1 + 3x + 4.5x2)(1−3x+7.5x2)

x2 term(s): 7.5x2 − 9x2 + 4.5x2 = 3x2 .

A1

Total

4

5

MFP3 - AQA GCE Mark Scheme 2013 January series

Q

Solution

Marks

Total

Comments

M1

3 PI: y  kx 2 e x PI

y ' PI  2kxe x  kx 2 e x y '' PI  2ke x  4kxe x  kx 2 e x

m1

Product rule used in finding both derivatives

m1

Subst. into DE

2k = 6 ; k = 3; y PI  3x 2 e x

A1

CSO

(GS: y =) ex(Ax+B) + 3 x 2 e x

B1F

5

E1

1

2ke x  4kxe x  kx 2 e x − 4kxe x  2kx 2 e x + kx 2 e x =6e

x

Total

4(a) Integrand is not defined at x = 0

5



…… =



1 0



lim a0



1 a

A1 x 4 ln x dx }

lim a 5 1 a5 – [ ln a  ] 25 25 a0 5 lim

But

a0

involving the ‘original’ ln x

A1

x5 x5 (+ c) ln x  5 25

x 4 ln x dx = {

= 

M1

x5 x5  1  ln x    dx 5 5  x

OE ...= kx 5 ln x   f x  , with f(x) not

(b) x 4 ln x dx =

ex(Ax+B) + kx 2 e x , ft c’s k .

Limit 0 replaced by a limiting process and F(1)−F(a) OE

M1

a 5 ln a = 0

E1

Accept

lim x0

x k ln x = 0 for

any k>0 So



1 0

x 4 ln x dx = 

1 25

A1

Total

6 7

Dep on M and A marks all scored

MFP3 - AQA GCE Mark Scheme 2013 January series

Q

Solution 5

(a)

Marks

Comments

dy sec x  y  tan x dx tan x IF is exp (



sec 2 x dx) tan x

M1

= eln(tanx) = tan x

(b)

Total

2

tan x

A1

2

AG Be convinced

dy  (sec 2 x) y  tan 2 x dx

d  y tan x  tan2x dx

M1



y tanx = tan 2 x dx



y tanx = tan x − x (+c)

 4

c2

 tan

 4

 4



 4

LHS as differential of y×IF PI

A1

 y tanx = (sec 2 x  1) dx

3 tan

and with integration attempted

c

so y tanx = tanx − x+ 2  y = 1 +(2 − x 

 4

m1

Using tan2x = ± sec2x ± 1 PI or other valid methods to integrate tan2x

A1

Correct integration of tan2x; condone absence of +c.

m1

Boundary condition used in attempt to find value of c

 4

) cotx

Total

A1 6

8

ACF

MFP3 - AQA GCE Mark Scheme 2013 January series

Q Solution 6(a)(i) y = ln e 3 x cos x = ln e3x + ln cos x = 3x + ln cos x dy 1 3  ( sin x) dx cos x dy  3  tan x dx



(ii)



d2 y d3 y 2 ;    2 sec x(sec x tan x) sec x dx 2 dx 3 d4 y  4 sec x(sec x tan x) tan x  2 sec 4 x 4 dx





y″(0)= −1;

1 2 0 3  2 4 x  x  x ... 2! 3! 4! 1 1 = 3x  x 2  x 4 2 12

(d)(i)

{ln(1+px)} = px 



1 2 2 p x 2

A1

Comments

Chain rule for derivative of ln cosx 3

CSO AG

M1 for d/dx{ [f(x)]2 }= 2f(x)f′(x) 3

ACF

M1

Mac. Thm with attempt to evaluate at least two derivatives at x=0

A1F

At least 3 of 5 terms correctly obtained. Ft one miscopy in (a)

A1

B1

3

CSO AG Be convinced

1

accept (px)2 for p2x2; ignore higher powers;



1  3x  x 2 {ln e cos x  ln(1  px)} =   1  1 2 1 2 2  4 3    2 3 x  x  O( x )   px  p x  O( x )  2 2    x  For

(ii)

M1

A1

ln e 3 x cos x = 0  3 x 

(c)

Total

B1; M1

(b) Maclaurin’s Thm: x 4 (iv ) x2 x2 y(0)+x y′(0)+ y″(0)+ y′′′(0) + y (0) 4! 2! 3!

y(0) = ln1 = 0; y′(0) =3; y′′′(0) = 0; y (iv ) (0) = −2

Marks B1

 1  e 3 x cos x   to exist, p = 3  ln x  0  x 2  1  px 

lim

lim

 3 p  1 p2 …… = ( )    O( x)  2 2 x0  x  1 2

Value of limit   

p2 = 4. 2

M1

Law of logs and expansions used;

A1

p=3 convincingly found

m1

Divide throughout by x2 before taking limit. (m1 can be awarded before or after the A1 above)

A1

Total

4

14

Must be convincingly obtained

MFP3 - AQA GCE Mark Scheme 2013 January series

Q

7(a)

Solution Solving

Total

Auxl. Eqn. m2 −6m + 10 = 0 (m − 3)2 + 1= 0

M1

m=3±i

A1

CF (yCF =) e3t(A cos t + B sin t)

M1

OE Condone x for t here; ft c’s 2 non-real values for ‘m’.

For PI try (yPI =) ke2t

M1

Condone x for t here

1 4k  12k  10k  1  k  2

A1

1 2t e 2

B1F

PI Completing sq or using quadratic formula to find m.

6

M1

OE

dy dy  2x dx dt

A1

OE

d2 y d2 y dy 4  2 t 2 2 dt dx dt

A1

3 3 1  d2 y dy  dy 2 2 + 40t y  4t 2 e 2t t 4t 2  2  − (12t  1)2t dt dt   dt

M1

1 2

4t

3 d2 y 2{ 2

dt

Product rule OE used dep on previous M1 being awarded at some stage 5

CSO

A.G.

Subst. using (b) into given DE to eliminate all x

3

dy − 6 + 10 y}  4t 2 e 2t dt 3 2

t≠0 so divide by 4t gives

2

Chain rule

d dt d ( f (t ) )  ( f (t ) ) OE dx dx dt d dx d eg ( g( x) ) ( g( x) )  dt dt dx

M1

m1

(d)

CF +PI with 2 arb. constants and both CF and PI functions of t only

dy d t dy = dx dx dt

d 2 y d  dy  dt d  dy  dy =  2 x  = (2x)   +2 2 dx  dt  dx dt  dt  dt dx 2 dy d y = (2x)(2x) +2 2 dt dt

(c)

Comments

d y dy 6  10 y  e 2t (*) 2 dt dt

GS of (*) is (yGS =) e3t(A cos t + B sin t) +

(b)

Marks

2

d2 y dy 6  10 y  e 2t (*) 2 dt dt

y = e 3 x ( A cos x 2  B sin x 2 ) 

1 2 x2 e 2

Total

A1

2

CSO

B1

1

OE Must include y=

14

A.G.

MFP3 - AQA GCE Mark Scheme 2013 January series

Q

Solution

8(a)(i)

r  sin

2 3

(2 

1 3 9 3 3  = cos ) =  2 3 2 4 4

(ii) x = ON = (3√3)/8

Marks

Total

M1; A1

2

Polar eqn of PN is r cos = ON 3 3 r= sec 8

M1

(iii) Area Δ ONP = 0.5 × rN × rP × sin (/3)

M1

=

(b)(i)

A1

1 3 3 3 3 3 27 3    = 2 8 4 2 128

 sin

n

 cos  d =

u

n

A1

(+c)

AG Be convinced OE With correct or ft from (a)(i) (ii), values for rP and rN .

2

M1

du

sin n 1  n 1

=

A1

2

Comments

Be convinced PI

2

(ii) Area of shaded region bounded by line OP

1 and arc OP = 2 1 2





2

3

1   sin 2  2  cos   d 2   2





2

(1  cos 4 ) d +

3

1 4



2

4 sin 2  cos 2  cos  d

2

B1

Correct limits

M1

2 sin 2 2  1  cos 4

B1

sin22 cos = 4sin2 cos2 cos

A1

Correct integration of 0.5(1−cos4)

3





Use of



  sin 4  2 =   + 8   2

(sin 2   sin 4  ) cos  d

3

m1

Writing 2nd integrand in a suitable form to be able to use (b)(i) OE PI

A1

Last two terms OE

3

  sin 4 sin  sin   2 =      8 3 5  2 3 3

=

 12



1 2 r d 2

M1

5

21 3 2  160 15

A1

Total TOTAL

8 16 75

CSO

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