A2AS MATH Past Papers Mark Schemes Standard MayJune Series 2014 14531

8/10/2019 A2AS MATH Past Papers Mark Schemes Standard MayJune Series 2014 14531

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ADVANCEDGeneral Certificate of Education

2014

Mathematics

Assessment Unit C4assessing

Module C4: Core Mathematics 4

[AMC41]

THURSDAY 22 MAY, MORNING

8454.01F

MARK

SCHEME


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8454.01F 2

GCE ADVANCED/ADVANCED SUBSIDIARY (AS) MATHEMATICS

Introduction

The mark scheme normally provides the most popular solution to each question. Other solutions given

by candidates are evaluated and credit given as appropriate; these alternative methods are not usually

illustrated in the published mark scheme.

The marks awarded for each question are shown in the right-hand column and they are prexed by theletters M,Wand MWas appropriate. The key to the mark scheme is given below:

M indicates marks for correct method.

W indicates marks for working.

MW indicates marks for combined method and working.

The solution to a question gains marks for correct method and marks for an accurate working based on

this method. Where the method is not correct no marks can be given.

A later part of a question may require a candidate to use an answer obtained from an earlier part of the

same question. A candidate who gets the wrong answer to the earlier part and goes on to the later part is

naturally unaware that the wrong data is being used and is actually undertaking the solution of a parallel

problem from the point at which the error occurred. If such a candidate continues to apply correct method,

then the candidates individual working must be followed through from the error. If no further errors are

made, then the candidate is penalised only for the initial error. Solutions containing two or more working

or transcription errors are treated in the same way. This process is usually referred to as follow-through

marking and allows a candidate to gain credit for that part of a solution which follows a working or

transcription error.

Positive marking:

It is our intention to reward candidates for any demonstration of relevant knowledge, skills or

understanding. For this reason we adopt a policy of following throughtheir answers, that is, having

penalised a candidate for an error, we mark the succeeding parts of the question using the candidates

value or answers and award marks accordingly.

Some common examples of this occur in the following cases:

(a) a numerical error in one entry in a table of values might lead to several answers being incorrect, butthese might not be essentially separate errors;

(b) readings taken from candidates inaccurate graphs may not agree with the answers expected butmight be consistent with the graphs drawn.

When the candidate misreads a question in such a way as to make the question easier only a proportion of

the marks will be available (based on the professional judgement of the examining team).


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AVAILABLE

MARKS

8454.01F 3

1 (i)( ) ( )x x

x

x

A

x

B

5 12 7

5 1++

+--

=-

MW1

2 7 ( 1) (5 )x A x B x+ +- = - M1 W1

1: 9 6B=x = - - 3

B2

=-

M1 W1

: 6x A5 3= = A 21

= MW1

(ii)( )( )

dx x

xx

5 12 7

+--

=( ) ( )

dx x

x2 5

12 1

3+-

-c m M1 ln lnx x c5 12

123 + +- - - MW3 10

2 (i) AB =AO + OB MW1

AB 3 4 7 5i j i j+ + += -

AB 4 9i j+= W1

(ii) direction vector is 4 9i j+ M1

vector equation of line: ( )3 2 4 9i j i jr + +m= - M1 W1

(iii) r= (3 +4)i+ (9 2)j MW1

3 + 4= 11 = 2 MW1

9 2 = 16 = 2 W1

The point with position vector (11i+ 16j) lies on this line. W1 9


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8454.01F 4

AVAILABLE

MARKS3 u x3 5= - 3ddx

u= MW1

6 2 ( 5) 2 10x u u+ += = MW1

x 2= ,u x1 3= = u 4= MW1

3

26 dx x x3 5- =

4

1

( ) du u u3

2 10 21

+

=

9

4 (i) V r= 1

0( 1)e dxx 2+ M2 W1

V r= 1

0( 2 1)e e dxx x2 + + MW1

1eeV x

2 2

xx

2

r= + +0

; E MW2

e eV2

2 121 2 0

2

+ + + +r= -d dn n( 2 24.0 (3 . .)e e s f V

2 4 32 +

r= - =^ h MW1

(ii) The bowl has negligible thickness.

The bowl has a flat base. MW1 8

M1 W1

4

1du u u

32

310

23

21

+` j MW1u u

154

920

25

23

= +1

4

; E MW2

23.8 (3 . .)s f45

1072= = MW1


5/7

AVAILABLE

MARKS

8454.01F 5

5 dV

V1

2

1 = k dt M2 W1

2V kt c21

+= MW2

When 0, 64 (given)t V= =

2 (8) (0)k c+= c= 16 M1 W1

2 16V kt21

+=

When 1, 48 (given)t V= =

2 16k48 += 2.1435fk =- MW1

2.1435 ) 16tf +2 (V21

= -

When 0, . 7.46 ( . .)s fV t 2 1435

16

3f

= = = MW1

Ice has completely melted at 7.28 pm. MW1 10

6 (i) ,sin cosR R24 7a a= =

( ) 24 7sin cosR2 2 2 2 2+ +a a =

R 25= M1 W1

cos

sintan

RR

a

aa=

tan7

24a = . ( . .)s f1 29 3a = M1 W1

7 24 25 .sin cos sinx x x 1 29- = -^ h

(ii) 625sin cos sinx x x7 24 2 2

a- = -^ ^h h MW1

sin cos dx x x7 241

2-=^ h cosec dx x625

1 2a-^ h M1

1.29 . .cot tan s f x c

625

1

7

243

1a a=

-- + = =

-^ ^h h

1.cot x

625 1 29= -

- ^ h+ c MW1 7


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8454.01F 6

AVAILABLE

MARKS7 (i) : tanf x x1"

-1- M1 W1

Domain: x x1 1R! G G- MW1

Range: ( )f x4 4

G Gr r1- MW1

(ii) : tan tangf x x x" " M1 W1

: tangf x x" MW1

Range: 0 ( ) 1gf xG G MW1

(iii)

0

1

y

x

MW3 11

tany x=

, 14

r-` j , 14r` j

4

4


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AVAILABLE

MARKS

8454.01F 7

8 2 2 2cosddt

xt= - M1 W1

4 sin td

d

t

y= - MW1

d

d

d

d

dd

x

y

t

y

t

x'= M1

sin t4cos t2 2 2= -

- W1

1 2 sin

sint

t

2 24

2- --^ h M1

1

sin t=

- W1

d

d

x

y2= M1

1

sint2

=-

t 4

5r

= MW1

( )

1sinx4

2 5

2

5

2

5r r r= - = - MW1

2cosy 445

422

2r= = - =-d n MW1 11

Alternative Solution

2 2 2cosddt

xt= - M1 W1

4sintddt

y= - W1

d

d

d

d

dd

x

y

t

y

t

x'= M1

sin t4cos t2 22

2--

= W1 M1

0 2 2 2 4cos sint t2 2 += - MW1

0 2 2 (1 2 ) 4sin sint t2 2 2 += - - M1

0 4 (1 )sin sint t2+= W1

0sin sinort t2

1= = -

ort t t4

5

4

5"r

r r= = = MW1

(

1)

sinx4

2 5

25

25

= - = - MW1

24 4cosy4

5

2

22

r= = - = -f p MW1

Total 75

A2AS MATH Past Papers Mark Schemes Standard MayJune Series 2014 14531

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