solomon c4 a-l ans

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FOR EDEXCEL

GCE ExaminationsAdvanced Subsidiary

Core Mathematics C4

Paper A

MARKING GUIDE

This guide is intended to be as helpful as possible to teachers by providingconcise solutions and indicating how marks could be awarded. There are

obviously alternative methods that would also gain full marks.

Method marks (M) are awarded for knowing and using a method.

Accuracy marks (A) can only be awarded when a correct method has been used.

(B) marks are independent of method marks.

Written by Shaun Armstrong

Solomon Press

These sheets may be copied for use solely by the purchasers institute.


2/48

Solomon PressC4A MARKS page 2

C4 Paper A Marking Guide

1. 2x(2 +y) +x2 d

d

y

x 2y

d

d

y

x= 0 M2 A2

d

d

y

x=

2

2 (2 )

2

x y

y x

+

M1 A1 (6)

2. (a) f( 110

) =1

10

3

1=

910

3=

3

10

3

( )= 10 M1 A1

(b) = 3(1 12)x

= 3[1 + ( 1

2 )(x) +

312 2

( )( )

2

(x)2 +

3 512 2 2

( )( )( )

3 2

(x)3 + ] M1

= 3 + 32x + 9

8x

2+ 15

16x

3+ A2

(c) 10 = f( 110

) 3 + 320

+ 9800

+ 1516000

= 3.1621875 (8sf) B1

(d) =10 3.1621875

10

100% = 0.003% (1sf) M1 A1 (8)

3. (a) 1 + 3= 5 = 2 M1p= 9 p = 7 A15 + 2= 11 q = 2 A1

(b) 1 + 3= 25 = 8 M1when = 8, r = (i + 7j 5k) + 8(3ij + 2k) = (25ij + 11k) (25, 1, 11) lies on l A1

(c) OC = [(1 + 3)i + (7 )j + (5 + 2)k] [(1 + 3)i + (7 )j + (5 + 2)k].(3ij + 2k) = 0 M1

3 + 9 7 + 10 + 4= 0 A1

= 1 OC = (4i + 6j 3k), C(4, 6, 3) M1 A1(d) A : = 2, B : = 8, C: = 1 AC: CB = 3 : 7 M1 A1 (11)

4. (a) 1

( 6)( 3)x x dx = 2 dt M1

1

( 6)( 3)x x

6

A

x +

3

B

x

1 A(x 3) +B(x 6) M1x = 6 A = 1

3, x = 3 B = 1

3A2

13 (

1

6x 1

3x ) dx = 2 dt

lnx 6 lnx 3 = 6t+ c M1 A1t= 0, x = 0 ln 6 ln 3 = c, c = ln 2 M1 A1x = 2 ln 4 0 = 6t+ ln 2 M1t= 1

6ln 2 = 0.1155 hrs = 0.1155 60 mins = 6.93 mins 7 mins A1

(b) ln6

2( 3)

x

x

= 6t, t= 16

ln6

2( 3)

x

x

as x 3, t cannot make 3 g B2 (12)


3/48


5. (a) x 0 0.5 1 1.5 2

y 0 1.716 1.472 1.093 1.083 B2

area 12

0.5 [0 + 1.083 + 2(1.716 + 1.472 + 1.093)] = 2.41 (3sf) B1 M1 A1

(b) volume = 2

0 16x e2x dx M1

u = 16x, u = 16, v = e2x, v = 12

e2x M1

I = 8x e2x 8e

2x dx A2

= 8x e2x 4e2x + c A1

volume = [8x e2x 4e2x] 20

= {(16e4 4e4) (0 4)} M1= 4(1 5e4) A1 (12)

6. (a) = (cosx cos 5x) dx M1 A1= sinx 1

5sin 5x + c M1 A1

(b) u2

=x + 1 x = u2 1,d

d

x

u= 2u M1

x = 0 u = 1, x = 3 u = 2 B1

I =2

12 2

( 1)u

u

2u du =

2

1 (2u4 4u2 + 2) du M1 A1

= [ 25

u5 4

3u

3+ 2u] 21 M1 A1

= ( 645

323

+ 4) ( 25

43

+ 2) = 115

5 M1 A1 (12)

7. (a) cos 2t= 12

, 2t= 3

, t= 6

M1 A1

(b)d

d

x

t= 2 sin 2t,

d

d

y

t= cosec tcot t M1

d

d

y

x=

cosec cot

2sin2

t t

t

M1 A1

t= 6

, y = 2, grad = 2

y 2 = 2(x 12

) M1

y = 2x + 1 A1

(c) x = 0 t= 4

B1

area =

6

4 cosec t (2 sin 2t) dt M1

=

6

4 cosec t 4 sin tcos t dt =

4

6 4 cos t dt M1 A1

(d) = [4 sin t]

4

6

= 2 2 2 = 2( 2 1) M2 A1 (14)

Total (75)


4/48


Performance Record C4 Paper A

Question no. 1 2 3 4 5 6 7 Total

Topic(s) differentiation binomialseries

vectors differential

equation,partial

fractions

trapezium

rule,integration

integration parametric

equations

Marks 6 8 11 12 12 12 14 75

Student


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FOR EDEXCEL

GCE Examinations

Advanced Subsidiary

Core Mathematics C4

Paper B

MARKING GUIDE







Solomon Press



6/48

Solomon PressC4B MARKS page 2

C4 Paper B Marking Guide

1. u =x2, u = 2x, v = sinx, v = cosx M1

I = x2 cosx 2x cosx dx = x2

cosx + 2x cosx dx A2u = 2x, u = 2, v = cosx, v = sinx M1

I = x2 cosx + 2x sinx 2 sinx dx A1= x2 cosx + 2x sinx + 2 cosx + c A1 (6)

2. 21

ydy = x dx M1

y1 =322

3x + c M1 A1

x = 1, y = 2 12

= 23

+ c, c = 16

M1 A1

1y

=322

3x 1

6,

1

y= 1

6

322

3x = 1

6(1

324x ) M1

y =32

6

1 4xA1 (7)

3. 8x 2y 2xd

d

y

x 2y

d

d

y

x= 0 M1 A2

(1, 3) 8 + 6 + 2d

d

y

x+ 6

d

d

y

x= 0,

d

d

y

x= 1

4M1 A1

grad of normal = 4 M1y + 3 = 4(x + 1) [y = 4x 7 ] M1 A1 (8)

4. (a) = 1 + (3)(ax) + ( 3)( 4)2

(ax)

2+

( 3)( 4)( 5)

3 2

(ax)3

+ M1 A1

= 1 3ax + 6a2

x

2

10a3

x

3

+ A1

(b)3

6

(1 )

x

ax

+= (6 x)( 1 3ax + 6a2x2 + )

coeff. ofx2

= 36a2

+ 3a = 3 M1

12a2

+ a 1 = 0 A1(4a 1)(3a + 1) = 0 M1a = 1

3 , 1

4A1

(c) a = 13

3

6

(1 )

x

ax

+= (6 x)( + 2

3x

2+ 10

27x

3+ ) M1

coeff. ofx3

= (6 1027

) + (1 23

) = 209

23

= 149

A1 (9)

5. (a) =5

11

3 1x +dx = [

122

3(3 1)x + ] 51 M1 A1

= 23

(4 2) = 43

M1 A1

(b) = 5

11

3 1x +dx M1

= [ 13

ln3x + 1] 51 M1 A1

= 13

(ln 16 ln 4) = 13

ln 4 = 23

ln 2 [ k= 23

] M1 A1 (9)


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6. (a) 15 17xA(1 3x)2 +B(2 +x)(1 3x) + C(2 +x)x = 2 49 = 49A A = 1 B1

x = 13

283

= 73

C C= 4 B1

coeffsx2 0 = 9A 3B B = 3 M1 A1

(b) =0

1 (1

2 x++

3

1 3x+

2

4

(1 3 )x) dx

= [ln2 +x ln1 3x + 43 (1 3x)1] 01 M1 A3

= (ln 2 + 0 + 43

) (0 ln 4 + 13

) M1

= 1 + ln 8 M1 A1 (11)

7. (a) x = 1 1 + 4 cos = 1, cos = 12

, = 3

, 53

M1

y > 0 sin > 0 = 3

A1

(b)d

d

x

= 4 sin ,

d

d

y

= 2 2 cos M1

d

d

y

x =2 2 cos

4sin

M1 A1

atP, grad = 12

3

2

2 2

4

= 2

2 3M1

grad of normal =2 3

2 2

2= 6 A1

y 6 = 6 (x 1) M1

y = 6x, when x = 0,y = 0 passes through origin A1

(c) cos =1

4

x +, sin =

2 2

yM1

2

( 1)

16

x ++

2

8

y= 1 M1 A1 (12)

8. (a) AB = (7ij + 12k) (3i + 3j + 2k) = (10i 4j + 10k) M1 r = (3i + 3j + 2k) + (5i 2j + 5k) A1

(b) OC = [i + (5 2)j + (7 + 7)k]

AC = OC OA = [(3 + )i + (2 2)j + (9 + 7)k] M1 A1

BC = OC OB = [(7 +)i + (6 2)j + (19 + 7)k] A1

AC.BC = (3 + )(7 +)+(2 2)(6 2)+(9 + 7)(19 + 7) = 0 M12 4+ 3 = 0 A1( 1)( 3) = 0 M1

= 1, 3 OC = (i + 3j) or (3ij + 14k) A2

(c) AC= 16 0 4+ + = 2 5 , BC= 36 16 144+ + = 14 M1

area = 12

2 5 14 = 14 5 M1 A1 (13)

Total (75)


8/48


Performance Record C4 Paper B

Question no. 1 2 3 4 5 6 7 8 Total

Topic(s) integration differentialequation

differentiation binomial

series

integration partial

fractions

parametric

equations

vectors

Marks 6 7 8 9 9 11 12 13 75

Student


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Advanced Subsidiary

Core Mathematics C4

Paper C

MARKING GUIDE







Solomon Press



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Solomon PressC4C MARKS page 2

C4 Paper C Marking Guide

1. u = lnx, u = 1x

, v =x, v = 12x

2M1

I = [ 12x

2lnx] 21

2

112x dx A1

= [

1

2x

2

lnx

1

4x

2

]

2

1 M1 A1= (2 ln 2 1) (0 1

4) = 2 ln 2 3

4M1 A1 (6)

2. x 0 0.5 1 1.5 2

arctanx 0 0.4636 0.7854 0.9828 1.1071 B2

(a) 12

1 [0 + 1.1071 + 2(0.7854)] = 1.34 (3sf) B1 M1 A1

(b) 12

0.5 [0 + 1.1071 + 2(0.4636 + 0.7854 + 0.9828)] = 1.39 (3sf) M1 A1 (7)

3. (a) 6x 2 +y +xd

d

y

x+ 2y

d

d

y

x= 0 M1 A2

(1, 3) 6 2 + 3 d

d

y

x+ 6

d

d

y

x= 0,

d

d

y

x= 1 M1 A1

grad of normal = 1y 3 = (x + 1) M1

y = 2 x A1

(b) sub. 3x2 2x +x(2 x) + (2 x)2 11 = 0 M13x

2 4x 7 = 0 A1

(3x 7)(x + 1) = 0 M1

x = 1 (atP) or 73

( 73

, 13

) A1 (11)

4. (a) AB =10

15

5

, r =3

9

7

+ 2

3

1

M1 A1

(b) 3 + 2= 9 = 3 M1

when = 3, r =3

9

7

+ 32

3

1

=9

0

4

(9, 0, 4) lies on l A1

(c) OD =3 2

9 3

7

+

+

3 2

9 3

7

+

+

.2

3

1

= 0 M1

6 + 4 27 + 9 7 + = 0 A1

= 2 OD = 735

, D (7, 3, 5) M1 A1

(d) AB = 100 225 25+ + = 350 , OD = 49 9 25+ + + 83 M1

area = 12

350 83 = 85.2 (3sf) M1 A1 (11)


11/48


5. (a)d

dt

= k( 20) B2

(b) 1

20d = k dt M1

ln 20 = kt+ c M1 A1t= 0, = 37 c = ln 17 M1

ln 2017

= kt, = 20 + 17ekt A1

t= 4, = 36 36 = 20 + 17e4k M1

k= 14

ln 1617

= 0.01516 A1

t= 10, = 20 + 17e0.01516 10

= 34.6C (3sf) A1

(c) 33 = 20 + 17e0.01516t

M1

t= 10.01516

ln 1317

= 17.70 minutes = 17 mins 42 secs M1 A1 (13)

6. (a) x = 0 t= 0 at O B1

y = 0 t= 0 (at O) or 2

t= 2

atA B1

(b) = volume when region abovex-axis is rotated through 2d

d

x

t= 3 cos t M1

volume =

2

0 (2sin 2t)2 3cos t dt=

2

0 12 sin2

2tcos t dt M1 A1

(c) t= 0 u = 0, t= 2

u = 1,d

d

u

t= cos t B1

sin2

2t= 4sin2tcos

2t= 4sin

2t(1 sin2t) M1

=1

0 12 4u2(1 u2) du M1

= 481

0 (u2u4) du A1

= 48[ 13

u3 1

5u

5] 10 M1 A1

= 48[( 13

15

) (0)] = 325

M1 A1 (13)

7. (a)8

(1 )(2 )

x

x x

+

1

A

x++

2

B

x

8 xA(2 x) +B(1 +x) M1x = 1 9 = 3A A = 3 A1

x = 2 6 = 3B B = 2 f(x) = 31 x+

+2

2 xA1

(b) =12

0 (3

1 x++

2

2 x) dx = [3 ln1 +x 2 ln2 x]

12

0 M1 A1

= (3 ln 32

2 ln 32

) (0 2 ln 2) M1

= ln 32

+ ln 4 = ln 6 M1 A1

(c) f(x) = 3(1 +x)1

+ 2(2 x)1(1 +x)

1= 1 x +x2x3 + B1

(2 x)1 = 21(1 12x)

1 M1

= 12

[1 + (1)( 12x) +

( 1)( 2)

2

( 1

2x)

2+

( 1)( 2)( 3)

3 2

( 12x)

3+ ] M1

= 12

(1 + 12x + 1

4x

2+ 1

8x

3+ ) A1

f(x) = 3(1 x +x2x3 + ) + (1 + 12x + 1

4x

2+ 1

8x

3+ ) M1

= 4 5

2x +

13

4 x

2

23

8 x

3

+ A1 (14)

Total (75)


12/48


Performance Record C4 Paper C


Topic(s) integration trapeziumrule

differentiation vectors differential

equation

parametric

equations

partial

fractions,binomial

series

Marks 6 7 11 11 13 13 14 75

Student


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Advanced Subsidiary

Core Mathematics C4

Paper D

MARKING GUIDE







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Solomon PressC4D MARKS page 2

C4 Paper D Marking Guide

1. (a) = 23

(1 32x)

3= 1

8(1 3

2x)

3B1

= 18

[1 + (3)( 32

x) + ( 3)( 4)2

( 3

2 x)2 + ( 3)( 4)( 5)

3 2

(

32

x)3 + ] M1

= 18

+ 916

x + 2716

x2

+ 13532

x3

+ A3

(b) x < 23

B1 (6)

2. (a) 2x + 3y + 3xd

d

y

x 4y

d

d

y

x= 0 M1 A2

d

d

y

x=

2 3

4 3

x y

y x

+

M1 A1

(b) grad =6 6

8 9

= 0 M1

normal parallel toy-axis x = 3 M1 A1 (8)

3. (a) 2x3 5x2 + 6 (Ax +B)x(x 3) + C(x 3) +Dx M1

x = 0 6 = 3C C= 2x = 3 15 = 3D D = 5 A1coeffsx

3 A = 2 B1

coeffsx2 5 =B 3A B = 1 M1 A1

(b) =2

1 (2x + 1 2

x+

5

3x ) dx

= [x2

+x 2 lnx + 5 lnx 3] 21 M1 A2

= (4 + 2 2 ln 2 + 0) (1 + 1 + 0 + 5 ln 2) M1= 4 7 ln 2 A1 (10)

4. (a) x dx = k(5 t) dt M112x

2= k(5t 1

2t2) + c M1 A1

t= 0, x = 0 c = 0 B1t= 2, x = 96 4608 = 8k, k= 576 M1 A1

t= 1 12x

2= 576 9

2, x = 5184 = 72 M1 A1

(b) 3 hours 5 mins t= 3.0833, x = 12284 = 110.83 M1 A1

d

d

x

t=

576(5 3.0833)

110.83

= 9.96,

d

d

x

t< 10 so she should have left M1 A1 (12)


15/48

Solomon PressC4D MARKS page 3

5. (a)1

4

5

.3

a

b

= 0 3 + 4a + 5b = 0 M1 A1

(b) 4 +s = 3 + 3t (1)1 + 4s = 1 + at (2)

1 + 5s = 6 + bt (3) B1(1) s = 3t 7 M1

sub. (2) 1 + 4(3t 7) = 1 + at12t 28 = at, t(12 a) = 28, t= 28

12 aM1 A1

sub. (3) 1 + 5(3t 7) = 6 + bt

15t 28 = bt, t(15 b) = 28, t= 2815 b

A1

28

12 a=

28

15 b, 12 a = 15 b, b = a + 3 M1

sub (a) 3 + 4a + 5(a + 3) = 0, a = 2, b = 1 M1 A1

(c) t= 2 r =3

1

6

+ 23

2

1

=3

3

4

, (3, 3, 4) M1 A1 (12)

6. (a) u2

= 1 x x = 1 u2,d

d

x

u= 2u M1

x = 0 u = 1, x = 1 u = 0 B1

area =1

0 1x x dx =0

1 (1 u2) u (2u) du M1

=1

0 (2u2 2u4) du A1

= [ 23

u3 2

5u

5] 10 M1 A1

= ( 23

25

) (0) = 415

M1 A1

(b) = 1

0

x2(1 x) dx M1

= 1

0 (x2x3) dx

= [ 13x

3 1

4x

4]10 M1 A1

= {( 13

14

) (0)} = 112

M1 A1 (13)

7. (a)d

d

x

t= 6 cos t (sin t),

d

d

y

t= 2 cos 2t M1 A1

d

d

y

x=

2cos2

6cos sin

t

t t=

2cos2

3sin 2

t

t= 2

3 cot 2t M1 A1

(b) 23 cot 2t= 0 2t=

2 , 3

2 t=

4 , 3

4 M1 A1

( 32

, 1), ( 32

, 1) A1

(c) t= 6

, x = 94

, y = 32

, grad = 23 3

B1

y 32

= 23 3

(x 94

) M1

6 3y 9 = 4x + 9

2x + 3 3y = 9 A1

(d) y2

= sin2

2t= 4 sin2tcos

2t= 4(1 cos2t)cos2t M2

cos2t=

3

x y2 = 4(1

3

x)

3

x, y

2= 4

9x(3 x) M1 A1 (14)

Total (75)


16/48


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Core Mathematics C4

Paper E

MARKING GUIDE







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Solomon PressC4E MARKS page 2

C4 Paper E Marking Guide

1. = (cosec2

2x 1) dx M1 A1

= 12

cot 2xx + c M1 A1 (4)

2. (a) 4 sinx + (2 cosy)d

d

y

x= 0 M1 A2

d

d

y

x=

4sin

2cos

x

y=

2sin

cos

x

y= 2 sinx secy M1 A1

(b) grad = 2 32

23

= 2 B1

y 6

= 2(x 3

) M1

6y = 12x 44x 2y = A1 (8)

3. (a)2

2 20

1 2 8

x

x x

+

+

=2 20

(1 2 )(1 4 )

x

x x

+ +

1 2

A

x

+

1 4

B

x+

B1

2 + 20xA(1 + 4x) +B(1 2x) M1x = 1

2 12 = 3A A = 4 A1

x = 14

3 = 32B B = 2

2

2 20

1 2 8

x

x x

+

+ 4

1 2x 2

1 4x+A1

(b)2

2 20

1 2 8

x

x x

+

+ = 4(1 2x)1 2(1 + 4x)1

(1 2x)1 = 1 + (1)(2x) + ( 1)( 2)2

(2x)2 + ( 1)( 2)( 3)

3 2

(2x)3 + M1

= 1 + 2x + 4x2

+ 8x3

+ A1

(1 + 4x)1

= 1 + (1)(4x) + ( 1)( 2)2

(4x)

2+

( 1)( 2)( 3)

3 2

(4x)3

+

= 1 4x + 16x2 64x3 + A1

22 20

1 2 8

x

x x+

+ = 4(1 + 2x + 4x2 + 8x3 + ) 2(1 4x + 16x

2 64x3 + ) M1

= 2 + 16x 16x2 160x3 + A1 (9)

4. (a) PQ = (2i 9j + k) (i 8j + 3k) = (3ij 2k) M1

r = (i 8j + 3k) + (3ij 2k) A1

(b) 6 += 2 = 4 M1a + 4= 9 a = 7 A1b= 1 b = 3 A1

(c) = cos1

3 1 ( 1) 4 ( 2) ( 1)

9 1 4 1 16 1

+ +

+ + + +M1 A1

= cos1

1

14 18= 86.4 (1dp) M1 A1 (9)

5. (a) dy = ke0.2t dt M1

y = 5ke0.2t

+ c A1

t= 0, y = 2 2 = 5k+ c, c = 2 5k M1y = 5ke0.2t 5k+ 2 A1

(b) t= 2, y = 1.6 1.6 = 5ke0.4 5k+ 2 M1

k=0.4

0.4

5e 5

= 0.2427 (4sf) M1 A1

(c) as t, yh (in metres) M1 h = 5k+ 2 = 0.787 m = 78.7 cm h = 79 M1 A1 (10)


19/48


6. (a) x = 0 t2 = 2

t 0 t= 2 (0, 2 + 2 ) M1 A1y = 0 t(t+ 1) = 0t 0 t= 0 (2, 0) M1 A1

(b)d

d

x

t= 2t M1

area = 02 t(t+ 1) (2t) dt A1

=2

0 (2t3

+ 2t) dt

= [ 12

t4

+ 23

t3]

20 M1 A1

= (2 + 43

2 ) (0) = 2 + 43

2 M1 A1 (10)

7. (a) let y = ax, lny =x ln a M1

1

y

d

d

y

x= ln a M1

d

d

y

x=y ln a = a

xln a

d

dx(a

x) = a

xln a A1

(b)d

d

y

x= 4

xln 4 2x 1 ln 2 M1 A1

x = 0, y = 32

, grad = ln 4 12

ln 2 = 32

ln 2 M1

y = ( 32

ln 2)x + 32

, 2y = 3x ln 2 + 3, 3x ln 2 2y + 3 = 0 M1 A1

(c) 4x

ln 4 2x 1 ln 2 = 0(2

x)

2 2 ln 2 1

2(2

x) ln 2 = 0 M1

12

(2x) ln 2[4(2

x) 1] = 0 M1

2x = 14

, x = 2 (2, 1516 ) A2 (12)

8. (a) x 0 0.5 1 1.5 2 2.5 3y 0 0.5774 0.7071 0.7746 0.8165 0.8452 0.8660 B2

(i) 12

1 [0 + 0.8660 + 2(0.7071 + 0.8165)] = 1.96 (3sf) B1 M1 A1

(ii) 12

0.5[0+0.8660+2(0.5774+0.7071+0.7746+0.8165+0.8452)]

= 2.08 (3sf) M1 A1

(b) = 3

0 1x

x +dx M1

=

3

0

1 1

1

x

x

+

+ dx =

3

0

(1

1

1x + ) dx M1

= [x lnx + 1] 30 M1 A1

= {(3 ln 4) (0)} = (3 ln 4) M1 A1 (13)

Total (75)


20/48


Performance Record C4 Paper E


Topic(s) integration differentiation partialfractions,binomial

series


equation

parametric

equations

differentiation trapezium

rule,integration

Marks 4 8 9 9 10 10 12 13 75

Student


21/48

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary

Core Mathematics C4

Paper F

MARKING GUIDE







Solomon Press



22/48

Solomon PressC4F MARKS page 2

C4 Paper F Marking Guide

1. 4x +y +xd

d

y

x 2y

d

d

y

x= 0 M1 A2

SP:d

d

y

x= 0 4x +y = 0, y = 4x M1 A1

sub. 2x2 4x2 16x

2 + 18 = 0 M1x

2= 1, x = 1 (1, 4), (1, 4) A2 (8)

2. x = 2 tan u d

d

x

u= 2 sec

2u M1

x = 0 u = 0, x = 2 u = 4

B1

I =

4

02

2

4tan

4sec

u

u 2 sec2u du =

4

0 2 tan2u du A1

=

4

0 (2 sec2u 2) du M1

= [2 tan u 2u]

40 M1 A1

= (2 2

) (0) = 12

(4 ) M1 A1 (8)

3. (a) =1225

24( )

= 24

25= 4

256 = 2

56 [ k= 2

5] M1 A1

(b) = 1 + ( 12

)( 12x) +

312 2

( )( )

2

( 1

2x)

2+

3 512 2 2

( )( )( )

3 2

( 1

2x)

3+ M1

= 1 14x + 3

32x

2 5

128x

3+ A3

(c) x = 112

(1 +121

2)x

=

121

24(1 )

= 2

56

x = 112 (1 +1

212 )x

1 14 ( 112 ) + 332 ( 112 )2 5128 ( 112 )

3 M1

= 0.97979510

6 52

0.97979510 = 2.44949 (5dp) M1 A1 (9)

4. (a) 4s = 7 3t (1)7 3s = 1 (2)4 +s = 8 + 2t (3) M1(2) s = 2, sub. (1) t= 5 B1 M1check (3) 4 + 2 = 8 10, true intersect A1intersect at (7j 4k) + 2(4i 3j + k) = (8i +j 2k) A1

(b) = cos1 4 ( 3) ( 3) 0 1 216 9 1 9 0 4 + +

+ + + +M1 A1

= cos1

10

26 13

= 57.0 (1dp) M1 A1 (9)


23/48


5. (a)d

d

x

t=

2

1 (2 ) ( 1)

(2 )

t t

t

=

2

2

(2 )t,

d

d

y

t= (1 + t)2 M1 B1

d

d

y

x=

2

1

(1 )t+

2

2

(2 )t=

2

2

(2 )

2(1 )

t

t

+= 1

2

22

1

t

t

+

M1 A1

(b) t= 1, x = 1, y = 12

, grad = 18

B1

grad of normal = 8

y 12 = 8(x 1) [y = 8x152

] M1 A1

(c) x(2 t) = t M1

2x = t(1 +x), t=2

1

x

x+A1

y =2

1

1

1 xx++

=1

(1 ) 2

x

x x

++ +

y = 11 3

x

x

++

M1 A1 (11)

6. (a) = (sec2x 1) dx M1

= tanxx + c M1 A1

(b) =

sin

cos

x

xdx, let u = cosx,

d

d

u

x= sinx M1

= 1

u (1) du =

1

udu A1

= lnu + c = lnu1 + c = lnsecx + c M1 A1

(c) volume =

3

0 x tan2x dx M1

u =x, u = 1, v = tan2x, v = tanxx M1

I =x(tanxx) (tanxx) dx A1=x tanxx2 lnsecx + 1

2x

2+ c A1

volume = [x tanx 12x

2 lnsecx]

30

= {( 13 3 118 2 ln 2) (0)} = 118

2( 6 3 ) ln 2 M1 A1 (13)

7. (a)d

d

V

t= kV,

d

d

V

h= 10hh2 B2

d

d

V

t=

d

d

V

h

d

d

h

t kV= (10hh2)

d

d

h

tM1

13

kh2(15 h) = h(10 h)d

d

h

t

kh(15 h) = 3(10 h)d

d

h

t

d

d

h

t= (15 )

3(10 )

kh h

h

M1 A1

(b)3(10 )

(15 )

h

h h

A

h + 15B

h , 3(10 h) A(15 h) +Bh M1

h = 0 A = 2, h = 15 B = 1 3(10 )(15 )

h

h h

2h

115 h

A2

(c) 3(10 )

(15 )

h

h h

dh = k dt, (2

h 1

15 h) dh = k dt M1

2 lnh + ln15 h = kt+ c M1 A1t= 0, h = 5 2 ln 5 + ln 10 = c, c = ln 250 M12 lnh + ln15 h ln 250 = kt

ln2(15 )

250

h h= kt,

2(15 )

250

h h= e

kt, h

2(15 h) = 250ekt M1 A1

(d) t= 2, h = 4 176 = 250e2k M1

k=

1

2

ln

176

250 = 0.175 (3sf) M1 A1 (17)

Total (75)


24/48


Performance Record C4 Paper F


Topic(s) differentiation integration binomialseries

vectors parametric

equations

integration differential

equation,partial

fractions

Marks 8 8 9 9 11 13 17 75

Student


25/48

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary

Core Mathematics C4

Paper G

MARKING GUIDE







Solomon Press



26/48

Solomon PressC4G MARKS page 2

C4 Paper G Marking Guide

1. 2x + 2y2

+ 2x 2yd

d

y

x+

d

d

y

x= 0 M2 A2

d

d

y

x=

22 2

4 1

x y

xy

++

M1 A1 (6)

2. u =x2, u = 2x, v = ex, v = ex M1 A1

I = x2 ex 2x ex dx = x2 ex + 2x e

x dx A2

u = 2x, u = 2, v = ex, v = ex M1

I = x2 ex 2x ex 2ex

dx A1

= x2 ex 2x ex 2ex + c A1 (7)

3. (a) (1 + ax)n

= 1 + nax +( 1)

2

n n(ax)

2+ B1

an = 4,2 ( 1)

2

a n n= 24 B1

a = 4n

, sub.

2

16

n ( 1)

2

n n= 24 M1 A1

8(n 1) = 24n, n = 12

, a = 8 M1 A1

(b) (1 +128 )x

= +

3 512 2 2

( )( )( )

3 2

(8x)

3+ M1

k= 516

512 = 160 A1 (8)

4. x 0 0.75 1.5 2.25 3y 2.7183 2.0786 1.0733 0.5336 0.3716 B2

(a) = 1

2

1.5 [2.7183 + 0.3716 + 2(1.0733)] = 3.93 (3sf) B1 M1 A1

(b) = 12

0.75 [2.7183 + 0.3716 + 2(2.0786 + 1.0733 + 0.5336)] M1

= 3.92 (3sf) A1

(c) curve must be above top of trapezia in some places and below in othershence position of ordinates determines whether estimate is high or low B2 (9)

5. (a) AB = (4j + k) (9i 8j + 2k) = (9i + 12jk) M1 r = (9i 8j + 2k) + (9i + 12jk) A1at C, 2 = 1, = 3 M1 A1

OC = (9i 8j + 2k) + 3(9i + 12jk) = (18i + 28jk) A1

(b) AC = 3(9i + 12jk), AC= 3 81 144 1+ + = 45.10 M1 A1 distance = 200 45.10 = 9020 m = 9.02 km (3sf) M1 A1 (9)


27/48


6. (a) 1

PdP = 0.05e

0.05tdt M1

lnP = e0.05t+ c M1 A1t= 0, P= 9000 ln 9000 = 1 + c, c = 1 + ln 9000 M1lnP = 1 + ln 9000 e0.05t A1t= 10 lnP = 1 + ln 9000 e0.5 = 9.498 M1

P= e9.498

= 13339 = 13300 (3sf) A1

(b) t, lnP 1 + ln 9000 M1P e1 + ln 9000 = 9000e = 24465 = 24500 (3sf) M1 A1 (10)

7. (a) x = 2 t= 1, x = 9 t= 2 B1

d

d

x

t= 3t

2M1

area =2

12

t 3t2 dt =

2

1 6t dt A1

= [3t2] 21 = 3(4 1) = 9 M1 A1

(b) = 2

1 (

2

t

)2 3t2 dt =

2

1 12 dt M1

= [12t] 21 = 12(2 1) = 12 M1 A1

(c) t=2

y x = ( 2

y)

3+ 1 =

3

8

y+ 1 M1

y3 = 81x

, y =3

2

1x M1 A1 (11)

8. (a) u = sinx d

d

u

x= cosx B1

I = 26cos

cos (2 sin )

x

x xdx = 2

6cos

(1 sin )(2 sin )

x

x x dx M1

= 26

(1 )(2 )u u du M1 A1

(b)6

(1 )(1 )(2 )u u u+

1

A

u++

1

B

u+

2

C

u

6 A(1 u)(2 u) +B(1 + u)(2 u) + C(1 + u)(1 u) M1u = 1 6 = 6A A = 1 A1u = 1 6 = 2B B = 3 A1u = 2 6 = 3C C= 2 A1

2

6

(1 )(2 )u u 1

1 u++

3

1 u 2

2 u

(c) x = 0 u = 0, x = 6

u = 12

M1

I =12

0 (1

1 u++

3

1 u 2

2 u) du

= [ln1 + u 3 ln1 u + 2 ln2 u]12

0 M1 A2

= (ln 32

3 ln 12

+ 2 ln 32

) (0 + 0 + 2 ln 2) M1

= 3 ln 32

+ 3 ln 2 2 ln 2

= 3 ln 3 3 ln 2 + ln 2 = 3 ln 3 2 ln 2 M1 A1 (15)

Total (75)


28/48


Performance Record C4 Paper G


Topic(s) differentiation integration binomialseries

trapezium

rule


equation

parametric

equations

partial

fractions,integration

Marks 6 7 8 9 9 10 11 15 75

Student


29/48

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary

Core Mathematics C4

Paper H

MARKING GUIDE







Solomon Press



30/48

Solomon PressC4H MARKS page 2

C4 Paper H Marking Guide

1. (a) = 1 + ( 32

)(4x) +3 12 2

( )( )

2(4x)

2+

3 1 12 2 2

( )( )( )

3 2

(4x)

3+ M1

= 1 + 6x + 6x2 4x3 + A3

(b) x < 14

B1 (5)

2. u = 1 + sinx d

d

u

x= cosx M1

x = 0 u = 1, x = 2

u = 2 B1

I =2

1 u3

du A1

= [ 14

u4] 21 M1

= 4 14

= 154

M1 A1 (6)

3. (a) 11( 4)( 3)

xx x

++ 4

Ax +

+3

Bx

x + 11 A(x 3) +B(x + 4) M1x = 4 7 = 7A A = 1 A1x = 3 14 = 7B B = 2 A1

11

( 4)( 3)

x

x x

++

23x

14x +

(b) =2

0 (2

3x 1

4x +) dx

= [2 lnx 3 lnx + 4] 20 M1 A1

= (0 ln 6) (2 ln 3 ln 4) M1

= ln

2

27 M1 A1 (8)

4. =

2

6 (2 sinx + cosecx)

2dx M1

=

2

6 (4 sin

2x + 4 + cosec

2x) dx A1

=

2

6 (2 2 cos 2x + 4 + cosec

2x) dx M1

= [6x sin 2x cotx]

2

6

M1 A2

= {(3 + 0 + 0) ( 32

3 )} M1

= (2 + 32

3 ) = 12

(4 + 3 3 ) A1 (8)

5. (a) 2x 3y 3xd

d

y

x 2y

d

d

y

x= 0 M1 A2

d

d

y

x=

2 3

3 2

x y

x y

+

M1 A1

(b) grad = 5 M1

y + 2 = 5(x 2) [y = 5x 12 ] M1 A1 (8)


31/48


6. (a) =1 6 5 3 ( 1) ( 6)

1 25 1 36 9 36

+ +

+ + + +M1 A1

=27

27 81=

27

9=

3 3

9= 1

33 M1 A1

(b) sin (AOB) = 213

1 ( 3) = 23

M1 A1

area = 12 3 3 9 23 = 272 2 M1 A1

(c) = OA sin (AOB) = 3 3 23

= 3 2 M1 A1 (10)

7. (a)d

d

x

t= 2t 1,

d

d

y

t=

2

4 (1 ) 4 ( 1)

(1 )

t t

t

=

2

4

(1 )tB1 M1

d

d

y

x=

2

4

(2 1)(1 )t t M1 A1

(b) t= 1, x = 2, y = 2, grad = 13

M1

y + 2 =

13 (x

2) M1

3y + 6 = x + 2x + 3y + 4 = 0 A1

(c) t(t 1) + 3 41

t

t+ 4 = 0 M1

t(t 1)2 + 12t+ 4(1 t) = 0t3 2t2 7t 4 = 0 A1

t= 1 is a solution (t+ 1) is a factor M1(t+ 1)(t

2 3t 4) = 0 M1

(t+ 1)(t+ 1)(t 4) = 0 A1

t= 1 (atP) or t= 4 Q (12, 163

) M1 A1 (14)

8. (a) 1

PdP = k dt M1

lnP = kt+ c A1t= 0, P= 300 ln 300 = c M1lnP = kt+ ln 300

ln300

P = kt,300

P= e

kt, P= 300e

ktM1 A1

(b) t= 1, P= 360 360 = 300ek M1

k= ln 65

= 0.182 (3sf) A1

(c) P= 300e0.1823t

when t= 2, P= 432; when t= 3, P= 518 B1model does not seem suitable as data diverges from predictions B1

(d) 1

PdP = (0.4 0.25 cos 0.5t) dt M1

lnP = 0.4t 0.5 sin 0.5t+ c A1t= 0, P= 300 ln 300 = c

ln300

P = 0.4t 0.5 sin 0.5t [P= 300e0.4t 0.5 sin 0.5t] M1 A1

(e) second model: t= 1, 2, 3 P= 352, 438, 605 M1 A1the second model seems more suitable as it fits the data better B1 (16)

Total (75)


32/48


Performance Record C4 Paper H


Topic(s) binomialseries

integration partial

fractions

integration differentiation vectors parametric

equations

differential

equations

Marks 5 6 8 8 8 10 14 16 75

Student


33/48

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary

Core Mathematics C4

Paper I

MARKING GUIDE







Solomon Press



34/48

Solomon PressC4I MARKS page 2

C4 Paper I Marking Guide

1. 3x2

+ 2y + 2xd

d

y

x 2y

d

d

y

x= 0 M1 A2

(2, 4) 12 8 + 4d

d

y

x+ 8

d

d

y

x= 0,

d

d

y

x= 1

3 M1 A1

grad of normal = 3 M1y + 4 = 3(x 2) M1

y = 3x 10 A1 (8)

2. (a) =1 12 21

44 (1 )x = 2(1

121

4)x B1

= 2[1 + ( 12

)( 14x) +

1 12 2

( )( )

2

( 1

4x)

2+ ] = 2 1

4x 1

64x

2+ M1 A2

(b) x < 4 B1

(c) x = 0.01 (4 12)x = 3.99 = 399

100= 1

10399 M1

x = 0.01 (4 12)x 2 1

400

1640000

= 1.997498438 M1

399 10 1.997498438 = 19.9749844 (9sf) M1 A1 (9)

3. (a) 0.9959, 0.6931, 0.2569 (4dp) B2

(b) (i) = 12

(1.0986 + 0) = 1.726 (3dp) B1 M1 A1

(ii) = 12

2

[1.0986 + 0 + 2(0.6931)] = 1.952 (3dp) M1 A1

(iii) = 12

4

[1.0986 + 0 + 2(0.9959 + 0.6931 + 0.2569)]

= 1.960 (3dp) A1

(c) 1.96; large change from 1 to 2 strips but from 2 to 4 strips the change isless than 0.01 so the error in 4 strip value is likely to be less than 0.005 B2 (10)

4. (a) x = 1 = 4

, x = 1 = 4

B1

d

d

x

= sec

2 M1

volume =

4

4 (cos

2)

2 sec2 d =

4

4 cos

2 d A1

=

4

4 (

12

+ 12

cos 2 ) d M1

= [ 12 + 1

4sin 2]

4

4

M1 A1

= [(

8+ 1

4) (

8 14 )] M1

= ( 4

+ 12

) = 14

( + 2) A1

(b) y = cos2 =

2

1

sec =

2

1

1 tan + y =

2

1

1 x+M2 A1 (11)


35/48


5. (a) AB = (5i 4j) (2ij + 6k) = (3i 3j 6k) B1

AC = (7i 6j 4k) (2ij + 6k) = (5i 5j 10k) = 53

AB M1

AC is parallel to AB , also common point single straight line A1

(b) 3 : 2 B1

(c) AD = (3i +j + 4k) (2ij + 6k) = (i + 2j 2k) B1

BD = (3i +j + 4k) (5i 4j) = (2i + 5j + 4k) B1AD .BD = 2 + 10 8 = 0 perpendicular M1 A1

(d) = 12

1 4 4+ + 4 25 16+ + = 12

3 3 5 = 92

5 M2 A1 (11)

6. (a) x = 2 sin u d

d

x

u= 2 cos u M1

x = 0 u = 0, x = 3 u = 3

B1

I =

3

01

2cos u 2 cos u du =

3

0 1 du A1

= [u]

30 = 3 0 =

3 M1 A1

(b) u =x, u = 1, v = cosx, v = sinx M1

I = [x sinx]

20 sinx dx A2

= [x sinx + cosx]

20 M1

= ( 2

+ 0) (0 + 1) = 2

1 M1 A1 (11)

7. (a) when x = 14

,d

d

x

t= 3

4 6 = 1

8M1 A1

ddxt

= kx(1 x) 18 = k 14 34 , k= 23 ddxt =23x(1 x) M1 A1

(b) 1

(1 )x xdx = 23 dt M1

1

(1 )x x A

x+

1

B

x, 1 A(1 x) +Bx M1

x = 0 A = 1 A1x = 1 B = 1 A1

(1

x+

1

1 x) dx = 23 dt

lnx ln1 x = 23

t+ c M1 A1

t= 0, x = 14

ln 14

ln 34

= c, c = ln 13

M1 A1

t= 3 lnx ln1 x = 2 + ln 13

ln3

1

x

x= 2,

3

1

x

x= e

2M1

3x = e2(1 x), x(e2 + 3) = e2 M1

x =2

2

e

e 3+ % destroyed =

2

2

e

e 3+ 100% = 71.1% (3sf) A1 (15)

Total (75)


36/48


Performance Record C4 Paper I


Topic(s) differentiation binomialseries

trapezium

rule

parametric

equations

vectors integration differential

equation,partial

fractions

Marks 8 9 10 11 11 11 15 75

Student


37/48

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary

Core Mathematics C4

Paper J

MARKING GUIDE







Solomon Press



38/48

Solomon PressC4J MARKS page 2

C4 Paper J Marking Guide

1. x(x 2) = 0, x = 0, 2 crossesx-axis at (0, 0) and (2, 0)

volume = 2

0 (x2 2x)2 dx M1

= 2

0 (x4 4x3 + 4x2) dx A1

= [ 15x

5x4 + 4

3x

3] 20 M1 A1

= {( 325

16 + 323

) (0)} = 1615

M1 A1 (6)

2. u = 1 12x x = (1 u)2,

d

d

x

u= 2(1 u) = 2u 2 M1 A1

I = 1

u (2u 2) du = (2

2

u) du A1

= 2u 2 lnu + c M1 A1

= 2(1 12x ) 2 ln1

12x + c A1 (6)

3. (a) 4 cos 2x sec2yd

d

y

x= 0 M1 A2

d

d

y

x= 4 cos 2x cos

2y M1 A1

(b) grad = 4 12

14

= 12

B1

y 3

= 12

(x 6

) M1

y 3

= 12x

12

y = 12x +

4A1 (8)

4. (a)d

d

x

t=

121

2at

,

d

d

y

t= a(1 2t) M1

d

d

y

x=

121

2

(1 2 )a t

at

= 2 t (1 2t) M1 A1

(b) y = 0 t= 0 (at O) or 1 (atA) B1t= 1, x = a, y = 0, grad = 2 M1y 0 = 2(xa) A1atB, x = 0 y = 2a M1area = 1

2a 2a = a2 M1 A1 (9)

5. (a)d

d

y

x= k y

12y

dy = k dx M1

122y = kx + c M1 A1

(0, 4) 4 = c M1

2 y = kx + 4 A1

(b) (2, 9) 6 = 2k+ 4, k= 1 M1 A1

2 y =x + 4, y = 12

(x + 4) M1

y = 14

(x + 4)2

A1 (9)


39/48


6. (a) let radius = r, tan 30 = 13

=r

h, r=

3

hM1

V= 13

r2h = 13

h2

3

h= 1

9h3 M1 A1

(b) (i)d

d

V

t= 120,

d

d

V

h= 1

3h2 B1

ddVt

= ddVh

ddht

, 120 = 13 h

2 ddht

, ddht

=2

360h

M1 A1

when h = 6,d

d

h

t= 3.18 cm s

1(2dp) M1 A1

(ii) V= 8 120 = 960 = 19

h3 h = 3 9 960

= 14.011 M1

d

d

h

t= 0.58 cm s1 (2dp) A1 (10)

7. (a) AB =3

6

1

4

1

3

=1

5

2

r =4

1

3

+ 1

5

2

M1 A1

(b) 4 + = 3 + 2 (1)1 + 5 = 7 3 (2)3 2 = 9 + (3) B12 (1) + (3): 5 = 15 + 5, = 4, = 1 M1 A1sub. (2): 1 5 = 7 + 12, not true do not intersect M1 A1

(c) OC =3 2

7 3

9

+

+

, BC = OC OB =6 2

13 3

8

+

+

M1 A1

1

5

2

.6 2

13 3

8

+

+

= 0, 6 + 2 65 15 16 2= 0 M1 A1

= 5 OC =7

8

4

M1 A1 (13)

8. (a) x(3x 7) A(1 x)(1 3x) +B(1 3x) + C(1 x) M1x = 1 4 = 2B B = 2 A1x = 1

3 2 = 2

3C C= 3 A1

coeffsx2 3 = 3A A = 1 A1

(b) =14

0 (1 +2

1 x 3

1 3x) dx = [x 2 ln1 x + ln1 3x]

14

0 M1 A1

= ( 1

4

2 ln 34

+ ln 1

4

) (0) M1

= 14

+ ln 169

+ ln 14

= 14

+ ln 49

M1 A1

(c) f(x) = 1 + 2(1 x)1 3(1 3x)1(1 x)1 = 1 +x +x2 +x3 + B1(1 3x)1 = 1 + 3x + (3x)2 + (3x)3 + = 1 + 3x + 9x2 + 27x3 + M1 A1 f(x) = 1 + 2(1 +x +x2 +x3 + ) 3(1 + 3x + 9x2 + 27x3 + ) M1

= 7x 25x2 79x3 + A1 (14)

Total (75)


40/48


Performance Record C4 Paper J


Topic(s) integration integration differentiation parametricequations

differential

equation

connected

rates

vectors partial

fractions,binomial

series

Marks 6 6 8 9 9 10 13 14 75

Student


41/48

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary

Core Mathematics C4

Paper K

MARKING GUIDE







Solomon Press



42/48

Solomon PressC4K MARKS page 2

C4 Paper K Marking Guide

1. = 3

12

(3 1)x

x

+dx M1

= 3

129 6 1x x

x

+ +dx =

3

1 (9x + 6 +1

x) dx A1

= [92x

2

+ 6x + lnx]31 M1 A1

= {( 812

+ 18 + ln 3) ( 92

+ 6 + 0)} M1

= (48 + ln 3) A1 (6)

2. (a) (1 3x)2 = 1 + (2)(3x) + ( 2)( 3)2

(3x)2 + ( 2)( 3)( 4)

3 2

(3x)3 + M1

= 1 + 6x + 27x2

+ 108x3

+ A3

(b)2

2

1 3

x

x

= (2 x)2(1 3x)2 = (4 4x +x2)(1 + 6x + 27x2 + 108x3 + ) M1

= 4 + 24x + 108x2

+ 432x3 4x 24x2 108x3 +x2 + 6x3 + A1

for smallx,

22

1 3

x

x

= 4 + 20x + 85x

2

+ 330x3

A1 (7)

3. (a)2

2

7 3 2

(1 2 )(1 )

x x

x x

+ +

+

1 2

A

x+

1

B

x++

2(1 )

C

x+

7 + 3x + 2x2A(1 +x)2 +B(1 2x)(1 +x) + C(1 2x)

x = 12

9 = 94A A = 4 B1

x = 1 6 = 3C C= 2 B1coeffsx

2 2 =A 2B B = 1 M1

f(x) = 41 2x

+1

1 x++

2

2

(1 )x+A1

(b) =2

1 (4

1 2x+ 1

1 x++

2

2

(1 )x+) dx

= [2 ln1 2x + ln1 +x 2(1 +x)1] 21 M1 A3

= (2 ln 3 + ln 3 23

) (0 + ln 2 1) M1

= ln 3 ln 2 + 13

= 13

ln 6 [p = 13

, q = 6 ] M1 A1 (11)

4. (a) 4 = 6 + 14 (1)

3 2 = 3 + 2 (2) B1(1) + 2 (2): 6 = 12 + 18, = 1, = 2 M1 A1

r =

7

03

2

5

42

=

3

81

M1 A1

(b) a (5) = 3, a = 8 M1 A1

(c) cos =5 ( 5) 4 14 ( 2) 2

25 16 4 25 196 4

+ +

+ + + +M1 A1

=27

45 15=

9

3 5 5=

3

5 5= 3

255 M1 A1 (11)


43/48


5. (a) 2x 4y 4xd

d

y

x+ 4y

d

d

y

x= 0 M1 A2

d

d

y

x=

2 4

4 4

x y

x y

=2

2 2

x y

x y

M1 A1

(b) grad = 32

M1

y 2 = 32

(x 1) M1

2y 4 = 3x 33x 2y + 1 = 0 A1

(c)2

2 2

x y

x y

= 32

M1

2(x 2y) = 3(2x 2y), y = 2x A1sub. x2 8x2 + 8x2 = 1 M1

x2

= 1, x = 1 (atP) or 1 Q (1, 2) A1 (12)

6. (a)d

d

N

t= kN B1

(b) 1

NdN = k dt M1

lnN = kt+ c M1 A1t= 0, N=N0 lnN0 = c M1

lnN = kt+ lnN0, ln0

N

N = kt M1

0

N

N= e

kt, N=N0e

ktA1

(c) 2N0 =N0e6k

M1

k= 16

ln 2 = 0.116 (3sf) M1 A1

(d) 10N0 =N0e0.1155t

M1

t=1

0.1155 ln 10 = 19.932 hours = 19 hours 56 mins M1 A1 (13)

7. (a) x +1

x= sec + tan +

1

sec tan +=

2(sec tan ) 1

sec tan

+ ++

M1

=2 2

sec 2sec tan tan 1

sec tan

+ + ++

=2

2sec 2sec tan

sec tan

++

M1 A1

=2sec (sec tan )

sec tan

++

= 2 sec M1 A1

(b)2

1x

x

+=

2

cos cos =

2

2

1

x

x +M1

21y

y

+=

2

sin sin =

2

2

1

y

y +

2

2 2

4

( 1)

x

x ++

2

2 2

4

( 1)

y

y += 1 M1 A1

(c)d

d

x

= sec tan + sec2 M1

= sec (tan + sec ) =2

1

2

x

x

+x = 1

2(x

2+ 1) M1 A1

(d)d

d

y

= cosec cot cosec2 M1

= cosec (cot + cosec ) = 2

1

2

y

y

+y = 1

2 (y2 + 1) A1

d

d

y

x=

2

2

1

1

y

x

+

+M1 A1 (15)

Total (75)


44/48


Performance Record C4 Paper K


Topic(s) integration binomialseries

partial

fractions

vectors differentiation differential

equation

parametric

equations

Marks 6 7 11 11 12 13 15 75

Student


45/48

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary

Core Mathematics C4

Paper L

MARKING GUIDE







Solomon Press



46/48

Solomon PressC4L MARKS page 2

C4 Paper L Marking Guide

1. (a)d

d

n

t= 0 e0.5t= 5 M1

t= 2 ln 5 = 3.219 mins = 3 mins 13 secs M1 A1

(b) dn = (e0.5t

5) dt

n = 2e0.5t

5t+ c M1 A1t= 0, n = 20 20 = 2 + c, c = 18 M1n = 2e

0.5t 5t+ 18 A1

(c) as tincreases, n rapidly becomes very large not realistic B1 (8)

2. 6x +y +xd

d

y

x 4y

d

d

y

x= 0 M1 A2

(1, 4) 6 + 4 +d

d

y

x 16

d

d

y

x= 0,

d

d

y

x= 2

3M1 A1

grad of normal = 32

M1

y 4 = 32

(x 1) M1

2y 8 = 3x + 33x + 2y 11 = 0 A1 (8)

3. (a) u = 2 x2 d

d

u

x= 2x M1

I = 1

u ( 1

2 ) du = 1

2

1

udu A1

= 12

lnu + c = 12

ln2 x2 + c M1 A1

(b) =

4

0 (12 sin 4x + 12 sin 2x) dx M1 A1

= [ 18

cos 4x 14

cos 2x]

40 M1 A1

= ( 18

0) ( 18

14

) = 12

M1 A1 (10)

4. (a) x 1 2 3y 0 1.665 3.144 B1

area 12

1 [0 + 3.144 + 2(1.665)] = 3.24 (3sf) B1 M1 A1

(b) volume = 3

1 x2

lnx dx M1

u = lnx, u = 1x, v =x

2, v = 13x3

I = 13x

3lnx 13x

2dx M1 A2

= 13x

3lnx 1

9x

3+ c A1

volume = [ 13x

3lnx 1

9x

3] 31

= {(9 ln 3 3) (0 19

)} M1

= (9 ln 3 269

) A1 (11)


47/48

Solomon PressC4L MARKS page 3

5. (a)2

5 8

(1 2 )(1 )

x

x x

+

1 2

A

x++

1

B

x+

2(1 )

C

x

5 8xA(1 x)2 +B(1 + 2x)(1 x) + C(1 + 2x)(1 x) M1

x = 12

9 = 94A A = 4 A1

x = 1 3 = 3C C= 1 A1coeffsx

2 0 =A 2B B = 2 M1 A1

f(x) =4

1 2x+ +2

1 x 21

(1 )x

(b) f(x) = 4(1 + 2x)1

+ 2(1 x)1 (1 x)2

(1 + 2x)1

= 1 + (1)(2x) + ( 1)( 2)2

(2x)

2+

( 1)( 2)( 3)

3 2

(2x)3

+ M1

= 1 2x + 4x2 8x3 + A1(1 x)1 = 1 +x +x2 +x3 + B1

(1 x)2 = 1 + (2)(x) + ( 2)( 3)2

(x)2 + ( 2)( 3)( 4)

3 2

(x)3 +

= 1 + 2x + 3x2

+ 4x3

+ A1

f(x) = 4(1 2x + 4x2 8x3) + 2(1 +x +x2 +x3) (1 + 2x + 3x2 + 4x3) M1= 5 8x + 15x2 34x3 + A1

(c) x < 12

A1 (12)

6. (a)d

d

x

t= 1 + cos t,

d

d

y

t= cos t M1

d

d

y

x=

cos

1 cos

t

t+M1 A1

(b)cos

1 cos

t

t+= 0, cos t= 0, t=

2M1 A1

( 2

+ 1, 1) A1

(c) =

0

sin t (1 + cos t) dt =

0

(sin t+ 1

2

sin 2t) dt M1 A1

= [cos t 14

cos 2t] 0 M1 A1

= (1 14

) (1 14

) = 2 M1 A1 (12)

7. (a) AB = (8j 6k) (3i + 6j 8k) = (3i + 2j + 2k) M1 r = (3i + 6j 8k) + (3i + 2j + 2k) A1

(b) 3 3 = 2 + 7 (1)6 + 2 = 10 4 (2)8 + 2 = 6 + 6 (3) B1(3) (2): 14 = 4 + 10, = 1, = 4 M1 A1

check (1) 3 12 = 2 7, true intersect B1(c) r = (2i + 10j + 6k) (7i 4j + 6k) (9, 14, 0) M1 A1

(d) OC = [(2 + 7)i + (10 4)j + (6 + 6)k]

AC = OC OA = [(5 + 7)i + (4 4)j + (14 + 6)k] M1 A1 [(5 + 7)i + (4 4)j + (14 + 6)k].(3i + 2j + 2k) = 0 M1

15 21+ 8 8+ 28 + 12= 0 A1

= 3 OC = (19i 2j + 24k) M1 A1 (14)

Total (75)


48/48

Performance Record C4 Paper L


Topic(s) differentialequation

differentiation integration trapezium

rule,integration

partial

fractions,binomial

series

parametric

equations

vectors

Marks 8 8 10 11 12 12 14 75

Student

solomon c4 a-l ans

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