core mathematics c4 - mr. samuel...

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary Core Mathematics C4 Paper A

MARKING GUIDE

This guide is intended to be as helpful as possible to teachers by providing concise 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 purchaser’s institute.

Solomon Press C4A MARKS page 2

C4 Paper A – Marking Guide 1. 2x(2 + y) + x2 d

dyx

− 2y ddyx

= 0 M2 A2

ddyx

= 22 (2 )2x yy x

+−

M1 A1 (6)

2. (a) f( 110 ) =

110

3

1− =

910

3 = 310

3( )

= 10 M1 A1

(b) = 3(1 − 12)x − = 3[1 + ( 1

2− )(−x) + 31

2 2( )( )2

− − (−x)2 + 3 51

2 2 2( )( )( )3 2

− − −× (−x)3 + …] M1

= 3 + 32 x + 9

8 x2 + 1516 x3 + … A2

(c) 10 = f( 110 ) ≈ 3 + 3

20 + 9800 + 15

16000 = 3.1621875 (8sf) B1

(d) = 10 3.162187510

− × 100% = 0.003% (1sf) M1 A1 (8)

3. (a) 1 + 3λ = −5 ∴ λ = −2 M1 p − λ = 9 ∴ p = 7 A1 −5 + 2λ = 11 ∴ q = 2 A1

(b) 1 + 3λ = 25 ∴ λ = 8 M1 when λ = 8, r = (i + 7j − 5k) + 8(3i − j + 2k) = (25i − j + 11k) ∴ (25, −1, 11) lies on l A1

(c) OC = [(1 + 3λ)i + (7 − λ)j + (−5 + 2λ)k] ∴ [(1 + 3λ)i + (7 − λ)j + (−5 + 2λ)k].(3i − j + 2k) = 0 M1 3 + 9λ − 7 + λ − 10 + 4λ = 0 A1 λ = 1 ∴ OC = (4i + 6j − 3k), C (4, 6, −3) M1 A1

(d) A : λ = −2, B : λ = 8, C : λ = 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

Ax −

+ 3

Bx −

1 ≡ A(x − 3) + B(x − 6) M1 x = 6 ⇒ A = 1

3 , x = 3 ⇒ B = − 13 A2

13 ∫ ( 1

6x − − 1

3x −) dx = ∫ 2 dt

lnx − 6 − lnx − 3 = 6t + c M1 A1 t = 0, x = 0 ∴ ln 6 − ln 3 = c, c = ln 2 M1 A1 x = 2 ⇒ ln 4 − 0 = 6t + ln 2 M1 t = 1

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

(b) ln 62( 3)

xx−−

= 6t, t = 16 ln 6

2( 3)xx−−

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


5. (a) x 0 0.5 1 1.5 2 y 0 1.716 1.472 1.093 1.083 B2 area ≈ 1

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

(b) volume = π2

0∫ 16x e−2x dx M1

u = 16x, u′ = 16, v′ = e−2x, v = 12− e−2x M1

I = −8x e−2x − ∫ −8e−2x dx A2

= −8x e−2x − 4e−2x + c A1 volume = π[−8x e−2x − 4e−2x] 2

0 = π{(−16e−4 − 4e−4) − (0 − 4)} M1 = 4π(1 − 5e−4) A1 (12)

6. (a) = ∫ (cos x − cos 5x) dx M1 A1

= sin x − 15 sin 5x + c M1 A1

(b) u2 = x + 1 ⇒ x = u2 − 1, dd

xu

= 2u M1

x = 0 ⇒ u = 1, x = 3 ⇒ u = 2 B1

I = 2

1∫2 2( 1)u

u− × 2u du =

2

1∫ (2u4 − 4u2 + 2) du M1 A1

= [ 25 u5 − 4

3 u3 + 2u] 21 M1 A1

= ( 645 − 32

3 + 4) − ( 25 − 4

3 + 2) = 1155 M1 A1 (12)

7. (a) cos 2t = 12 , 2t = π3 , t = π6 M1 A1

(b) ddxt

= −2 sin 2t, ddyt

= −cosec t cot t M1

ddyx

= cosec cot2sin 2

t tt

−−

M1 A1

t = π6 , y = 2, grad = 2

∴ y − 2 = 2(x − 12 ) M1

y = 2x + 1 A1

(c) x = 0 ⇒ t = π4 B1

∴ area = π6π4∫ cosec t × (−2 sin 2t) dt M1

= −π6π4∫ cosec t × 4 sin t cos t dt =

π4π6∫ 4 cos t dt M1 A1

(d) = [4 sin t]π4π6

= 2 2 − 2 = 2( 2 − 1) M2 A1 (14)

Total (75)


Performance Record – C4 Paper A

Question no. 1 2 3 4 5 6 7 Total

Topic(s) differentiation binomial series

vectors differential equation,

partial fractions

trapezium rule,

integration

integration parametric equations

Marks 6 8 11 12 12 12 14 75

Student

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary Core Mathematics C4 Paper B

MARKING GUIDE



Solomon Press


Solomon Press C4B MARKS page 2

C4 Paper B – Marking Guide 1. u = x2, u′ = 2x, v′ = sin x, v = −cos x M1 I = −x2 cos x − ∫ −2x cos x dx = −x2 cos x + ∫ 2x cos x dx A2

u = 2x, u′ = 2, v′ = cos x, v = sin x M1 I = −x2 cos x + 2x sin x − ∫ 2 sin x dx A1

= −x2 cos x + 2x sin x + 2 cos x + c A1 (6)

2. ∫ 21y

dy = ∫ x dx M1

−y−1 = 322

3 x + c M1 A1

x = 1, y = −2 ⇒ 12 = 2

3 + c, c = 16− M1 A1

− 1y

= 322

3 x − 16 , 1

y = 1

6 − 322

3 x = 16 (1 −

324x ) M1

y = 32

6

1 4x− A1 (7)

3. 8x − 2y − 2x ddyx

− 2y ddyx

= 0 M1 A2

(−1, −3) ⇒ −8 + 6 + 2 ddyx

+ 6 ddyx

= 0, ddyx

= 14 M1 A1

grad of normal = −4 M1 ∴ y + 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 + 6a2x2 − 10a3x3 + … A1

(b) 36

(1 )x

ax−

+ = (6 − x)( 1 − 3ax + 6a2x2 + …)

coeff. of x2 = 36a2 + 3a = 3 M1 12a2 + a − 1 = 0 A1 (4a − 1)(3a + 1) = 0 M1 a = 1

3− , 14 A1

(c) a = 13− ∴ 3

6(1 )

xax−

+ = (6 − x)(… + 2

3 x2 + 1027 x3 + …) M1

coeff. of x3 = (6 × 1027 ) + (−1 × 2

3 ) = 209 − 2

3 = 149 A1 (9)

5. (a) = 5

1∫1

3 1x + dx = [

122

3 (3 1)x + ] 51 M1 A1

= 23 (4 − 2) = 4

3 M1 A1

(b) = π5

1∫1

3 1x + dx M1

= π[ 13 ln3x + 1] 5

1 M1 A1

= 13 π(ln 16 − ln 4) = 1

3 π ln 4 = 23 π ln 2 [ k = 2

3 ] M1 A1 (9)


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

3 ⇒ 283 = 7

3 C ⇒ C = 4 B1

coeffs x2 ⇒ 0 = 9A − 3B ⇒ B = 3 M1 A1

(b) = 0

1−∫ ( 12 x+

+ 31 3x−

+ 24

(1 3 )x−) dx

= [ln2 + x − ln1 − 3x + 43 (1 − 3x)−1] 0

1− M1 A3

= (ln 2 + 0 + 43 ) − (0 − ln 4 + 1

3 ) M1 = 1 + ln 8 M1 A1 (11)

7. (a) x = 1 ∴ −1 + 4 cos θ = 1, cos θ = 12 , θ = π3 , 5π

3 M1

y > 0 ∴ sin θ > 0 ∴ θ = π3 A1

(b) dd

xθ

= −4 sin θ, dd

yθ

= 2 2 cos θ M1

∴ ddyx

= 2 2 cos4sin

θθ−

M1 A1

at P, grad = −12

32

2 2

4

×

× = − 2

2 3 M1

grad of normal = 2 32

× 22

= 6 A1

∴ y − 6 = 6 (x − 1) M1 y = 6 x, when x = 0, y = 0 ∴ passes through origin A1

(c) cos θ = 14

x + , sin θ = 2 2

y M1

∴ 2( 1)

16x + +

2

8y = 1 M1 A1 (12)

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

(b) OC = [µ i + (5 − 2µ)j + (−7 + 7µ)k] AC = OC − OA = [(3 + µ)i + (2 − 2µ)j + (−9 + 7µ)k] M1 A1 BC = OC − OB = [(−7 + µ)i + (6 − 2µ)j + (−19 + 7µ)k] A1 AC . BC = (3 + µ)(−7 + µ)+(2 − 2µ)(6 − 2µ)+(−9 + 7µ)(−19 + 7µ) = 0 M1 µ 2 − 4µ + 3 = 0 A1 (µ − 1)(µ − 3) = 0 M1 µ = 1, 3 ∴ OC = (i + 3j) or (3i − j + 14k) A2

(c) AC = 16 0 4+ + = 2 5 , BC = 36 16 144+ + = 14 M1 area = 1

2 × 2 5 × 14 = 14 5 M1 A1 (13) Total (75)


Performance Record – C4 Paper B

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

Topic(s) integration differential equation

differentiation binomial series

integration partial fractions

parametric equations

vectors

Marks 6 7 8 9 9 11 12 13 75

Student

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary Core Mathematics C4 Paper C

MARKING GUIDE



Solomon Press


Solomon Press C4C MARKS page 2

C4 Paper C – Marking Guide 1. u = ln x, u′ = 1

x, v′ = x, v = 1

2 x2 M1

I = [ 12 x2 ln x] 2

1 − 2

1∫12 x dx A1

= [ 12 x2 ln x − 1

4 x2] 21 M1 A1

= (2 ln 2 − 1) − (0 − 14 ) = 2 ln 2 − 3

4 M1 A1 (6) 2. x 0 0.5 1 1.5 2 arctan x 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 + x ddyx

+ 2y ddyx

= 0 M1 A2

(−1, 3) ⇒ −6 − 2 + 3 − ddyx

+ 6 ddyx

= 0, ddyx

= 1 M1 A1

grad of normal = −1 ∴ y − 3 = −(x + 1) M1 y = 2 − x A1

(b) sub. ⇒ 3x2 − 2x + x(2 − x) + (2 − x)2 − 11 = 0 M1 3x2 − 4x − 7 = 0 A1 (3x − 7)(x + 1) = 0 M1 x = −1 (at P) or 7

3 ∴ ( 73 , 1

3− ) A1 (11)

4. (a) AB = 10155

−

, ∴ r = 397

−

+ λ 23

1

−

M1 A1

(b) 3 + 2λ = 9 ∴ λ = 3 M1

when λ = 3, r = 397

−

+ 323

1

−

= 904

−

∴ (9, 0, −4) lies on l A1

(c) OD = 3 29 37

λλλ

+ − − +

∴ 3 29 37

λλλ

+ − − +

.23

1

−

= 0 M1

6 + 4λ − 27 + 9λ − 7 + λ = 0 A1

λ = 2 ∴ OD = 735

−

, D (7, 3, −5) M1 A1

(d) AB = 100 225 25+ + = 350 , OD = 49 9 25+ + + 83 M1 area = 1

2 × 350 × 83 = 85.2 (3sf) M1 A1 (11)


5. (a) ddtθ = −k(θ − 20) B2

(b) ∫ 120θ −

dθ = ∫ −k dt M1

lnθ − 20 = −kt + c M1 A1 t = 0, θ = 37 ⇒ c = ln 17 M1

ln 2017

θ − = −kt, θ = 20 + 17e−kt A1

t = 4, θ = 36 ⇒ 36 = 20 + 17e−4k M1 k = − 1

4 ln 1617 = 0.01516 A1

t = 10, θ = 20 + 17e−0.01516 × 10 = 34.6°C (3sf) A1

(c) 33 = 20 + 17e−0.01516t M1 t = − 1

0.01516 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 π2 ∴ t = π2 at A B1

(b) = volume when region above x-axis is rotated through 2π

ddxt

= 3 cos t M1

∴ volume = ππ2

0∫ (2 sin 2t)2 × 3 cos t dt = π2

0∫ 12π sin2 2t cos t dt M1 A1

(c) t = 0 ⇒ u = 0, t = π2 ⇒ u = 1, ddut

= cos t B1

sin2 2t = 4 sin2 t cos2 t = 4 sin2 t (1 − sin2 t) M1

∴ = 1

0∫ 12π × 4u2(1 − u2) du M1

= 48π1

0∫ (u2 − u4) du A1

= 48π[ 13 u3 − 1

5 u5] 10 M1 A1

= 48π[( 13 − 1

5 ) − (0)] = 325 π M1 A1 (13)

7. (a) 8(1 )(2 )

xx x

−+ −

≡ 1

Ax+

+ 2

Bx−

8 − x ≡ A(2 − x) + B(1 + x) M1 x = −1 ⇒ 9 = 3A ⇒ A = 3 A1

x = 2 ⇒ 6 = 3B ⇒ B = 2 ∴ f(x) = 31 x+

+ 22 x−

A1

(b) = 12

0∫ ( 31 x+

+ 22 x−

) dx = [3 ln1 + x − 2 ln2 − x]120 M1 A1

= (3 ln 32 − 2 ln 3

2 ) − (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 + x2 − x3 + … B1 (2 − x)−1 = 2−1(1 − 1

2 x)−1 M1

= 12 [1 + (−1)(− 1

2 x) + ( 1)( 2)2

− − (− 12 x)2 + ( 1)( 2)( 3)

3 2− − −

× (− 12 x)3 + …] M1

= 12 (1 + 1

2 x + 14 x2 + 1

8 x3 + …) A1

∴ f(x) = 3(1 − x + x2 − x3 + …) + (1 + 12 x + 1

4 x2 + 18 x3 + …) M1

= 4 − 52 x + 13

4 x2 − 238 x3 + … A1 (14)

Total (75)


Performance Record – C4 Paper C


Topic(s) integration trapezium rule

differentiation vectors differential equation


partial fractions, binomial

series

Marks 6 7 11 11 13 13 14 75

Student

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary Core Mathematics C4 Paper D

MARKING GUIDE



Solomon Press


Solomon Press C4D MARKS page 2

C4 Paper D – Marking Guide 1. (a) = 2−3(1 − 3

2 x)−3 = 18 (1 − 3

2 x)−3 B1

= 18 [1 + (−3)( 3

2− x) + ( 3)( 4)2

− − ( 32− x)2 + ( 3)( 4)( 5)

3 2− − −

× ( 32− x)3 + …] M1

= 18 + 9

16 x + 2716 x2 + 135

32 x3 + … A3

(b) x < 23 B1 (6)

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

− 4y ddyx

= 0 M1 A2

ddyx

= 2 34 3

x yy x

+−

M1 A1

(b) grad = 6 68 9−

− − = 0 M1

∴ normal parallel to y-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 = −2 x = 3 ⇒ 15 = 3D ⇒ D = 5 A1 coeffs x3 ⇒ A = 2 B1 coeffs x2 ⇒ −5 = B − 3A ⇒ B = 1 M1 A1

(b) = 2

1∫ (2x + 1 − 2x

+ 53x −

) 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 M1

12 x2 = k(5t − 1

2 t2) + c M1 A1 t = 0, x = 0 ⇒ c = 0 B1 t = 2, x = 96 ⇒ 4608 = 8k, k = 576 M1 A1 t = 1 ⇒ 1

2 x2 = 576 × 92 , x = 5184 = 72 M1 A1

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

∴ ddxt

= 576(5 3.0833)110.83

− = 9.96, ddxt

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


5. (a) 145

.3ab

= 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 + at 12t − 28 = at, t(12 − a) = 28, t = 28

12 a− M1 A1

sub. (3) ⇒ 1 + 5(3t − 7) = −6 + bt 15t − 28 = bt, t(15 − b) = 28, t = 28

15 b− A1

2812 a−

= 2815 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

16

− −

+ 232

1

−

= 334

− −

, ∴ (3, −3, −4) M1 A1 (12)

6. (a) u2 = 1 − x ⇒ x = 1 − u2, dd

xu

= −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

5 u5] 10 M1 A1

= ( 23 − 2

5 ) − (0) = 415 M1 A1

(b) = π1

0∫ x2(1 − x) dx M1

= π1

0∫ (x2 − x3) dx

= π[ 13 x3 − 1

4 x4] 10 M1 A1

= π{( 13 − 1

4 ) − (0)} = 112 π M1 A1 (13)

7. (a) ddxt

= 6 cos t × (−sin t), ddyt

= 2 cos 2t M1 A1

ddyx

= 2cos 26cos sin

tt t−

= 2cos 23sin 2

tt−

= 23− cot 2t M1 A1

(b) 23− cot 2t = 0 ⇒ 2t = π2 , 3π

2 ⇒ t = π4 , 3π4 M1 A1

∴ ( 32 , 1), ( 3

2 , −1) A1

(c) t = π6 , x = 94 , y = 3

2 , grad = − 23 3

B1

∴ y − 32 = − 2

3 3(x − 9

4 ) M1

6 3 y − 9 = −4x + 9 2x + 3 3 y = 9 A1

(d) y2 = sin2 2t = 4 sin2 t cos2 t = 4(1 − cos2 t)cos2 t M2

cos2 t = 3x ∴ y2 = 4(1 −

3x )

3x , y2 = 4

9 x(3 − x) M1 A1 (14)

Total (75)


Performance Record – C4 Paper D


Topic(s) binomial series

differentiation partial fractions

differential equation

vectors integration parametric equations

Marks 6 8 10 12 12 13 14 75

Student

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary Core Mathematics C4 Paper E

MARKING GUIDE



Solomon Press


Solomon Press C4E MARKS page 2

C4 Paper E – Marking Guide 1. = ∫ (cosec2 2x − 1) dx M1 A1

= 12− cot 2x − x + c M1 A1 (4)

2. (a) −4 sin x + (2 cos y) ddyx

= 0 M1 A2

ddyx

= 4sin2cos

xy

= 2sincos

xy

= 2 sin x sec y M1 A1

(b) grad = 2 × 32 × 2

3 = 2 B1

∴ y − π6 = 2(x − π3 ) M1

6y − π = 12x − 4π 4x − 2y = π A1 (8)

3. (a) 22 20

1 2 8x

x x+

+ − = 2 20

(1 2 )(1 4 )x

x x+

− + ≡

1 2A

x− +

1 4B

x+ B1

2 + 20x ≡ A(1 + 4x) + B(1 − 2x) M1 x = 1

2 ⇒ 12 = 3A ⇒ A = 4 A1

x = 14− ⇒ −3 = 3

2 B ⇒ B = −2 22 20

1 2 8x

x x+

+ − ≡ 4

1 2x− − 2

1 4x+ A1

(b) 22 20

1 2 8x

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 8x

x x+

+ − = 4(1 + 2x + 4x2 + 8x3 + …) − 2(1 − 4x + 16x2 − 64x3 + …) M1

= 2 + 16x − 16x2 − 160x3 + … A1 (9)

4. (a) PQ = (2i − 9j + k) − (−i − 8j + 3k) = (3i − j − 2k) M1 ∴ r = (−i − 8j + 3k) + λ(3i − j − 2k) A1

(b) 6 + µ = 2 ∴ µ = −4 M1 a + 4µ = −9 ∴ a = 7 A1 b − µ = 1 ∴ b = −3 A1

(c) = cos−1 3 1 ( 1) 4 ( 2) ( 1)9 1 4 1 16 1

× + − × + − × −+ + × + +

M1 A1

= cos−1 114 18×

= 86.4° (1dp) M1 A1 (9)

5. (a) ∫ dy = ∫ −ke−0.2t dt M1

y = 5ke−0.2t + c A1 t = 0, y = 2 ⇒ 2 = 5k + c, c = 2 − 5k M1 ∴ y = 5ke−0.2t − 5k + 2 A1

(b) t = 2, y = 1.6 ⇒ 1.6 = 5ke−0.4 − 5k + 2 M1

k = 0.40.4

5e 5−−

− = 0.2427 (4sf) M1 A1

(c) as t → ∞, y → h (in metres) M1 ∴ “h” = −5k + 2 = 0.787 m = 78.7 cm ∴ h = 79 M1 A1 (10)


6. (a) x = 0 ⇒ t2 = 2 t ≥ 0 ∴ t = 2 ∴ (0, 2 + 2 ) M1 A1 y = 0 ⇒ t(t + 1) = 0 t ≥ 0 ∴ t = 0 ∴ (2, 0) M1 A1

(b) ddxt

= −2t M1

area = 0

2∫ t(t + 1) × (−2t) dt A1

= 2

0∫ (2t3 + 2t) dt

= [ 12 t4 + 2

3 t3] 20 M1 A1

= (2 + 43 2 ) − (0) = 2 + 4

3 2 M1 A1 (10) 7. (a) let y = ax, ∴ ln y = x ln a M1

1y

ddyx

= ln a M1

ddyx

= y ln a = ax ln a ∴ ddx

(ax) = ax ln a A1

(b) ddyx

= 4x ln 4 − 2x − 1 ln 2 M1 A1

x = 0, y = 32 , grad = ln 4 − 1

2 ln 2 = 32 ln 2 M1

∴ y = ( 32 ln 2)x + 3

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

(c) 4x ln 4 − 2x − 1 ln 2 = 0 (2x)2 × 2 ln 2 − 1

2 (2x) ln 2 = 0 M1

12 (2x) ln 2[4(2x) − 1] = 0 M1

2x = 14 , x = −2 ∴ (−2, 15

16 ) A2 (12) 8. (a) x 0 0.5 1 1.5 2 2.5 3 y 0 0.5774 0.7071 0.7746 0.8165 0.8452 0.8660 B2 (i) ≈ 1

2 × 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) = π3

0∫ 1x

x + dx M1

= π3

0∫1 1

1x

x+ −

+ dx = π

3

0∫ (1 − 11x +

) dx M1

= π[x − lnx + 1] 30 M1 A1

= π{(3 − ln 4) − (0)} = π(3 − ln 4) M1 A1 (13) Total (75)


Performance Record – C4 Paper E


Topic(s) integration differentiation partial fractions, binomial

series

vectors differential equation


differentiation trapezium rule,

integration

Marks 4 8 9 9 10 10 12 13 75

Student

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary Core Mathematics C4 Paper F

MARKING GUIDE



Solomon Press


Solomon Press C4F MARKS page 2

C4 Paper F – Marking Guide 1. 4x + y + x d

dyx

− 2y ddyx

= 0 M1 A2

SP: ddyx

= 0 ∴ 4x + y = 0, y = −4x M1 A1

sub. 2x2 − 4x2 − 16x2 + 18 = 0 M1 x2 = 1, x = ± 1 ∴ (−1, 4), (1, −4) A2 (8)

2. x = 2 tan u ⇒ dd

xu

= 2 sec2 u M1

x = 0 ⇒ u = 0, x = 2 ⇒ u = π4 B1

I = π4

0∫2

24 tan4sec

uu

× 2 sec2 u du = π4

0∫ 2 tan2 u du A1

= π4

0∫ (2 sec2 u − 2) du M1

= [2 tan u − 2u]π40 M1 A1

= (2 − π2 ) − (0) = 12 (4 − π) M1 A1 (8)

3. (a) = 1225

24( )− = 2425 = 4

25 6× = 25 6 [ k = 2

5 ] M1 A1

(b) = 1 + ( 12− )( 1

2 x) + 31

2 2( )( )2

− − ( 12 x)2 +

3 512 2 2( )( )( )

3 2− − −

× ( 12 x)3 + … M1

= 1 − 14 x + 3

32 x2 − 5128 x3 + … A3

(c) x = 112 ⇒ (1 +

121

2 )x − = 121

24(1 )− = 25 6

x = 112 ⇒ (1 +

121

2 )x − ≈ 1 − 14 ( 1

12 ) + 332 ( 1

12 )2 − 5128 ( 1

12 )3 M1 = 0.97979510 ∴ 6 ≈ 5

2 × 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 M1 check (3) −4 + 2 = 8 − 10, true ∴ intersect A1 intersect at (7j − 4k) + 2(4i − 3j + k) = (8i + j − 2k) A1

(b) = cos−1 4 ( 3) ( 3) 0 1 216 9 1 9 0 4× − + − × + ×

+ + × + + M1 A1

= cos−1 1026 13

−×

= 57.0° (1dp) M1 A1 (9)


5. (a) ddxt

= 21 (2 ) ( 1)

(2 )t t

t× − − × −

− = 2

2(2 )t−

, ddyt

= −(1 + t)−2 M1 B1

ddyx

= − 21

(1 )t+ ÷ 2

2(2 )t−

= −2

2(2 )2(1 )

tt

−+

= 12−

221

tt

− +

M1 A1

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

8− B1 grad of normal = 8 ∴ y − 1

2 = 8(x − 1) [ y = 8x − 152 ] M1 A1

(c) x(2 − t) = t M1

2x = t(1 + x), t = 21

xx+

A1

y = 21

11 x

x++ = 1

(1 ) 2x

x x+

+ + ∴ y = 1

1 3xx

++

M1 A1 (11)

6. (a) = ∫ (sec2 x − 1) dx M1

= tan x − x + c M1 A1

(b) = ∫ sincos

xx

dx, let u = cos x, ddux

= −sin x M1

= ∫ 1u

× (−1) du = − ∫ 1u

du A1

= −lnu + c = lnu−1 + c = lnsec x + c M1 A1

(c) volume = ππ3

0∫ x tan2 x dx M1

u = x, u′ = 1, v′ = tan2 x, v = tan x − x M1 I = x(tan x − x) − ∫ (tan x − x) dx A1

= x tan x − x2 − lnsec x + 12 x2 + c A1

volume = π[x tan x − 12 x2 − lnsec x]

π30

= π{( 13 3 π − 1

18 π2 − ln 2) − (0)} = 118 π2( 6 3 − π) − π ln 2 M1 A1 (13)

7. (a) ddVt

= −kV, ddVh

= 10πh − πh2 B2

ddVt

= ddVh

× ddht

∴ −kV = (10πh − πh2) ddht

M1

− 13 kπh2(15 − h) = πh(10 − h) d

dht

−kh(15 − h) = 3(10 − h) ddht

∴ ddht

= − (15 )3(10 )kh h

h−

− M1 A1

(b) 3(10 )(15 )

hh h

−−

≡ Ah

+ 15

Bh−

, 3(10 − h) ≡ A(15 − h) + Bh M1

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

hh h

−−

≡ 2h

− 115 h−

A2

(c) ∫ 3(10 )(15 )

hh h

−−

dh = ∫ −k dt, ∫ ( 2h

− 115 h−

) dh = ∫ −k dt M1

2 lnh + ln15 − h = −kt + c M1 A1 t = 0, h = 5 ⇒ 2 ln 5 + ln 10 = c, c = ln 250 M1 2 lnh + ln15 − h − ln 250 = −kt

ln 2(15 )

250h h− = −kt,

2 (15 )250

h h− = e−kt, h2(15 − h) = 250e−kt M1 A1

(d) t = 2, h = 4 ⇒ 176 = 250e−2k M1 k = 1

2− ln 176250 = 0.175 (3sf) M1 A1 (17)

Total (75)


Performance Record – C4 Paper F


Topic(s) differentiation integration binomial series

vectors parametric equations

integration differential equation,

partial fractions

Marks 8 8 9 9 11 13 17 75

Student

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary Core Mathematics C4 Paper G

MARKING GUIDE



Solomon Press


Solomon Press C4G MARKS page 2

C4 Paper G – Marking Guide 1. 2x + 2y2 + 2x × 2y d

dyx

+ ddyx

= 0 M2 A2

ddyx

= −22 2

4 1x yxy+

+ M1 A1 (6)

2. u = x2, u′ = 2x, v′ = e−x, v = −e−x M1 A1 I = −x2 e−x − ∫ −2x e−x dx = −x2 e−x + ∫ 2x e−x dx A2

u = 2x, u′ = 2, v′ = e−x, v = −e−x M1 I = −x2 e−x − 2x e−x − ∫ −2e−x dx A1

= −x2 e−x − 2x e−x − 2e−x + c A1 (7)

3. (a) (1 + ax)n = 1 + nax + ( 1)2

n n− (ax)2 + … B1

∴ an = −4, 2 ( 1)

2a n n− = 24 B1

⇒ a = 4n

− , sub. ⇒ 216n

× ( 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 3 y 2.7183 2.0786 1.0733 0.5336 0.3716 B2

(a) = 12 × 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 others hence position of ordinates determines whether estimate is high or low B2 (9)

5. (a) AB = (4j + k) − (9i − 8j + 2k) = (−9i + 12j − k) M1 ∴ r = (9i − 8j + 2k) + λ(−9i + 12j − k) A1 at C, 2 − λ = −1, λ = 3 M1 A1 ∴ OC = (9i − 8j + 2k) + 3(−9i + 12j − k) = (−18i + 28j − k) A1

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


6. (a) ∫ 1P

dP = ∫ 0.05e−0.05t dt M1

lnP = −e−0.05t + c M1 A1 t = 0, P = 9000 ⇒ ln 9000 = −1 + c, c = 1 + ln 9000 M1 lnP = 1 + ln 9000 − e−0.05t A1 t = 10 ⇒ lnP = 1 + ln 9000 − e−0.5 = 9.498 M1 P = e9.498 = 13339 = 13300 (3sf) A1

(b) t → ∞, lnP → 1 + ln 9000 M1 ∴ P → e1 + ln 9000 = 9000e = 24465 = 24500 (3sf) M1 A1 (10) 7. (a) x = 2 ⇒ t = 1, x = 9 ⇒ t = 2 B1

ddxt

= 3t2 M1

∴ area = 2

1∫2t

× 3t2 dt = 2

1∫ 6t dt A1

= [3t2] 21 = 3(4 − 1) = 9 M1 A1

(b) = π2

1∫ ( 2t

)2 × 3t2 dt = π2

1∫ 12 dt M1

= π[12t] 21 = 12π(2 − 1) = 12π M1 A1

(c) t = 2y

∴ x = ( 2y

)3 + 1 = 38y

+ 1 M1

∴ y3 = 81x −

, y = 32

1x − M1 A1 (11)

8. (a) u = sin x ⇒ ddux

= cos x B1

I = ∫ 26cos

cos (2 sin )x

x x− dx = ∫ 2

6cos(1 sin )(2 sin )

xx x− −

dx M1

= ∫ 26

(1 )(2 )u u− − du M1 A1

(b) 6(1 )(1 )(2 )u u u+ − −

≡ 1

Au+

+ 1

Bu−

+ 2

Cu−

6 ≡ A(1 − u)(2 − u) + B(1 + u)(2 − u) + C(1 + u)(1 − u) M1 u = −1 ⇒ 6 = 6A ⇒ A = 1 A1 u = 1 ⇒ 6 = 2B ⇒ B = 3 A1 u = 2 ⇒ 6 = −3C ⇒ C = −2 A1 ∴ 2

6(1 )(2 )u u− −

≡ 11 u+

+ 31 u−

− 22 u−

(c) x = 0 ⇒ u = 0, x = π6 ⇒ u = 12 M1

I = 12

0∫ ( 11 u+

+ 31 u−

− 22 u−

) du

= [ln1 + u − 3 ln1 − u + 2 ln2 − u]120 M1 A2

= (ln 32 − 3 ln 1

2 + 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)


Performance Record – C4 Paper G


Topic(s) differentiation integration binomial series

trapezium rule

vectors differential equation


partial fractions,

integration

Marks 6 7 8 9 9 10 11 15 75

Student

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary Core Mathematics C4 Paper H

MARKING GUIDE



Solomon Press


Solomon Press C4H MARKS page 2

C4 Paper H – Marking Guide 1. (a) = 1 + ( 3

2 )(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 + sin x ⇒ ddux

= cos x M1

x = 0 ⇒ u = 1, x = π2 ⇒ u = 2 B1

I = 2

1∫ u3 du A1

= [ 14 u4] 2

1 M1

= 4 − 14 = 15

4 M1 A1 (6)

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

xx x

++ −

≡ 4

Ax +

+ 3

Bx −

x + 11 ≡ A(x − 3) + B(x + 4) M1 x = −4 ⇒ 7 = −7A ⇒ A = −1 A1 x = 3 ⇒ 14 = 7B ⇒ B = 2 A1

11( 4)( 3)

xx x

++ −

≡ 23x −

− 14x +

(b) = 2

0∫ ( 23x −

− 14x +

) 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. = ππ2π6∫ (2 sin x + cosec x)2 dx M1

= π π2π6∫ (4 sin2 x + 4 + cosec2 x) dx A1

= ππ2π6∫ (2 − 2 cos 2x + 4 + cosec2 x) dx M1

= π[6x − sin 2x − cot x]π2π6

M1 A2

= π{(3π + 0 + 0) − (π − 32 − 3 )} M1

= π(2π + 32 3 ) = 1

2 π(4π + 3 3 ) A1 (8)

5. (a) 2x − 3y − 3x ddyx

− 2y ddyx

= 0 M1 A2

ddyx

= 2 33 2

x yx y

−+

M1 A1

(b) grad = 5 M1 ∴ y + 2 = 5(x − 2) [ y = 5x − 12 ] M1 A1 (8)


6. (a) = 1 6 5 3 ( 1) ( 6)1 25 1 36 9 36× + × + − × −+ + × + +

M1 A1

= 2727 81×

= 279

= 3 39

= 13 3 M1 A1

(b) sin (∠AOB) = 2131 ( 3)− = 2

3 M1 A1

area = 12 × 3 3 × 9 × 2

3 = 272 2 M1 A1

(c) = OA × sin (∠AOB) = 3 3 × 23 = 3 2 M1 A1 (10)

7. (a) ddxt

= 2t − 1, ddyt

= 24 (1 ) 4 ( 1)

(1 )t t

t× − − × −

− = 2

4(1 )t−

B1 M1

ddyx

= 24

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

(c) t(t − 1) + 3 × 41

tt−

+ 4 = 0 M1

−t(t − 1)2 + 12t + 4(1 − t) = 0 t3 − 2t2 − 7t − 4 = 0 A1 t = −1 is a solution ∴ (t + 1) is a factor M1 (t + 1)(t2 − 3t − 4) = 0 M1 (t + 1)(t + 1)(t − 4) = 0 A1 t = −1 (at P) or t = 4 ∴ Q (12, 16

3− ) M1 A1 (14)

8. (a) ∫ 1P

dP = ∫ k dt M1

lnP = kt + c A1 t = 0, P = 300 ⇒ ln 300 = c M1 lnP = kt + ln 300

ln300P = kt,

300P = ekt, P = 300ekt M1 A1

(b) t = 1, P = 360 ⇒ 360 = 300ek M1 k = ln 6

5 = 0.182 (3sf) A1

(c) P = 300e0.1823t when t = 2, P = 432; when t = 3, P = 518 B1 model does not seem suitable as data diverges from predictions B1

(d) ∫ 1P

dP = ∫ (0.4 − 0.25 cos 0.5t) dt M1

lnP = 0.4t − 0.5 sin 0.5t + c A1 t = 0, P = 300 ⇒ ln 300 = c

ln300P = 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 A1 the second model seems more suitable as it fits the data better B1 (16) Total (75)


Performance Record – C4 Paper H


Topic(s) binomial series

integration partial fractions

integration differentiation vectors parametric equations

differential equations

Marks 5 6 8 8 8 10 14 16 75

Student

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary Core Mathematics C4 Paper I

MARKING GUIDE



Solomon Press


Solomon Press C4I MARKS page 2

C4 Paper I – Marking Guide 1. 3x2 + 2y + 2x d

dyx

− 2y ddyx

= 0 M1 A2

(2, −4) ⇒ 12 − 8 + 4 ddyx

+ 8 ddyx

= 0, ddyx

= 13− M1 A1

grad of normal = 3 M1 ∴ y + 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 )(− 1

4 x) + 1 12 2( )( )

2− (− 1

4 x)2 + …] = 2 − 14 x − 1

64 x2 + … M1 A2

(b) x < 4 B1

(c) x = 0.01 ⇒ (4 − 12)x = 3.99 = 399

100 = 110 399 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 × π × (1.0986 + 0) = 1.726 (3dp) B1 M1 A1

(ii) = 12 × π2 × [1.0986 + 0 + 2(0.6931)] = 1.952 (3dp) M1 A1

(iii) = 12 × π4 × [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 is less than 0.01 so the error in 4 strip value is likely to be less than 0.005 B2 (10)

4. (a) x = −1 ⇒ θ = − π4 , x = 1 ⇒ θ = π4 B1

dd

xθ

= sec2 θ M1

volume = ππ4π4−∫ (cos2 θ)2 × sec2 θ dθ = π

π4π4−∫ cos2 θ dθ A1

= ππ4π4−∫ ( 1

2 + 12 cos 2θ ) dθ M1

= π[ 12 θ + 1

4 sin 2θ ]π4π4−

M1 A1

= π[( π8 + 1

4 ) − (− π8 − 1

4 )] M1

= π( π4 + 1

2 ) = 14 π(π + 2) A1

(b) y = cos2 θ = 21

sec θ = 2

11 tan θ+

∴ y = 21

1 x+ M2 A1 (11)


5. (a) AB = (5i − 4j) − (2i − j + 6k) = (3i − 3j − 6k) B1 AC = (7i − 6j − 4k) − (2i − j + 6k) = (5i − 5j − 10k) = 5

3 AB M1

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

(b) 3 : 2 B1

(c) ADuuur

= (3i + j + 4k) − (2i − j + 6k) = (i + 2j − 2k) B1 BD

uuur = (3i + j + 4k) − (5i − 4j) = (−2i + 5j + 4k) B1

ADuuur

. BDuuur

= −2 + 10 − 8 = 0 ∴ perpendicular M1 A1

(d) = 12 × 1 4 4+ + × 4 25 16+ + = 1

2 × 3 × 3 5 = 92 5 M2 A1 (11)

6. (a) x = 2 sin u ⇒ dd

xu

= 2 cos u M1

x = 0 ⇒ u = 0, x = 3 ⇒ u = π3 B1

I = π3

0∫1

2cosu × 2 cos u du =

π3

0∫ 1 du A1

= [u]π30 = π3 − 0 = π3 M1 A1

(b) u = x, u′ = 1, v′ = cos x, v = sin x M1

I = [x sin x]π20 − ∫ sin x dx A2

= [x sin x + cos x]π20 M1

= ( π2 + 0) − (0 + 1) = π2 − 1 M1 A1 (11)

7. (a) when x = 14 , d

dxt

= 34 ÷ 6 = 1

8 M1 A1

ddxt

= kx(1 − x) ∴ 18 = k × 1

4 × 34 , k = 2

3 ∴ ddxt

= 23 x(1 − x) M1 A1

(b) ∫ 1(1 )x x−

dx = ∫ 23 dt M1

1(1 )x x−

≡ Ax

+ 1

Bx−

, 1 ≡ A(1 − x) + Bx M1

x = 0 ⇒ A = 1 A1 x = 1 ⇒ B = 1 A1

∴ ∫ ( 1x

+ 11 x−

) dx = ∫ 23 dt

lnx − ln1 − x = 23 t + c M1 A1

t = 0, x = 14 ⇒ ln 1

4 − ln 34 = c, c = ln 1

3 M1 A1

t = 3 ⇒ lnx − ln1 − x = 2 + ln 13

ln 31

xx−

= 2, 31

xx−

= e2 M1

3x = e2(1 − x), x(e2 + 3) = e2 M1

x = 2

2e

e 3+ ∴ % destroyed =

2

2e

e 3+ × 100% = 71.1% (3sf) A1 (15)

Total (75)


Performance Record – C4 Paper I


Topic(s) differentiation binomial series

trapezium rule


vectors integration differential equation,

partial fractions

Marks 8 9 10 11 11 11 15 75

Student

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary Core Mathematics C4 Paper J

MARKING GUIDE



Solomon Press


Solomon Press C4J MARKS page 2

C4 Paper J – Marking Guide 1. x(x − 2) = 0, x = 0, 2 ∴ crosses x-axis at (0, 0) and (2, 0)

volume = π2

0∫ (x2 − 2x)2 dx M1

= π2

0∫ (x4 − 4x3 + 4x2) dx A1

= π[ 15 x5 − x4 + 4

3 x3] 20 M1 A1

= π{( 325 − 16 + 32

3 ) − (0)} = 1615 π M1 A1 (6)

2. u = 1 − 12x ⇒ x = (1 − u)2, d

dxu

= −2(1 − u) = 2u − 2 M1 A1

I = ∫ 1u

× (2u − 2) du = ∫ (2 − 2u

) du A1

= 2u − 2 lnu + c M1 A1

= 2(1 − 12x ) − 2 ln1 −

12x + c A1 (6)

3. (a) 4 cos 2x − sec2 y ddyx

= 0 M1 A2

ddyx

= 4 cos 2x cos2 y M1 A1

(b) grad = 4 × 12 × 1

4 = 12 B1

∴ y − π3 = 12 (x − π6 ) M1

y − π3 = 12 x − π

12

y = 12 x + π4 A1 (8)

4. (a) ddxt

= 121

2 at− , ddyt

= a(1 − 2t) M1

ddyx

= 121

2

(1 2 )a t

at−

− = 2 t (1 − 2t) M1 A1

(b) y = 0 ⇒ t = 0 (at O) or 1 (at A) B1 t = 1, x = a, y = 0, grad = −2 M1 ∴ y − 0 = −2(x − a) A1 at B, x = 0 ∴ y = 2a M1 area = 1

2 × a × 2a = a2 M1 A1 (9)

5. (a) ddyx

= 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 = 1

2 (x + 4) M1

y = 14 (x + 4)2 A1 (9)


6. (a) let radius = r, ∴ tan 30° = 13

= rh

, r = 3

h M1

V = 13 πr2h = 1

3 πh × 2

3h = 1

9 πh3 M1 A1

(b) (i) ddVt

= 120, ddVh

= 13 πh2 B1

ddVt

= ddVh

× ddht

, 120 = 13 πh2 d

dht

, ddht

= 2360πh

M1 A1

when h = 6, ddht

= 3.18 cm s−1 (2dp) M1 A1

(ii) V = 8 × 120 = 960 = 19 πh3 ∴ h = 3 9 960

π× = 14.011 M1

∴ ddht

= 0.58 cm s−1 (2dp) A1 (10)

7. (a) AB = 3

61

−

− 4

13

−

= 152

−

∴ r = 4

13

−

+ λ 152

−

M1 A1

(b) −4 + λ = 3 + 2µ (1) 1 + 5λ = −7 − 3µ (2) 3 − 2λ = 9 + µ (3) B1 2 × (1) + (3): −5 = 15 + 5µ, µ = −4, λ = −1 M1 A1 sub. (2): 1 − 5 = −7 + 12, not true ∴ do not intersect M1 A1

(c) OC = 3 27 39

µµ

µ

+ − − +

, BC = OC − OB = 6 213 38

µµ

µ

+ − − +

M1 A1

∴ 152

−

.6 213 38

µµ

µ

+ − − +

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

µ = −5 ∴ OC = 7

84

−

M1 A1 (13)

8. (a) x(3x − 7) ≡ A(1 − x)(1 − 3x) + B(1 − 3x) + C(1 − x) M1 x = 1 ⇒ −4 = −2B ⇒ B = 2 A1 x = 1

3 ⇒ −2 = 23 C ⇒ C = −3 A1

coeffs x2 ⇒ 3 = 3A ⇒ A = 1 A1

(b) = 14

0∫ (1 + 21 x−

− 31 3x−

) dx = [x − 2 ln1 − x + ln1 − 3x]140 M1 A1

= ( 14 − 2 ln 3

4 + ln 14 ) − (0) M1

= 14 + ln 16

9 + ln 14 = 1

4 + 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)


Performance Record – C4 Paper J


Topic(s) integration integration differentiation parametric equations

differential equation

connected rates

vectors partial fractions, binomial

series

Marks 6 6 8 9 9 10 13 14 75

Student

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary Core Mathematics C4 Paper K

MARKING GUIDE



Solomon Press


Solomon Press C4K MARKS page 2

C4 Paper K – Marking Guide 1. = π

3

1∫2(3 1)x

x+ dx M1

= π3

1∫29 6 1x x

x+ + dx =

3

1∫ (9x + 6 + 1x

) dx A1

= π[ 92 x2 + 6x + lnx] 3

1 M1 A1

= π{( 812 + 18 + ln 3) − ( 9

2 + 6 + 0)} M1 = π(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) 22

1 3xx

− −

= (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 small x, 22

1 3xx

− −

= 4 + 20x + 85x2 + 330x3 A1 (7)

3. (a) 2

27 3 2

(1 2 )(1 )x x

x x+ +

− + ≡

1 2A

x− +

1B

x+ + 2(1 )

Cx+

7 + 3x + 2x2 ≡ A(1 + x)2 + B(1 − 2x)(1 + x) + C(1 − 2x) x = 1

2 ⇒ 9 = 94 A ⇒ A = 4 B1

x = −1 ⇒ 6 = 3C ⇒ C = 2 B1 coeffs x2 ⇒ 2 = A − 2B ⇒ B = 1 M1 ∴ f(x) = 4

1 2x− + 1

1 x+ + 2

2(1 )x+

A1

(b) = 2

1∫ ( 41 2x−

+ 11 x+

+ 22

(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 = 1

3 − ln 6 [ p = 13 , q = 6 ] M1 A1 (11)

4. (a) 4λ = 6 + 14µ (1) −3 − 2λ = 3 + 2µ (2) B1 (1) + 2 × (2): −6 = 12 + 18µ, µ = −1, λ = −2 M1 A1

r = 703

−

− 2542

−

= 38

1

− −

M1 A1

(b) a − (−5) = −3, a = −8 M1 A1

(c) cos θ = 5 ( 5) 4 14 ( 2) 225 16 4 25 196 4× − + × + − ×

+ + × + + M1 A1

= 2745 15×

= 93 5 5×

= 35 5

= 325 5 M1 A1 (11)


5. (a) 2x − 4y − 4x ddyx

+ 4y ddyx

= 0 M1 A2

ddyx

= 2 44 4

x yx y

−−

= 22 2x yx y−−

M1 A1

(b) grad = 32 M1

∴ y − 2 = 32 (x − 1) M1

2y − 4 = 3x − 3 3x − 2y + 1 = 0 A1

(c) 22 2x yx y−−

= 32 M1

2(x − 2y) = 3(2x − 2y), y = 2x A1 sub. ⇒ x2 − 8x2 + 8x2 = 1 M1 x2 = 1, x = 1 (at P) or −1 ∴ Q (−1, −2) A1 (12)

6. (a) ddNt

= kN B1

(b) ∫ 1N

dN = ∫ k dt M1

lnN = kt + c M1 A1 t = 0, N = N0 ⇒ lnN0 = c M1 lnN = kt + lnN0, ln

0

NN

= kt M1

0

NN

= ekt, N = N0ekt A1

(c) 2N0 = N0e6k M1 k = 1

6 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 + 1x

= sec θ + tan θ + 1sec tanθ θ+

= 2(sec tan ) 1

sec tanθ θ

θ θ+ +

+ M1

= 2 2sec 2sec tan tan 1

sec tanθ θ θ θ

θ θ+ + +

+ =

22sec 2sec tansec tan

θ θ θθ θ+

+ M1 A1

= 2sec (sec tan )sec tanθ θ θ

θ θ+

+ = 2 sec θ M1 A1

(b) 2 1xx+ = 2

cosθ ⇒ cos θ = 2

21

xx +

M1

2 1yy+ = 2

sinθ ⇒ sin θ = 2

21

yy +

∴ 2

2 24

( 1)x

x + +

2

2 24

( 1)y

y + = 1 M1 A1

(c) dd

xθ

= sec θ tan θ + sec2 θ M1

= sec θ (tan θ + sec θ ) = 2 12

xx+ × x = 1

2 (x2 + 1) M1 A1

(d) dd

yθ

= −cosec θ cot θ − cosec2 θ M1

= −cosec θ (cot θ + cosec θ ) = −2 12

yy+ × y = 1

2− (y2 + 1) A1

∴ ddyx

= −2

211

yx

++

M1 A1 (15)

Total (75)


Performance Record – C4 Paper K


Topic(s) integration binomial series

partial fractions

vectors differentiation differential equation


Marks 6 7 11 11 12 13 15 75

Student

FOR EDEXCEL

GCE Examinations

Advanced Subsidiary Core Mathematics C4 Paper L

MARKING GUIDE



Solomon Press


Solomon Press C4L MARKS page 2

C4 Paper L – Marking Guide 1. (a) d

dnt

= 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 A1 t = 0, n = 20 ⇒ 20 = 2 + c, c = 18 M1 n = 2e0.5t − 5t + 18 A1

(c) as t increases, n rapidly becomes very large ∴ not realistic B1 (8)

2. 6x + y + x ddyx

− 4y ddyx

= 0 M1 A2

(1, 4) ⇒ 6 + 4 + ddyx

− 16 ddyx

= 0, ddyx

= 23 M1 A1

grad of normal = 32− M1

∴ y − 4 = 32− (x − 1) M1

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

3. (a) u = 2 − x2 ⇒ ddux

= −2x M1

I = ∫ 1u

× ( 12− ) du = 1

2− ∫ 1u

du A1

= 12− lnu + c = 1

2− ln2 − x2 + c M1 A1

(b) = π4

0∫ ( 12 sin 4x + 1

2 sin 2x) dx M1 A1

= [ 18− cos 4x − 1

4 cos 2x]π40 M1 A1

= ( 18 − 0) − ( 1

8− − 14 ) = 1

2 M1 A1 (10) 4. (a) x 1 2 3 y 0 1.665 3.144 B1 area ≈ 1

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

(b) volume = π3

1∫ x2 ln x dx M1

u = ln x, u′ = 1x

, v′ = x2, v = 13 x3

I = 13 x3 ln x − ∫ 1

3 x2 dx M1 A2

= 13 x3 ln x − 1

9 x3 + c A1

volume = π[ 13 x3 ln x − 1

9 x3] 31

= π{(9 ln 3 − 3) − (0 − 19 )} M1

= π(9 ln 3 − 269 ) A1 (11)


5. (a) 25 8

(1 2 )(1 )x

x x−

+ − ≡

1 2A

x+ +

1B

x− + 2(1 )

Cx−

5 − 8x ≡ A(1 − x)2 + B(1 + 2x)(1 − x) + C(1 + 2x)(1 − x) M1 x = 1

2− ⇒ 9 = 94 A ⇒ A = 4 A1

x = 1 ⇒ −3 = 3C ⇒ C = −1 A1 coeffs x2 ⇒ 0 = A − 2B ⇒ B = 2 M1 A1 f(x) = 4

1 2x+ + 2

1 x− − 2

1(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) ddxt

= 1 + cos t, ddyt

= cos t M1

ddyx

= cos1 cos

tt+

M1 A1

(b) cos1 cos

tt+

= 0, cos t = 0, t = π2 M1 A1

∴ ( π2 + 1, 1) A1

(c) = π

0∫ sin t × (1 + cos t) dt = π

0∫ (sin t + 12 sin 2t) dt M1 A1

= [−cos t − 14 cos 2t] π0 M1 A1

= (1 − 14 ) − (−1 − 1

4 ) = 2 M1 A1 (12)

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

(b) 3 − 3λ = −2 + 7µ (1) 6 + 2λ = 10 − 4µ (2) −8 + 2λ = 6 + 6µ (3) B1 (3) − (2): −14 = −4 + 10µ, µ = −1, λ = 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 + 7µ)i + (10 − 4µ)j + (6 + 6µ)k] AC = OC − OA = [(−5 + 7µ)i + (4 − 4µ)j + (14 + 6µ)k] M1 A1 ∴ [(−5 + 7µ)i + (4 − 4µ)j + (14 + 6µ)k].(−3i + 2j + 2k) = 0 M1 15 − 21µ + 8 − 8µ + 28 + 12µ = 0 A1 µ = 3 ∴ OC = (19i − 2j + 24k) M1 A1 (14) Total (75)


Performance Record – C4 Paper L


Topic(s) differential equation

differentiation integration trapezium rule,

integration

partial fractions, binomial

series


vectors

Marks 8 8 10 11 12 12 14 75

Student

core mathematics c4 - mr. samuel...

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