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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)
Solomon Press C4A MARKS page 3
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)
Solomon Press C4A MARKS page 4
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
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 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)
Solomon Press C4B MARKS page 3
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)
Solomon Press C4B MARKS page 4
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
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 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)
Solomon Press C4C MARKS page 3
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)
Solomon Press C4C MARKS page 4
Performance Record – C4 Paper C
Question no. 1 2 3 4 5 6 7 Total
Topic(s) integration trapezium rule
differentiation vectors differential equation
parametric equations
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
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 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)
Solomon Press C4D MARKS page 3
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)
Solomon Press C4D MARKS page 4
Performance Record – C4 Paper D
Question no. 1 2 3 4 5 6 7 Total
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
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 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)
Solomon Press C4E MARKS page 3
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)
Solomon Press C4E MARKS page 4
Performance Record – C4 Paper E
Question no. 1 2 3 4 5 6 7 8 Total
Topic(s) integration differentiation partial fractions, binomial
series
vectors differential equation
parametric equations
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
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 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)
Solomon Press C4F MARKS page 3
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)
Solomon Press C4F MARKS page 4
Performance Record – C4 Paper F
Question no. 1 2 3 4 5 6 7 Total
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
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 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)
Solomon Press C4G MARKS page 3
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)
Solomon Press C4G MARKS page 4
Performance Record – C4 Paper G
Question no. 1 2 3 4 5 6 7 8 Total
Topic(s) differentiation integration binomial series
trapezium rule
vectors differential equation
parametric equations
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
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 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)
Solomon Press C4H MARKS page 3
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)
Solomon Press C4H MARKS page 4
Performance Record – C4 Paper H
Question no. 1 2 3 4 5 6 7 8 Total
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
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 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)
Solomon Press C4I MARKS page 3
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)
Solomon Press C4I MARKS page 4
Performance Record – C4 Paper I
Question no. 1 2 3 4 5 6 7 Total
Topic(s) differentiation binomial series
trapezium rule
parametric equations
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
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 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)
Solomon Press C4J MARKS page 3
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)
Solomon Press C4J MARKS page 4
Performance Record – C4 Paper J
Question no. 1 2 3 4 5 6 7 8 Total
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
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 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)
Solomon Press C4K MARKS page 3
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)
Solomon Press C4K MARKS page 4
Performance Record – C4 Paper K
Question no. 1 2 3 4 5 6 7 Total
Topic(s) integration binomial series
partial fractions
vectors differentiation differential equation
parametric equations
Marks 6 7 11 11 12 13 15 75
Student
FOR EDEXCEL
GCE Examinations
Advanced Subsidiary Core Mathematics C4 Paper L
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 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)
Solomon Press C4L MARKS page 3
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)