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.

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 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)

1
2. ∫ y2
dy = ∫ x dx M1
3
−y−1 = 2
3
x2 + c M1 A1
x = 1, y = −2 ⇒ 1
2
= 2
3
+ c, c= − 16 M1 A1
3 3 3
1 2 1 1 1 2 1
− = 3
x − 2
6
, = 6
− 3
x =
2
6
(1 − 4x ) 2
M1
y y
6
y= 3
A1 (7)
1 − 4x 2

dy dy
3. 8x − 2y − 2x − 2y =0 M1 A2
dx dx
dy dy dy
(−1, −3) ⇒ −8 + 6 + 2 +6 = 0, = 1
4
M1 A1
dx dx dx
grad of normal = −4 M1
∴ y + 3 = −4(x + 1) [ y = −4x − 7 ] M1 A1 (8)

( −3)( −4) ( −3)( −4)( −5)

4. (a) = 1 + (−3)(ax) + 2
(ax)2 + 3× 2
(ax)3 + … M1 A1
2 2 3 3
= 1 − 3ax + 6a x − 10a x + … A1
6− x
(b) = (6 − x)( 1 − 3ax + 6a2x2 + …)
(1 + ax)3
coeff. of x2 = 36a2 + 3a = 3 M1
12a2 + a − 1 = 0 A1
(4a − 1)(3a + 1) = 0 M1
a = − 13 , 14 A1

6− x
(c) a = − 13 ∴ = (6 − x)(… + 2
3
x2 + 10
27
x3 + …) M1
(1 + ax)3
coeff. of x3 = (6 × 10
27
) + (−1 × 2
3
)= 20
9
− 2
3
= 14
9
A1 (9)

5 1 1
5. (a) = ∫1 3x + 1
dx = [ 23 (3x + 1) 2 ] 15 M1 A1
2 4
= 3
(4 − 2) = 3
M1 A1
5 1
(b) = π∫ dx M1
1 3x + 1
= π[ 13 ln3x + 1] 15 M1 A1
1 1 2 2
= 3
π(ln 16 − ln 4) = 3
π ln 4 = 3
π ln 2 [k= 3
] M1 A1 (9)

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6. (a) 15 − 17x ≡ A(1 − 3x)2 + B(2 + x)(1 − 3x) + C(2 + x)
x = −2 ⇒ 49 = 49A ⇒ A=1 B1
x= 31
⇒ 28
3
7
= 3C ⇒ C=4 B1
coeffs x2 ⇒ 0 = 9A − 3B ⇒ B=3 M1 A1
0 1 3 4
(b) = ∫ −1 (
2+ x
+
1 − 3x
+
(1 − 3x)2
) dx

= [ln2 + x − ln1 − 3x + 4

3
(1 − 3x)−1] 0−1 M1 A3
4
= (ln 2 + 0 + 3
) − (0 − ln 4 + 13 ) M1
= 1 + ln 8 M1 A1 (11)

7. (a) x = 1 ∴ −1 + 4 cos θ = 1, cos θ = 1

2
, θ = π
3
, 5π
3
M1
y > 0 ∴ sin θ > 0 ∴ θ = π
3
A1

dx dy
(b) = −4 sin θ, = 2 2 cos θ M1
dθ dθ
dy
∴ = 2 2 cos θ M1 A1
dx −4sin θ
2 2 × 12 2
at P, grad = − 3
=− M1
4× 2 3
2

2 3 2
grad of normal = × = 6 A1
2 2
∴ y− 6 = 6 (x − 1) M1
y= 6 x, when x = 0, y = 0 ∴ passes through origin A1
x +1 y
(c) cos θ = , sin θ = M1
4 2 2
( x + 1)2 y2
∴ + =1 M1 A1 (12)
16 8

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
1
area = 2
× 2 5 × 14 = 14 5 M1 A1 (13)

Total (75)

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 Performance Record – C4 Paper B

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

Topic(s) integration differential differentiation binomial integration partial parametric vectors
equation series fractions equations

Marks 6 7 8 9 9 11 12 13 75
Student

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