1. No category

FOR OCR
GCE Examinations
Advanced / 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 using a valid 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.
C4 Paper B – Marking Guide
1.
u = x, u′ = 1, v′ = e3x, v =
1
3
e3x
M1 A1
∫
I =
=
1
1 3x
x e3x −
e dx
3
3
3x
1
1 3x
xe − 9e +c
3
2.
M1
A1
x2 + 0 x + 1
x4 + x3 − 5 x2 + 0 x
x4 + x3 − 6 x2
x2 + 0 x
x2 + x
− x
2
x +x−6
− 9
M2
− 9
− 6
− 3
∴ quotient = x2 + 1, remainder = −x − 3
3.
4.
A2
(i)
= −cosec2 x2 × 2x = −2x cosec2 x2
M1 A1
(ii)
= cos x × (3 + 2 cos x) − sin2x × (−2sin x)
M1 A1
2
2
= 3cos x + 2 cos x +22sin x = 3cos x + 2 2
M1 A1
(3 + 2 cos x)
(3 + 2 cos x )
(i)
(3 + 2 cos x)
(1 − 3x)−2 = 1 + (−2)(−3x) +
( −2)( −3)
2
2
3
(−3x)2 +
( −2)( −3)( −4)
3× 2
(−3x)3 + …
= 1 + 6x + 27x + 108x + …
(ii)
 2− x 


 1 − 3x 
2
∴ for small x,
(i)
(ii)
6.
(i)
(ii)
 2− x 


 1 − 3x 
2
= 4 + 20x + 85x2 + 330x3
dx
dy
−1
= 12 at 2 ,
= a(1 − 2t)
dt
dt
dy
a(1 − 2t )
=
= 2 t (1 − 2t)
1
1 at − 2
dx
2
A3
A1
A1
M1 A1
4s
= −7 − 3t (1)
7 − 3s = 1
(2)
−4 + s = 8 + 2t
(3)
(2) ⇒ s = 2, sub. (1) ⇒ t = −5
check (3) −4 + 2 = 8 − 10, true ∴ intersect
intersect at (7j − 4k) + 2(4i − 3j + k) = (8i + j − 2k)
M1
B1 M1
A1
A1
4 × (−3) + (−3) × 0 + 1× 2
= 57.0° (1dp)
M1 A1
 Solomon Press
C4B MARKS page 2
(8)
M1 A1
16 + 9 + 1 × 9 + 0 + 4
−10
26 × 13
(7)
M1
B1
M1
A1
M1
A1
= cos−1
(6)
M1
y = 0 ⇒ t = 0 (at O) or 1 (at A)
t = 1, x = a, y = 0, grad = −2
∴ y − 0 = −2(x − a)
at B, x = 0 ∴ y = 2a
area = 12 × a × 2a = a2
= cos−1
(4)
= (2 − x)2(1 − 3x)−2 = (4 − 4x + x2)(1 + 6x + 27x2 + 108x3 + …) M1
= 4 + 24x + 108x2 + 432x3 − 4x − 24x2 − 108x3 + x2 + 6x3 + …
5.
(4)
(9)
7.
(i)
∫
∫
dy =
−ke−0.2t dt
M1
−0.2t
y = 5 ke
+c
t = 0, y = 2 ⇒ 2 = 5k + c,
∴ y = 5ke−0.2t − 5k + 2
(ii)
t = 2, y = 1.6 ⇒ 1.6 = 5ke−0.4 − 5k + 2
k=
(iii)
8.
(i)
A1
M1
A1
c = 2 − 5k
−0.4
= 0.2427 (4sf)
5e −0.4 − 5
M1 A1
as t → ∞, y → h (in metres)
∴ “h” = −5k + 2 = 0.787 m = 78.7 cm ∴ h = 79 (2sf)
dy
dy
+ 4y
=0
dx
dx
2x − 4y − 4x
grad =
M1 A1
3
2
M1
∴ y−2=
3
2
(x − 1)
M1
2y − 4 = 3x − 3
3x − 2y + 1 = 0
(iii)
x − 2y
2x − 2 y
A1
3
2
=
M1
2(x − 2y) = 3(2x − 2y),
y = 2x
sub. ⇒ x2 − 8x2 + 8x2 = 1
x2 = 1, x = 1 (at P) or −1
∴ Q (−1, −2)
9.
(i)
(ii)
A1
M1
A1
du
= cos x
dx
u = sin x ⇒
6cos x
I =
∫
cos 2 x (2 − sin x)
=
∫
(1 − u 2 )(2 − u )
6
6
(1 + u )(1 − u )(2 − u )
≡
dx =
6cos x
∫
(1 − sin 2 x)(2 − sin x)
dx
du
A
1+ u
(iii)
6
≡
(1 − u 2 )(2 − u )
1
1+ u
x = 0 ⇒ u = 0, x =
I =
1
2
∫0
(
1
1+ u
+
3
1− u
π
6
−
B
1− u
+
C
2−u
+
+
3
1− u
⇒ u=
2
2−u
= (ln
− 3 ln
M1
A1
A1
A1
2
2−u
−
1
2
M1
) du
1
1
2
M1
M1 A1
= [ln1 + u − 3 ln1 − u + 2 ln2 − u] 02
3
2
(11)
B1
6 ≡ A(1 − u)(2 − u) + B(1 + u)(2 − u) + C(1 + u)(1 − u)
u = −1
⇒
6 = 6A
⇒
A=1
u=1
⇒
6 = 2B
⇒
B=3
u=2
⇒
6 = −3C ⇒
C = −2
∴
(9)
M1 A1
dy
2x − 4 y
x − 2y
=
=
4x − 4 y
2x − 2 y
dx
(ii)
M1
M1 A1
3
2
+ 2 ln ) − (0 + 0 + 2 ln 2)
M1 A1
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
(14)
Total
(72)
 Solomon Press
C4B MARKS page 3
Performance Record – C4 Paper B
Question no.
Topic(s)
1
2
3
4
5
6
7
8
9
Total
integration algebraic differentiation binomial parametric vectors differential differentiation integration,
division
series equations
equation
partial
fractions
Marks
4
4
6
7
8
Student
 Solomon Press
C4B MARKS page 4
9
9
11
14
72