```FOR OCR
GCE Examinations
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 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 J – Marking Guide
1.
2.
3
= [ 13 (x2 − 4) 2 ] 42
3
2
=
1
3
(12
=
1
3
× ( 2 3 )3 =
(i)
=
M1 A1
− 0)
M1
1
3
×8× 3 3 = 8 3
(2 x − 3)( x + 3)
(2 x − 3)( x − 2)
(ii)
M1 A1
x+3
x−2
=
(5)
M1 A1
2 x2 + 0 x + 4
4
3
2
x − 2 2x + 0x + 0x + 0x − 1
4
3
2 x + 0 x − 4 x2
4 x2 + 0 x − 1
4 x2 + 0 x − 8
7
2
∴ quotient = 2x2 + 4, remainder = 7
3.
(i)
dy
=0
dx
4 cos 2x − sec2 y
4.
(i)
∴ y−
π
3
1
2
×
=
1
2
y−
π
3
=
1
2
y=
1
2
x+
1
4
(x −
x−
M1 A1
=
1
2
B1
π
6
)
M1
π
12
π
4
A1
y
− 12
dy =
∫
k dx
2 y = kx + c
M1 A1
(0, 4) ⇒ 4 = c
∴ 2 y = kx + 4
M1
A1
(2, 9) ⇒ 6 = 2k + 4, k = 1
∴ 2 y = x + 4,
y = 12 (x + 4)
M1
M1
y=
(i)
(ii)
(7)
M1
1
2
5.
(6)
dy
= k y
dx
∫
(ii)
A2
M1 A1
dy
= 4 cos 2x cos2 y
dx
(ii)
M2
1
4
(x + 4)2
x = 0 ⇒ t2 = 2
t≥0 ∴ t= 2
∴ (0, 2 +
y = 0 ⇒ t(t + 1) = 0
t≥0 ∴ t=0
∴ (2, 0)
dx
dy
= −2t,
= 2t + 1
dt
dt
dy
2t + 1
=−
2t
dx
A1
2)
M1 A1
A1
M1
M1 A1
t = 2, x = −2, y = 6, grad = − 54
M1
∴ y − 6 = − 54 (x + 2)
M1
4y − 24 = −5x − 10
5x + 4y − 14 = 0
A1
 Solomon Press
C4J MARKS page 2
(8)
(9)
6.
(i)
(ii)
1 + 3x ≡ A(1 − 3x) + B(1 − x)
x=1
⇒
4 = −2A ⇒
1
x= 3
⇒
2 = 23 B ⇒
=
1
4
∫0
(
3
1 − 3x
−
2
1− x
M1
A1
A1
A = −2
B=3
) dx
1
= [−ln1 − 3x + 2 ln1 − x] 04
1
4
= (−ln
= ln
(iii)
7.
(i)
(ii)
9
16
+ 2 ln
− ln
1
4
3
4
) − (0)
M1
9
4
A1
= ln
f(x) = 3(1 − 3x)−1 − 2(1 − x)−1
(1 − x)−1 = 1 + x + x2 + x3 + …
(1 − 3x)−1 = 1 + 3x + (3x)2 + (3x)3 + … = 1 + 3x + 9x2 + 27x3 + …
∴ f(x) = 3(1 + 3x + 9x2 + 27x3 + …) − 2(1 + x + x2 + x3 + …)
= 1 + 7x + 25x2 + 79x3 + …
1 3
   
 4 .  a
 5  b 
   
= 0 ∴ 3 + 4 a + 5b = 0
sub. (3) ⇒
sub (a) ⇒
t(15 − b) = 28,
12 − a = 15 − b,
,
 −3 
 
1
 −6 
 
 3
 3
 
 −3  ,
 −4 
 
+ 2  −2  =
t=2 ∴ r=
(i)
u = x2, u′ = 2x, v′ = e 2 , v = 2 e 2
1
 
1
1
x
2
= 2 x2 e
(ii)
M1 A1
t=
28
15 − b
A1
b=a+3
∫
−
4x e
1
x
2
− [8x e
1
x
2
1
x
2
8e
− 8x e + 16 e
du
u = sin t ⇒
= cos t
dt
2
2
π
2
1
x
2
1
x
2
1
x
2
I =
∫0
=4∫
2
M1 A1
(12)
M1
d x]
A1
+ c or 2 e
1
x
2
(x2 − 4x + 8) + c
A1
M1
2
B1
2
sin 2t = 4 sin t cos t = 4 sin t (1 − sin t)
1
M1 A1
M1
⇒ u=1
2
M1
A2
, v = 2e
∫
1
x
2
t = 0 ⇒ u = 0, t =
x
dx
1
x
2
−
∴ (3, −3, −4)
1
x
1
x
2
u = 4x, u′ = 4, v′ = e
I = 2 x2 e
28
12 − a
3 + 4a + 5(a + 3) = 0, a = −2, b = 1
(iii)
I = 2 x2 e
t=
1 + 5(3t − 7) = −6 + bt
28
15 − b
=
(12)
B1
M1
t(12 − a) = 28,
15t − 28 = bt,
28
12 − a
B1
M1 A1
M1
A1
M1 A1
4 + s = −3 + 3t
(1)
(2)
1 + 4s = 1 + at
1 + 5s = −6 + bt
(3)
(1) ⇒ s = 3t − 7
sub. (2) ⇒ 1 + 4(3t − 7) = 1 + at
12t − 28 = at,
8.
M1 A1
M1
2
4u (1 − u ) du
1
0
( u 2 − u 4) d u
= 4[ 13 u3 −
= 4[( 13 −
1
5
1
5
A1
u5] 10
) − (0)] =
M1
8
15
M1 A1
(13)
Total
(72)
 Solomon Press
C4J MARKS page 3
Performance Record – C4 Paper J
Question no.
1
Topic(s)
integration
Marks
5
2
3
4
6
rational differentiation differential parametric
partial
expressions,
equation equations fractions,
algebraic
integration,
division
binomial
series
6
7
8
Student
 Solomon Press
C4J MARKS page 4
5
9
12
7
8
Total
vectors integration
12
13
72
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