FOR OCR
GCE Examinations
Advanced / Advanced Subsidiary
Core Mathematics C3
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 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.
C3 Paper D – Marking Guide
1.
(i)
(ii)
LHS = sin x cos 30 + cos x sin 30 + sin x cos 30 − cos x sin 30
= 2 sin x cos 30
= 3 sin x
[a= 3]
M1 A1
let x = 45,
M1
sin 75 + sin 15 =
3 sin 45
3×
=
2.
6
M1
A1
(ii)
3e2y − 16ey + 5 = 0
(3ey − 1)(ey − 5) = 0
ey = 13 , 5
M1
M1
A1
1
3
(i)
, ln 5
M1 A1
dy
6
= 2ex −
x
dx
x = 1, y = 2e, grad = 2e − 6
∴ y − 2e = (2e − 6)(x − 1)
[ y = (2e − 6)x + 6 ]
1
2
area =
(i)
2
×6
1
2x − 1
∫1
=
1
2
=
A1
M1 A1
−6
2e − 6
3
×
3− e
=
=
3
3− e
9
3− e
M1 A1
M1 A1
(ln 3 − 0) =
= π∫
2
1
(2 x − 1)2
1
1
2
M1 A1
ln 3
M1 A1
dx
= π[ − 16 − ( − 12 )] =
M1 A1
1
3
π
M1 A1
( 3π
, 5)
2
B2
O
(
(
π
, −1)
2
3π
, −5)
2
adding,
(iii)
(8)
y
( π2 , 1)
(ii)
(8)
dx
= π[ − 12 (2x − 1)−1] 12
(i)
(7)
M1
= [ 12 ln2x − 1] 12
(ii)
(6)
x=0 ⇒ y=6
y = 0 ⇒ (2e − 6)x + 6 = 0
x=
5.
M1 A1
2x − 3 = e
x = 12 (e + 3)
(ii)
4.
=
1
2
(i)
y = ln
3.
1
2
A1
x
⇒ −1 = a + b
⇒ −5 = a − b
B1
−6 = 2a ∴ a = −3, b = 2
−3 + 2 cosec x = 0
cosec x = 32 ,
sin x =
2
3
M1
x = 0.73, π − 0.7297
x = 0.73, 2.41 (2dp)
A2
 Solomon Press
C3D MARKS page 2
M1 A1
(8)
6.
(i)
2 cos 2 x
sin x
+
sin 2 x
cos x
cos 2 x
sin x
+
sin x cos x
cos x
LHS ≡
≡
(ii)
7.
(i)
(ii)
M1
≡
cos 2 x + sin 2 x
sin x cos x
A1
≡
(cos 2 x − sin 2 x ) + sin 2 x
sin x cos x
M1
≡
cos 2 x
sin x cos x
≡ cot x ≡ RHS
A1
cot x = 1 + cot2 x − 7
(cot x + 2)(cot x − 3) = 0
M1
M1
A1
M1
≡
cos x
sin x
cot x = cosec2 x − 7,
cot2 x − cot x − 6 = 0,
cot x = −2 or 3
tan x = − 12 or 13
x = π − 0.4636 or 0.32
x = 0.32, 2.68 (2dp)
A2
f(x) > 0
B1
y = 3e
x = 1 + ln
y
3
y
3
M1
x
3
f −1(x) = 1 + ln , x ∈
, x>0
(iii)
f(ln 2) = 3eln 2 − 1 = 3e−1eln 2 = 6e−1
gf(ln 2) = g(6e−1) = 30e−1 − 2
(iv)
f −1g(x) = f −1(5x − 2) = 1 + ln
∴ 1 + ln
x=
(i)
1
5
5x − 2
3
A2
M1 A1
A1
5x − 2
3
5x − 2
= e3
3
= 4,
M1 A1
M1
(3e3 + 2)
dy
= 2x −
dx
1
(4 + ln
2
A1
x)
x = 1, y = −1, grad =
∴ y+1=
7
4
− 12
×
1
x
= 2x −
1
2 x 4 + ln x
7
4
A1
(x − 1)
SP:
2x −
1
2 x 4 + ln x
let f(x) = 2x −
M1
A1
=0
M1
1
2 x 4 + ln x
f(0.3) = −0.40, f(0.4) = 0.088
sign change, f(x) continuous ∴ root
(iii)
2x −
1
2 x 4 + ln x
=0
⇒
M1
A1
2x =
1
2 x 4 + ln x
x2 =
1
4 4 + ln x
=
1
(4 + ln
4
x)
x=
(iv)
(11)
M1 A1
4y + 4 = 7x − 7
7x − 4y = 11
(ii)
(11)
x−1
x − 1 = ln
8.
M1
1
(4 + ln
4
− 12
=
x)
− 12
1
(4 + ln
2
x1 = 0.381512, x2 = 0.378775, x3 = 0.378999,
∴ α = 0.37898 (5dp)
x4 = 0.378981, x5 = 0.378982,
M1
x)
− 14
A1
M1 A1
A1
(13)
Total
(72)
 Solomon Press
C3D MARKS page 3
Performance Record – C3 Paper D
Question
no.
Topic(s)
1
2
3
4
5
6
7
8
Total
trigonometry exponentials differentiation integration functions, trigonometry functions differentiation,
and
trigonometry
numerical
logarithms
methods
Marks
6
7
8
8
8
Student
 Solomon Press
C3D MARKS page 4
11
11
13
72