FOR EDEXCEL
GCE Examinations
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 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.
© Science Exam Papers
C3 Paper D – Marking Guide
1.
(a)
(b)
= f(2) = 2 + ln 4
y = 2 + ln (3x − 2),
x = 13 (2 + ey − 2)
f −1(x) =
2.
M1 A1
1
3
3x − 2 = e
y−2
M1
(2 + ex − 2)
M1 A1
3(cosec2 x − 1) − 4 cosec x + cosec2 x = 0
4 cosec2 x − 4 cosec x − 3 = 0
(2 cosec x + 1)(2 cosec x − 3) = 0
cosec x = − 12 or 32
sin x = −2 (no solutions) or
M1
M1
A1
2
3
M1
x = 0.73, π − 0.7297
x = 0.73, 2.41 (2dp)
3.
(a)
(i)
=
y
ln 2
y
ln 2
=
= ln x − ln e = 2 ln x − 1 = 2y − 1
M1 A1
= 4 − (2y − 1)
M1
y = (5 − 2y)ln 2
y(2 ln 2 + 1) = 5 ln 2
y=
M1
5ln 2
2ln 2 + 1
A1
y
x = e = 4.27 (2dp)
4.
(a)
LHS ≡
≡
(b)
5.
(a)
(b)
A1
2sin( x + y ) cos( x − y )
2 cos( x + y ) cos( x − y )
M1 A1
sin( x + y )
cos( x + y )
M1 A1
≡ tan (x + y) ≡ RHS
let x = 30°, y = 22.5° ∴ tan (30 + 22.5) =
tan 52.5 =
3
+ 1
2
2
1+ 1
2
2
=
=
3+ 2
1+ 2
×
=
f(x) = 3 −
3+ 2
1+ 2
+
=
M1
M1
6 −
3 −
2 +2
x + 11
(2 x + 1)( x − 3)
A1
M1 A1
=
4 x 2 − 13x + 3
(2 x + 1)( x − 3)
M1 A1
(4 x − 1)( x − 3)
(2 x + 1)( x − 3)
f ′(x) = 4 × (2 x + 1) − (42x − 1) × 2 =
(2 x + 1)
x = −2 ⇒ y = 3, grad =
2
3
(9)
B1
3(2 x 2 − 5 x − 3) − ( x − 1)(2 x + 1) + ( x + 11)
(2 x + 1)( x − 3)
=
(8)
B1 A1
1− 2
1− 2
3− 6 + 2−2
1− 2
x −1
x−3
sin 60 + sin 45
cos 60 + cos 45
=
∴ y−3=
(6)
M1 A1
2
(ii)
(b)
A2
ln x
ln 2
(5)
=
4x −1
2x + 1
6
M1 A1
(2 x + 1)2
2
3
A1
(x + 2)
M1
3y − 9 = 2x + 4
2x − 3y + 13 = 0
A1
(10)
 Solomon Press
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6.
(a)
(b)
(c)
dy
= 3e3x × cos 2x + e3x × (−2 sin 2x) = e3x(3 cos 2x − 2 sin 2x)
dx
d2 y
dx 2
= 3e3x × (3 cos 2x − 2 sin 2x) + e3x(−6 sin 2x − 4 cos 2x)
M1 A1
= e3x(5 cos 2x − 12 sin 2x)
A1
3x
e (3 cos 2x − 2 sin 2x) = 0
3 cos 2x = 2 sin 2x
tan 2x = 32
SP:
2x = 0.98279,
7.
d y
when x = 0.491,
(a)
dx
M1
M1
x = 0.491 (3sf)
2
(d)
M1 A1
2
= −31.5,
2
d y
dx 2
M1 A1
< 0 ∴ maximum
M1 A1
(11)
y
(0, 4a2)
y = 2x − a
y = 4 a 2 − x2
(0, a)
O ( 12 a, 0) (2a, 0)
(−2a, 0)
(b)
B3
B3
x
4 − x2 = 2 x − 1
x2 + 2x − 5 = 0,
x>
1
2
∴ x = −1 +
M1
x=
−2 ± 4 + 20
2
=
−2 ± 2 6
2
M1
6
A1
2
4 − x = −(2x − 1)
x2 − 2 x − 3 = 0
(x + 1)(x − 3) = 0
x < 12 ∴ x = −1,
8.
(a)
x = −1, −1 +
M1
A1
6
dy
3
6
=2−
×2=2−
2x + 5
2x + 5
dx
grad = −4, grad of normal = 14
∴ y+4=
(b)
M1
1
4
7
4
x−
x+
7
2
7
2
1
4
(x + 2)
[y=
M1
A1
1
4
x−
7
2
]
M1 A1
= 2x − 3 ln (2x + 5)
− 3 ln (2x + 5) = 0,
M1
let f(x) =
7
4
x+
7
2
− 3 ln (2x + 5)
f(1) = −0.59, f(2) = 0.41
sign change, f(x) continuous ∴ root
(c)
(d)
7
4
x+
7
2
(12)
M1
A1
− 3 ln (2x + 5) = 0
7x + 14 − 12 ln (2x + 5) = 0
7x = 12 ln (2x + 5) − 14
x = 127 ln (2x + 5) − 2
M1
A1
x1 = 1.5648, x2 = 1.5923, x3 = 1.6039, x4 = 1.6087, x5 = 1.6107
q = 1.61 (3sf)
f(1.605) = −0.0073, f(1.615) = 0.0029
sign change, f(x) continuous ∴ root ∴ q = 1.61 (3sf)
M1 A1
A1
M1
A1
(14)
Total
(75)
 Solomon Press
C3D MARKS page 3
© Science Exam Papers
Performance Record – C3 Paper D
Question
no.
Topic(s)
Marks
1
2
3
4
5
6
7
8
Total
differentiation functions differentiation,
rational
functions trigonometry exponentials trigonometry
numerical
expressions,
and
methods
differentiation
logarithms
5
6
8
9
10
11
12
14
75
Student
 Solomon Press
C3D MARKS page 4
© Science Exam Papers