FOR EDEXCEL
GCE Examinations
Advanced Subsidiary
Core Mathematics C3
Paper L
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 L – Marking Guide
1.
(a)
2( x − 2) + 1
x−2
f(x) =
=2+
1
x−2
M1
x > 2 ∴ f(x) > 2
(b)
ff(x) = f(
=
2(2 x − 3) − 3( x − 2)
(2 x − 3) − 2( x − 2)
=
2.
2x − 3
)
x−2
A1
2( 2xx−−23 ) − 3
2 x −3 − 2
x−2
(a)
4x − 3 = ln 2
x = 14 (3 + ln 2)
(b)
ln (2y − 1) − ln (3 − y) = ln
y=
(a)
(b)
B1
2y −1
3− y
=1
M1
A1
y(e + 2) = 3e + 1
M1
A1
dy
6
= 2ex −
x
dx
x = 1, y = 2e, grad = 2e − 6
∴ y − 2e = (2e − 6)(x − 1)
(7)
M1
[ y = (2e − 6)x + 6 ]
A1
M1 A1
x=0 ⇒ y=6
y = 0 ⇒ (2e − 6)x + 6 = 0
−6
2e − 6
3
×
3− e
1
2
×6
=
=
3
3− e
9
3− e
M1 A1
M1 A1
=
( x − 10)(2 x − 1) − ( x − 8)( x + 4)
( x − 3)( x + 4)(2 x − 1)
M1 A1
=
x 2 − 17 x + 42
( x − 3)( x + 4)(2 x − 1)
A1
=
( x − 14)( x − 3)
( x − 3)( x + 4)(2 x − 1)
x − 14
( x + 4)(2 x − 1)
= 1,
=
x − 14
( x + 4)(2 x − 1)
x − 14 = 2x2 + 7x − 4
M1
x + 3x + 5 = 0
b2 − 4ac = 9 − 20 = −11
b2 − 4ac < 0 ∴ no real roots
tan x + tan 45
− tan x
1 − tan x tan 45
tan x + 1
= 4 + tan x
1 − tan x
A1
M1
A1
=4
(9)
M1
A1
tan x + 1 = (4 + tan x)(1 − tan x)
tan2 x + 4 tan x − 3 = 0
M1
A1
tan x = −4 ± 16 + 12 = −2 ±
M1
2
(8)
M1 A1
2
5.
(6)
M1
M1 A1
3e + 1
e+2
area =
(b)
M1 A1
=e
x=
(a)
=x
2x − 3
x−2
f −1(x) =
2y − 1 = e(3 − y),
4.
4 x − 6 − 3x + 6
2x − 3 − 2x + 4
(c)
2y −1
3− y
3.
=
M1
7
x = 180 − 77.9, −77.9 or 32.9, −180 + 32.9
x = −147.1, −77.9, 32.9, 102.1 (1dp)
B1 M1
A2
(9)
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© Science Exam Papers
6.
(a)
y
y = 3x + 5a
B3
y = x − a
(0, 5a)
B3
O
( − 53
a, 0)
(a, 0)
(0, −a)
(−a, 0)
(b)
x
⇒ x = − 32 a
−x − a = 3x + 5a
M1 A1
−x − a = −(3x + 5a) ⇒ x = −2a,
7.
(a)
x
2
cos x ≡ cos2
x
2
− sin2
cos x ≡ (1 − sin2
cos x ≡ 1 − 2 sin2
LHS ≡
tan
x
2
2 tan2
tan
x
2
8.
(a)
x
2
x
2
x
2
M1 A1
(10)
) − sin2
M1
x
2
M1
x
2
A1
1 − (1 − 2sin 2 2x )
M1
2sin 2x cos 2x
2sin 2
≡
(c)
x = −2a, − a
cos (A + B) ≡ cos A cos B − sin A sin B
let A = B =
(b)
3
2
2sin
x
2
x
cos 2x
2
= 2 sec2
x
2
− tan
= −1 or
x
2
x
2
sin 2x
≡ tan
x
2
≡ RHS
− 5,
tan
x
2
= 2(1 + tan2
− 3 = 0,
(2 tan
≡
cos 2x
x
2
M1 A1
− 3)(tan
x
2
x
2
)−5
M1
+ 1) = 0
M1
3
2
A1
= 135 or 56.310
B1
x = 112.6° (1dp), 270°
A2
dy
= 2 × e−x + (2x + 3) × (−e−x) = −(2x + 1)e−x
dx
SP: −(2x + 1)e−x = 0
1
x = − 12 ∴ ( − 12 , 2e 2 )
(12)
M1 A1
M1 A1
(b)
x = 0, y = 3, grad = −1, grad of normal = 1
∴ y=x+3
(c)
x + 3 = (2x + 3)e−x
x + 3 − (2x + 3)e−x = 0
let f(x) = x + 3 − (2x + 3)e−x
f(−2) = 8.4, f(−1) = −0.72
sign change, f(x) continuous ∴ root
M1
A1
(d)
x1 = −1.1619, x2 = −1.2218, x3 = −1.2408, x4 = −1.2465 = −1.25 (2dp)
M1 A2
(e)
f(−1.255) = 0.026, f(−1.245) = −0.016
sign change, f(x) continuous ∴ root
M1
A1
(14)
Total
(75)
M1
A1
M1
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© Science Exam Papers
Performance Record – C3 Paper L
Question
no.
Topic(s)
Marks
1
2
3
4
5
6
7
8
Total
functions exponentials differentiation rational trigonometry functions trigonometry differentiation,
expressions
numerical
and
methods
logarithms
6
7
8
9
9
10
12
14
75
Student
 Solomon Press
C3L MARKS page 4
© Science Exam Papers