FOR EDEXCEL
GCE Examinations
Advanced Subsidiary
Core Mathematics C3
Paper A
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.
C3 Paper A – Marking Guide
1.
2.
dx
dy
= 2 sec y × sec y tan y + sec2 y = sec2 y(2 tan y + 1) = 2 tan 2y + 1
dy
dx
=1÷
cos y
dx
dy
cos 2 y
2 tan y + 1
=
(a)
= f( 12 ) = − 52
(b)
gf(x) =
∴
M1 A1
=
2
3x − 1
M1 A1
=6
2 = 6(3x − 1)
x = 49
3.
e2y − x + 2 = 0
sub.
⇒
e2y = x − 2
2y = ln (x − 2)
ln (x + 3) − ln (x − 2) − 1 = 0
x+3
x−2
x+3
=
x−2
d
dx
M1
e
A1
x + 3 = e(x − 2)
3 + 2e = x(e − 1)
M1
M1
2e + 3
e −1
(tan x) =
=
=
(b)
dy
dx
x=
, y=
∴ y−
π
2
(a)
1
2
ln (
2e + 3
e −1
− 2) = 0.53 (2dp)
d  sin x 


dx  cos x 
cos x × cos x − sin x × (− sin x)
cos 2 x
cos 2 x + sin 2 x
cos 2 x
π
2
=
1
cos 2 x
= sec2 x
, grad = 2 + π
= (2 + π)(x −
at P, x = 0
∴ y = π2 −
5.
= 4.91 (2dp), y =
= 2 × tan x + 2x × sec2 x = 2 tan x + 2x sec2 x
π
4
π
4
π
4
A2
(8)
M1 A1
M1 A1
M1 A1
B1
)
M1
(2 + π) = − 14 π2
M1 A1
(10)
3 cos x + sin x = R cos x cos α + R sin x sin α
R cos α = 3, R sin α = 1
∴ R = 32 + 12 = 10
tan α = 13 , α = 18.4 (3sf)
∴ 3 cos x° + sin x° =
(b)
(6)
M1
A1
=1
x=
(a)
M1
A1
⇒
ln
4.
(4)
M1 A1
2
(3x − 4) + 3
2
3x − 1
M1 A1
M1 A1
M1 A1
10 cos (x − 18.4)°
2
6 cos x + 2 sin x cos x = 0
2 cos x(3 cos x + sin x) = 0
cos x = 0 or 3 cos x + sin x = 10 cos (x − 18.4) = 0
x = 90, 270 or x − 18.4 = 90, 270
x = 90, 108.4 (1dp), 270, 288.4 (1dp)
 Solomon Press
C3A MARKS page 2
M1
M1
A1
B1 M1
A1
(10)
6.
(a)
(i)
y
(ii)
y
y=2
(0, q)
y=4
(p, 0)
O
M1 A1
M2 A2
x
(p − 1, 0)
x=1
O
x=0
(b)
y = 0 ⇒ 2x − 1 = 0 ⇒ x =
1
2
∴p=
x
1
2
M1 A1
x=0 ⇒ y=1 ∴q=1
(c)
y=
2x − 1
,
x −1
y(x − 1) = 2x − 1
x(y − 2) = y − 1,
(a)
(b)
M1
y −1
y−2
x=
M1
x −1
x−2
∴ f −1(x) =
7.
B1
A1
(i)
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 ]
(ii)
let x = 45, sin 75 + sin 15 =
3 sin 45 =
3×
1
2
=
1
2
6
2(cosec2 y − 1) + 5 cosec y + cosec2 y = 0
3 cosec2 y + 5 cosec y − 2 = 0,
(3 cosec y − 1)(cosec y + 2) = 0
cosec y = −2 or 13 (no solutions)
(12)
M1 A1
A1
M2 A1
M1
M1
A1
sin y = − 12
y = 180 + 30, 360 − 30
y = 210, 330
8.
(a)
2
x +x−6
x2 + 0 x + 1
x4 + x3 − 5 x2 + 0 x
x4 + x3 − 6 x2
x2 + 0 x
x2 + x
− x
∴ f(x) = x2 + 1 +
= x2 + 1 −
(b)
B1 M1
A1
M1 A1
− 9
− 6
− 3
A1
= x2 + 1 −
1
x−2
[A = 1, B = −1, C = −2]
M1 A1
y
y = x2 + 1
y=
1
x−2
O
f(x) = 0 ⇒ x2 + 1 =
(c)
− 9
−x − 3
x2 + x − 6
x+3
( x − 2)( x + 3)
(12)
B2
x
1
x−2
, graphs intersect once ∴ exactly one real root B1
e.g. x0 = 3,
x1 = 2.1, x2 = 2.1848, x3 = 2.1732, x4 = 2.1747
∴ root = 2.17 (3sf)
f(2.165) = −0.37, f(2.175) = 0.016
sign change, f(x) continuous ∴ root
M1 A1
A1
M1
A1
(13)
Total
(75)
 Solomon Press
C3A MARKS page 3
Performance Record – C3 Paper A
Question
no.
Topic(s)
1
2
3
4
5
6
7
8
Total
differentiation functions exponentials differentiation trigonometry functions trigonometry
rational
and
expressions,
logarithms
numerical
methods
Marks
4
6
8
10
Student
 Solomon Press
C3A MARKS page 4
10
12
12
13
75
FOR EDEXCEL
GCE Examinations
Advanced Subsidiary
Core Mathematics C3
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 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.
C3 Paper B – Marking Guide
1.
( x + 3)( x + 4)
(2 x + 1)( x + 4)
x+3
2x + 1
(a)
=
(b)
ln (x2 + 7x + 12) − ln (2x2 + 9x + 4) = 1,
x+3
2x + 1
ln
=
x+3
2x + 1
= 1,
x + 3 = e(2x + 1),
(a)
x 2 + 7 x + 12
2x2 + 9x + 4
1
A1
3 − e = x(2e − 1)
M1
×3=
M1 A1
A1
3
4 5
∴ y− 2 5 =
(x − 3)
M1
4 5 y − 40 = 3x − 9
3x − 4 5 y + 31 = 0
(b)
A1
y− 2 5 =−4
normal:
5
3
(x − 3)
at Q, x = 0 ∴ y − 2 5 = 4 5 ,
3.
(a)
y= 6 5
adding,
P + Q = 2A ⇒ A =
subtracting,
P − Q = 2B ⇒ B =
P+Q
2
2 sin 3x cos 2x = 0
sin 3x = 0 or cos 2x = 0
3x = 0, π, 2π or 2x = π2 ,
π
4
x = 0,
4.
M1
sin (A + B) ≡ sin A cos B + cos A sin B
sin (A − B) ≡ sin A cos B − cos A sin B
adding, sin (A + B) + sin (A − B) ≡ 2 sin A cos B
let P = A + B, Q = A − B
∴ sin P + sin Q ≡ 2 sin
(b)
(a)
(4, 0)
(b)
dy
dx
2π
3
,
cos
P+Q
2
P−Q
2
P−Q
2
3
2
x × ln
− 18
∴ y−0=
x
4
M1
A1
M1
3π
2
A2
5
2
+ x ×
1
x
=
1
2
3
2
x (5 ln
x
4
+ 2)
area =
ln
1
2
×
3
2
x (5 ln
x
4
x
4
1
2
x = 4e
M1 A1
M1
1
2
A1
×4=1
A1
+ 2) = 0
= − 52
M1
− 52
M1 A1
 Solomon Press
C3B MARKS page 2
(9)
A1
(x − 4)
at Q, x = 0, y =
1
2
M1 A1
A1
grad = 8, grad of normal = − 18
(c)
(9)
M1
3π
4
,
M1 A1
B1
5
2
=
π
3
,
(7)
B1
1
−
3
(3x + 11) 2
2
3
4 5
grad =
M1
=e
20 = 2 5
−
1
(3x + 11) 2
2
=
=1
A1
x = 3, y =
dy
dx
ln
3−e
2e − 1
x=
2.
M1 A2
(10)
5.
(a)
= 2[x2 + 2x] + 2 = 2[(x + 1)2 − 1] + 2
= 2(x + 1)2
M1
A1
(b)
translation by 1 unit in negative x direction
stretch by scale factor of 2 in y direction (either first)
B3
(c)
y
2
y = 2(x + 1)2,
x+1=±
y
2
x = −1 ±
y
2
(d)
= (x + 1)2
M1
M1
(domain ⇒ +), ∴ f −1(x) = −1 +
x
2
, x∈
, x≥0
A2
y
y = f(x)
B3
y = f −1(x) is reflection of
y = f(x) in line y = x
−1
y = f (x)
O
6.
x
(13)
(a)
f(x) > −2
B1
(b)
x = 0, y = e − 2 ∴ P (0, e − 2)
y = 0, 0 = e3x + 1 − 2
3x + 1 = ln 2
x = 13 (ln 2 − 1) ∴ Q ( 13 (ln 2 − 1), 0)
B1
M1
M1
A1
(c)
f ′(x) = 3e3x + 1
at P, grad = 3e
∴ y − (e − 2) = 3e(x − 0)
y = 3ex + e − 2
M1
A1
M1
A1
(d)
at Q, grad = 6
tangent at Q: y − 0 = 6(x −
1
3
(ln 2 − 1))
y = 6x − 2 ln 2 + 2
intersect: 3ex + e − 2 = 6x − 2 ln 2 + 2
x(3e − 6) = 4 − e − 2 ln 2
4 − e − 2 ln 2
3e − 6
x=
7.
B1
(a)
arccos θ =
(b)
π
3
,
θ = cos
= −0.0485 (3sf)
π
3
=
1
2
A1
(13)
y
−2
(d)
M1
M1 A1
x+2
B2
y = arccos (x − 1)
B3
y=
(c)
B1
M1
O
2
x
let f(x) = arccos (x − 1) − x + 2
f(0) = 1.7, f(1) = −0.16
sign change, f(x) continuous ∴ root
M1 A1
A1
x1 = 0.83944, x2 = 0.88598, x3 = 0.87233,
x4 = 0.87632, x5 = 0.87515, x6 = 0.87549
∴ α = 0.875 (3dp)
M1 A2
A1
(14)
Total
(75)
 Solomon Press
C3B MARKS page 3
Performance Record – C3 Paper B
Question
no.
Topic(s)
1
2
3
4
rational
differentiation trigonometry differentiation
expressions,
exponentials
and
logarithms
Marks
7
9
9
10
Student
 Solomon Press
C3B MARKS page 4
5
functions
13
6
7
Total
exponentials trigonometry,
and
numerical
logarithms,
methods
differentiation
13
14
75
FOR EDEXCEL
GCE Examinations
Advanced Subsidiary
Core Mathematics C3
Paper C
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.
C3 Paper C – Marking Guide
1.
(a)
(b)
2.
(a)
=
x+4
(2 x + 1)( x + 1)
=
( x + 4) − 2( x + 1)
(2 x + 1)( x + 1)
M1
=
2− x
(2 x + 1)( x + 1)
A1
2− x
(2 x + 1)( x + 1)
=
2
2x + 1
−
M1
1
2
2(2 − x) = 2x2 + 3x + 1
2x2 + 5x − 3 = 0
(2x − 1)(x + 3) = 0
x = −3, 12
M1
A1
if θ =
M1
π
2
M1
, sin θ = 1, cosec θ = 1
∴ cosec θ − sin θ = 1 − 1 = 0
∴ statement is false
(b)
1 − sin θ = 2 sin θ
sin2 θ + 2 sin θ − 1 = 0
M1
2 (no solutions) or −1 +
2
5.
2
M1 A1
θ = 0.4271, π − 0.4271
θ = 0.43, 2.71 (2dp)
M1
A1
(a)
2x − 3 = e
x = 12 (e + 3)
M1
M1 A1
(b)
3e2y − 16ey + 5 = 0
(3ey − 1)(ey − 5) = 0
ey = 13 , 5
M1
M1
A1
y = ln
4.
A1
2
sin θ = −2 ± 4 + 4 = −1 −
3.
1
3
, ln 5
×3=
M1 A1
(a)
=
1
3x − 2
(b)
=
2 × (1 − x ) − (2 x + 1) × (−1)
3
3x − 2
(c)
=
(a)
(i)
(1 − x)2
3
2
3
1
(6)
(7)
(8)
M1 A1
=
x 2 × e2x + x 2 × 2e2x =
3
M1 A2
(1 − x )2
1
2
1
x 2 e2x(3 + 4x)
y
M1 A2
(ii)
(8)
y
3
2
( − , 6)
O
x
B3
(−2, −4) (2, −4)
O
x
M2 A2
(1, −12)
(b)
a = 4, b = 2
B2
 Solomon Press
C3C MARKS page 2
(9)
6.
(a)
4 − ln 3x = 0,
(b)
ln 3x = 4,
x=
1
3
e4
y
( 13 e4, 0)
O
(c)
7.
(a)
B2
x
y = 4 − ln 3x
ln 3x = 4 − y
x = 13 e4 − y
∴ f −1(x) =
(d)
M1 A1
M1
M1
e4 − x
1
3
A1
2−x
fg(x) = 4 − ln 3e
= 4 − (ln 3 + ln e2 − x)
= 4 − ln 3 − (2 − x)
= x + 2 − ln 3
[a = 2, b = 3]
M1
M1
A1
(10)
4 sin x + 3 cos x = R sin x cos α + R cos x sin α
R cos α = 4, R sin α = 3
∴ R = 42 + 32 = 5
tan α = 34 , α = 0.644 (3sf)
M1 A1
M1 A1
∴ 4 sin x + 3 cos x = 5 sin (x + 0.644)
(b)
(c)
8.
(a)
minimum = −5
occurs when x + 0.6435 =
3π
2
5 sin (2θ + 0.6435) = 2
sin (2θ + 0.6435) = 0.4
2θ + 0.6435 = π − 0.4115, 2π + 0.4115
2θ = 2.087, 6.051
θ = 1.04, 3.03 (2dp)
dy
dx
=
1
2
x
− 12
− 4e1 − 4x
∴ y−
SP:
3
2
=
1
2
− 12
x
1
8 x
(c)
(d)
(e)
1
3
(x −
1
4
M1
B1 M1
M1
A2
)
1
3
A1
[ 4x − 12y + 17 = 0 ]
− 4e1 − 4x = 0,
1
2 x
M1 A1
= 4e1 − 4x
= e1 − 4x
M1
8 x = e4x − 1
4x − 1 = ln 8 x
x = 14 (1 + ln 8 x )
M1
A1
x1 = 0.7699, x2 = 0.7372, x3 = 0.7317, x4 = 0.7308 = 0.731 (3dp)
let f(x) =
1
2
x
− 12
(13)
M1
grad = −3, grad of normal =
(b)
, x = 4.07 (3sf)
B1
M1 A1
M1 A2
− 4e1 − 4x
f(0.7305) = −0.00025, f(0.7315) = 0.0017
sign change, f(x) continuous ∴ root
M1
A1
x1 = 6.304, x2 = 1.683 × 1019
diverges rapidly away from root
B2
(14)
Total
(75)
 Solomon Press
C3C MARKS page 3
Performance Record – C3 Paper C
Question
no.
Topic(s)
1
2
3
4
5
6
7
8
Total
rational
trigonometry exponentials differentiation functions functions trigonometry differentiation,
expressions
and
numerical
logarithms
methods
Marks
6
7
8
8
Student
 Solomon Press
C3C MARKS page 4
9
10
13
14
75
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.
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
=
4x −1
2x + 1
6
(2 x + 1)2
2
3
M1 A1
A1
(x + 2)
M1
3y − 9 = 2x + 4
2x − 3y + 13 = 0
A1
 Solomon Press
C3D MARKS page 2
(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)
(10)
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
Performance Record – C3 Paper D
Question
no.
Topic(s)
Marks
1
2
3
4
5
6
7
8
Total
functions trigonometry exponentials trigonometry
rational
differentiation functions differentiation,
and
expressions,
numerical
logarithms
differentiation
methods
5
6
8
9
10
Student
 Solomon Press
C3D MARKS page 4
11
12
14
75
FOR EDEXCEL
GCE Examinations
Advanced Subsidiary
Core Mathematics C3
Paper E
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.
C3 Paper E – Marking Guide
1.
2.
=
x 2 (2 x + 1)
( x + 2)( x − 2)
=
x2
( x + 2)( x − 3)
(a)
x−2
(2 x + 1)( x − 3)
×
M1 A2
M1 A1
sin x
cos x
LHS ≡ 2 sin x cos x −
M1
2
≡ 2sin x cos x − sin x
M1 A1
cos x
2
sin x
≡ sin x(2 cos x − 1) ≡
× cos 2x ≡ tan x cos 2x ≡ RHS
cos x
(b)
3.
M1
A1
B1
M1 A1
(a)
f(1) = 2.30, f(1.5) = −18.5
sign change, f(x) continuous ∴ root
M1
A1
(b)
x2 + 5x − 2 sec x = 0 ⇒
2
cos x
2
x2 + 5x =
cos x =
(a)
(b)
(a)
A1
M1
1
−
1
(1 − cos x) 2
2
× sin x =
=
(ii)
= 3x2 × ln x + x3 ×
dx
=
dy
1
x
1× (3 − 2 y ) − ( y + 1) × (−2)
(3 − 2 y )2
1
5
sin x
2 1 − cos x
(11)
M1 A2
= x2(3 ln x + 1)
M1 A2
5
=
M1
A1
M1 A2
(3 − 2 y )2
(3 − 2y)2
M1 A1
(11)
3 sin θ + cos θ = R sin θ cos α + R cos θ sin α
tan α =
(c)
2
x2 + 5x
f ′(x) = 2x + 5 − 2 sec x tan x
SP: 2x + 5 − 2 sec x tan x = 0
f ′(1.05345) = 0.00046, f ′(1.05355) = −0.0022
sign change, f ′(x) continuous ∴ root ∴ x-coord of P = 1.0535 (5sf)
R cos α =
(b)
∴ g(x) =
x2 + 5x
M1 A2
dy
dx
=1÷
=
dy
dx
5.
M1
x1 = 1.3119, x2 = 1.3269, x3 = 1.3302, x4 = 1.3310 = 1.331 (3dp)
(i)
(10)
M1
x2 + 5x
2
x = arccos
4.
M1 A1
cos x
tan x cos 2x = 2 cos 2x
cos 2x (tan x − 2) = 0
cos 2x = 0 or tan x = 2
2x = 90, 270 or x = 63.4
x = 45°, 63.4° (1dp), 135°
(c)
(5)
1
3
3 , R sin α = 1
, α=
π
6
maximum = 2
occurs when θ +
π
6
2 sin (θ +
3 = 0,
θ +
π
6
π
6
)+
= − π3 , −π +
∴
=
π
3
π
2
∴ R=
3 +1 = 2
3 sin θ + cos θ = 2 sin (θ +
, θ =
)
M1 A1
B1
M1 A1
π
3
sin (θ +
π
6
)=−
3
2
= − π3 , − 2π
3
M1
B1 M1
θ = − 5π6 , − π2
A2
 Solomon Press
C3E MARKS page 2
M1 A1
π
6
(12)
6.
(a)
f(x) ≤ 3
B1
(b)
y
y = f(x)
y = f −1(x)
O
(c)
B3
x
y = 3 − x2
x2 = 3 − y
x = ± 3− y
M1
f −1(x) =
M1 A2
3− x , x ∈ , x ≤ 3
4
3
(d)
= f( ) =
(e)
3− x =
3−x=
11
9
M1 A1
8
3− x
64
M1
(3 − x )2
(3 − x)3 = 64
3−x=4
x = −1
7.
(a)
M1
A1
10k
M1
⇒
A=
sub (2)
⇒
7 = 13e10k × e−60k
7
e−50k = 13
A1
= 0.0124 (3sf)
M1 A1
1
− 50
7
ln 13
10 × 0.01238
∴ A = 13e
(c)
13
(1)
(2)
(1)
∴ k=
(b)
18 = 5 + Ae−10k
12 = 5 + Ae−60k
⇒
⇒
t = 10, T = 18
t = 60, T = 12
(13)
e−10 k
= 13e
= 14.7 (3sf)
M1
A1
T = 5 + 14.71e−0.01238t
dT
= −0.01238 × 14.71 e−0.01238t = −0.1822e−0.01238t
dt
dT
= −0.1822e−0.01238 × 20 = −0.142
when t = 20,
dt
∴ temperature decreasing at rate of 0.142 °C per minute (3sf)
−0.01238(t − 60)
T = 5 + 14.71e
= 5 + 14.71e0.7428 − 0.01238t
= 5 + 14.71e0.7428 × e−0.01238t
B = 30.9 (3sf)
= 5 + 30.9e−0.01238t,
M1 A1
M1
A1
M1
M1
A1
(13)
Total
(75)
 Solomon Press
C3E MARKS page 3
Performance Record – C3 Paper E
Question
no.
Topic(s)
Marks
1
2
rational
trigonometry
expressions
5
10
3
4
6
numerical differentiation trigonometry functions
methods,
differentiation
11
11
Student
 Solomon Press
C3E MARKS page 4
5
12
13
7
Total
exponentials
and
logarithms,
differentiation
13
75
FOR EDEXCEL
GCE Examinations
Advanced Subsidiary
Core Mathematics C3
Paper F
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.
C3 Paper F – Marking Guide
1.
2.
3.
3
sin θ
= −8 cos θ
M1
3 = −8 sin θ cos θ = −4 sin 2θ
sin 2θ = − 34
M1
A1
2θ = 180 + 48.590, 360 − 48.590 = 228.590, 311.410
θ = 114.3, 155.7 (1dp)
M1
A2
(a)
g(x) = (x + a)2 − a2 + 2
∴ g(x) ≥ 2 − a2
M1 A1
A1
(b)
gf(3) = g(1 − 3a) = (1 − 3a)2 + 2a(1 − 3a) + 2
∴ 1 − 6a + 9a2 + 2a − 6a2 + 2 = 7, 3a2 − 4a − 4 = 0
(3a + 2)(a − 2) = 0
a = − 23 , 2
M1
A1
M1
A1
(a)
3x + 1 = e2
x = 13 (e2 − 1)
M1
M1 A1
(b)
consider ln (3x2 + 5x + 3) ≥ 0
⇒
3x2 + 5x + 3 ≥ 1
3x2 + 5x + 2 ≥ 0
(3x + 2)(x + 1) ≥ 0
−1 − 23
M1
A1
∴ if (e.g.) x = − 34 , ln (3x2 + 5x + 3) = ln
∴ if x = − 34 , ln (3x2 + 5x + 3) < 0
4.
(a)
dx
=1×
dy
1 − 2y + y ×
1 − 2y −
=
y
1− 2 y
1
−
1
(1 − 2 y ) 2
2
(1 − 2 y ) − y
1− 2y
=
15
16
= −0.0645...
M1
∴ statement is false
A1
× (−2)
=
1− 3y
1− 2 y
M1
M1 A1
y = −1, x = − 3 , grad =
∴ y+1=
1
4
3 (x +
4y + 4 =
1
4
(a)
3
B1
3)
M1
3x+3
3 x − 4y − 1 = 0
5.
[ p = −4, q = −1 ]
A1
(8)
y
( π6 , 3)
( 7π
, 1)
6
O
x
x=
(b)
(8)
M1 A1
dy
dx
1− 2 y
=
=1÷
1− 3y
dy
dx
(b)
(7)
M1
2
3
x ≤ −1 or x ≥ −
(6)
2 + sec (x −
x−
π
6
=π−
x=
5π
6
3π
2
,
2π
3
x=
π
6
) = 0,
π
3
,π+
sec (x −
π
3
=
2π
3
,
M2 A3
5π
3
π
6
) = −2,
cos (x −
4π
3
) = − 12
M1
B1 M1
A2
 Solomon Press
C3F MARKS page 2
π
6
(10)
6.
(a)
y
(−3, 6)
(3, 6)
(0, 4)
O
(b)
x
y
B3
(c)
y
(−3, 3)
(0, 8)
(0, 2)
O ( 32 , 0) x
M2 A2
( − 92 , 2)
O
7.
(a)
(b)
f(x) = 1 +
y=
x
4x
2x − 5
15
(2 x − 5)( x − 1)
−
=
2 x 2 − 7 x + 5 + 4 x( x − 1) − 15
(2 x − 5)( x − 1)
=
6 x 2 − 11x − 10
(2 x − 5)( x − 1)
3x + 2
,
x −1
M2 A2
=
B1
M1 A1
(3x + 2)(2 x − 5)
(2 x − 5)( x − 1)
=
3x + 2
x −1
M1 A1
y(x − 1) = 3x + 2
M1
x(y − 3) = y + 2
M1
y+2
y−3
x=
x+2
x−3
3( x − 1) + 5
=
x −1
∴ f −1(x) =
f(x) =
A1
3+
5
x −1
M1
x < 1 ∴ f(x) < 3 ∴ domain of f −1(x) is x ∈
8.
, x<3
A1
−1
(c)
f(x) = 2 ⇒ x = f (2) = −4
(a)
dy
= 2x −
dx
1
(4 + ln
2
x)
x = 1, y = −1, grad =
∴ y+1=
7
4
− 12
×
M1 A1
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
1
2 x 4 + ln x
M1
, f(0.3) = −0.40, f(0.4) = 0.088
M1
sign change, f(x) continuous ∴ root
(c)
2x −
1
2 x 4 + ln x
=0
⇒
A1
2x =
1
2 x 4 + ln x
x2 =
1
4 4 + ln x
=
1
(4 + ln
4
x)
x=
(d)
(12)
M1 A1
4y + 4 = 7x − 7
7x − 4y = 11
(b)
(11)
1
(4 + ln
4
− 12
=
x)
− 12
1
(4 + ln
2
M1
x)
x1 = 0.38151, x2 = 0.37877, x3 = 0.37900, x4 = 0.37898 (5dp)
− 14
A1
M1 A2
(13)
Total
(75)
 Solomon Press
C3F MARKS page 3
Performance Record – C3 Paper F
Question
no.
Topic(s)
1
2
3
4
5
6
7
8
Total
trigonometry functions exponentials differentiation trigonometry functions
rational differentiation,
and
expressions,
numerical
logarithms,
functions
methods
proof
Marks
6
7
8
8
10
Student
 Solomon Press
C3F MARKS page 4
11
12
13
75
FOR EDEXCEL
GCE Examinations
Advanced Subsidiary
Core Mathematics C3
Paper G
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.
C3 Paper G – Marking Guide
1.
(a)
(b)
2.
(a)
(b)
dy
= 3(3x − 5)2 × 3 = 9(3x − 5)2
dx
grad = 9
∴ y − 1 = 9(x − 2)
[ y = 9x − 17 ]
A1
M1 A1
9(3x − 5)2 = 9
3x − 5 = ± 1
x = 2 (at P), 43
M1
A1
∴ Q ( 43 , −1)
A1
cos (A + B) ≡ cos A cos B − sin A sin B
cos (A − B) ≡ cos A cos B + sin A sin B
adding, 2 cos A cos B ≡ cos (A + B) + cos (A − B)
M1 A1
2 cos (x +
π
2
) cos (x +
cos (2x +
2π
3
) + cos
π
3
=1
M1
cos (2x +
2π
3
)=1−
1
2
=
1
2
A1
, 2π +
π
3
2π
3
2x +
2x = π,
x=
3.
π
2
,
1
cos x
π
3
= 2π −
π
6
)=1
=
5π
3
,
7π
3
B1
M1
5π
6
A2
× (−sin x) = −tan x
M1 A2
(a)
=
(b)
= 2x × sin 3x + x2 × 3 cos 3x = 2x sin 3x + 3x2 cos 3x
(c)
=
d
−1
[ 6(2 x − 7) 2 ]
dx
(a)
M1 A2
B1
×2=−
22 + 32 =
tan α =
(c)
(9)
6
M1 A2
3
(10)
(2 x − 7) 2
2 sin x − 3 cos x = R sin x cos α − R cos x sin α
R cos α = 2, R sin α = 3
∴ R=
(b)
− 32
(7)
M1
5π
3
= −3(2 x − 7)
4.
M1
3
2
13
M1 A1
, α = 56.3 (3sf)
M1 A1
∴ 2 sin x° − 3 cos x° =
13 sin (x − 56.3)°
cosec x° + 3 cot x° = 2
⇒
1
sin x
⇒
⇒
1 + 3 cos x = 2 sin x
2 sin x° − 3 cos x° = 1
+
3cos x
sin x
=2
B1
13 sin (x − 56.31) = 1
sin (x − 56.31) =
1
13
M1
x − 56.31 = 16.10, 180 − 16.10 = 16.10, 163.90
x = 72.4, 220.2 (1dp)
 Solomon Press
C3G MARKS page 2
B1 M1
A2
(10)
5.
(a)
(b)
let f(x) = 2x3 − x2 + 4x + 15
f( − 32 ) = − 27
− 94 − 6 + 15 = 0 ∴ (2x + 3) is a factor
4
x2 − 2x
2x + 3 2x3 − x2
2x3 + 3x2
2
− 4x
2
− 4x
M1 A1
+ 5
+ 4x + 15
M1 A1
+ 4x
− 6x
10x + 15
10x + 15
∴ f(x) = (2x + 3)(x2 − 2x + 5)
∴
(c)
2x2 + x − 3
3
2 x − x 2 + 4 x + 15
dy
=
dx
SP:
=
(2 x + 3)( x − 1)
(2 x + 3)( x 2 − 2 x + 5)
1× ( x 2 − 2 x + 5) − ( x − 1)(2 x − 2)
2
( x − 2 x + 5)
− x2 + 2 x + 3
( x 2 − 2 x + 5)2
2
M1 A1
− x2 + 2 x + 3
M1 A2
( x 2 − 2 x + 5)2
−(x + 1)(x − 3) = 0
∴ (−1, − 14 ), (3, 14 )
M1
A2
(a)
P = 30 + 50e0.002 × 30 = 83.1
∴ population = 83 100 (3sf)
M1
A1
(b)
30 + 50e0.002t > 84
e0.002t > 54
50
M1
A1
t>
(c)
(d)
7.
x −1
x2 − 2 x + 5
=0
−x2 + 2x + 3 = 0,
x = −1, 3
6.
=
=
(a)
1
0.002
ln 54
, t > 38.5 ∴ 2018
50
30 + 50e0.002t = 26 + 50e0.003t,
e0.003t − e0.002t = 0.08,
e0.001t − 1 = 0.08e−0.002t
0.001t = ln (1 + 0.08e−0.002t)
t = 1000 ln (1 + 0.08e−0.002t)
M1 A1
50e0.003t − 50e0.002t = 4
e0.002t(e0.001t − 1) = 0.08
y
(b, 0)
A1
M1 A2
A1
(ii)
O
M1 A2
( 13 a, 0)
O
(0, 2b)
(c)
(13)
y
x
(0, a)
(b)
M1
M1
t1 = 69.887, t2 = 67.251, t3 = 67.595
∴ 2047
(i)
(12)
x = 0 ⇒ y = −1 ∴ b = −1
y = 0 ⇒ 2 − x+9 = 0
x = 2 2 − 9 = −5 ∴ a = −5
y = 2 − x+9 ,
x+9 = 2 − y
2
x + 9 = (2 − y)
x = (2 − y)2 − 9
∴ f −1(x) = (2 − x)2 − 9
f(−9) = 2 ∴ domain of f −1(x) is x ∈
x
M1 A2
B1
M1 A1
M1
, x≤2
M1 A1
M1 A1
(14)
Total
(75)
 Solomon Press
C3G MARKS page 3
Performance Record – C3 Paper G
Question
no.
Topic(s)
1
2
3
4
differentiation trigonometry differentiation trigonometry
Marks
7
9
10
10
Student
 Solomon Press
C3G MARKS page 4
5
6
7
Total
rational
exponentials functions
expressions,
and
differentiation logarithms,
numerical
methods
12
13
14
75
FOR EDEXCEL
GCE Examinations
Advanced Subsidiary
Core Mathematics C3
Paper H
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.
C3 Paper H – Marking Guide
1.
(a)
(b)
= f(2) = −2
gf(x) = g(2 − x2) =
6 − 3x 2
∴
5(sec2 2θ
5 sec2 2θ
(5 sec 2θ
sec 2θ =
=
3 − 2 x2
x2 =
2.
M1 A1
9
4
1
2
2
3(2 − x )
2(2 − x 2 ) − 1
6 − 3x
=
2
M1 A1
3 − 2 x2
2(6 − 3x2) = 3 − 2x2
,
x = ± 32
,
M1 A1
− 1) − 13 sec 2θ = 1
− 13 sec 2θ − 6 = 0
+ 2)(sec 2θ − 3) = 0
− 52 or 3
M1
M1
A1
cos 2θ = − 52 (no solutions) or
1
3
2θ = 70.529, 360 − 70.529, 360 + 70.529, 720 − 70.529
= 70.529, 289.471, 430.529, 649.471
θ = 35.3°, 144.7°, 215.3°, 324.7° (1dp)
3.
(2 x − 3)( x + 3)
(2 x − 3)( x − 2)
(a)
=
(b)
ln (2x2 + 3x − 9) − ln (2x2 − 7x + 6) = 2,
ln
x+3
x−2
= 2,
2
B1 M1
A2
x+3
x−2
=
x+3
x−2
4.
(a)
2 x2 + 3x − 9
ln
2x2 − 7 x + 6
=2
= e2
M1
A1
e2 − 1
( 3π
, 5)
2
B3
x
( π2 , −1)
(
3π
2
⇒ −1 = a + b
, −5) ⇒ −5 = a − b
adding,
B1
−6 = 2a ∴ a = −3, b = 2
−3 + 2 cosec x = 0, cosec x =
x = 0.73, π − 0.7297,
5.
3
2
,
sin x =
M1 A1
2
3
x = 0.73, 2.41 (2dp)
⇒ 18 000 = 2000e3k, e3k = 9
k = 13 ln 9 = 0.732 (3sf)
M1
A2
(a)
t = 3, N = 18 000
(b)
4000 = 2000e0.7324t
1
t = 0.7324
ln 2 = 0.9464 hours
M1
M1 A1
∴ doubles in 57 minutes (nearest minute)
A1
(c)
(7)
y
O
(c)
M1
A1
2e 2 + 3
( π2 , 1)
(b)
(7)
M1 A2
x + 3 = e (x − 2)
3 + 2e2 = x(e2 − 1)
x=
(6)
N = 2000e0.7324t,
when t = 3,
dN
= 0.7324 × 2000e0.7324t = 1465e0.7324t
dt
M1
M1 A1
M1 A1
dN
= 13 200 ∴ increasing at rate of 13 200 per hour (3sf) A1
dt
 Solomon Press
C3H MARKS page 2
(9)
(10)
6.
(a)
d
d
(sec x) =
[(cos x)−1]
dx
dx
= −(cos x)−2 × (−sin x)
=
sin x
cos 2 x
=
1
cos x
×
M1 A1
sin x
cos x
= sec x tan x
(b)
(c)
7.
A1
dy
= 2e2x × sec x + e2x × sec x tan x = e2x sec x (2 + tan x)
dx
x = 0, y = 1, grad = 2
∴ y = 2x + 1
e2x sec x (2 + tan x) = 0
tan x = −2
x = −1.11 (2dp)
SP:
M1 A1
M1
A1
M1
M1
A1
(11)
(a)
f(x) = (x − 1)2 − 1 + 5 = (x − 1)2 + 4
M1 A1
(b)
f(x) ≥ 4
B1
(c)
y = (x − 1)2 + 4
(x − 1)2 = y − 4
x − 1 = ± y−4
M1
x=1±
y−4
−1
x−4
f (x) = 1 +
(d)
(e)
M1 A1
translation by 4 units in negative x direction
translation by 1 unit in negative y direction (either first)
dy
−1
= 12 ( x − 4) 2
dx
x = 8, y = 3, grad =
(a)
(b)
(c)
1
4
A1
[ y = 35 − 4x ]
dy
= −e2x−2 + ex
dx
M1 A1
M1
A1
e2, grad =
M1
x = 2, y =
y=
3
2
3
4
3
2
2
e =
3
4
(12)
M1 A1
SP: −e2x−2 + ex = 0
let f(x) = −e2x−2 + ex
f(1.3) = −0.70, f(1.4) = 0.29
sign change, f(x) continuous ∴ root
∴ y−
(d)
B2
M1
∴ grad of normal = −4
∴ y − 3 = −4(x − 8)
8.
M1
3
4
e2
2
e (x − 2)
M1
M1 A1
2
ex
∴ x = 0 ⇒ y = 0 so passes through origin
A1
x1 = −1.125589, x2 = −1.125803, x3 = −1.125804 (7sf)
∴ x-coordinate of B = −1.1258 (5sf)
M1 A2
A1
(13)
Total
(75)
 Solomon Press
C3H MARKS page 3
Performance Record – C3 Paper H
Question
no.
Topic(s)
Marks
1
2
functions trigonometry
6
7
3
4
5
7
8
Total
rational
functions, exponentials differentiation functions, differentiation,
expressions, trigonometry
and
differentiation numerical
exponentials
logarithms,
methods
and
differentiation
logarithms
7
9
10
Student
 Solomon Press
C3H MARKS page 4
6
11
12
13
75
FOR EDEXCEL
GCE Examinations
Advanced Subsidiary
Core Mathematics C3
Paper I
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.
C3 Paper I – Marking Guide
1.
=
=
2x
2 x + 3x − 5
2x
(2 x + 5)( x − 1)
=
2.
x2 − x
×
2
×
M1
x3
x( x − 1)
M1 A1
x3
2
x(2 x + 5)
M1 A1
(a)
A (0, 5), B (0, e2)
∴ AB = e2 − 5
B2
B1
(b)
3 + 2ex = ex + 2 = e2ex
3 = ex(e2 − 2)
M1
ex =
3
e2 − 2
,
x = ln
∴ y = e2ex = e2 ×
3.
(a)
(b)
3
2
e −2
3
M1 A1
e2 − 2
=
3e2
M1 A1
2
e −2
2
2
f ′(x) = 2 x × (4 x + 1) − ( x2 + 3) × 4 = 4 x + 2 x −212
(4 x + 1)
4 x 2 + 2 x − 12
(4 x + 1)2
≥0
M1
2(2x − 3)(x + 2) ≥ 0
x ≤ −2 or x ≥
(a)
3
2
dy
2
= 2x − 5 +
x
dx
5
3
−2
M1
A1
grad of normal = − 35
M1
∴ y + 6 = − 35 (x − 3)
M1
5y + 30 = −3x + 9
3x + 5y + 21 = 0
A1
2x − 5 +
SP:
2
x
=0
M1
M1
A1
(a)
= g(5) = log2 16 = 4
M1 A1
(b)
y = log2 (3x + 1)
3x + 1 = 2y
x = 13 (2y − 1)
M1
g−1(x) =
M1 A1
(c)
1
3
(8)
A1
2x2 − 5x + 2 = 0
(2x − 1)(x − 2) = 0
x = 12 , 2
5.
M1 A1
3
2
M1
x = 3, y = −6, grad =
(b)
(8)
M1 A2
(4 x + 1)
for x ≠ − 14 , (4x + 1)2 > 0 ∴ 4x2 + 2x − 12 ≥ 0
4.
(5)
(2x − 1)
fg−1(x) = f [ 13 (2x − 1)] = 2(2x − 1) − 1 = 2(2x) − 3
(8)
M1
x
∴ 2(2 ) − 3 = 2
2x = 52
x=
ln 52
ln 2
or
A1
ln 5 − ln 2
ln 2
M1 A1
 Solomon Press
C3I MARKS page 2
(9)
6.
(a)
cos (A + B) ≡ cos A cos B − sin A sin B
cos (A − B) ≡ cos A cos B + sin A sin B
subtracting, cos (A + B) − cos (A − B) ≡ −2 sin A sin B
let P = A + B, Q = A − B
(b)
7.
(a)
P+Q
2
P−Q
⇒ B=
2
P+Q
P−Q
sin
2
2
adding,
P + Q = 2A ⇒ A =
subtracting,
P − Q = 2B
∴ cos P − cos Q ≡ −2 sin
M1 A1
M1
A1
(cos 5x − cos x) + sin 3x = 0
−2 sin 3x sin 2x + sin 3x = 0
sin 3x(1 − 2 sin 2x) = 0
sin 3x = 0 or sin 2x = 12
3x = 0, 180, 360 or 2x = 30, 150
x = 0, 15, 60, 75, 120
(i)
y
M1
M1
A1
B1
M1 A2
(ii)
(11)
y
B3
O (0, 0)
(0, 0) O (2a, 0)
(b)
= f(3a2) = 9a4 − 6a3
(2a, 0) x
B3
M1 A1
2
gf(x) = 3a(x − 2ax)
∴ 3a(x2 − 2ax) = 9a3
x2 − 2ax − 3a2 = 0
(x + a)(x − 3a) = 0
x = −a, 3a
M1
(a)
f(0.7) = −0.25, f(0.8) = 0.23
sign change, f(x) continuous ∴ root
M1
A1
(b)
f ′(x) = 2 + cos x + 3 sin x
x = 0, y = −3, grad = 3
∴ y = 3x − 3
M1
A1
M1 A1
(c)
cos x + 3 sin x = b cos x cos c + b sin x sin c
b cos c = 1, b sin c = 3
(c)
8.
(−2a, 0)
x
A1
M1
A1
∴ b = 12 + 32 = 10
tan c = 3, c = 1.25 (3sf)
∴ a = 2, b = 10 , c = 1.25
(d)
SP:
2+
(12)
M1
M1
A2
10 cos (x − 1.249) = 0
cos (x − 1.249) = −
2
10
M1
x − 1.249 = π − 0.8861, π + 0.8861 = 2.256, 4.028
x = 3.50, 5.28 (2dp)
M1
A2
(14)
Total
(75)
 Solomon Press
C3I MARKS page 3
Performance Record – C3 Paper I
Question
no.
Topic(s)
1
2
3
4
5
6
7
8
Total
rational exponentials differentiation differentiation functions trigonometry functions
numerical
expressions
and
methods,
logarithms
differentiation,
trigonometry
Marks
5
8
8
8
Student
 Solomon Press
C3I MARKS page 4
9
11
12
14
75
FOR EDEXCEL
GCE Examinations
Advanced Subsidiary
Core Mathematics C3
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 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.
C3 Paper J – Marking Guide
1.
(a)
cos2 x = ( 3 − 1)2 = 3 − 2 3 + 1 = 4 − 2 3
M1
2
(b)
cos 2x = 2 cos x − 1 = 2(4 − 2 3 ) − 1 = 7 − 4 3
M1 A1
2(cos y cos 30 − sin y sin 30) =
M1 A1
3
2
3 cos y − sin y =
3
2
3 cos y =
3
2
tan y =
2.
(a)
f(x) = (x −
∴ f(x) ≥
(b)
(c)
3.
5
2
sin y −
3 (sin y cos 30 − cos y sin 30)
3 cos y
B1
sin y
3 ÷
5
2
3 2
)
2
19
4
−
3
5
=
9
4
3
M1 A1
+ 7 = (x −
3 2
)
2
+
19
4
A1
M1 A1
2
fg(x) = f(2x − 1) = (2x − 1) − 3(2x − 1) + 7
∴ 4x2 − 4x + 1 − 6x + 3 + 7 = 17
2 x2 − 5 x − 3 = 0
(2x + 1)(x − 3) = 0
x = − 12 , 3
x2 + 4x −
4
x4 + x3 − 13x2
x4 − 3x3 + 3x2
4x3 − 16x2
4x3 − 12x2
2
− 4x
2
− 4x
2
x − 3x + 3
∴ f(x) = x2 + 4x − 4 +
M1
A1
M1
A1
+
+
+
+
2x − 5
x 2 − 3x + 3
M1
26x
12x
14x − 17
12x − 12
2x − 5
, A = 4, B = −4, C = 2, D = −5
(a)
x = 1 ⇒ y = −2, grad = 5
∴ grad of normal = − 15
M1
− 15
M1
(x − 1)
dx
=
dy
1
2
sec
y
2
tan
0 ≤ y < π ∴ tan
y
2
A1
y
2
dy
=
dx
x=
π
3
≥0 ∴
1
−
1
(3 + 2 cos x) 2
2
dx
=
dy
1
2
sec
y
2
sec 2
y
2
−1 =
1
2
x x2 − 1
× (−2 sin x) = −
π
3
M1 A1
M1 A1
sin x
3 + 2 cos x
, y = 2, grad = − 14 3
∴ y − 2 = − 14 3 (x −
)
M1 A1
M1 A1
[ 3 3 x + 12y − 24 − π 3 = 0 ]
 Solomon Press
C3J MARKS page 2
(10)
M1
dy
dx
2
=1÷
=
dy
dx
x x2 −1
(b)
A3
M1 A2
( x − 3x + 3)
5y + 10 = −x + 1
x + 5y + 9 = 0
4.
(9)
+ 26x − 17
2
f ′(x) = 2x + 4 + 2 × ( x − 3 x +23) − (2 x −25) × (2 x − 3)
∴ y+2=
(8)
M1 A1
= g(11) = 21
(a)
(b)
1
2
M1 A1
(11)
5.
(a)
(b)
6.
f(x) > 5
B1
2x − 3
y=5+e
2x − 3 = ln (y − 5)
x = 12 [3 + ln (y − 5)]
M1
M1
∴ f −1(x) =
1
2
[3 + ln (x − 5)], x ∈ , x > 5
A2
(c)
x = f −1(7) =
1
2
(3 + ln 2)
M1 A1
(d)
f ′(x) = 2e2x − 3
grad = 4
∴ y − 7 = 4[x −
(a)
LHS ≡
≡
1
2
[ y = 4x + 1 − 2 ln 2 ]
(3 + ln 2)]
2cos 2 x
sin x
+
sin 2 x
cos x
cos 2 x
sin x
+
sin x cos x
cos x
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
cos x
sin x
≡ cot x ≡ RHS
A1
cot x = cosec2 x − 7
cot x = 1 + cot2 x − 7
cot2 x − cot x − 6 = 0
(cot x + 2)(cot x − 3) = 0
cot x = −2 or 3
tan x = − 12 or 13
M1
A1
M1
x = π − 0.4636 or 0.32
x = 0.32, 2.68 (2dp)
A2
(a)
f(x) ≥ 0
B1
(b)
= f(0) = 5
M1 A1
(c)
fg(x) = f [ln (x + 3)] = 2 ln (x + 3) − 5
∴ 2 ln (x + 3) − 5 = 3
2 ln (x + 3) = 2, 8
ln (x + 3) = 1, 4
x = e − 3, e4 − 3
M1
(b)
(11)
M1
≡
≡
7.
M1
A1
M1 A1
M1
(11)
M1
A1
M1 A1
let h(x) = f(x) − g(x)
h(3) = −0.79, f(4) = 1.1
sign change, h(x) continuous ∴ root
M1
A1
(e)
x1 = 3.396, x2 = 3.428, x3 = 3.430, x4 = 3.431
M1 A2
(f)
h(3.4305) = −0.000052, f(3.4315) = 0.0018
sign change, h(x) continuous ∴ root ∴ α = x4 to 4sf
M1
A1
(15)
Total
(75)
(d)
 Solomon Press
C3J MARKS page 3
Performance Record – C3 Paper J
Question
no.
Topic(s)
Marks
1
2
trigonometry functions
8
9
3
4
6
7
Total
rational
differentiation functions,
trigonometry functions,
expressions,
differentiation
exponentials
differentiation
and logs,
numerical
methods
10
11
Student
 Solomon Press
C3J MARKS page 4
5
11
11
15
75
FOR EDEXCEL
GCE Examinations
Advanced Subsidiary
Core Mathematics C3
Paper K
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.
C3 Paper K – Marking Guide
1.
(a)
(b)
arctan (x − 2) = − π3
M1
x − 2 = tan (− π3 ) = − 3
M1
x=2−
A1
3
2
1 − 2 sin θ − sin θ − 1 = 0
2 sin2 θ + sin θ = 0
sin θ (2 sin θ + 1) = 0
sin θ = 0 or − 12
θ = 0 or − π6 , −π +
M1
M1
A1
π
6
θ = − 5π6 , − π6 , 0
2.
(a)
(b)
=
4x
( x + 3)( x − 3)
=
4 x − 2( x − 3)
( x + 3)( x − 3)
=
2x + 6
( x + 3)( x − 3)
=
2
x−3
A2
2
x+3
−
(8)
M1
M1
2( x + 3)
( x + 3)( x − 3)
=
M1
A1
23 − 8 = 0 ∴ (x − 2) is a factor of (x3 − 8)
2
x−2
x + 2x
x3 + 0 x2
x3 − 2 x2
2 x2
2 x2
B1
+ 4
+ 0x − 8
M1 A1
+ 0x
− 4x
4x − 8
4x − 8
∴ x3 − 8 = (x − 2)(x2 + 2x + 4)
∴
3.
(a)
3x − 8 x + 4
=
( x − 2)( x 2 + 2 x + 4)
(3x − 2)( x − 2)
=
−x
−x
2
= 2x × e + x × (−e ) = xe (2 − x)
(c)
=
(b)
cos x × (3 + 2 cos x) − sin x × (−2sin x)
(3 + 2 cos x )2
=
M1 A2
3cos x + 2
(3 + 2 cos x)2
M1 A1
(ex − 3)(ex − 5) = 0
ex = 3, 5
x = ln 3, ln 5
M1 A1
assume log2 3 is rational
B1
∴ log2 3 =
p
q
(9)
M1 A1
where p and q are integers and q ≠ 0
p
⇒
2q = 3
⇒
2p = 3q
2 and 3 are co-prime ∴ only solution is p = q = 0
but q ≠ 0 ∴ contradiction ∴ log2 3 is irrational
 Solomon Press
C3K MARKS page 2
(9)
M1 A1
(3 + 2 cos x)2
3cos x + 2 cos 2 x + 2 sin 2 x
M1 A1
M1 A1
−x
(b)
(a)
x2 + 2 x + 4
3x − 2
= −cosec2 x2 × 2x = −2x cosec2 x2
=
4.
x3 − 8
2
M1
M1
A1
M1
A1
(10)
5.
(a)
(b)
f(x) > 0
y = 3e
B1
x−1
x − 1 = ln
,
x = 1 + ln
x
3
, x>0
M1 A2
(c)
f(ln 2) = 3eln 2 − 1 = 3e−1eln 2 = 6e−1
gf(ln 2) = g(6e−1) = 30e−1 − 2
(d)
f −1g(x) = f −1(5x − 2) = 1 + ln
∴ 1 + ln
x=
(a)
1
5
5x − 2
3
= 4,
M1 A1
A1
5x − 2
3
5x − 2
= e3
3
M1 A1
M1
(3e3 + 2)
A1
2x2 + 3 ln (2 − x) = 0 ⇒
3 ln (2 − x) = −2x2
ln (2 − x) = − 23 x2
2−x= e
− 23 x 2
x=2− e
− 23 x 2
M1
[ k = − 23 ]
x1 = 1.90988, x2 = 1.91212, x3 = 1.91262, x4 = 1.91273
∴ α = 1.913 (3dp)
f(1.9125) = 0.0070, f(1.9135) = −0.020
sign change, f(x) continuous ∴ root
(c)
f ′ ( x) = 4 x +
3
2− x
3
2− x
× (−1) = 4x −
= 0,
4x2 − 8x + 3 = 0,
x = 12 , 32
(a)
(i)
4x =
A1
M1 A1
A1
M1
A1
3
2− x
3
2− x
,
M1 A1
4x(2 − x) = 3
M1
(2x − 3)(2x − 1) = 0
y
(12)
M1
(b)
∴ 4x −
7.
M1
y
3
f −1(x) = 1 + ln , x ∈
6.
y
3
(ii)
M1
A1
(13)
y
(−45, 15)
O
(−135, −1)
B3
x
(135, −1)
O
x
M1 A2
(135, −1)
(b)
2 2 cos x − 2 2 sin x = R cos x cos α − R sin x sin α
R cos α = 2 2 , R sin α = 2 2 ,
tan α = 1, α = 45
∴ f(x) = A + 4 cos (x + 45)°
(c)
(d)
∴ R=
8+8 = 4
3
3 + 4 cos (x + 45) = 0,
M1 A1
B1
B1
cos (x + 45) = −
3
4
x + 45 = 180 − 41.4, 180 + 41.4 = 138.6, 221.4
x = 93.6, 176.4 (1dp)
M1
M1
A2
(14)
Total
(75)
 Solomon Press
C3K MARKS page 3
Performance Record – C3 Paper K
Question
no.
Topic(s)
Marks
1
trigonometry
8
2
3
4
6
7
Total
rational
differentiation exponentials functions
numerical
functions,
expressions
and
methods,
trigonometry
logarithms,
differentiation
proof
9
9
10
Student
 Solomon Press
C3K MARKS page 4
5
12
13
14
75
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.
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)
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
3
3− e
9
3− e
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)
1
2
×6
x − 14
( x + 4)(2 x − 1)
= 1,
=
=
=
M1 A1
x − 14
( x + 4)(2 x − 1)
x − 14 = 2x2 + 7x − 4
2
x + 3x + 5 = 0
b2 − 4ac = 9 − 20 = −11
b2 − 4ac < 0 ∴ no real roots
5.
(6)
M1
M1 A1
3e + 1
e+2
area =
(b)
M1 A1
B1
=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
tan x + tan 45
− tan x
1 − tan x tan 45
tan x + 1
= 4 + tan x
1 − tan x
M1 A1
M1
A1
M1
A1
=4
A1
M1
A1
tan x = −4 ± 16 + 12 = −2 ±
M1
7
x = 180 − 77.9, −77.9 or 32.9, −180 + 32.9
x = −147.1, −77.9, 32.9, 102.1 (1dp)
 Solomon Press
C3L MARKS page 2
(9)
M1
tan x + 1 = (4 + tan x)(1 − tan x)
tan2 x + 4 tan x − 3 = 0
2
(8)
B1 M1
A2
(9)
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
− 3)(tan
M1 A1
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
M1
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)
 Solomon Press
C3L MARKS page 3
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,
and
expressions
numerical
logarithms
methods
6
7
8
9
9
Student
 Solomon Press
C3L MARKS page 4
10
12
14
75