EDEXCEL CORE MATHEMATICS C3 (6665)
Question
number
MOCK PAPER MARK SCHEME
Scheme
Marks
2x2 +7x +6 = (x + 2)(2x + 3)
1.
M1 A1
7(3  2 x)
3x 2

(2  x)(3  2 x)
3x 5
=
7
(2  x) x 3
some correct algebraic cancelling
M1 A1
(4)
(4 marks)
2.
x ℝ
(a)
f –1(x) =
(b)
gf –1(x) = g( 12 x) =
(c)
Range gf1(x)  2
1
2
x,
3
4
x2 + 2
B1 B1
(2)
M1 A1
(2)
B1
(1)
(5 marks)
3.
(i)
e2x + 3 = 6
2x + 3 = ln 6
x=
(ii)
M1
1
(ln 6  3)
2
M1 A1
(3)
ln (3x + 2) = 4
3x + 2 = e4
x=
M1
1 4
(e  2)
3
M1 A1
(3)
(6 marks)
1
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EDEXCEL CORE MATHEMATICS C3 (6665)
Question
number
4.
(i)
MOCK PAPER MARK SCHEME
Scheme
u = x3
v = e3x
Marks
du
 3x2
dx
dv
 3e3x
dx
dy
 3x2 e3x + x33e3x or equiv
dx
(ii)
u = 2x
M1 A1 A1 (3)
du
2
dx
v = cos x
dv
 –sin x
dx
dy
2 cos x  2 x sin x

or equiv
dx
cos 2 x
(iii)
(iv)
u = tan x
du
 sec2 x
dx
y = u2
dy
 2u
du
M1 A1 A1 (3)
dy
 2u sec2 x
dx
M1
dy
 2 tan x sec2 x
dx
A1
u = y2
du
 2y
dy
x = cos u
dx
  sin u
du
(2)
M1
dx
  2y sin y2
dy
A1
dy
1

dx 2 y sin y 2
M1 A1
(4)
(12 marks)
2
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EDEXCEL CORE MATHEMATICS C3 (6665)
Question
number
5. (a) (i)
(ii)
(b)
Scheme
Marks
sin (A + B) – sin (A – B)
= sin A cos B + sin B cos A – sin A cos B + sin B cos A
M1
= 2 sin B cos A
A1 cso
(*)
(2)
cos (A – B) – cos (A + B)
= cos A cos B + sin A sin B – cos A cos B + sin A sin B
M1
= 2 sin A sin B
A1 cso
(*)
sin  A  B   sin  A  B  2 sin B cos A
=
cos A  B   sin  A  B 
2 sin A sin B
=
(2)
M1
cos A
sin A
= cot A
(c)
MOCK PAPER MARK SCHEME
A1
(*)
A1 cso
Let A = 75 and B = 15
B1
sin 90 o  sin 60 o
 cot 75 o
o
o
cos 60  cos 90
M1
3
2 =2 3
cot 75 =
1
0
2
M1 A1
(3)
1
(4)
(11 marks)
3
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EDEXCEL CORE MATHEMATICS C3 (6665)
Question
number
(a)
6.
MOCK PAPER MARK SCHEME
Scheme
Marks
y
(0.4, 4)
(–0.5, 2)
O
x
1
x<0
B1 shape
0 < x < 1 B1 shape
x>1
B1 shape
B1 points
(b)
(4)
y
3
x
4
(2.5, –2)
(3.4, –4)
M1
any translation
M1 correct
direction,
translation
B1 points
B1 asymptotes
(4)
y
(c)
–1
0
x
1
B1 shape > 0
B1 shape < 0
(–0.4, –4) (0.4, –4)
B1 points
B1 asymptotes
(4)
(12 marks)
4
© Science Exam Papers
EDEXCEL CORE MATHEMATICS C3 (6665)
Question
number
7.
MOCK PAPER MARK SCHEME
Scheme
(a)
Marks
y
y = ln x
B1 x-intercept
labelled
x
1
O
B1 shape
(2)
(b)
dy 1
1
 so tangent line to (e, 1) is y = x + C
dx x
e
the line passes through (e, 1) so 1 = e
1
+ C and C = 0
e
The line passes through the origin.
y=
y
O
(c)
M1
A1
(3)
y = ln x
x
1
All lines y = mx passing through the origin and having a gradient > 0 lie
above the x-axis.
1
Those having a gradient < will lie below the line.
e
y=
(d)
x
e
M1
x
so it cuts y = ln x between x =1 and x = e.
e
B1
B1
(2)
x0 = 1.86
xn
x1 = e 3 = 1.859
M1
x2 = 1.858
A1
x3 = 1.858
x4 = 1.858
x5 = 1.857
(e)
When x = 1.8575, ln x –
A1
1
x = 0.000 064 8… > 0
3
(3)
M1
When x = 1.8565, ln x = –0.000 140… < 0
A1
Change of sign implies there is a root between.
A1
(3)
(13 marks)
5
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EDEXCEL CORE MATHEMATICS C3 (6665)
Question
number
8.
(a)
MOCK PAPER MARK SCHEME
Scheme
Marks
4 sin  – 3 cos  = R sin  cos   R cos  sin 
sin  terms give 4 = R cos 
cos  terms give 3 = R sin 
(b)
tan  = 0.75
M1
 = 36.9o
A1
R2 = 42 + 32 = 25  R = 5
M1 A1
(4)
5 sin ( – 36.9o) = 3
sin ( – 36.9o) = 0.6
M1
 – 36.9o = 36.9o , 143.1
A1 M1
 = 73.7o , 180
awrt 74 A1 A1
(c)
Max value 5
B1
(d)
sin ( –36.9o) = 1
M1
(5)
(1)
  36.9 = 90
 = 90 + 36.9 = 126.9
A1
(2)
(12 marks)
6
© Science Exam Papers
EDEXCEL CORE MATHEMATICS C3 (6665)
Question
MOCK PAPER MARK SCHEME
AO1
AO2
Q1
Specification
Section
1.1
AO3
AO4
AO5
2
2
4
Q2
1.2
3
2
5
Q3
3.1, 3.2
2
4
6
Q4
4.1, 4.2, 4.3
5
6
Q5
2.3
3
5
Q6
1.3, 1.4
7
5
Q7
3.2, 5.2
2
3
2
3
3
13
Q8
2.1, 2.3
2
2
3
2
3
12
TOTAL
26
29
5
8
7
75
1
3
Totals
12
11
12
7
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