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6665/01
Edexcel GCE
Core Mathematics C3
Gold Level (Hard) G1
Time: 1 hour 30 minutes
Materials required for examination
Mathematical Formulae (Green)
Items included with question papers
Nil
Candidates may use any calculator allowed by the regulations of the Joint
Council for Qualifications. Calculators must not have the facility for symbolic
algebra manipulation, differentiation and integration, or have retrievable
mathematical formulas stored in them.
Instructions to Candidates
Write the name of the examining body (Edexcel), your centre number, candidate number, the
unit title (Core Mathematics C3), the paper reference (6665), your surname, initials and
signature.
Information for Candidates
A booklet ‘Mathematical Formulae and Statistical Tables’ is provided.
Full marks may be obtained for answers to ALL questions.
There are 8 questions in this question paper. The total mark for this paper is 75.
Advice to Candidates
You must ensure that your answers to parts of questions are clearly labelled.
You must show sufficient working to make your methods clear to the Examiner. Answers
without working may gain no credit.
Suggested grade boundaries for this paper:
Gold 1
A*
A
B
C
D
E
68
59
50
43
37
31
This publication may only be reproduced in accordance with Edexcel Limited copyright policy.
©2007–2013 Edexcel Limited.
1.
The point P lies on the curve with equation
y = 4e2x + 1.
The y-coordinate of P is 8.
(a) Find, in terms of ln 2, the x-coordinate of P.
(2)
(b) Find the equation of the tangent to the curve at the point P in the form y = ax + b, where a
and b are exact constants to be found.
(4)
June 2008
2.
A curve C has equation
y=
3
,
(5  3 x ) 2
x≠
5
.
3
The point P on C has x-coordinate 2.
Find an equation of the normal to C at P in the form ax + by + c = 0, where a, b and c are
integers.
(7)
June 2010
3.
A curve C has equation y = x2ex.
(a) Find
dy
, using the product rule for differentiation.
dx
(3)
(b) Hence find the coordinates of the turning points of C.
(3)
d2 y
(c) Find
.
dx 2
(2)
(d) Determine the nature of each turning point of the curve C.
(2)
June 2007
Gold 1: 9/12
2
4.
(i) Given that y =
dy
ln( x 2  1)
, find
.
dx
x
(4)
(ii) Given that x = tan y, show that
dy
1
=
.
dx 1  x 2
(5)
January 2010
5.
Figure 1
Figure 1 shows a sketch of the curve C with the equation y = (2x2 − 5x + 2)e−x.
(a) Find the coordinates of the point where C crosses the y-axis.
(1)
(b) Show that C crosses the x-axis at x = 2 and find the x-coordinate of the other point where C
crosses the x-axis.
(3)
(c) Find
dy
.
dx
(3)
(d) Hence find the exact coordinates of the turning points of C.
(5)
June 2010
Gold 1 :9/12
3
6.
(a) Express 3 sin x + 2 cos x in the form R sin (x + α) where R > 0 and 0 < α <

.
2
(4)
(b) Hence find the greatest value of (3 sin x + 2 cos x)4.
(2)
(c) Solve, for 0 < x < 2π, the equation 3 sin x + 2 cos x = 1,
decimal places.
giving your answers to 3
(5)
June 2007
7.
(a) Express 4 cosec2 2θ − cosec2 θ in terms of sin θ and cos θ.
(2)
(b) Hence show that
4 cosec2 2θ − cosec2 θ = sec2 θ .
(4)
(c) Hence or otherwise solve, for 0 < θ < ,
4 cosec2 2θ − cosec2 θ = 4
giving your answers in terms of .
(3)
June 2012
8.
(a) Starting from the formulae for sin (A + B) and cos (A + B), prove that
tan (A + B) =
tan A  tan B
.
1  tan A tan B
(4)
(b) Deduce that
1   3 tan 


tan    =
.
6
 3  tan 

(3)
(c) Hence, or otherwise, solve, for 0  θ  π,
1 + √3 tan θ = (√3 − tan θ) tan (π − θ).
Give your answers as multiples of π.
(6)
January 2012
TOTAL FOR PAPER: 75 MARKS
END
Gold 1: 9/12
4
Question
Number
1.
Scheme
Marks
e2 x1  2
(a)
2x  1  ln 2
x
M1
1
 ln 2  1
2
A1 (2)
dy
 8e 2 x 1
dx
(b)
x
1
 ln 2  1 
2
B1
dy
 16
dx
B1
1


y  8  16  x   ln 2  1 
2


M1
y  16 x  16  8ln 2
A1 ( 4)
(6 marks)
Question
Number
2.
Scheme
At P, y  3
B1

18 
dy
3
 3(2)  5  3x  (3) or
3 
dx
 (5  3x) 
dy
18

  18
dx (5  3(2))3
1
1
m(N) =
or
18
18
1
N: y  3  18 ( x  2)
N:
Marks
M1A1
M1
M1
M1
A1
x  18 y  52  0
[7]
Gold 1 :9/12
5
Question
Number
3.
Scheme
dy
 x 2 e x  2 xe x
dx
dy
 0 , ex(x2 + 2x) = 0
(b) If
dx
[ex  0]
x(x + 2) = 0
(x=0)
x = 0, y = 0
Marks
(a)
(c)
d2 y
 x 2 e x  2 xe x  2 xe x  2e x
2
dx
M1,A1,A1 (3)
setting (a) = 0
and
M1
x = –2
x = –2, y = 4e–2 ( = 0.54…)
A1
A1 √
(3)
  ( x 2  4 x  2)e x 
M1, A1
(2)
d2 y
d2 y
>
0
(=2)
x
=
–2,
< 0 [ = –2e–2 ( = –0.270…)]
2
2
dx
dx
(d)
M1: Evaluate, or state sign of, candidate’s (c) for at least one of candidate’s
x value(s) from (b)
x = 0,
minimum
Gold 1: 9/12
maximum
6
M1
A1 (cso) (2)
(10 marks)
Question
Number
Q4
(i)
Scheme
y 
Marks
ln( x 2  1)
x
u  ln( x 2  1)
M1
du
2x
 2
dx x  1

 u  ln( x 2  1)

Apply quotient rule:  du
2x


 dx x 2  1
A1
v x


dv
 1
dx

 2x 
2
 2
  x   ln( x  1)
dy
x 1
 
dx
x2
M1
A1
(4)
 dy

2
1
 2
 2 ln( x 2  1) 

( x  1)
x
 dx

(ii)
x  tan y
dx
 sec2 y
dy
dy
1

dx
sec2 y
M1*
A1
 cos y
dM1*
2
dy
1

dx
1  tan 2 y
Hence,
dM1*
dy
1

, (as required)
dx
1  x2
A1 AG
(5)
[9]
Gold 1 :9/12
7
Question
Number
5.
(a)
Scheme
Marks
y
 0, 5
M1A1
O
 52 , 0
x
(2)
x  20
(b)
B1
2 x  5   (15  x) ;  x  
M1;A1 oe.
10
3
(3)
(c)
fg(2)  f (3)  2(3)  5 ;  11  11
M1;A1
(d)
g( x)  x2  4x  1  ( x  2)2  4  1  ( x  2)2  3 . Hence g min   3
Either g min   3 or g( x) …  3
or g(5)  25  20  1  6
 3 „ g( x) „ 6 or  3 „ y „ 6
M1
(2)
B1
A1
(3)
[10]
Question
Number
6.
(a)
Scheme
Marks
Complete method for R: e.g. R cos   3 , R sin   2 , R  (3 2  2 2 )
R  13
A1
or 3.61 (or more accurate)
2
3
Complete method for tan  
[Allow tan   ]
3
2
 = 0.588
(Allow 33.7°)
(b) Greatest value =
sin( x  0.588) 
(c)
M1
A1
 13  = 169
4
1
13
(4)
M1, A1
sin(x + their ) =
( = 0.27735…)
( x + 0.588)
(x + 0.588)
1
their R
= 0.281( 03… ) or 16.1°
=  – 0.28103…
Must be  – their 0.281 or 180° – their 16.1°
or (x + 0.588)
= 2 + 0.28103…
Must be 2 + their 0.281 or 360° + their 16.1°
x = 2.273 or x = 5.976 (awrt)
Both
(radians only)
If 0.281 or 16.1° not seen, correct answers imply this A mark
Gold 1: 9/12
M1
8
(2)
M1
A1
M1
M1
A1
(5)
(11 marks)
Question
Number
7. (a)
Scheme
Marks
4
1
 2
2
sin 2 sin 
4
1

 2
2
(2sin  cos ) sin 
4cos ec 2 2  cos ec 2 
B1 B1
(2)
(b)
4
1
4
1
 2 
 2
2
2
2
(2sin  cos ) sin  4sin  cos  sin 
1
cos 2 

sin 2  cos 2  sin 2  cos 2 
sin 2 

sin 2  cos 2 
1

 sec 2 
cos 2 

Using 1  cos 2   sin 2 
M1
M1
M1A1*
(4)
(c)
sec 2   4  sec  2  cos  

 2
3
,
3
1
2
M1
A1,A1
(3)
(9 marks)
Gold 1 :9/12
9
Question
number
Scheme
Marks
M1A1
8. (a)
M1
(
A1 *
(4)
M1
(b)
M1
A1 *
(3)
M1
(c)
M1
M1 A1
θ=
M1
A1
θ=
(6)
(13 marks)
Gold 1: 9/12
10
Statistics for C3 Practice Paper G1
Mean score for students achieving grade:
Qu
1
2
3
4
5
6
7
8
Max
score
6
7
10
9
10
11
9
13
75
Gold 1 :9/12
Modal
score
Mean
%
69
76
73
60
61
62
57
51
62
ALL
4.15
5.30
7.34
5.38
6.05
6.84
5.09
6.63
46.78
A*
6.72
8.91
8.65
12.08
A
B
C
D
E
U
5.31
6.24
9.12
7.70
7.66
9.40
6.98
9.66
62.07
4.48
5.74
7.87
6.17
6.22
7.47
5.04
7.53
50.52
3.77
5.12
6.78
4.83
4.88
5.74
3.59
5.97
40.68
2.97
4.22
5.51
3.74
3.75
3.99
2.44
4.35
30.97
2.12
2.99
4.09
2.42
2.77
2.44
1.60
3.19
21.62
1.01
1.55
2.18
1.21
1.66
0.99
0.76
1.60
10.96
11