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Paper 1
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ADDITIONAL MATHEMATICS
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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
General Certificate of Education Ordinary Level
4037/01, 0606/01
October/November 2009
MARK SCHEME
Maximum Mark : 80
IMPORTANT NOTICE
Mark Schemes have been issued on the basis of one copy per Assistant examiner and two copies per Team
Leader.
[Turn over
November 2009 4037/1 Markscheme
1(i) 2a 3 − 7a 2 + 7 a 2 + 16 = 0
leading to a 3 = −8 , a = −2
Page 1 of 4
M1 for use of x = a
M1
A1
[2]
(ii)
3
2 (i)
⎛6
⎜
⎜3
⎜2
⎜⎜
⎝1
M1 for substitution of x = −
2
⎛ 1⎞
⎛ 1⎞
⎛ 1⎞
2 ⎜ − ⎟ − 7 ⎜ − ⎟ − 14 ⎜ − ⎟ + 16
⎝ 2⎠
⎝ 2⎠
⎝ 2⎠
= 21
3
2
5
2
1
4
5
2
(ii)
2 ⎞ ⎛ 5 ⎞ ⎛ 43 ⎞
⎟⎜ ⎟ ⎜ ⎟
3 ⎟ ⎜ 3 ⎟ ⎜ 32 ⎟
=
0 ⎟ ⎜ 2 ⎟ ⎜ 35 ⎟
⎟⎜ ⎟ ⎜ ⎟
7 ⎟⎠ ⎜⎝ 1 ⎟⎠ ⎜⎝ 22 ⎟⎠
M1
1
2
A1
[2]
B1,B1
[2]
B1 for each matrix, must be in
correct order
B2,1,0
-1 for each error
[2]
3 4 ( 2k + 1) = 4 ( k + 2 )
2
4k 2 + 3k − 1 = 0
1
leading to k = ,−1
4
M1
A1
M1 for use of ‘ b 2 − 4ac ’
Correct quadratic equation
M1
A1
M1 for correct attempt at solution
A1 for both l values
[4]
4 (13 − 3 y ) + 3 y 2 = 43
2
(13 − x )
+
M1 for eliminating one variable
A1
A1 for correct quadratic
M1
M1 for correct attempt at solving
quadratic
A1,A1
A1 for each correct pair
2
= 43 )
3
6 ( 2 y 2 − 13 y + 21) = 0
(or x
2
M1
(or 2 ( 2 x − 13x + 20 ) = 0 )
2
( 2 y − 7 )( y − 3) = 0
(or ( 2 x − 5 ) ( x − 4 ) = 0
7 ⎛
5
⎞
⎜ x = or 4 ⎟
2 ⎝
2
⎠
5 ⎛
7
⎞
( or x = 4 or
⎜ y = or 3 ⎟ )
2 ⎝
2
⎠
y = 3 or
(
) (
2
)
2
5 (i) 3 + 2 + 3 − 2 = 22
AC =
22
[5]
M1 for use of Pythagoras
M1
A1
[2]
3− 2
(ii) tan A =
3+ 2
(
(
)
)
(3 − 2 ) 3 − 2
=
(3 + 2 ) 3 − 2
11 − 6 2
7
M1
M1 for correct ratio
M1, A1
M1 for rationalising
[3]
November 2009 4037/1 Markscheme
6 (i)
3x 2 − 10 x − 8 = 0
(3x + 2)(x − 4) = 0
2
critical values − ,4
3
2
A = {x : − ≤ x ≤ 4}
3
(ii)
B = {x : x ≥ 3}
A ∩ B = {x : 3 ≤ x ≤ 4}
7 (i)
Page 2 of 4
M1
M1 for attempt to solve quadratic
A1
A1 for critical values
A1
[3]
B1
B1
[2]
M1, A1
[2]
C8 = 1287
13
(ii) 7 teachers, 1 student : 6
6 teachers, 2 students 7C6 ×6 C2 :105
5 teachers, 3 students 7C5 ×6 C3 :420
531
B1 for values of x that define B.
B1 for attempt to combine the sets
correctly and correct use of notation
M1 for correct C notation
B1
B1
B1
B1
[4]
8 (i) When t = 0, N = 1000
B1
[1]
dN
= −1000ke − kt
dt
dN
when t = 0,
= −20 leading to
dt
1
k=
50
(iii) 500 = 1000e − kt
(ii)
t = −50 ln
1
leading to 34.7 mins
2
9 (i) 20 × −2 (1 − 2 x )
19
M1 for differentiation
M1
B1
B1 for
A1
[3]
M1
M1 for attempt to formulate
equation using half life
M1 for a correct attempt at solution
M1
A1
[3]
B1,B1
[2]
(ii) x 2
1
+ 2 x ln x
x
dN
= −20 stated
dt
B1 for 20 and (1 − 2 x )
B1 for -2
19
M1 for attempt to differentiate a
product.
1
B1 for
x
M1
B1
A1
[3]
(iii)
x ( 2sec ( 2 x + 1) ) − tan ( 2 x + 1)
2
x2
M1 for attempt to differentiate a
quotient.
B1 for differentiation of tan ( 2 x + 1)
M1
B1
A1
[3]
November 2009 4037/1 Markscheme
dy
= 9x2 − 4x + 2
dx
at P grad = 7
tangent y − 3 = 7( x − 1)
10 (i)
Page 3 of 4
M1
A1
M1
A1
M1 for differentiation and attempt
to find gradient
M1 for attempt to find tangent
equation, allow unsimplified
[4]
(ii) at Q , 7 x − 4 = 3 x 3 − 2 x 2 + 2 x
leading to 3 x3 − 2 x 2 − 5 x + 4 = 0
(x − 1)(3x 2 + x − 4) − 0
M1
B1,M1
(x − 1)(3x + 4)(x − 1) = 0
M1
4
40
leading to x = − , y = −
3
3
A1
sin θ cos θ
+
cos θ sin θ
sin 2 θ + cos 2 θ
=
cos θ sin θ
1
=
cos θ sin θ
11 (a) tan θ + cot θ =
= cos ecθ sec θ
(b)(i) tan x = 3sin x
sin x
= 3sin x
cos x
sin x − 3sin x cos x = 0
1
leading to cos x = , sin x = 0
3
o
o
x = 70.5 , 289.5 and x = 180o
[5]
M1
2 cos ec 2 y + 3cos ecy − 2 = 0
( 2 cos ecy − 1) ( cos ecy + 2 ) = 0
1
7π 11π
leading to sin y = − , y = ,
2
6 6
M1 for attempt to get in terms of sin
and cos and attempt to get one
fraction
M1 for use of identity
M1
A1
[3]
M1
M1 for use of tan x =
A1√A1
B1
√A1 on their x = 70.5o
B1 for x = 180o
sin x
cos x
and correct attempt to solve
[4]
(ii) 2 cot 2 y + 3 cos ecy = 0
2 ( cos ec 2 y − 1) + 3cos ecy = 0
M1 for equating tangent and curve
equations
B1 for realising ( x – 1) is a factor
and attempt to factorise
M1 for factorisation and attempt to
solve quadratic
A1 for both
M1
M1
M1
A1,A1
[5]
M1 for use of correct identity
M1 for attempt to solve quadratic
M1 for dealing with cosec
November 2009 4037/1 Markscheme
12 EITHER
(i) πr 2 h = 1000 , leading to
1000
h=
πr 2
(ii) A = 2 πrh + 2 πr 2
leading to given answer
2000
A = 2 πr 2 +
r
dA
2000
(iii)
= 4πr − 2
dr
r
dA
2000
when
= 0, 4πr = 2
dr
r
leading to r = 5.42
d2 A
4000
= 4π + 3
2
dr
r
Page 4 of 4
M1
M1 for attempt to use volume
A1
[2]
M1
A1
M1for attempt to use surface area
GIVEN ANSWER
[2]
M1
A1
DM1
M1 for attempt to differentiate and
set to 0
DM1 for solution
A1
[4]
M1
M1 for second derivative method or
gradient method’
+ ve when r = 5.42 so min value
A1
Amin=554
A1 for minimum, can be given if r
incorrect but + ve
A1
12 OR (i) y = x + cos 2 x
dy
= 1 − 2 sin 2 x
dx
dy
1
when
= 0 , sin 2 x =
dx
2
π 5π
leading to x = ,
12 12
M1
(iv)
5π
12
(ii) Area =
∫ x + cos 2 x.dx
π
12
[3]
M1 for attempt to differentiate
A1
M1
M1
A1,A1
[6]
M1 for setting to 0 and attempt to
solve
M1 for correct order of operations
M1
M1 for attempt to integrate
A1,A1
DM1
A1for each term correct
DM1 for correct use of limits
(Trig terms cancel out)
5π
⎡ x2 1
⎤ 12
= ⎢ + sin 2 x ⎥
⎣2 2
⎦π
12
=
π2
12
A1
[5]