1. No category

SPEC/5/MATME/SP1/ENG/TZ0/XX/M
MARKSCHEME
SPECIMEN
MATHEMATICS
Standard Level
Paper 1
11 pages
–5–
SPEC/5/MATME/SP1/ENG/TZ0/XX/M
SECTION A
1.
(a)
1
AE ! AD
2
A1
attempt to find AD
M1
e.g. AB ! BD , u ! v
1
AE # (u ! v )
2
1 !
" 1
"# u ! v#
2 %
$ 2
A1
N2
[3 marks]
(b)
1
ED # AE # (u ! v )
2
A1
DC # 3v
A1
attempt to find EC
1
e.g. ED ! DC, (u ! v ) ! 3v
2
M1
1
7 " 1
!
EC # u ! v " # (u ! 7v ) #
2
2 $ 2
%
A1
N2
[4 marks]
Total [7 marks]
2.
(a)
min value of r is &1 , max value of r is 1
(b)
C
(c)
linear, strong negative
A1A1
N2
[2 marks]
A1
N1
[1 mark]
A1A1
N2
[2 marks]
Total [5 marks]
–6–
3.
SPEC/5/MATME/SP1/ENG/TZ0/XX/M
(a)
4 (ms 1 )
A1
(b)
recognising that acceleration is the gradient
4!0
e.g. a (1.5) "
2!0
M1
a " 2 (ms 2 )
(c)
A1
recognizing area under curve
e.g. trapezium, triangles, integration
M1
correct substitution
6
1
e.g. (3 # 6) 4, $ v(t ) dt
0
2
A1
distance " 18 (m)
A1
N1
[1 mark]
N1
[2 marks]
N2
[3 marks]
Total [6 marks]
4.
(a)
(b)
(i)
new mean is 20 # 10 " 30
A1
N1
(ii)
new sd is 6
A1
N1
[2 marks]
(i)
new mean is 20 % 10 " 200
A1
N1
(ii)
METHOD 1
A1
A1
N2
variance is 36
new variance is 36 % 100 " 3600
METHOD 2
new sd is 60
new variance is 602 " 3600
A1
A1
N2
[3 marks]
Total [5 marks]
–7–
5.
(a)
SPEC/5/MATME/SP1/ENG/TZ0/XX/M
attempt to use substitution or inspection
du
" ex
e.g. u # 1 ! e x so
dx
M1
correct working
A1
du
" ln u
e.g. #
u
ln (1 ! e x ) ! C
(b)
A1
N3
[3 marks]
METHOD 1
attempt to use substitution or inspection
e.g. let u " sin 3x
du
" 3cos3 x
dx
1
1 u2
u du " $ ! C
#
3
3 2
sin 2 3 x
# sin 3x cos3xdx " 6 ! C
M1
A1
A1
A1
N2
[4 marks]
METHOD 2
attempt to use substitution or inspection
e.g. let u " cos 3 x
du
" %3sin 3x
dx
1
1 u2
% # u du " % $ ! C
3
3 2
cos 2 3 x
# sin 3x cos 3xdx " 6 ! C
M1
A1
A1
A1
N2
[4 marks]
METHOD 3
recognizing double angle
correct working
1
e.g. sin 6 x
2
% cos 6 x
!C
6
1
cos 6 x
# 2 sin 6 xdx " % 12 ! C
# sin 6 xdx "
M1
A1
A1
A1
N2
[4 marks]
Total [7 marks]
–8–
6.
(a)
recognizing double angle
e.g. 3 ! 2sin x cos x , 3sin 2 x
M1
A1A1
a " 3, b " 2
(b)
substitution 3sin 2 x "
3
2
N3
[3 marks]
M1
1
2
finding the angle
#
$#
e.g. , 2 x "
6
6
A1
sin 2 x "
x"
SPEC/5/MATME/SP1/ENG/TZ0/XX/M
A1
5#
12
A1
N2
Note: Award A0 if other values are included.
[4 marks]
Total [7 marks]
7.
(a)
1 &
%
) or ( 2 '
x
(
)
2&
3 %
f ''( x) " 2 x ) or 3 '
( x )
6 &
%
f '''( x) " (6 x 4 ) or ( 4 '
x )
(
% 24 &
f (4) ( x) " 24 x 5 ) or 5 '
( x )
f '( x) " ( x
2
A1
N1
A1
N1
A1
N1
A1
N1
[4 marks]
(b)
f ( n ) ( x) "
((1) n n !
or ((1)n n!* x
xn 1
( n 1)
+
A1A1A1
N3
[3 marks]
Total [7 marks]
–9–
SPEC/5/MATME/SP1/ENG/TZ0/XX/M
SECTION B
8.
(a)
f ( x) ! 3( x 2 " 2 x " 1) # 12
A1
2
A1
AG
! 3 x " 6 x " 3 # 12
! 3x 2 " 6 x # 9
(b)
N0
[2 marks]
(i)
vertex is (#1, # 12)
A1A1
N2
(ii)
y ! #9, or (0, # 9)
A1
N1
(iii)
evidence of solving f ( x) ! 0
e.g. factorizing, formula
M1
correct working
A1
e.g. 3( x " 3)( x # 1) ! 0, x !
#6 $ 36 " 108
6
x ! #3, x ! 1, or (#3, 0), (1, 0)
A1A1
N2
[7 marks]
A1A1A1
N3
(c)
Note: Award A1 for a parabola opening upward,
A1 for vertex in approximately correct position,
A1 for intercepts in approximately correct positions.
Scale and labelling not required.
[3 marks]
(d)
% p & % #1 &
' (!'
(, t ! 3
) q * ) #12 *
A1A1A1
N3
[3 marks]
Total [15 marks]
– 10 –
9.
(a)
(i)
(ii)
number of ways of getting X ! 6 is 5
5
P ( X ! 6) !
36
number of ways of getting X , 6 is 21
P ( X , 6) !
(iii)
21 " 7 !
"! #
36 $ 12 %
P ( X ! 7 | X , 6) !
6 " 2!
"! #
21 $ 7 %
SPEC/5/MATME/SP1/ENG/TZ0/XX/M
A1
A1
N2
A1
A1
N2
A2
N2
[6 marks]
(b)
attempt to find P ( X # 6)
5 21
e.g. 1 $ $
36 36
10
36
fair game if E (W ) ! 0 (may be seen anywhere)
attempt to substitute into E ( X ) formula
P ( X # 6) !
M1
A1
R1
M1
" 5 ! " 21 !
" 10 !
e.g. 3 " # % 1" # $ k " #
$ 36 % $ 36 %
$ 36 %
correct substitution into E (W ) ! 0
A1
" 5 ! " 21 !
" 10 !
e.g. 3 " # % 1" # $ k " # ! 0
$ 36 % $ 36 %
$ 36 %
work towards solving
e.g. 15 % 21 $ 10k ! 0
M1
36 ! 10k
36
k!
(! 3.6)
10
A1
A1
N4
[8 marks]
Total [14 marks]
– 11 –
10.
(a)
(b)
SPEC/5/MATME/SP1/ENG/TZ0/XX/M
A1A1
f !( x) " # sin x $ 3 cos x
(i)
at A, f !( x) " 0
R1
correct working
e.g. sin x " 3 cos x
A1
tan x " 3
% 4%
x" ,
3 3
attempt to substitute their x into f ( x)
A1
N2
[2 marks]
A1
M1
& 4% '
& 4% '
e.g. cos ( ) $ 3 sin ( )
3
* +
* 3 +
(ii)
correct substitution
&
1
3'
e.g. # $ 3 (( #
))
2
* 2 +
A1
correct working that clearly leads to #2
1 3
e.g. # #
2 2
A1
q " #2
AG
correct calculations to find f !( x) either side of x "
4%
3
N0
A1A1
e.g. f !( ! " " # #$ f !%& ! " " $ #
f !( x) changes sign from negative to positive
so A is a minimum
(c)
max when x "
e.g.
N0
[10 marks]
R1
%
into f ( x)
3
A1
& 3'
1
$ 3 ((
))
2
* 2 +
max value is 2
(d)
AG
%
3
correctly substituting x "
R1
r " 2, a "
%
3
A1
N1
[3 marks]
A1A1
N2
[2 marks]
Total [17 marks]