HL MATH – Monday, 5/4/15
–3–
Spring Break: Paper 1, problems 2, 3, 7 (no calculator)
2.
[Maximum mark: 7]
SPEC/5/MATHL/HP1/ENG/TZ0/XX
9 x 3 45 x 2 74 x 40 0 .
[1 mark]
,
3.
±
–4–
SPEC/5/MATHL/HP1/ENG/TZ0/XX
[6 marks]
[Maximum mark: 6]
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .[3
. . marks]
.
7.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .–. 8. .–. . . . . . . . . . . .SPEC/5/MATHL/HP1/ENG/TZ0/XX
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[3 marks]
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[Maximum mark: 8]
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. ..1.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .x. ..+.. y.. ..+.. z.. .=
. ..3.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. 2.. ..x..+.. 3.. y.. ..+.. z.. .=
x 3y z
.. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
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. . .marks]
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. . .marks]
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HL MATH – Monday, 5/4/15
–8–
Spring Break: Paper 2, problems 6 (with calculator)
6.
[Maximum mark: 6]
f is of the form f ( x)
x
4 and y
x+a
,x
bx + c
c
b
f has
2
,1
3
2
–9–
a , b and c .
7.
SPEC/5/MATHL/HP2/ENG/TZ0/XX
SPEC/5/MATHL/HP2/ENG/TZ0/XX
[Maximum mark: 9]
............................................................................
1
with
. . . .a.constant
. . . . . . .velocity
. . . . . . .of
. . . . . . . . . . . . . . . . . .1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
it moves with a constant velocity of
in a direction
............................................................................
............................................................................
– 13 –
SPEC/5/MATHL/HP2/ENG/TZ0/XX[4 marks]
............................................................................
Do NOT. .write
. . . .solutions
. . . . . . .on
. . this
. . . page.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .[5
. . marks]
..
11.
. . . . . . . .mark:
. . . . .13]
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[Maximum
............................................................................
. . . . . . . . . . . . . .$P
.........................................................
interest
. . . . . rate
. . . .of. .I. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . .$R
............................................
. . $. S. n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . n. th. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Let
.......................................................................
............................................................................
. . . . . . . . . . . . . . . . . . . . .S.1 .and
. . .show
. . . . that
.........................................
............................................................................
. . . . . . . . . . . . . . . . . . . . . . . . . . I. . .2 . . . . . . . . . . .I. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
S2
P 1
R 1
1
.
. . . . . . . . . . . . . . . . . . . . . . . . .100
. . . . . . . . . . . . . .100
................................
.......................................................................
S n . Hence show that
.......................................................................
n
n
. . . . . . . . . . . . . . . S. n. . .P. .1. . . I. . . . . 100
. . . R. . . 1. . . .I . . . . .1. .. . . . . . . . . . . . . . . .[7
. .marks]
........
100
I
100
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in 5 years (i.e. 60 months).
.......................................................................
[6 marks]
& !
"2#
3
[3 marks]
HL MATH – Monday, 5/4/15
Solutions
2.
(a)
sum "
Total [6 marks]
45
40
, product "
9
9
A1
[1 mark]
(b)
it follows that 3! "
solving, ! "
45
40
and ! (! 2 # " 2 ) "
9
9
5
3
A1
5 $ 25
40
2%
& #" !"
3" 9
# 9
1
" " ( %)
3
the other two roots are 2,
A1A1
M1
A1
4
3
A1
[6 marks]
Total [7 marks]
–6–
SPEC/5/MATHL/HP1/ENG/TZ0/XX/M
HL MATH – Monday, 5/4/15
3.
(a)
P (no heads from n coins tossed) ! 0.5n
1 1 1 1 1 1
P (no head) ! " # " # "
3 2 3 4 3 8
7
!
24
(A1)
M1
A1
[3 marks]
(b)
P (2 coins and no heads)
P (no heads)
1
! 12
7
24
2
!
7
P (2 | no heads) !
M1
A1
A1
[3 marks]
Total [6 marks]
4.
(a)
1
E ( X ) ! % 12 x 3 (1 $ x)dx
M1
0
1
& x 4 x5 '
! 12 ( $ )
* 4 5 +0
!
A1
3
5
A1
[3 marks]
(b)
f !( x) ! 12(2 x $ 3x 2 )
at the mode f !( x) ! 12(2 x $ 3 x 2 ) ! 0
therefore the mode !
2
3
A1
M1
A1
[3 marks]
Total [6 marks]
5.
(a)
f ($ x) ! 2cos ( $ x) # ($ x)sin ($ x)
! 2cos x # x sin x
therefore f is even
"!
f ( x) #
M1
A1
A1
[3 marks]
(b)
f !( x) ! $2sin x # sin x # x cos x (! $ sin x # x cos x)
f !!( x) ! $ cos x # cos x $ x sin x (! $ x sin x)
so f !!(0) ! 0
A1
A1
AG
[2 marks]
continued !!
[4 marks]
Total [7 marks]
HL MATH – Monday, 5/4/15
7.
(a)
using row operations,
M1
to obtain 2 equations in the same 2 variables
A1A1
for example y + z " 1
2 y + 2z " ! +1
the fact that one of the left hand sides is a multiple of the other left hand side
indicates that the equations do not have a unique solution, or equivalent
R1AG
[4 marks]
(b)
(i)
! "3
A1
(ii)
put z " "
then y " 1 # "
and x " +2" or equivalent
M1
A1
A1
[4 marks]
Total [8 marks]
–6–
5.
(a)
(i)
3
displacement ! " v dt
0
! 0.703 (m)
(ii)
3
total distance ! " v dt
0
! 2.05 (m)
SPEC/5/MATHL/HP2/ENG/TZ0/XX/M
(M1)
A1
(M1)
A1
[4 marks]
(b)
solving the equation
t
0
cos (u 2 ) du ! 1
(M1)
"
t ! 1.39 (s)
A1
[2 marks]
Total [6 marks]
6.
vertical asymptote x ! #4 $ #4b % c ! 0
1
horizontal asymptote y ! #2 $ ! #2
b
1
b ! # and c ! #2
2
2
%a
1! 3
1 2
# & #2
2 3
a ! #3
M1
M1
A1A1
M1
A1
[6 marks]
HL MATH – Monday, 5/4/15
–7–
SPEC/5/MATHL/HP2/ENG/TZ0/XX/M
7.
(a)
let the interception occur at the point P, t hrs after 12:00
then, SP ! 20t and MP ! 30t
using the sine rule,
SP 2
sin !
! !
MP 3 sin135
whence ! ! 28.1
A1
M1A1
A1
[4 marks]
(b)
using the sine rule again,
MP
sin135
!
MS sin (45 " 28.1255...)
sin135
30t ! 10 #
sin16.8745...
t ! 0.81199...
the interception occurs at 12:49
M1A1
M1
A1
A1
[5 marks]
Total [9 marks]
– 10 –
SPEC/5/MATHL/HP2/ENG/TZ0/XX/M
HL MATH – Monday, 5/4/15
11.
(a)
(i)
I "
!
S1 # P &1 $
!%R
" 100 #
A1
2
I "
I "
!
!
S2 # P &1 $
! % R &1 $
!%R
" 100 #
" 100 #
M1A1
2
I "
! !
I ""
!
# P &1 $
! % R &1 $ &1 $
!!
" 100 #
" " 100 # #
(ii)
AG
extending this,
n
n
! !
I "
I "
I "
!
!
Sn # P &1 $
! % R && 1 $ & 1 $
! $ ... $ & 1 $
!
" 100 #
" 100 #
" " 100 #
n
!!
"
I "
R
1
$
% 1!
&
&
!
n
& " 100 #
!
I "
!
"
#
# P &1 $
! %
I
" 100 #
100
n
n
"
I " 100 R ! !
I "
!
%
1
$
% 1!
= P &1 $
&
!
&
!
!
I &" " 100 #
" 100 #
#
1
"
!!
#
M1A1
M1A1
AG
[7 marks]
(b)
(i)
M1
putting S60 # 0, P # 5000, I # 1
60
60
A1
A1
5000 $ 1.01 # 100 R (1.01 % 1)
R # ($)111.22
(ii)
putting n # 20, P # 5000, I # 1, R # 111.22
20
20
S20 # 5000 $ 1.01 % 100 $ 111.22(1.01 % 1)
# ($)3652
which is the outstanding amount
M1
A1
A1
[6 marks]
Total [13 marks]
HL MATH – Monday, 5/4/15