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8/18/2019 Jawapan MT Kertas 2
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SULIT
3472/2
Additional
MathematicsPaper 2
September
2008
S!T"# P$%U#USA$ A!A&MI!
'A(ATA$ PLA'A#A$ PA)A$%
PP#I!SAA$ P#*U(AA$ SPM
TA)U$ 2008
A&&ITI"$AL MAT)MATI*S
Paper 2
MA#!I$% S*)M
This marking scheme consists of 12 printed pages
8/18/2019 Jawapan MT Kertas 2
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+USTI"$,"#!I$% / S"LUTI"$ MA#!S T"TAL
13 x + 2 y +1 =8
x2 + 9 x - y = 8
P1
7 3
2
x y
−= or
7 2
3
y x
−=
P1
Substitute
7 3
2
x
y
−= or
7 2
3
y
x
−=
into non inear e!uation
2 7 39 " # 8
2
x x x
−+ − = or
$1
27 2 7 2" # 9" # 8
3 3
y y y
− −+ − =
"2%+23#"%-1#=& or "'(-83#"(-2# =& or so)e !uadratic
e!uation using formua or competing the s!uares
$1
% = -11*, % =1 1
( = 2&*7, ( =2 1 -
2 "a# PR = 2 2i j− 1
"b# QS QR RS = + or QS QR QP = +
QS = ' 'i j+
. P R k RS = . P R = 3 ' + 2i j i j− + + +
. 2 / 2" 3 # P R i j i j= + = + =0 . 2 P R RS = =0coinear
P1
1
$1
1
"c#
. 2 1 P R RS = 1 -
3"a# 2
sin cos
2sin cos 1 "2cos 1#
A A
A A A+
+ −,
use of sin2 = 2 sin cos or cos 2 = 2 cos 2 -1
P1
1 1 1
2cos 2cos cos A A A+ = $1
= sec 1
"b# 2 " sec2% -1# = sec % +1, use of 1 + tan2 % = sec2 % P1
" 2 sec % 3# " sec % +1 # =& $1
sec % =
3
2 or sec % = -1 $1
cos % =2
3 or cos % = -1 $1
4 = '8*19 ,18& ,311*81o o o or
=. .'8 11,18& ,311 '9o o o
1 8
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5uestion 6orking Soution arks Tota
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'"a#"i#
"ii#
22"18# 27"22# 32"29# 37" # '2"2#32*7
18 22 29 2
p
p
+ + + +=
+ + + + p = 2/
median =/& '&
29* " #"3'* 29*#29
−+ −
= 32*9
$1
1
$1
1'"b# eight of the bars proportiona to the fre!uenc( or
:abe the o;er and upper boundariesmid
pointscass inter)a correct(*
<orrect ;a( of finding the )aue of mode*
oda mass = 33
$1
$1
17
"a# 23 'dy
x xdx
= +
"1,-1# , 7dy
dx
=
radient of norma =1
7−
>!uation of norma
(-"-1# =1
7− " %-1#
7(+% + /=&
$1
$1
$1
"b# '2'dy
xdx
−= −
'
2'"1*98 2#
2
y
x
δ
δ
−≈ −
&*&3≈
$1
$1
1
-
/"a# =8&&&
8&&&, 8&&&"&*9#, 8&&&"&*9#2,?r =&*9
8&&&
1 &*9 sα = − = 8&,&&&
P1
$1
1
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5uestion 6orking Soution arks Tota
8"c# Ase of distance formua for P or P<
P = 2 2" 1# " 1&# x y or − + − 2 2" 3# " 2# PC x y= + + − $1
Ase 2P< = P, $1
2 2 2 22 " 3# " 2# " 1# " 1&# x y x y+ + − = − + −
2 23 3 2/ ' '9 & x y x y+ + + − = 1 .0
9"a# ( =3% , ( ='-%2
"%+'#"%-1#=& so)e simutaneous e!uation
% =1, % = ''-%2 =&, %= 2± or find the imits of
integration
$1
Ase area of triange =1 3
"1#"3#2 2
= or
Dntegrate
1
&
"3 # x dx∫ or
2
2
1
"' # x dx−∫ 1 22 3
& 1
3'
2 3
x xor x
−
$1
Substitution,8 1
8 '3 3
− − −
$1
=2
13
unit2
dd up 2 area,3 2
12 3
+$1
213
/unit
1
"b#Eoume of cone =
21"3# "1# 3
3π π = or
13
&
9
3
xπ
$1
3π 2
2 2
1
"' # x dxπ −∫
=
23 +
1
8
1/ 3 +
x x
xπ
− +
$1
=
3 + 3 38"2# 2 8"1# 1F 1/"2# 1/"1#
3 + 3 +π
− + − − +
G
$1
5uestion 6orking Soution arks Tota
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=8
31+
π
Eoume generated = 3π +8
31+
π 1
=8
/1
π 1 .0
1&"a# p = &*3 or ! =&*7 P1
i* P"4=#=3& + 2+
+ "&*3# "&*7#C Ase of P"4=r#=
" # " #n r n r
r C p q −
$1
=&*&'/'' 1
ii* P"4@2# = P"4=&# + P"4=1#3& & 3& 3& 1 29
& 1"&*3# "&*7# "&*3# "&*7#C C + $1
=&*&&&9//& or '9*//& 1&−× 1
"b#i P " H @ c # =&*2&2< =-2*& 1
22*&3
µ −− = , use of H = X µ σ − $1
31*1mm µ = 1
iiP"
3& 31*1 32 31*1#
3 3 z
− −< <
=1-&*3&9-&*388 $1
=&*2/&7 1 .0
11"a# rc ength B = 8"&*92# $1
=7*3/
&*92 rad = 2*71 o 2 o '2I
sin "2*71 o # = 8
AD $1
=/*3/'
cos "2*71 o #=8
OD $1
='*8'8
CB = 8- '*8'8 = 3*12
Perimeter =3*12 + /*3/'+7*3/ " add a the sides# $1
=1/*88 1
11"b#tan "2*71 o#=
8
AC $1
=1&*&
rea of J< =1
"8#"1&*
$1
= '2*&1
5uestion 6orking Soution arks Tota
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11"b#rea of sector JB =
21"8# "&*92#
2 use of =
21
2r θ
$1
=29*''
rea of the shaded region 5,
='2*&1-29*'' $1
= 12*7 1 .0
12 "a# '=v P1"b# &,ma% =av
&+2 =−= t a $1
st 2
+=
'2
++
2
+ 2
ma% +
−
=v $1
1
ma%'
12
−−= msv 1
"c# used )@&
2
' &t t − + <
( ) ( ) &'1 <−− t t $1
1 't < < 1
"d# ( )∫ +−= dt t t s '+2
1
&
23
'2
+
3
+−= t
t t +
'
1
23
'2
+
3
+− t
t t
$1$1
Substitute the )aues of t $1
m3
223=
1 .0
13"a# use 1&&
&
1 × P
P
/&*3,1+2,/2*1 === z y x $1111
used∑∑
=w
Iw I $1
( ) ( ) ( ) ( )
1'
312+21+&+1+2'12& +++= I
$1"used I #
79*13/= 1
5uestion 6orking Soution arks Tota
"c# 18&,1'' == R P I I P1
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( ) ( ) ( ) ( )
1'
312+21&8+1+2''*19' +++= I $1
= 1'7*93 1 .0
1' "a# D +&&≤+ y x 1
DD 3 y x≤ 1
DDD 2&&≥ y 1
<annot ha)e sign K=I"b# Jne of graph of straight ine is correct $1
the graph of straight ine are correct $1 The shaded region of L is correct 1
"c# "i# 2&& 1
"ii# ma%imum point "3&&,2&&# based on the
raph
1
( ) ( )2 3&& 2& 2&&+
- substitute an( number based on the )aue in shadedregion
$1
11&& 1 .0
1 "a# used cosine rue
( ) ( ) ( ) ( ) &2228&cos+*12+*1&2+*12+*1& −+=QS $1
QS = 1'*8/ cm 1"b# used sine rue
+*9
3+sin
8/*1'
sin=
R $1
sin L = &*89719
&21*11/=∠QRS 1
Q Q'S
P
"i# <an see an(;here in the diagram
1
"ii# Mind PQS ∠ , used sine rue , hence find.QPQ∠
8/*1'
8&sin
+*12
sin o PQS =∠
o PQS 93*++=∠ ,oQPQ 1'*/8
. =∠
$1
+estion ,orin1 / Soltion Mars Total
Mind area of PQS ∆ or area of . PQQ∆ $1 1
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( )( ) o PQS ara 8&sin+*12+*1&2
1=∆
= /'*/3cm2
( ) ( ) o PQQara 1'*/8sin+*1&+*1&2
1.=∆
= 1*1/ cm2
Mind the area of PS Q .∆
1/*+1/3*/' −= $1
= 13*'7 1 .0
Jr an( other methods
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$mber o l11a1e
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0
-
20
22
24
2-
8
8
%raph or +estion 4b5$mber o l11a1e
.6 24 26 34 36 44
Modal mass9 33
Mass15
!.
$.
$.
.0 . 20 2 30
:
;20
;.
;.0
;0
0
0
.0
.
1&og x
:
:
:
%
%
%
$o7a5
3 40
;2
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raph for 5uestion1'"b#
00
00
L