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



+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

SULIT3



'"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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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


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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&*&#2

$1

= '2*&1


SULIT7



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


"c# 18&,1'' == R P I I P1

SULIT8



( ) ( ) ( ) ( )

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



SULIT11

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



SULIT12

raph for 5uestion1'"b#

00

00

L

jawapan mt kertas 2

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