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3472/1 NO. KAD PENGENALAN

Matematik

Tambahan

Kertas 1 ANGKA GILIAN

Okt!ber

2""42 jam

#A$ATAN PENDIDIKAN %A$A&

'#IAN %IMILA 2""4

MATEMATIK TAM$A&AN

Kertas 1

Dua jam

#ANGAN $'KA KETA% ('E%TION% INI

%E&INGGA GI)EN T&ATTA&'

1. Tuliskan angka giliran and nombor kad pengenalan anda pada ruang yang

disediakan.

2. Calon dikehendaki membaca arahan

di halaman 2.

Kertas questions ini mengandungi 16 halaman bercetak.

Kod Pemeriiksa

Questions Marks

Penuh

Marks

Diperoleh

1 2

2 3

3

3

! 3

6 3

" 3

#

$ 3

1%

11 3

12 3

13 31

1!

16 2

1" 3

1#

1$ 2

2% 3

21

22 2

23

2 32!

&umlah

3472/1 ' 2%% (ak )ipta &abatan Pendidikan *abah

*ON+IDENTIAL

*ON+IDENTIAL 3472/1

,Lihat sebe-ah



MAKL'MAT 'NT'K *ALON

1. Kertas questions ini mengandungi 25 questions.

2. Answer a-- questions.

3. Bagi setiap questions berikan %AT' answer sahaa.

. Answer hendaklah ditulis dengan elas dalam ruang yang disediakan dalamkertas questions.

!. Tunukkan langkah!langkah penting dalam kera mengira anda. "ni boleh membantu

anda untuk mendapatkan marks.

6. #ekiranya anda hendak menukarkan answer$ batalkan kera mengira yang telah

dibuat. Kemudian tuliskan answer yang baru.

". %aah yang mengiringi questions tidak dilukiskan mengikut skala kecuali dinyatakan.

#. &arks yang diperuntukkan bagi setiap questions and ceraian questions ditunukkan

dalam kurungan.

$. #atu senarai rumus disediakan di halaman ' hingga 5.

1%. #ebuah buku si(ir matematik empat angka disediakan.

11. Anda dibenarkan menggunakan kalkulator sainti(ik yang tidak boleh diprogram.

12 Kertas questions ini hendaklah diserahkan di akhir peperiksaan.

*ON+IDENTIAL 2 3472/1



+umus,rumus berikut boleh membantu anda men-nser questions. *imbol,simbol /ang gi0en

that adalah /ang biasa digunakan.

he olloing ormulae ma/ be helpul in ansering the questions. he s/mbols gi0en are the

ones commonl/ used.

ALGE$A

1.a

acbb )

2

2 −±−= #.

a

bb

c

ca

log

loglog =

2. am an 4 am5n $. n 4 a 5 n 7 18d

3. am ÷ an 4 am 7 n 1%. 9812:2

d nan

# n −+=

. am8n 4 amn 11. n 4 ar n71

!. logamn 4 logam 5 logan

6. nmn

m

aaa logloglog −= 12. 1;1

81

1

81

≠−−

=−−

= r r

r a

r

r a

#

nn

n

". mnm an

a loglog = 13. 1;1

<−

= r r

a# α

KALK'L'% *AL*'L'%

1. / 4 u0;d)

du*

d)

d*u

d)

dy+= . <uas di baah cur0e

-rea under a cur0e8

2.*

u y = ;

2*

d)

d*u

d)

du*

d)

dy −

= 4 ∫ b

a

yd) or or8

4 ∫ b

a

)dy

3.d)

du

du

dy

d)

dy×= !. =sipadu janaan >olume o re0olution8

4 d) y

b

a

∫ 2

π or or8

4 ∫ b

a

dy )2π

,Lihat sebe-ah


3472/1

3472/1

*ON+IDENTIAL



%TATI%TIK %TATI%TI*%8

1. +

) ) ∑= ". ∑

∑=i

ii

,

" , "

2. ∑∑= (

()

) #. 8?

?

r n

n P r

n

−=

3. 2

228

) +

)

+

) )−=

−= ∑∑

σ $.?8?

?

r r n

nC r

n

−=

.2

228

) (

()

(

) ) ( −=

−=

∑∑

∑∑

σ 1%. P-∪@8 4 P-8 5 P@8 7P-∩@8

11. PA 4 r8 4 r nr r

n q pC − ; p 5 q 4 1

!.C

(

- + .m

m

−+= 2

1

12. min Mean8 4 np

13. npq=σ

6. 1%%1 ×=o

/

/ " 1.

σ

µ −= 0

1

GEOMETI GEOMET08

1. &arak Distance8 . <uas segitiga -rea o a triangle8 4

4 221

221 88 y y ) ) −+− 882

1312312133221 y ) y ) y ) y ) y ) y ) ++−++

2. itik tengah Midpoint8

++

=2

;2

8;6 2121 y y ) ) y ) !. 22 y )r +=

3. itik /ang membahagi suatu tembereng 6. 22B

y )

y )ir

+

+=

garis

- point di0iding a segment o a line8

+

+

+

+=

nm

myny

nm

m)n) y ) 2121 ;8;

TIGONOMETI TIGONOMET0

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1. Panjang lengkok; s 4 jθ #. B A B A B A sinkoskossin8sin ±=±

length o an arc; s 4 r θ B A B A B A sincoscossin8sin ±=±

2. <uas sector; θ 2

2

1 . = $. B A B A B A sinsinkoskos8kos =±

-rea o a sector; θ 2

2

1r A = B A B A B A sinsincoscos8cos =±

3. sin2- 5 kos2- 4 1 1%. B A

B A B A

tantan1

tantan8tan

±=±

sin2- 5 cos2- 4 1

11. A

A A

2tan1

tan22tan

−=

. sek 2- 4 1 5 tan2-

sec2- 4 1 5 tan2- 12.C

c

B

b

A

a

sinsinsin==

!. kosek 2- 4 1 5 kot2-

cosec2- 4 1 5 cot2- 13. a2 4 b2 5 c2 7 2bc kos-

a2 4 b2 5 c2 7 2bc cos-6. sin2- 4 2sin-kos-

sin2- 4 2sin-cos-

1. <uas segitiga -rea o a triangle8

". kos2- 4 kos2- 7 sin2- 4 C absin2

1

4 2kos2- 7 1

4 1 7 2sin2-

cos2- 4 cos2- 7 sin2-4 2cos2- 7 1

4 1 7 2sin2-

Answe a-- questions.

,Lihat sebe-ah*ON+IDENTIAL 6 3472/1

3472/1

*ON+IDENTIALCor eaminer

use



1. Ei0en that a unction ( is deined as ( F ) → 1 5 )2. Cind the 0alue o ( 71 )8.

:2 marks9

1

Answer F GGGGGGGGGGGGG

2. Ei0en that ( )8 4 ) 5 1 and g )8 4 2 ) 5 182 7 ". Cind 821− g( . :3 marks9

2

Answer F GGGGGGG...GGGGG

3. Cind the possible 0alues o m i a straight line y 4 m ) 5 2 is a tangent to thecur0e y 4 )2 5 2 ) 5 6. : marks9

3

Answer F GGGGGGGGGGGGG.

. Ei0en that y 4 )2 5 # ) 7 !. Hsing the completing square method; determinethe maimum or minimum point or the quadratic unction.

:3 marks9

3

2

*ON+IDENTIAL " 3472/1

3472/1

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Cor eamineruse



Answer F GGGGG..GGGGGGGG

!. Cind the range o the 0alue o ) such that 3 )2 5 ! ) < 12. :3 marks9

!

Answer F GGGGGGGGGG...

6. *ol0e #2 ).2 ) 41%2

1. :3 marks9

6

Answer F GGGGGGGGGGGGG.. 3

3

3

3472/1

*ON+IDENTIAL

,Lihat sebe-ah



". *impli/#1log

2"log

2

2 to the simplest orm. :3 marks9

"

Answer F GGGGGGGGGGGGG...

#. Ei0en that point P; #8 and point Q72; !8. Cind the equation o a straightline hich is perpendicular to the straight line PQ and passing through point

6; %8. : marks9

#


$. Diagram 1 shos a triangle -@). = the area o triangle -@) is " unit2.

Cind the 0alue o a. :3 marks9

$

Answer F a 4 .GGGGGGGGGG..

3

3

*ON+IDENTIAL # 3472/1

3472/1 *ON+IDENTIAL

Cor eaminer

use

Diagram1



1%. Ei0en that ;81

82

3

)

) ) (

−= ind the 0alue o 8.1I ( : marks9

1%


11. =n Diagram 2; the minor arc PQ is subtended at an angle o %o at the centre

J ith radii 1# cm. Cind the length o major arc PQ in terms o π . :3marks9

11

11

3

*ON+IDENTIAL $ 3472/1 Cor

aminer use

Diagram 2



Answer F GGGGGGGGGGGG...

12. able 1 shos the mass distribution o a group o students. = the mean othe mass or this distribution is #.3 kg; ind the 0alue o m. :3 marks9

mass kg8 36 7 % 1 7 ! 6 7 !% !1 7 !! !6 7 6% 6% 7 6!

Lumber o

students

6 12 m 1% 3

12

Answer F m 4 .GGGGGGGGGGG

13. Ei0en that lg !; lg 2! and lg 12! are three consecuti0e terms in an -rithmeticProgression. Cind the common dierence and tenth term in this -rithmetic

progression. :3 marks9

,Lihat sebe-ah

3

3

*ON+IDENTIAL 1% 3472/1

3472/1

*ON+IDENTIAL

Cor eamineruse

able 1



13


GGGGGGGGGGGGG..

1. nth term o a geometric progression is 2' 7 n. Cind theF

a8 irst and second term

b8 common ratio and

c8 sum o the number o terms here n is large enough such that %nr » .

: marks9

1

Answer F a8 GGGGGGGGGG.G

b8 GGGGGGGGGGG

c8 GGGGGGGGGGG

1!.

J

Diagram 3 shos part o graph log 3 y against log 3 ). Corm an equationhich relate y ith ). : marks9

log 3 )

(3;"8

log 3 y

K%;728

3

*ON+IDENTIAL 11 3472/1Cor eaminer

use

Diagram 3

3472/1 *ON+IDENTIAL



1!


16. 0aluate ∫ − d) )

) 28

3 :2 marks9

16

Answer F GGGGGGGGGGG.......

1". he gradient unction o a cur0e hich passes through point

1; 38 is 2

32

) ) + . Cind the equation o cur0e.

:3 marks9

,Lihat sebe-ah

2


3472/1 *ON+IDENTIAL

Cor eaminer

use



1"

Answer F GGGGGGGGGGG...

1#. Ei0en that position 0ectors 4P and 4/ are∼∼

+ i 3 and∼∼

+− i 26

respecti0el/. Cind P/ and unit 0ector in the direction o P/ . : marks9

1#

Answer F GGGGGGGGGGGG..

GGGGGGG.....................

1$.

@

J

-

+ *P

Q

∼b

∼a

3


3472/1 *ON+IDENTIAL

Cor eaminer

use

Diagram



Diagram shos 0ector ∼

= a4A and 0ector ∼

= b4B . press the olloing

0ector in terms o∼a and

∼b . :2 marks9

i8 P/

ii8 %# .

1$

Answer F i8 GGGGGGGGGGG..

ii8 GGGGGGGGGGG.

2%. Ei0en that cos ) 4 a and %o N ) N $%o; ind an epression in terms o a or

i8 tan )

ii8 sin7 )8. :3 marks9

2%


21. *ketch the graph y 4 2kos2) 71 or %o N ) N 36%o. : marks9

Answer F

,Lihat sebe-ah

2

3


3472/1 *ON+IDENTIAL

Cor eaminer

use



21

22. (o man/ dierent arrangements o alphabets rom a ord *EMELANG

can be ormed in a ro. :2 marks9

22

Answer F GGGGGGGGGGGGG.

23.

Diagram ! shos a target o to circles ith one centre P and Q hich radii

is r and 2r. Jne participant can score 3 and 1 points i he hits the area P and

Q respecti0el/. = the probabilit/ that participant hits the target is!

; ind

the probabilit/ he scoresa8 % point

b8 3 points

c8 1 point in a shot. : marks9

P Q

2

*ON+IDENTIAL 1! 3472/1

3472/1 *ON+IDENTIAL

Cor eamineruse

Diagram !



23 Answer F a8GGGGGGGGGGG

b8GGGGGGGGGGG

c8.GGGGGGGGGGG

2. he probabilit/ o one student in orm ! o a school able to urther stud/ to

Corm 6 is %.". Crom number o 2"6 students in Corm !; ind the

a8 number o studentsb8 standard de0iation o that number o students ho are able to urther

their studies to Corm 6. :3 marks9

2

Answer F a8 GGGGGGGGGGGG

b8GGGGGGGGGGGG

2!. he height o *abah ootball team pla/ers are normall/ distributed ith the

mean 1.# m and 0ariance %.%$ m2. Cind the probabilit/ o height o a chosen pla/er ithin 1."! m and 1.$3 m. : marks9

,Lihat sebe-ah

3


3472/1 *ON+IDENTIAL

Cor eaminer

use



2!


END O+ ('E%TION PAPE

#A$ATAN PENDIDIKAN %A$A&

'#IAN %IMILA OKTO$E 2""4

ADDITIONAL MAT&EMATI*%

P-P+ 1

N!ta &ana sat ari5aa P1 !r P2 !r P3 !r P4 b!-eh i5er!-ehi.

(esti!ns %!-ti!ns 6 markin s8heme Marks

1. ) y =+ 21 ; or y ) =+ 21 :P19

181

−=−

) ) ( :P29

2

2 1821 =− ( :P19

"812812 −+= g :P29

4 2 :P39

3

3 m) 5 2 4 )2 5 2 ) 56 :P19 )2 5 2 7 m8 ) 5 4 % :P29

2 7 m82 7 188 4 % :P39

m 4 72; m 4 6 :P9

!

112

222

−−++= ) ) y 8 :P19

$81:2 −+= ) y 9 :P29

-nserF 71; 7$8 :P39

3

!. 3 )2 5 ! ) 712 N %

3 ) 7 8 ) 5 38 N % logical actor8 :P19

3

3472/1 *ON+IDENTIAL




3

= ) ; ) 4 73 :P29

-nserF 73 N ) N3

:P39

6. 23

82

2

4 2 71%

:P19

26 5 4 2 71% or " ) 4 71% :P29

) 4"

1%− :P39

3


".

2

2

3

2

3log83log :P19

3log

3log2

3

2

2

= :P29

4#

3 :P39

3

#.Eradient o a straight line PQ 4

2

!#

+−

or2

1 :P19

Eradient o perpendicular line ith PQ 4 72 :P292

6

%−=

−−

)

y or similar 9P39

y 4 72 ) 512 :P9

$."

313

%22%

2

1=

−

a :P19

4 "2662%2

1=−++− a or similar :P29

4 72 :P39

3

1%.

322

81

8812.83818I )

) ) ) ) ) ( −

−−−−= or similar :P29

2

83

83#83381I

−−+−

= ( or similar :P39

42"

1 :P9

3472/1

*ON+IDENTIAL




11.%

$

2

1#%

π π = or 32%o :P19

4 8$

221#

π π − or 81#2

36%

32%π ×

o

o

:P29

4 32π :P39

3


12.

atau

m

m

atau

m

m

31%126

1#$232!3%#!1622#3.#

31%126

363!#1%!3#12363#3.#

++++++++++=

+++++×+×+×++×+×

=

:P19

m

m

++

=3!

#16$!3.# :P29

m 4 1! :P39

3

13. lg !; lg 2!; lg 12!

T 4 lg !; T 2 4 lg 2! 4 2lg ! :P19

d 4 2lg ! 7 lg ! 4 lg ! :P29

T 1% 4 lg ! 5 $lg !

4 1%lg ! :P39

3

1. a8 T n 4 23 7 n

T 4 2371 4 22 4

T 2 4 2372 4 21 4 2 :P19

b8 !.%

2 ==r :P19

c8!.%1

−=

α # :P19

#!.%

== :P29

1!. 3;"8 %;728

33

$

%3

82"==

−−−

=m :P19

c 4 72

log 3 y 4 3log 3 ) 7 2 :P292

1%

3

1%1% 1%logloglog −+= ) y :P39

3472/1

*ON+IDENTIAL





21. / 4 2cos2) 71 %o N ) N 36%o

*hape o graph cos2) :P19

*hape o graph 2cos 2) :P29

line y 4 71 :P39

Cinalise graph :P9

22. )M+<-LE

Lumber o a/s 4!

!

2

$:P19

4 1#1% :P29

2

23.a8 probabilit/ o getting % point 4

!

1:P19

b8 ratio o area o small circle F area o bull e/e 4 1 F :P19

the probabilit/ o getting 3 points 4

1

!

×

4 !

1

:P29

c8 probabilit/ o getting 1 point 4!

38

11

!

=−× :P19

2. Probabilit/ 4 %." total o students 4 2"6a8 number o students ho urther stud/ to Corm 6 4 2"6%." :P19

4 1$3 person :P29

b8 3.%".%2"6 ) )=σ

3

%

y

)

71

36%o1#%o

*ON+IDENTIAL ! 3472/1



4 ".613 :P19

2!. µ 4 1.# m

σ2 4 %.%$ m2

σ 4 %.3 m :P19

p1."!≤ 0 ≤1.$38 :P29

4 83.%

#.1$3.1

3.%

#.1"!.1 −≤≤− 1 p

4 833.%16".% ≤≤− 1 p :P39

4 1 7 %.336 7 %.332! 4 %.233$ :P9

'#IAN %IMILA

ADDITIONAL MAT&EMATI*% %PM

Pene-aras # Lian 0ee Lin

%ekt!r Penrsan Akaemik9 #abatan Peniikan %abah

Ah-i Pane-

1. *he!n N!k Tai9 %MK P Likas9 K!ta Kinaba-

2. *hn % Ki!n9 %M#K Ti!n &a9 %anakan

3. Ph!!n *hie: +n9 %MK Ten!m9 Ten!m

4. %hirne *ha9 %MK &!- Trinit9 Ta:a

;. %rianih %e:an9 %MK Tam5ar-i9 Tam5ar-i

<. Theresa Lee *h!!n M!i9 %MK %t. Mi8hae- Penam5an9 Penam5an

7. Tin Kai *h9 Maktab %abah9 K!ta Kinaba-

=. 0e!h %ek G!h9 %MK Aaseh9 Laha Dat

3472/1

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3472/1

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