Mate Tambahan > Skema Spm M3t Kertas 2 Set 1

SULIT 3472/2 Matematik Tambahan Kertas 2 Tahun 2005

JABATAN PELAJARAN PERAK

LEARNING TO SCORE

MATEMATIK TAMBAHAN Kertas 2 Set 1 SKEMA JAWAPAN

Skema jawapan ini mengandungi 13 halaman bercetak.

Matematik Tambahan SPM Kertas 2 Set 1 1.

4m + 3p = 2 m2 + 3m + 5p = 15 dari (1), 4m = 2 – 3p

m=

------------------ (1) ------------------ (2)

2 − 3p ------------------ (3) 4

Ganti (3) ke dalam (2) 2

 2 − 3p   2 − 3p   + 3  + 5p = 15   4   4 

[1]

4 − 12 p + 9 p 2 6 −9p + + 5 p = 15 16 4 9 p 2 −12 p + 4 24 − 36 p 80 p + + 16 16 16

= 15

9 p 2 − 12 p − 36 p + 80 p + 4 + 24 = 15 16 9 p 2 + 32 p + 28 = 240 9 p 2 + 32 p − 212 = 0 p =

2.

− 32 ±

(32)2

[1]

− 4(9)(−212)

2(9)

=

− 32 ± 8656 18

[1]

p = 3.39, -6.95 m = -2.04, 5.71

[1] [1]

(a)

x y + =1 8 8

[1]

(b)

x=

5(0) + 3(8) 3+5

x=3 C(3,5) (c)

Kecerunan AB =

y=

5(8) + 3(0) 3+5

[1]

y=5 [1]

−8 = -1 8

kec. CD = 1 [mAB mCD = -1] Persamaan CD : y - 5 = 1(x - 3) y=x+2 bila y = 0, x = -2 pintasan-x = -2

Learning To Score 2005 Unit Kurikulum Jabatan Pelajaran Perak

[1] [1] [1]

Halaman

2

Matematik Tambahan SPM Kertas 2 Set 1

y

3.

2 1 y = kos 3x O

30°

60° 90° 120° 150° 180°

x

-1 y = 2 sin 2x -2 (a)

Bentuk graf kos x Graf 1.5 kalaan Semua betul dan lengkap seperti rajah

[1] [1] [1]

(b)

Bentuk graf sin x Graf 1 kalaan bilangan penyelesaian = 3

[1] [1] [1]

∑ x = 40

4. (a)

Min = x = Varians =

∑x

dan

∑x

=

2

= 1600

40 =8 5

[1]

N 1600 − (8) 2 5

[1]

= 256 (b)

[1]

Katakan nombor itu ialah y

∑x − y =8 N 40 − y =8 4

[1]

y=8

[1]

1600-(8)2 = 1536 Sisihan piawai =

1536 − (8) 2 4

[1]

= 17.89 Learning To Score 2005 Unit Kurikulum Jabatan Pelajaran Perak

[1] Halaman

3

Matematik Tambahan SPM Kertas 2 Set 1 5.

(a)

dy − 18 = = −18(2 x + 1) − 4 4 dx (2 x + 1) − 18(2 x + 1) −4+1 −4 y = ∫ − 18(2 x + 1) dx = +c (−4 + 1)(2)

[1]

= 3(2 x + 1) −3 + c 3 = +c (2 x + 1) 3 3 +c Melalui titik (−1, 5), 5 = (2(−1) + 1) 3 5 = −3 + c c=8 3 +8 Persamaan lengkung : y = (2 x + 1) 3 (b)

Titik (0, 11)

3 + 8 = 11 (2(0) + 1) 3

[1]

− 18 dy = dx (2(0) + 1) 4

[1]

= -18

[1]

1 18

[1]

Kecerunan garis normal = Persamaan normal : y = (a)

[1]

Memotong paksi-y : x = 0

y=

6.

[1]

1 x + 11 18

[1]

D  2

Perimeter semibulatan pertama = D + π 

=D+

πD 2

πD  D + 2 + 2+π =D+ 2  2 

Perimeter semibulatan kedua = ( D + 2) + π 

[1]

πD  D + 4 + 4 + 2π =D+ 2  2 

Perimeter semibulatan ketiga = ( D + 4) + π  Beza sepunya 1 = D + Beza sepunya 2 = D +

Learning To Score 2005 Unit Kurikulum Jabatan Pelajaran Perak

πD

πD   + 2+π −D +  = 2+π 2 2  

πD

πD   + 4 + 2π −  D + + 2+π  = 2+π 2 2  

Halaman

4

Matematik Tambahan SPM Kertas 2 Set 1 Oleh kerana beza sepunya 1 = beza sepunya 2, maka ditunjukkan ia

6. (b) (i)

membentuk janjang aritmetik.

[1]

Maka beza sepunya = 2 + π

[1]

Mempunyai perimeter melebihi 100 cm bermaksud Tn > 100

a + (n − 1)d > 100 D+

πD 2

+ (n − 1)(2 + π ) > 100

 22   7  7 22 7+ + (n − 1)(2 + ) > 100 2 7

18 + (n − 1)

[1]

36 > 100 7

n − 1 > (100 − 18) n > 16

7 36

17 18

[1]

n = 17

[1]

Diameter semibulatan terkecil = 7 + 16(2) = 39 cm.

(ii)

Diameter semibulatan yang kelapan T8 = 7 + (8 – 1)(2) = 21

1  21  Luas semibulatan yang kelapan = π   2  2

2

[1]

= 173.25 cm2 7.

(a)

[1]

x

1

2

3

4

5

6

log10 y

−0.17

−0.05

0.09

0.22

0.34

0.47

[1]

Rujuk graf Lampiran A Paksi betul dan skala seragam

[1]

6 titik ditanda betul

[1]

Garis lurus penyuaian terbaik

[1]

Learning To Score 2005 Unit Kurikulum Jabatan Pelajaran Perak

Halaman

5

Matematik Tambahan SPM Kertas 2 Set 1 7.

(b)

log10 y = x log10 m − log10 n (i)

log10 m = kecerunan graf =

0.28 − 0.02 4.5 − 2.5

[1]

= 0.13

[1]

m = 1.349 (ii)

− log10 n = pintasan paksi- log10 y

→

8.

(a)

(i)

[1]

= −0.3

[1]

n = 1.995

[1]

→

→

AC = AO + OC = −2 x + 10 y ~

→

(ii)

[1]

→

[1]

~

→

OB = OA + AB 2  = 2 x + 10 y  ~ 5 ~

[1]

 →

OB = 2 x + 4 y ~

→

(b)

(i)

→

OR = h OB = h  2 x + 4 y  = 2h x + 4h y ~ ~ ~  ~ →

(ii)

[1]

~

[1]

→

AR = k AC = k  − 2 x + 10 y  = −2k x + 10k y ~ ~ ~ ~ 

→

→

→

OR = OA + A R = 2 x +  − 2k x + 10k ~ ~  = (2 − 2k ) x + 10k y ~

[1]

y  ~ [1]

~

→

(c)

OR = = 2h x + 4h y = (2 − 2k ) x + 10k y ~

~

~

~

2h = 2 − 2k

------------------(1)

[1]

4h = 10k

------------------(2)

[1]

h=

5 7

[1]

k=

2 7

[1]

Learning To Score 2005 Unit Kurikulum Jabatan Pelajaran Perak

Halaman

6

Matematik Tambahan SPM Kertas 2 Set 1 9.

(a)

∠BOC = ∠POQ

s = jθ 8 = 6θ 8 4 θ= = 6 3 sudut ∠BOC = (b)

[1]

4 radian 3

[1]

LOPQ =

1 2 4 (6)   = 24 2 3

[1]

LOBC =

1 2 4 (3)   = 6 2 3

[1]

Luas kawasan berlorek = 24 − 6

(c)

∠AOB = ∠COD = π −

4  s = jθ = 3 π −  3 

[1]

= 18 cm2

[1]

4 3

[1]

4  S AB = S CD = jθ = 3 π −  3 

[1]

= 3π − 4

[1]

Perimeter seluruh rajah = 6 + 3 + 3 + 8 + 2 (3π − 4) = 20 + 2 (3π − 4) = (12 + 6π) cm = 6(2 + π) cm 10.

(a)

Titik pusingan di A :

[1]

dy = 6 − 2x = 0 dx

[1]

x=3 y = 6(3) − (3)2 = 9

A(3, 9) [1]

Persamaan garis lurus OA ialah y = 3x (b)

Luas rantau =

∫ (6 x − x ) dx 3

 2 x3  = 3 x −  3   = (27 − 9 ) − =4 Learning To Score 2005 Unit Kurikulum Jabatan Pelajaran Perak

−

2

0

[1]

1 (3)(9) 2

[1]

3

−

27 2

[1]

0

27 27 = 18 − 2 2

1 unit 2 2

[1] Halaman

7

Matematik Tambahan SPM Kertas 2 Set 1

10.

(c)

Isipadu janaan = π =π

∫ (6 x − x ) 3

2 2

0

∫ (36 x 3

2

0

1 π (9 )2 (3) 3

dx −

[1]

)

− 12 x 3 + x 4 dx − 81π 3

 x5  = π 12 x 3 − 3x 4 +  − 81π 5 0 

[1]

=129.6π

[1]

− 81π

= 48.6π unit3 11.

(a)

(i)

6p = 2.4 p=

(ii)

(b)

(i)

(ii)

12.

(a)

I =

[1] [1]

2.4 = 0.4 6

[1]

Kebarangkalian = 1 − P(X = 1) − P(X = 0) = 1 − 6 C1 (0.4)1 (0.6) 5 − 6 C 0 (0.4) 0 (0.6) 6

[1]

= 1 − 0.186624 − 0.046656

[1]

= 0.7667

[1]

 

Kebarangkalian = P(X < 650) = P Z <

650 − 800   60 

[1]

= P( Z < -2.5 )

[1]

= 0.0062

[1]

Bilangan pelajar yang akan menerima biasiswa = np

∑W I ∑W

i i

= 1500( 0.0062)

[1]

= 9.3 = 9 orang

[1]

= 126

i

130(6) + 115(m) + 110(2) + 140(5) + 120(3) = 126 6+ m+2+5+3

[1] [1] [1]

m=4 (b)

x × 100 = 110 4.50 x = 4.95 Harga mangkuk pada tahun 2005 = RM4.95

Learning To Score 2005 Unit Kurikulum Jabatan Pelajaran Perak

[1] [1] [1]

Halaman

8

Matematik Tambahan SPM Kertas 2 Set 1 12.

13.

(c)

80 = 96 100 130(6) + 115(4) + 110(2) + 140(5) + 96(3) 20 = 122.4 120 ×

(a)

[1] [1] [1] [1]

D b

c = 4 cm 135° B

C

d =7 cm

b2 = c2 + d2 - 2cd kos b

b2 = (4)2 + (7)2 − 2(4)(7) kos 135°

[1]

2

b = 16 + 49 - (56)(-0.7071)

b2 = 104.5976

[1]

b = 10.23 oleh itu, panjang CD = 10.23

[1]

D

(b)

b = 8 cm

a = 4 cm 45°

A

d

B

sin A sin B = a b sin A sin 45 o = 4 8 0.7071 sin A = ×4 8

[1] [1]

sin A = 0.3536 A = 20.71°

= 20° 42´

Learning To Score 2005 Unit Kurikulum Jabatan Pelajaran Perak

[1]

Halaman

9

Matematik Tambahan SPM Kertas 2 Set 1 13.

c)

Bagi ∆ ABD, ∠D = 180 - 45° - 20.71°

∠D = 114.29° / 114° 18’ 1 Luas ABD = (4)(8) sin 114 o18 ' 2 = 16 (0.9115) = 14.58 Luas ∆ BCD =

[1]

[1]

1 (4)(7) sin 135 o 2

=14(sin 45°) =14(0.7071) = 9.8994 oleh itu, luas seluruh rajah = 14.5824 + 9.8994 = 24.4818 = 24.48 14

a)

b)

[1]

x + y ≤ 16

[1]

x≥5

[1]

x ≤ 2y

[1]

Rujuk graf lampiran B Paksi betul dan skala seragam Semua garis lurus betul Rantau R betul

c)

[1]

[1] [2,1,0] [1]

(i) Dari graf, bilangan bola tampar yang paling banyak dapat dibeli = 16 biji [1] (ii) 40x + 30y = k Dari graf, garis 40x + 30y melalui titik (5,0) Perbelanjaan minimum = 40(5) + 30(0) = RM200

15

a)

-3t2 + 10t + 8 = 0

[1] [1] [1]

2

3t - 10 t - 8 = 0 (3t+2)(t-4) =0 t=

−2 , 3

t=4

oleh itu, t = 4 sahaja

Learning To Score 2005 Unit Kurikulum Jabatan Pelajaran Perak

[1]

Halaman 10

Matematik Tambahan SPM Kertas 2 Set 1

15.

b)

a=

dv = -6t + 10 dt

bila a =0,

-6t + 10 = 0

10 5 = 6 3

t=

[1]

2

5 5 5 bila t = , v = −3  + 10  + 8 3  3 3

= 16

c)

2 m s −1 3

[1] [1]

s = ∫ v dt = ∫ − 3t 2 + 10t + 8 dt =

− 3t 3 10t 2 + + 8t + c = -t3 +5t2 + 8t + c 3 2

[1]

bila t = 0, s = 0 => c = 0 bila t = 3, s = -(3)3 + 5 (3)2 + 8(3) = 42 m

(d)

bila t = 2, s= -(2)3 + 5(2)2 + 8(2) = 28

[1]

oleh itu, 42 – 28 = 14 m

[1]

s = 0, -t3 + 5t2 + 8t = 0 -t(t2 - 5t - 8) = 0

t=0,

[1]

t2 - 5t - 8 = 0

t=

5 ± (−5) 2 − 4(1)(−8) 2(1)

= 6.275, -1.275 oleh itu, t = 6.275 s.

Learning To Score 2005 Unit Kurikulum Jabatan Pelajaran Perak

[1]

Halaman 11

Matematik Tambahan SPM Kertas 2 Set 1 Jawapan 7 (a)

Lampiran A

log 10 y 0.50

x

0.40 x 0.30 x

0.20

0.10

0

x

1

2 x

3

4

5

x

6

-0.10 x -0.20

-0.30

-0.40

-0.50

Learning To Score 2005 Unit Kurikulum Jabatan Pelajaran Perak

Halaman 12

Matematik Tambahan SPM Kertas 2 Set 1 Jawapan 14 (b)

Lampiran B

y

20

x=5

18

16

14

12

10

x

+

y

8

=

2 x=

16

y

6

4

R

2

0

2

4

x (5,0) 6

Learning To Score 2005 Unit Kurikulum Jabatan Pelajaran Perak

8

10

12

14

16

Halaman 13

18

20

x