Spm > Matematik Tambahan > M3t Kertas2 Set2

SULIT 3472/2 Matematik Tambahan Kertas 2 Tahun 2006

JABATAN PELAJARAN PERAK

LEARNING TO SCORE

MATEMATIK TAMBAHAN Kertas 2 Set 2 SKEMA JAWAPAN

Skema jawapan ini mengandungi 13 halaman bercetak.

Matematik Tambahan SPM Kertas 2 Set 2 BAHAGIAN A 1.

n = 8 – 2m m2 + 2mn = 20 m2 + 2m ( 8 – 2m ) = 20 3m2 – 16m + 20 = 0 ( 3m – 10 ) ( m – 2 ) = 0

[1] [1] [1]

10 m= , 2 3 10 4  10  Bila m = , n = 8 – 2  = 3 3 3 n = 8 – 2 (2) = 4

m=2 ,

2.

[1]

[1]

 3 x  1(1) 3 y  1(4)  ,  3 1   3 1 3x  1 3y  4 2 = , 3 = 4 4 16 x =3 , y= 3  16  B  3,   3

a) C (2, 3) = 

b)

mAB =

[1] [1]

[1]

3 4 7  2 1 3

mAQ = 

[1]

3 7

[1]

Persamaan PQ  y –

16 3 = –  x  3 3 7

[1]

21y + 9x = 139

3.

y 2

y  2 sin x

1  O -1 -2

60

90

120

y  kos



180

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3 x 2

x 240 270 300 360



(a) Bentuk graf kos Lengkap 1.5 kalaan

[1] [1]

(b) Tahu y = 2 sin x Bentuk graf sin (1 kalaan) Amplitud = 2

[1] [1] [1]

Bilangan Penyelesaian = 3

[1]

Halaman

2

Matematik Tambahan SPM Kertas 2 Set 2 4.

a)

(i)

m  2  m  3  m  4  2m   4m  9 5

8

9m  14  40 54 m 6 9

[1]

[1]

(ii) x 8 3 2 12 15

Sisihan piawai =

b)

5.

(a)

446 2 8 5 25.2  5.02

[1]

= (i) Min baru = 8(2) + 3 = 19

[1] [1]

(ii) Varians baru = 25.2 ( 22 ) = 100.8

[1] [1]

3 y  x  29 3 y  x  29 1 29 y  x 3 3 1 mnormal  3 dy  3 dx Pada titik A(2, 9),

(b)

x2 64 9 4 144 225 446

[1] [1]

dy p  4( 2)   3 dx ( 2) 2 p  8   3 4 p  4(3  8)  20

dy 20  4 x  2  4 x  20 x  2 dx x 2 4x 20 x 1 20 y   c  2x 2  c 2 1 x Melalui titik A(2, 9),

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[1]

[1]

9  2  2   2

9  8  10  c Persamaan Lengkung

[1]

20 c  2 

[1]

c = 9  8  10 = 9

[1]

20 y  2 x2  9 x

[1]

Halaman

3

Matematik Tambahan SPM Kertas 2 Set 2 6.

(a)

(i) 1, 2, 3, … , 50

82 = 4

Bilangan Baris = 50

(3)

Tinggi h = 50  3 = 150 cm

[1]

5

[1]

50 2(1)  (50  1)1 2

[1]

= 25(51) = 1275 segitiga

[1]

(ii) S50 

Sn  2000

(b)

n 2(1)  (n  1)1  2000 2

[1]

n (1 + n)  4000 n2 + n – 4000  0

[1]

63.75  n  62.75

63.75

62 baris boleh dibina.

[1]

62.75

[1]

BAHAGIAN B 7.

(a)

x3

1.00

8.00

27.00

42.88

64.00

xy

4.50

7.40

16.20

24.15

33.60

[1] [1]

Rujuk graf Lampiran A

(b)

Paksi betul dan skala seragam

[1]

5 titik ditanda betul

[1]

Garis lurus penyuaian terbaik

[1]

xy

1 n 3  x m m

[1]

(i)

1 = pintasan paksi- xy = 4 m

[1]

m= (ii)





1 4

[1]

n 15  8.5 = kecerunan graf = m 24  10

[1]

n = 0.4643 1   4

n  0.1161 Learning To Score 2006 Unit Kurikulum Jabatan Pelajaran Perak

[1] Halaman

4

Matematik Tambahan SPM Kertas 2 Set 2 8.

(a)

(i)

 1  1 (6 x ) = 3 x BC = AD = ~ ~ 2 2    AC = AB + BC = 2 y + 3 x

[1]

~

     BD = BA + AD =  AB + AD =  2 y + 6 x ~

(ii)

     CD = CA + AD =  AC + AD =  (2 y + 3 x ) + 6 x

(iii)

~

= 3x  2y ~

(b)

[1]

~

~

  AE = m AC  m  3 x  2 y   3m x  2m y ~  ~ ~ ~   BE = n BD  n  6 x  2 y   6 n x  2n y ~  ~ ~ ~

(i) (ii)

~

~

[1]

~

[1] [1]

   AE = AB + BE

(c)

3m x  2m y  2 y  6n x  2 n y ~

~

~

~

~

~

[1]

~

3m x  2m y  6n x  2  2n  y ~

~

3m = 6n --------------(1) m = 2n

[1]

2m = 2  2n

m = 1  n -------------(2)

Maka

[1]

2n = 1  n 3n = 1 n=

1 3

1  3

m = 2  

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[1]

2 3

[1]

Halaman

5

Matematik Tambahan SPM Kertas 2 Set 2

9.

a) sin x =

102 = 5

5 8

x 8

x = 38 41’  POQ = 2 x = 77o 22’ = 1.35 rad b)

c)

[1] [1]

Perimeter kaw. Berlorek = Panjang lengkok major PQ + Perentas PQ = 8 (2 ─ 1.35) + 10 = 39.47 + 10 = 49.47

1

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



Luas kaw. Berlorek = Luas bulatan   1.358  88 sin 7722' 2 2  2

1

=  (8)2 – [ 43.2 – 31.22] = 201.09 – 11.98 = 189.11

10.

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

a) y = 4x – x2 ------- (1) x – 2y = 0 ------- (2) Selesaikan persamaan (1) dan (2) 4x – x2 =

x 2

x ( -2x + 7 ) = 0 x = 0 atau x = y = 4x – x2

7 2

7 7  2 4

maka koordinat A  ,

[1]

dy  4  2x dx 7 = 4 – 2  = – 3 2 1 3 1 7 7 y– = x  3 2 4

[1]

mnormal =

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[1]  12 y = 4x + 7

[1]

Halaman

6

Matematik Tambahan SPM Kertas 2 Set 2 10.

b)

 4 x  x dx 4

Luas kaw. Berlorek =

2

7 2

 4x 2 x 3  =    3  2

4



7 2



 

[1]

49 16

  3   2 4 2  4  =  3   =  32 

1  7  7     2  2  4 

[1]

3  7       7 2  2     2    3   2  

64    24.5  14.29   3.0625 3 

       

4



49 16

7 2

= 10.67 – 10.21 + 3.0625 = 3.5225 unit2

c)

[1]

Isipadu kawasan berlorek = Isipadu kon + 

 4 x  x  4

7 2

2

2

dx



[1]



4 1 7 7 =         7 16 x 2  8 x 3  x 4 dx 3 4 2 2 2

343 =  96 =

343  96

16 x 3 x5  +    2x4   5  3

 512   16121      15   480 

4

7 2

+ 

= 4.1208 unit3

11.

a)

[1]

(i) P ( X = 1 ) = 10 C1 0.2  0.8 = 0.2684 1

(ii) P ( X = 0 ) =

10

9

[1] [1]

C 0 0.2  0.8 0

10

[1] [1]

= 0.1074

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[1]

Halaman

7

Matematik Tambahan SPM Kertas 2 Set 2 11.

b)

 50  60 X   65  60      4   4 = P 2.5  Z  1.25

(i) P 

[1]

= 1 – 0.1056 – 0.00621 = 0.88819

[1]

Bilangan pekerja kilang = 0.88819 x 500 = 444 orang.

(ii)

[1]

P ( X < m ) = 0.75 1 – P ( X  m ) = 0.75 P ( X  m ) = 0.25

 

m  60    0.25 4  m  60  0.674 4

P Z 

[1] [1]

m = 62.7

[1]

BAHAGIAN C 12.

a)

b)

300  100  110 P00 300  100 P00   RM 272.73 110 2(120 )  1(125)  x (111) i)  115 2 1 x 365  111x  115 3 x

[1] [1] [1]

365 + 111x = 345 + 115x 115x – 111x = 365 – 345 4x = 20 x=5

(ii)

[1]

4(110 )  2(120 )  1(125)  5(111) 1360  4  2 1 5 12  113.3 P04  100  113 .3 1800

[1] [1]

113.3  1800 = RM 2039.40 100  120 120  113.3 = 100

[1]

= 135.96

[1]

P04  c)

I 06 / 04

I 06 / 00

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[1]

[1]

Halaman

8

Matematik Tambahan SPM Kertas 2 Set 2 13.

(a)

AB 9.5  sin 42 sin 65

(i)

AB 

[1]

9.5 sin 42  7.01 cm sin 65

[1]

BD2 = (7.01)2 + (2)2  2(7.01)(2) kos 65°

(ii)

[1]

41.29  6.43 cm

BD =

[1]

7.01 6.43  sin ADB sin 65

(iii)

[1]

7.01sin 65  0.9881 6.43

sin ADB 

 ADB = 180  81.15 = 98.85

[1]

 ABC = 180  65  42 = 73

(iv)

Luas  ABC =

[1]

1 7.019.5 sin 73 2

[1]

= 31.84 cm2

[1]

B’

(b)

Jawapan (ii) 6.43 cm

Jawapan (i) 7.01 cm

[1] 65

81.15 D’

A’ 14

a)

x+y5

[1]

y>x

[1]

25x + 20y  400 5x + 4y  80

b)

[1]

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

[1] [2,1,0]

Rantau R betul c)

[1]

(i) y = x + 2

[1]

Dari graf, jumlah maksimum pemain = 8 + 10 = 18 orang

[1]

(ii) x = 2 Dari graf, 3 orang  bilangan pemain lelaki  17 orang Learning To Score 2006 Unit Kurikulum Jabatan Pelajaran Perak

[1] Halaman

9

Matematik Tambahan SPM Kertas 2 Set 2 15

a)

a = 4t  5

v

4t 2  5t  c 2

[1]

v = 2t2  5t + c Halaju awal = 3, t = 0, v =  3,  c =  3, b)

v = 2t2  5t  3

[1]

v<0 2t2  5t  3 < 0

[1]

(2t + 1)(t  3) < 0

 

[1]

3

1 2

1
0t<3

c)

s=

[1]

2t 3 5t 2 2 5   3t  c  t 3  t 2  3t  c 3 2 3 2

[1]

bila t = 0, s = 0, c = 0

s=

2 3 5 2 t  t  3t 3 2 ds = v = 2t2  5t  3 = 0 dt (2t + 1) (t  3) =0 t= 3

bila t = 3, s =

2 3 5 2 3  3  33  13.5 3 2

Kedua-dua bila t = 4, s =

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[1]

2 3 5 2 4   4   34   9.333 3 2

Halaman 10

Matematik Tambahan SPM Kertas 2 Set 2 15.

t=3

t=4

 9.333

 13.5

x t=0

Jumlah jarak yang dilalui = 13.5 + (13.5  9.333) = 17.67

[1]

(d) v 9

O

3 4

t

Bentuk graf

[1]

Titik (0, 3), (3, 0), (4, 9)

[1]

3

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Halaman 11

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

Lampiran A

xy

40

35 x

30

25

x

20 x

15

10 x

5 x

0

10

20

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30

40

50

60

70

Halaman 12

x3

Matematik Tambahan SPM Kertas 2 Set 2 Jawapan 14 (a)

Lampiran B

y 22

20

18

16

14

12

R

10

8

6

4

2

0

2

4

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6

8

10

12

14

16

Halaman 13

x