SKEMA PERCUBAAN SPM 2017

MATEMATIK TAMBAHAN KERTAS 2

SOALAN 1

1. a)

2

4 3 5

4 3( 1) 5

4 8

2

2 3 5

2(2) ( 1) 3(2)( 1) 5

3

x y

k

k

k

x py xy

p

p

b)

2

2

5 4

3

5 4 5 42 3 5 5

3 3

2 9 10 0

2 5 2 0

5, 2

2

5, 1

3

5 5,

2 3

xy

x xx x

x x

x x

x x

y

2.

2

Luas 28 224

8

1 22Perimeter 2 28 2 14 72

2 7

36 36

836 36

36 36 8 0

3 1 3 2 0

1 2,

3 3

24, 12

PQRS xy

yx

STR y x x

y x

xx

x x

x x

x x

y y

3. a)

2

2

2

(5 ) 5 1

5 5 1

4 4 0

( 4) ( 4) 4(1)(4)

2(1)

2

h h h

h h h

h h

h

h

b)

2 2

2 2

2 2

2 2

2

2

2 2 1

1 2 11

3 1

3 1 1 2(3 1) 11

4 2 2(9 6 1) 11

4 2 18 12 2 11

7 16 4 0

( 16) ( 16) 4(7)(4)

2(7)

2

7

2 13 1

7 7

x y x y

x y x y

x y

y y y y

y y y y

y y y y

y y

y

y

x

4. 2(y + 1)

2

32x – xy = 63

x + x + y + y = 14

y = 7 – x atau setara

2(7 – x + 1)

2

32x – x(7 – x) = 63

(3x – 13)(x – 3) = 0

x = 3, abaikan nilai 3

13x

y = 4, abaikan nilai 3

8y

5.

354

4

35

75.8

xy

xy

Luas

xy

yx

yx

Perimeter

6

6

1222

12

1

2

x Gantikan dalam

5.2,5.3

5.3,5.2

2

7,

2

5

07252

035244

035424

3564

2

2

yy

xx

xx

xx

xx

xx

xx

Ukuran Bilik, Panjang = 3.5 + 2 = 5.5 m

Lebar = 2.5 + 2 = 4.5 m

SOALAN 2

1. (a) a = 6000

r = 1.022

n = 2017 − 2011 + 1

= 7

T7 = 6000 × 1.0227 − 1

= RM6836.86

(b) Tn > 18000

6000 × 1.022n − 1

> 18000

1.022n − 1

> Error!

(n − 1) log10 1.022 > log10 3

n − 1 > Error!

n − 1 > 50.4842

n > 51.4842

∴ n = 52

(c) T5 = 6000 × 1.0225 − 1

= RM6545.68

Total interest earned

= 6545.68 − 6000

= RM546

2. a) 15 1 14(2) 29T

b) 20

10

1 19(2) 39

10(3 39) 210

2

T

S

c) 2(1) ( 1)(4) 435

2

15

2 15 30

nn

n

3. a) (2k + 1) – (k + 3) = (k + 3) – (k – 1)

2

1

k = 6 d = 4

b) a + (6 – 1)(4) = 5

a = –15

c) )4)(1()15(2

2 n

nSn

nnSn 172 2

4. a)

16

12

27623

30424

28

842

a

d

d

da

da

dadaa

cm

S

4000

12241622

2525

b)

292

12231624

T

cm

x

3

124

12

292

5. a) Number of sheep sold

d = 40, a = 40

for the first 21 month = 21 40 20(40)T

= 840

The number of sheep left after 21 month is 1000 – 840 = 160

b)

12

2 ( 1)2

122(1000) (12 1)( 40)

2

9360

9360 1 30

280800

n

nS a n d

S

1

2

SOALAN 3

1. a)

22

2

120

12 710

1510

x

x

x

b)

2 2

2

2

120 18 138

1510 18 1834

1834 1389.34

11 11

new

new

new

x

x

2. a)

2 2

2

2

15015

10

247215

10

22.2

xx

N

xx

N

b) i)

15016

11

26

xx

N

k

k

ii)

2 2

222472 26

1611

5.494

xx

N

3. a) 1

2

140 15

220.5 1014

24.07

m

N FL C

f

b) 22

228550 960

40 40

137.75

11.74

fxx

f

4. a) L = 19.5 atau F = 16 atau fm = 15 atau c = 5

)5(15

162

50

5.19

22.5

b)

81115106

)32(8)27(11)22(15)17(10)12(6

atau

81115106

)32(8)27(11)22(15)17(10)12(6 22222

2)5.22(50

27225

38.25

c) 41.25

5. a) i) 4

16

64

x

x

ii) 22

2

3 416

400

x

x

b)

2

22

16 12 28

64 84 148

400 480 880

1485.286

28

8805.286

28

3.487

N

x

x

x

SOALAN 4

1. a) i)

3 (1)( 6) 3 (1)(9)3,6 ,

3 1 3 1

6, 5

(6,5)

x y

x y

R

ii) 1(18 48 20 24) ( 24 36 48 10)

2

28

b) 2 2

2 2

( 2) ( 4) 2

4 8 19 0

x y

x x y y

2. a) 1(4 80 8) (20 8 16)

2

24

b) 1 21, 1

5 1( 7)

2

m m

y x

y x

c) 2 2

2 2

( 4) ( 8) 5

8 16 55 0

x y

x y x y

3. a) )0(1)5(0)4(2)2(5)1(4)0(0

2

1

3

b)

23

)1(3)2(2,

23

)5(3)0(2

5

7,3

c) AP = 2PC atau 2222 )1()5(2)2()0( yxyx

(x – 0)2 +(y – 2)

2 = 4[(x – 5)

2 + (y – 1)

2]

3x2 + 3y

2 – 40x – 4y + 100 = 0

4. a) Q(2, 0) or P(0, -6)

(2, 0) =

32

182,

32

02 yx

R = (5, 9)

b) 5)0()2( 22 yx

25)2( 22 yx

021422 xyx

5. a)

216

4082

1

0

0

2

8

5

4

0

0

2

1

unit

AOBLuas

b)

5

4,

5

16

23

5223,

23

4283

2:3:

C

CBAC

c)

0231267233

161642566442510168

28254

2

22

2222

2222

yxyx

yyxxyyxx

yxyx

PBPA

SOALAN 5

1. a)

b) 4 3cos 2

3 2

4 3cos 2

3 2

3 4 3cos 2

2 3 2

3 2 9cos 2

2 4

x x

x x

x x

x x

No of solution = 3

2. a)

b)

2cos 0

2cos

xx

xx

No. of solutions = 2

3. a) Bentuk graf –sin x

Amplitud 3

2

-2

0 2

Lakaran dalam julat 0 ≤ x ≤ 2π

b) Persamaan

xy 12

Garis lurus

xy 12 dalam julat 0 ≤ x ≤ 2π

3 penyelesaian

4.

5. a) xy 2sin2

Sine curve seen

One and a half cycle in

3π0

2x

Max value

32

, Min value

32

OR

Sketching the straight line from the

*equation involving x and y. K1

N1

2 9π 4

y x 2 9 3

sin2π 4 2

x x

or equivalent

y

xππ

2O 3π

2

3sin 2

2y x

32

32

π4

3π4

5π4

94

34

2 9π 4

y x

Number of solutions = 3

Curve and straight line sketched

correctly

b)

xy

xx

xx

1

2sin2

2sin2

Draw

xy 1 on the same axes

Number of solution = 4

SOALAN 6

1. a) 2

49

xdx

dy atau 0

49

2

x

3

2p dan

3

2q (kedua-dua jawapan betul)

b)

12,

3

2 dan

12,

3

2 (kedua-dua jawapan betul)

32

2 8

xdx

yd

27

3

2

83

dan 27

3

2

83

(kedua-duanya betul)

12,

3

2 adalah titik minimum

12,

3

2 adalah titik maksimum

2. a)

2

22

0

3

3 6 3

6 3 1

lim

6 1

x

y y x x x x

y y x x x x x x

yx x

x

dy yit

dx x

dyx

dx

b)

2

2

0

6 1 0

1 1,

6 12

1 1,

6 12

6 0

1 1,

6 12

dy

dx

x

x y

P

d y

dx

P

3. a)

2

32

312

3

16

xxdx

dy

xxy

b)

3

97,4sin

3

97

3

1464,4

3

1,0sin

3

1

3

1060,0

4,0

043

0312

0

2

2

ganPuTitik

yx

ganPuTitik

yx

xx

xx

xx

dx

dy

c)

3

97,4

0124612,4

2

1,0

0120612,0

612

2

2

2

2

2

2

dx

ydxWhen

dx

ydxWhen

xdx

yd

4. a)

2

2

2

12

12 9

1 12 9

3

dypx

dx

px

p

p

b)

2

3

3

3

3 12

12

4 1 12 1

15

12 15

dyx

dx

y x x c

c

c

y x x

5. 2

2

2 2

2

3 2

2 2

5

1

52 2

1

2 4 3

solvesimultaneous

2 4 3 3 2

2 0

0, 2

3, 3

(0,3)

2 3

(2,3)

2 7

y x x

dyx

dx

ym

x

yx

x

y x x

x x x x

x x

x x

y y

y x

y x

SOALAN 7

1. a) 0.6 0.4p q

i)

4 26

4( 4) 0.6 0.4

0.31104

P x C

ii)

4 2 5 1 6 06 6 6

4 5 6

( 4) ( 4) ( 5) ( 6)

0.6 0.4 0.6 0.4 0.6 0.4

0.54432

P x P x P x P x

C C C

b) 300 20

i)

285 320

285 300 320 300

20 20

0.75 1

1 ( 0.75) ( 1)

1 0.2266 0.1587

0.6147

P x

P z

P z

P z P z

ii) 18000.72

2500

( ) 0.72

3000.72

20

3000.583

20

288.34

P x y

yP z

y

y

2. a) i) 8 3 53( 3) (0.1) (0.9) 0.0331p x C

ii) 8 6 2 8 7 1 8 8 0

6 7 8

( 5) ( 6) ( 7) ( 8)

(0.9) (0.1) (0.9) (0.1) (0.9) (0.1)

0.9619

p y p y p y p y

C C C

b) i) 35 25( 35)

8

( 1.25)

0.1057

p x p z

p z

ii) 100( )

600

250.1667

8

250.967

8

17.26

p x t

tp z

t

t

3. a) i) 3 5

83

3 2( 3)

5 5

0.1239

P X C

ii) 0 8 1 7

8 80 1

( 2) ( 0) ( 1)

3 2 3 2

5 5 5 5

0.1064

P X P X P X

C C

b) i)

60 65( 60)

5

1

0.84134

P X P Z

P Z

ii)

75 65( 75)

5

2

0.97725

0.97725250

244.31

244

P X P Z

P Z

x

x

x

4. a) i) 15 13 2 15 14 15 15

13 14 15( 13) (0.75) (0.25) (0.75) (0.25) (0.75)

0.2361

P x C C C

ii) (0.75)(0.25) 10.2

554 / 555

n

n

b) i)

80 5880 ( )

15

0.03593

P x P z

ii) 58( ) 0.1

15

581.281

15

38.785

wP x

w

w

5. a) i) 5 1 4

1( 1) (0.6) (0.4)

0.0768

P X C

ii) 5 2 3

2( 2) (0.6) (0.4)

0.2304

P X C

b) i) 2 1.5( 2) 1

0.2

1 0.00621

0.9938

P X P

ii) ( ) 0.65

1.51 0.65

0.2

1.50.385

0.2

1.423

P x m

mP Z

m

m

SOALAN 8

1. a)

t

s 71.0 60.3 49.2 40.0 30.0 20.0

b) Skala pada kedua-dua paksi seragam

Semua titik diplot betul

Garis lurus penyuaian terbaik

c) i)

btat

s

Kelihatan garis menyentuh paksi-t

s= 80

a = 80

Keceruanan = 06

8020

b = –10

ii) 97.5

2. a)

x 1 2 3 4 5 6

1/y 3.70 5 6.25 7.69 9.09 10

Correct and uniform scale

All points correctly plotted

Line of best fit

b) i)

a

bx

ay

a

bx

y

11

1

gradienta

1

a

11.3

a = 0.7692

1.846

4.27692.0

int

b

b

erceptYa

b

ii) 0.1429

3. (a) x² 1 4 9 16 25 36

xy 89 82 72 58 38 15 2xy qx p

(b)

2.11 , 91

91

m q c p

p

(c) 2 20.25 , 9.56x y

4. (a) 10log x 0.13 0.21 0.37 0.48 0.58 0.75 0.87

10log y 0.98 0.64 1.18 1.26 1.34 1.48 1.58

10 10 10log log logy n x a

(b)

10

0.8

log 0.88

7.59

m n

c a

a

5. (a) p 1 2 3 4 5 6 7

p q 0.77 0.94 1.15 1.28 1.46 1.62 1.80

p q ap b

(b)

0.17

0.60

m a

c b

(c) 1.81 1.34 , 4.3p

SOALAN 9

1. a) yxTR 104 , yxSQ 510

b) )510( yxmSU atau ymxmSU 510

)104()10(5

2yxnxSU

ynxnxSU 1044 atau ynxnSU 10)44(

c) 4 – 4n = 10m dan 5m = 10n

6

1n ,

3

1m

d)

nTR

TR

TU

PQ 2

1

atau

yx

yx

1046

1

1042

1

3 : 2

2. a) i) 2 26 10

8

m

m

ii)

10 11

OQ OP PQ

i j

b)

2 2

3 3

2 (10 11 )

3

MN MR RN

QR RO

i j

c) i) ( )

3 4

PT PR RT PO OR RT

i j

ii) 6 8

3 4

2

OP i j

PT i j

OP PT

3. a) i)

1

2

12 ( 3 )

2

32

2

QL QR RL

QR QP

b a

b a

ii)

3 2

SN SR RN

PQ QR

a b

b) i)

32

2

32

2

QM hQL

h a b

h a hb

ii)

2

2(2 ) ( 3 2 )

3 (4 2 )

QM QN NM

QR kNS

b k a b

ka k b

Equating the coefficients of a and b

33 2 4 2

2

4 2,

3 3

hk h k

h k

c) 3 2PQ a PS b

Area of parallelogram PQRS

1

2 5.1 3 sin402

9.835

4. a) i) AC AB BC

p q

ii) 1

4

1

4

AE AD

q

iii)

1

4

BE BA AE

p q

b) i)

1

4

1

4

BF kBE

k q p

kq kp

ii)

AF hAC

h p q

hp hq

c)

1

4

11 ( )

4

AF AB BF

p kq kp

k p kq shown

5. a) i) 1

2

3

OF OQ

y

ii)

1

3

13 6

3

2 2

OE OP PE

OP PQ

OP x y

x y

b) i)

3 1 3

PG hPF

PO OG h PO OF

OG h PO OF PO

h x hy

ii)

2 2

2 2

OG kOE

k x y

kx ky

3 1 3 2 2

3 1 2 3 2

3 1 3

1

2

3

4

h x hy kx ky

h k h k

h h

h

k

SOALAN 10

1. a) )10(2

360

38 ATAU

90

19)10(

10 + 10 + )10(2360

38 ATAU 10 + 10 +

90

19)10(

atau 9

19205

45

194 atau

315

1678 atau 5.327

b) 6635.4

90

19327.5

5(4.6635)

5(4.6635) + 5 + 5

43.3125

c)

90

19)5(

2

1 2 atau

90

19)10(

2

1 2 ATAU 2)5(360

38 atau 2)10(

360

38

90

19)10(

2

1 2 -

90

19)5(

2

1 2 ATAU 2)10(360

38 - 2)5(

360

38

24.881

2. a)

b) Arc AB = K1 (arc AB or CD)

Arc CD

OE =

OD

Perimeter = 6 + 4.713 + 2 + 5.5644 + 4

c) Area of sector AOB =

Area of OCE

3. a) i) 4 1cos

12 3

70 32'

1.231

WPX

WPX

rad

ii)

180 90 70 32' 90

109 28'

1.911

WQY WQZ YQZ

rad

b) Perimeter of the shaded region

2 28(1.231) 12 4 4(1.911)

28.81

WX XY WY

cm

c) Area of the shaded region

2 2

2

1 1 18 4 128 8 1.231 4 1.911

2 2 2

13.20cm

4. a) tan 1

4

AB

OB

rad

b)

2 2

2

2 2

2.828

AB AB cm

OA

cm

c) 1

3

3 6

4

6 2.828

3.172

6 4.7134

OB

OQ

OQ OB cm

BQ cm

AP OP OA

PQ cm

Perimeter shaded region

3.172 2 4 4.713

13.89

AP AB BQ PQ

cm

d) Area of shaded region = Area sector OPQ – Area of AOB

2

2

1 16 2 2

2 4 2

12.14cm

5. a) = kos

-1 ( )

26

4

=81.15˚

= 1.417 rad

b) QP = 26.31 cm atau ∠QRY = 1.725 rad QR = 11 x 1.725 atau PR = 15 x 1.417

QR = 18.975 atau PR = 21.255

18.975 + 21.255 + 26.31

66.54

c) ½(11+15)(26.31) atau ½(112 )(1.725)

atau ½(152 )(1.417)

½(11+15)(26.31) - ½(112 )(1.725)-½(15

2 )(1.417)

181.755

SOALAN 11

1. a) 24 xxy

b) 244 xxx

1b

c) )3)(3(

2

1

4

1

2 )4( xx atau

3

)1()1(2

3

)4()4(2

32

32

3

)1()1(2

3

)4()4(2

32

32

- )3)(3(2

1

2

9

d)

1

0

22 )4( xx

5

)0()1(2

3

)0(16

5

)1()1(2

3

)1(16 54

354

3

15

53

2. a)

2

2

1

2

13 1

2

1 5

2 2

dyx

dx

m

y x

y x

b)

4 1

2

3

4

3

2

3

(4 )

4

31

2

2

3

y dy

y

c) 4

3

2 2

(4 )

4 3(4(4) ) (4(3)

2 2

1

2

y dy

3. a) 3

1

4

1

4

127.5 2

2127.5

4

1127.5

2 2

4

k

k

x dx

x

k

k

b) (i)

2

2

(0, 2)

2

2 0

2 1 0

2, 1

(1,1)

B

x x

x x

x x

x x

A

(ii)

1 2 2

0 1

1 22 3

2

0 1

2

4 22 3

1 8 18 8 4 2

2 3 3

5

6

y dy y dy

y yy y

4. a)

2

2

20 33

2

2 3 14 0

2 7 2 0

7, 2

2

2

xx

x x

x x

x x

k

b)

22

10

3

2

2

0

( 3)

26

3

1 14 497

2 3 3

26 49

3 3

25

33

L x dx

L

xx

c)

1

22

7

3

72

3

8

1(2) 3 4

3

8 4

12

( 3)

32

V

V

y dx

yy

5. a) 2

2

22

1

23

1

3 , 4

( 4)( 1) 0

4, 1

4 0

2

11 3 4

2

34

2 3

3 8 18 4

2 3 3

13

6

y x y x

x x

x x

x

x

x

xx

b) 2

2 2 2

1

23 5

1

3 5 3 3

1(3) (1) (4 )

3

83 16

3 5

8(2) 2 8(1) 13 16(2) 16(1)

3 5 3 5

86

15

x dx

x xx

SOALAN 12

1. a) 30

120

3 4 600

y

x y

x y

c) i) 150

ii) 0.5 0.6

30, 160

0.5(160) 0.6(30)

98

x y k

x y

2. a) 100

2

30

x y

y x

y x

c) i) y max = 80 y min = 50

ii)

(20,80)

40(20) 20(80)

2400

3. a)

3

50

1000

y x

y x

x y

c) i) RM720

ii) (25,75)

25 75

100

x y k

4. a) 2 5 30

3 2 24

2

x y

x y

x y

c) i) 4

ii) 200 250

200(5) 250(4)

2000

x y k

5. a)

165 150 6600

105 120 840

20

x y

x y

x y

c) i) 3 37y

ii) (0,44)

12 12

12(0) 12(44)

528

x y k

SOALAN 13

1. a) 150100

60.0

90.0 Xx

12010050.1

Xy

8010040.0

Xz

b)

100

)15(80)30(150)20(120)35(150

= 133.50

c) i) 50.133100

2000

2014 XP

26702014

RMP

ii) 50.133

100

115X

= 153.525

2. a) i) Price index I = Error!× 100

p = Error!× 100

= 126

ii) 87 = Error!× 100

q = 9.80

b) I,¯ = Error!

= Error!

= 113

c) i) I2010 = Error!× 113

= 207.92

ii) P2010 = Error!× 22.3

= RM46.37

3. a) i)

45.11

10610080.10

06

06

RMP

P

ii)

17.2

11510050.2

04

04

RMP

P

RM

b)

160

13395357795

95.13320352520

)20140()35()25115()20106(

M

M

M

I

I

I

c) i)

74.160

100

12013395

I

ii)

93.28

74.16010018

08

08

RMP

P

4. (a) 960100

120

800

m

(b) 10

09

100P

IP

x = 110 , y = 140

(c) 110(6) 125(4) 98(3) 110(2)

15

1686

15

111.6

(d) 4500111.6

100

5022

5. a) x =36

y = 54

z = 80

b) 5 8 15 4 100

5

125(5) 120(8) 105(3) 80(4)

20

111

n n n

n

c)

2000

2000

280100 111

252.25

P

P RM

d) 2008

2008

100 120280

336

336100

252.25

133.20

P

P

I

SOALAN 14

1. a) i) AC2 = (7)

2 + (6)

2 – 2(7)(6)kos(80)

8.3913

ii)

3913.8

35sin

12

sin

ACD

180 – 55.1093

124.8907

iii) 358907.124180 CAD

80sin)6)(7(2

1 atau 1093.20sin)12)(3913.8(

2

1

37.9911

b) i)

ii) 145°

2. a) i) 10sin 30

QT 20

QT

cm

ii) 3

20

81 22

cos

b) i) 12 9

sin sin50B

B = 113 11

ii) CD2 = 9

2 + 8.5

2 – 2 (9)(8.5) cos 47

0 49

’

CD = 7.107 cm.

iii) '1 1(8.5)(9)sin 46 49 (12)(9)sin16 49

2 2

42.730

3. a) 61

14

sin94 sin61

15.97

TSU

SU

SU

b) 2 2 26 12 15.97 2(12)(15.97)cos

18.70

RUS

RUS

c) 2 2 2

43.7

12 14 2(12)(14)cos43.7

9.853

RUT

RT

RT

d) 1 1(12)(15.97)sin18.7 (14)(15.97)sin25

2 2

77.97

4. a) i) 2 2 2

115

4 5 2(4)(5)cos115

7.610

QRS

SQ

SQ

ii) sin sin115

5 7.610

36.55

QSR

QSR

iii) 1 1(5)(5)sin50 (5)(4)sin115

2 2

18.64

b) i)

ii) ' ' ' 180 36.55 143.55Q S R

5. a) 15 4 sin 7

2

44.43

R

R

b) 2 2 24 5 (2 4 5cos44.43 )

3.526

QS

QS

c) 3.526 8

sin sin45

18.16

180 45 18.16

116.84

P

P

PQS

d) 13.526 8 sin116.84 7

2

12.58 7

19.58

SOALAN 15

S’

S R

Q

1. a)

max

16 8 0

2

4(2)(4 2) 16

dva t

dt

t

v

b) 2

2 3

2 3

2 3

16 4

48 0, 0

3

48

3

48(3) (3)

3

36

s t t dt

t t c t s

s t t

c) 2 3

2

48 0

3

48 0

3

0 6

s t t

t t

t t

d) 4 (4 ) 0

0 4

t t

t

2. a)

2

2

2

max

2

2 3

, 2 2 0

1

2 12 3

2

4

max

v t t c

v t t

v a t

t

v

b)

2

2

0

2 3 0

2 3 0

3 1 0

3

v

k k

k k

k k

k

c) 3

2

2

33

2

2

2 3

33

5

3

s t t dt

tt t

d) 3

2

2 3

2

0

3 03

3 9 0

3 9 0

4.854

s

tt t

t t t

t t

t

3. a) 15 3 0

0 5

t

t

b) 2

2

315

2

315(5) (5)

2

37.5

s t t

c) 2

2

315

2

315(8) (8)

2

24

2(37.5) 24 37.5 (37.5 24)

51

s t t

or

d)

4. a) 2

dva

dt

b) 0

2 8 0

4

v

t

t

c)

2

2

2

2 8

8

0, 0

8

2

2 8 2

3

3 8 3 15

15 3

12

s t dt

s t t c

t s

s t t

v

t

t

s

d) 4 7

0 4

25

v dt v dt

5. a)

2 3

2

0

3 0

3 0

0, 3

3

s

t t

t t

t t

t

b) 5 4

2 3 2 33(5) (5) 3(4) (4)

34

s s

c)

2

2

2 4

6 3

6 3 0

6 3 0

2

4 , 16

4 4 ( 16)

24

dst t

dt

t t

t t

t

s s

s

d)

2

6 6

6 6 0

1

6(1) 3( )

3

a t

t

t

v t