31
SKEMA JAWAPAN PEPERIKSAAN PERCUBAAN KERTAS 1 2017 SOALAN 1 BIL JAWAPAN SUB MARK MARKAH PENUH 1 1 ) 3 2 ( log : B1 5 ) 3 2 ( : B2 5 1 x x x x 2 5 x 3 3 2 1 3 1 2 x x 2 log x x 3 B2 B1 3 3 25 27 : 1 25 1 3 3 : 2 2 1 3 log log log log log a a a a a a B B n m 3 4 3 x 3y B2 : y x 3 2 1 2 3 ) 3 ( B1 : m = 9 x atau n = 3 3 y 3 3 5. 1 B2 : 3 1 3 log log 5 5 B1 : 3 3

# SKEMA JAWAPAN PEPERIKSAAN PERCUBAAN KERTAS 1 · skema jawapan peperiksaan percubaan kertas 1 2017 soalan 1 bil jawapan sub mark markah penuh 1 b1: log (2 3) 1 b2 : (2 3) 5 5 1 x x

others

• View
177

0

Embed Size (px)

Citation preview

SKEMA JAWAPAN

PEPERIKSAAN PERCUBAAN KERTAS 1

2017

SOALAN 1

BIL JAWAPAN SUB MARK MARKAH

PENUH

1

1)32(log:B1

5)32(:B2

5

1

xx

xx

2

5x

3 3

2 1

3

1

2

x

x

2log

x

x

3

B2

B1

3

3

2527:1

25133:2

213

logloglog

loglog

aaa

aa

aB

B

nm

3

4 3x – 3y

B2 : y

x

3

2

1

2

3

)3(

B1 : m = 9x atau n = 33

y

3 3

5. 1

B2 :

3

13

loglog

5

5

B1 :

3 3

SOALAN 2

1

1

7

7

7

7

1

7

777

7log

)753(log

7

1log

105log:B1

7log

7log5log3log:B2

1

1:B3

atau

kh

1kh

4

4

2

kh

hk

2

12

3log2log2

5log2log3log2

55

555

5log2log2log 55

2

5 or 3log2log 5

2

5 or

5log2log3log 12122

12

12log

90log

5

5 or 2 log 5 3 or 2 log52

4

B3

B2

B1

4

3 (a)

p

1

(b) 12

25

p

p

13log

23log2

5

m

m

m

m

m

m

9log

243log 2

1

3

B2

B1

4

4

2

23 xy

B3 : 2

log8loglog2 222 bc

B2: 4log

8loglog

2

2

2

2 bc

B1: bc 8loglog 4

2

4 @ 4log

8log

2

2

2b

c

4 4

5

q

pq

1

B3 : log2 3(1 – p) = pq

B2 : log2 3 = p(log2 4 + log2 3)

B1 : 12log

3log

2

2 = p

4 4

SOALAN 3

1 15

B2 : )2)(1()5)(1()42( nxnx

B1 : )5)(1()42( nx atau

)2)(1( nx

3 3

2 n = 42

87)2)(1(5 n

d = 2

3

B2

B1

3

3 1256

)168(2))8(4)200(2(2

55 S

)8(42005 T

3

B2

B1

3

4 a) k = 2h + 1

k -3h = (h + 2) – k

b) 9 – 6h

2

B1

1

3

5 a) d = 6

b) 5

B1 :

)5(283)6(283 )5()6(

22

1

2

3

SOALAN 4

1 7 1 3

4

B1 : 282 T atau 7

28

2

2

6

1

3

11

9

1

3

1r

3

B2

B1

3

3 3

2

6r dan p = 6

32

6r atau p = 6

p

p

p

ppp

18

2

36

6,0,6

2

3

B2

B1

3

4 x1 = 3 dan x2 = –1 B2 : (x – 3)(x + 1) = 0

B1 : 3x2 – 6x – 9 = 0

3

5 -24

B2 :

4

31

6

B1 : a = - 6 , 4

3r

3 3

SOALAN 5

1 (a) 5

B1 : )2(

)4(

)4(

)2(

x

x

x

x

2 4

(b)

2

243

B1 : a = T1 = 81 ...seen atau

3

11

81

2

2 (a)

3

2

5

4

27

8 or equivalent

2

3

1

2

T

T

T

T

(b) 3

4or equivalent

3

21

9

4

3 a) x = 6

B1 : 12

2412

x

b) 6096

B1 :

12

16

12

16 22310

1

1

4

4 atau 5.5

B3 :

2

111

2T

B2 : a = 11

B1 : 22

2

11

a

4 4

5 a) a = 8 , - 2

B1 :4

224

a

a

a

a

b) 16

2187 atau 136.6875

B1 :

2

38

9 3

16T

1

1

4

SOALAN 6

1 y = x3

3

3

B2 : 4 = 4 + c atau 12 = 12 + c atau c = 0

B1 : m = 1

2 p = 2 and 1q

p = 2 or 1q

04

)5(3

p or 55 q

qpxxy 52 2

3

B2

B1

3

3 s = 3 and t = 5

7 = 2s + 1 or t = 2(2) + 1

3

B1

3

4 r = 1, p = 0.001 (BOTH)

r = 1 or p = 0.001 (either one)

=-3 or r = or

3

5 m =

2

3 dan n = 3

B1 : –2 = m

3

y = –2x

2 + 3x

2

1

3

SOALAN 7

BIL JAWAPAN SUB MARK MARKAH

PENUH

B3 : 0)2)(2(

4

4

B2 : 1002

40

2

12

B1 : 40 rrr

2 (a) 842.1

5.65

(b) 23.025

)842.1()5(

2

1 2 * (candidate’s from a)

2

B1

2

B1

4

3 (a) 10.8 cm

(b) 19.44 cm2

)142.3

1809.0tan(612

2

19.012

2

1 2

6

9.0tanh

1

3

B2

B1

4

4 (a) θ = 0.842 rad

kos θ = 15

10

(b) 126.708 cm2

Area = 88)18.11)(10(

2

1)841.0()15(

2

1 2

2

B1

2

B1

4

SOALAN 8

BIL JAWAPAN SUB MARK MARKAH

PENUH

1 (a)

15

17

(b) 221

21

13

5

17

15

13

12

17

8 B1

1

2

3

2 (a) tan A =

12

5

(b) sin (A – B) = 65

56

5

3

13

12

5

4

13

5

1

2

B1

3

3 θ = 45o, 63.43

o, 206.57

o, 225

o

θ = 63.43o, 206.57

o OR θ = 45

o, 225

o

(tan θ - 2)(tan θ - 1) = 0

3

B2

B1

3

4

7

17 atau 2.4286

B2 :

)1)(5

12(1

15

12

atau tan (112.62 – 45)

B1 : tan = 5

12 atau tan 45 = 1

3 3

5 (a)

p

1

1

2

3

(b)p

p21

21 p seen

B1

SOALAN 9

BIL JAWAPAN SUB MARK MARKAH

PENUH

1 x = 0o, 30

o, 150

o, 180

o, 360

o (all correct)

B3 : 0o and 30

o (both)

B2 : sin x (-2sinx +1 )= 0

B1 : (1-2sin2x) + sin x - 1 = 0

4 4

2 0, 120, 180, 240, 360

B3 : 0, 180, 360 atau 120, 240

B2 : sin x(2 kos x + 1) = 0

B1 : 2 sin x kos x + sin x = 0

4 4

3 x =18.435o, 45

o,198.435

o, 225

o

B3 : 18.435o, 198.435

o or 45

o, 225

o

B2 : (3 tan x – 1 ) ( tan x – 1 ) = 0

3 tan2 x – 4 tanx +1 =0

B1 : 3 ( 1 + tan2x)- 4 tan x -2 = 0

4 4

4 x = 68o12’, 90

o, 248

o12’ , 270

o

B3 : cos x = 0 and x = 90o,270

o

OR

4 4

B3 : tan x = 2

5 and x = 68

o12’, 248

o12’

B2 : cos x (2 sin x - 5 cos x) = 0

B1 : 5 sin2x = 5 - 2sin x cos x

5 x = 35o 16’, 144

o44’ , 215

o16’, 324

o44’

B3 : 3

1sin x

B2 : 3(1 - 2sin2 x) = 1

B1 : 3 (cos2 x - sin2 x) = 1

4 4

SOALAN 10

BIL JAWAPAN SUB MARK MARKAH

PENUH

1 (a) 36.6

B1: (16.8 x 2) + 3

2

1

3

(b) 7

2 6

B2 : 22

10

1060a

a

B1 : Set 1 : x2 = 24 + 6a

2 atau Set 2 : x

2 = 36 + 4a

2

dengan a = min

3 3

3 a) 12/3 = 4

b)13.2665

1

2

3

4 m = 11

B1 : (m - 11)(m + 4) = 0

B1 : 143

7

3

25422

mm

3 3

5 1A and 323.1B

B1 : 22 2

8

462

8

40 BA OR

(b) Set A is to be preferred as the standard deviation of

set A is smaller than set B

2

1

3

SOALAN 11

BIL JAWAPAN SUB

MARK

MARKAH

PENUH

1 (a) 252 1

3 (b) 66

B1 : 4C3 ×

6C2 atau 60 atau

4C4 ×

6C1 atau 6

2

2 277 200

3 3

B2: 2

5

3

8

4

12 CCC

3 6270

B2 : 4C3 x

11C4 +

4C2 x

11C5 +

4C1 x

11C6 +

4C0 x

11C7

B1 : 4C3 x

11C4 atau

4C2 x

11C5 atau

4C1 x

11C6 atau

4C0 x

11C7

3

3

4 (a) 210 1

3 (b) 30

B1: 2

2

2

5

2

3 CCC

2

5 11

B2: 4

4

3

4

2

4 CCC

B1: seenCorCorC ...4

4

3

4

2

4

3

3

SOALAN 12

BIL JAWAPAN SUB

MARK

MARKAH

PENUH

1

(a) 10

3

1

3

(b) 10

1

B1 : n (M N) = 2 atau {3, 15}

2

2

15

1 or an equivalent single fraction

B2: 6

2

5

2

6

3

B1: 6

3 or

5

2 or

6

2

3

3

3 6

B2: 24x

x x

24

22 +

24

24

x x

24

2 =

4

1

B1: 24x

x x

24

22 atau

24

24

x x

24

2

3

3

4 (a)

10

3

B1: 2

1+ P(G) =

5

4

2

3

(b) 5

1

1

5 (a)

216

1 // 0.00463

1 3

(b) 72

25

B1: 6

1

6

5

6

5

6

5

6

1

6

5

6

5

6

5

6

1

2

SOALAN 13

BIL JAWAPAN SUB

MARK

MARKAH

PENUH

1 (a) 0.1573

B1: 0.5 – 0.3427

2

3

(b) 0.1573 1

2 (a) 60 2 3

B1: 5.012

54

X

(b) 30.85% 1

3 (a) 0.1977 1

3 (b) 31.75

B1: – 0.85 = 5

36x

2

4 (a) 0.9788 1

3 (b) 48

B1: 24

40

X

2

5 (a) 1.25

B1: 2.1

2.57.6 Z

2

3

(b) 0.1056 1

SOALAN 14

BIL JAWAPAN SUB

MARK

MARKAH

PENUH

1 (a) t = 1

B1: 23

24

t

2

4

(b) y = 2x – 4

B1: m2 = 2 atau 2 = 2(3) + c

2

2 (a) h = 4

B1:

(b) D (5, 9)

B1:

2

2 4

3 (10, 7)

B3: x = 10 atau y = 7

B2: 25

0

x atau 3

5

8

y

B1: AC : AB = 1 : 5 atau AC : CB = 1 : 4

4

4

4 9.7082

B3 : 22 )2(6()

2

16(

B2 : Koordinat titik pembahagi tembereng garis

dengan nisbah m : n; P(2

1, –2)

B1 :

31

)26(1)6(3,

31

)6(1)6(3

4

4

5 (a) h = 8

B1: 6

710

62

75

h

(b) (0, – 2)

B1: x = 0 atau y = – 2

atau

4

)1(6)3(2 atau

4

)1(7)3(5

2

2 4

SOALAN 15

BIL JAWAPAN SUB

MARK

MARKAH

PENUH

1 p = 1 dan q = ½

B2: p = 1 atau q = ½

B1: 3p + 4q – 5 = 0 atau p + 6q – 4 = 0

3

3

2 k =

5

6

B2 : 5 = 3m atau – 3 = – (3 + k) m

B1: 5a – 3b = m [3a – (3 + k)] b

3

3

3 h = 7

B2: 24 or 3 = )1(2

1h

B1:

4

)1

2

3 h

3

3

4 (a) – 3x

(b) 6x – 5y

B1 : QP = QR + RP

1

2 3

5 (a) 5w – 3u

(b) – 10w + 9u

B1: – 15w + 9u + 3u + 5w – 3u

1

2 3

SOALAN 16

BIL JAWAPAN SUB MARK MARKAH

PENUH

1 . a) = 5i – 8j

= 4i – 6j + i - 2j

PQ = PO + OQ

b) = 11i – 18j

= 15i -24j –(4i – 6j)

B1

B2

B1

3

OR = 15i – 24j –PO

PO + OR = 3(5i – 8j)

2 a) i – 4j

b) 17

4 ji

: 22 )4()1( atau 17

B1

2

B1

3

3 12.65

22 124

ji 124 or ji 124

3

B2

B1

3

4

53

27~~ji

53OC

~~~~34511 jiji

3

B2

B1

3

5 h =

7

8 k

–h + k = – + k and –8 = –7

AB =

8

khatau BC =

7

1 k

3

B2

B1

3

SOALAN 17

BIL JAWAPAN SUB MARK MARKAH

PENUH

1 = 2 18π = π h

r = 3 V = π h

V = π (6)

V = 2

= 4 πr

= x = x 2

=

2 r = 5

408 r

3.0

128

r

rdr

dA8

3

B2

B1

3

3 π1.2

3

3

)2.0()3(2

2.0dt

dratau r

dr

dA2

B2

B1

4 10

42

32

x or 4

2

23

2

2

23

xdx

dp

3

B2

B1

3

5

8

15

25

16h

2 = 9 x

4

1

dh

dV = 2

25

16h

3

B2

B1

3

SOALAN 18

BIL JAWAPAN SUB MARK MARKAH

PENUH

1 )(

8

1xf

)(4

1

2

1)(

2

1xfxdxxg

)(4

1)( xfdxxg

)()(4 xfdxxg

)(4)( xgxfdx

d

3

B2

B1

3

2

a) 2

2)(x

7

dx

dy

3

B2

3

2)2(

13222

x

xx

dx

dy

(b)c

2x

64x

B1

3 y =

3

3x+ x

2 + x + 4

1 = 3

)3( 3 +

2

)3(2 2 + (–3) + c

y = 3

3x +

2

2 2x + x + c

3

B2

B1

3

4

3

B2

B1

3

5 546 2 xxy

c = ‒ 5

c

)1(42

)1(62

2

cxx

y 42

6 2

y = dxx 46

46 xdx

dy

3

B2

B1

3

SOALAN 19

BIL JAWAPAN SUB MARK MARKAH

PENUH

1 = 18

= 10 + (2(4) -2(0))

= 2 5 + [2x]4

0

24 4

0 0( ) 2f x dx dx

3

B2

B1

3

2 22

2

2

pp

)42(22

2

p

p

p

xx

2

2

22

3

B2

B1

3

3 5

12 – 7

seenataudxxfdxxg 7)()(4

2

4

2

3

B2

B1

3

4 h = 3

3]5)3(2[

h –

3]5)2(2[

h = 7

32

3

2

3)(

)52(xorimitlcorrectthewith

x

h

3

B2

B1

3

5 −209

−11(19)

66 + ∫b,6f(x) dx = 85

3

B2

B1

3

SOALAN 20

BIL JAWAPAN SUB MARK MARKAH

PENUH

1 (a) 1.............................1

(b) 2..................................9

1................0))(1(4)6( 2

q

Bq

1

B1

2

3

2 p = -16 and k = 5

2(3)2+3p+30 = 0

2x2 -16x + 30 = 0

(x - 3)(x - 5) = 0

3 (a) x − 3 or x 1

0)1)(3( xx or

0322 xx

3

B2

B1

3

4 x ≤ −4 or x ≥ 7 3 3

- 3 1

(x + 4)(x − 7) ≤ 0

B2

B1

5

3

B2

B1

3

SOALAN 21

BIL JAWAPAN SUB

MARK

MARKAH

PENUH

1

6)1( 2 x

26)10(4 2 aor

.6)1(2)( 2 xxf

B1

B2

3

2 h = −1 and k = −10

(x − 1)2 – 10

2(x2 − 2x + (−1)

2 − (−1)

2 − 4)

3

B2

B1

3 a) 1

1/2 -1/ 3

b)

c)

1

1

4 (a) 1)1( 2 xy

2)2

2()

2

2(2 222 xx

(c) ( −1,1)

2

B1

1

5 610 p

0)6)(10( pp or

0)8)(2(4)2( 2 pp

3

B2

B1

SOALAN 22

BIL JAWAPAN SUB MARK MARKA

H

PENUH

1 (-2)2- 4(1)(3 – k) >0

4 – 12 + 4k > 0

k > 2

B1

B2

3

2 b2 − 4ac < 0 : Tiada punca/ No real roots:

−12q2

(2pq)2 − 4(p

2 + 3)(q

2)

3

B2

B1

-10 6

3

B1

B2

3

4

15 xy , pxxy 22

pxxx 2215

0)1(42 2 pxx

0)1)(2(4)4( 2 p

088 p

1p

B1

B2

3

5

1)3(2 xmx

01)3(2 xmx

Dua punca nyata, 042 acb

0)1)(1(4))3(( 2 m

0469 2 mm

0562 mm

0)5)(1( mm

5,1 mm

B1

B2

3

SOALAN 23

BIL JAWAPAN SUB

MARK

MARKAH

PENUH

1

B1

B2

3

2

n(x) = 1 - x

x=2

1

B1

2

1

3

3

2,

23

32

m

m

m

2)1

(

)1

(3

m

mm

m

)1

(m

mg

3

B2

B1

4

g(x)=2x+k

g2(x) = 2(2x+k)+k

= 4x+3k

∴ @ sebarang satu nilai

h /k (Both)

B1

B2

3

5 (a) b

(b) a

b

3

5

a

b

b

53

1

2

B1

SOALAN 24

BIL JAWAPAN SUB MARK MARKA

H

PENUH

a)

b)

1

B1

2

2 (a) k = 1

(b) t = 4, t = -3

0122 tt

1

2

B1

3 1. f(-1) = 1

p-q-5 =1

p-q = 6 p-q=6 / p+q =-4

f(1) = -9 +

p+q-5=-9

p+q = -4

sebarang satu set

2p=2

p=2

∴1+q=-4 both

q=-5

B1

B2

3

4 a) 1

b) x = 22.5, 112.5

2x = 45°, 225°

1

2

B1

5 x =

4

3

5x + 3 = 5

3x

f –1

= 5

3x

3

B2

B1

SOALAN 25

BIL JAWAPAN SUB

MARK

MARKAH

PENUH

1 a) 5

b)

c) Hubungan satu ke banyak

1

1

1

2 (a) or

(b) many to one

c)

1

1

1

3 a) banyak-banyak

b)

c)

1

1

1

4 a) Kodomain = { a, b, c, d, e }

c) Julat = { a, b, d, e }

1

1

1

5 a) {(5, g), (7, h), (9, h), (9, i), (11, g), (11, h), (11, i)}

b) Many-to-many relation Hubungan banyak dengan banyak

c) {g, h, i}

1

1

1