Spm > Matematik Tambahan > Kertas1 Set1 Jawapan

Matematik Tambahan SPM Skema Jawapan Kertas 1 Set 1 1.

(a) hk

–1

(b) kh

2. (a) p

p

–1

(– 4) = – 2

[1]

(– 2) = – 4

[1]

–1

3x 4x

(x) =

–1

3y

(y) = 4 y

3y 4 y

x =

x(4–y) = 3y 4x – xy = 3y y(x+3) = 4x

4x x3

y =

p(x)

(b)

=

p (5) 

=

3.

[1]

4x x3

,

x ≠ –3

[1]

4(5) 53

5 2

[1]

g(x) = 4 – 3x y = 4 – 3x 3x = 4–y x = g – 1(x) =

f(x) = fgg

4 y 3 4 x 3

–1

[1]

(x) =

4  x

fg   3 

= 2

4  x  3 5  

[1]

=

23  2 x 3

[1]

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Halaman 1

Matematik Tambahan SPM Skema Jawapan Kertas 1 Set 1

4.

y = 3 + 5x – 4x2

y = t ,

3 + 5x – 4x2 =

t

4x2–5x +t–3 = 0

[1]

Bersilang pada dua titik , guna (–5)2 – 4(4)(t–3) > 16t

5.

b

0

<

73

t <

73 16

2

– 4ac > 0 [1]

[1]

5(2x–1) = 3(x2 – 2) 3 x 2 – 10 x – 1 = 0 Guna

x =

[1]

 b  b 2  4ac 2a

x =

 (10)  (10) 2  4(3)(1) 2(3)

x =

10  112 6

x =

3.43,

x = – 0. 0972

[1]

[ 1 ] kedua – dua

g (x) = 4 – 2 ( x + h ) 2

6.

x+h = 0 ,

x = –h

(–h , 4 )

= (3 , k )

;

y = 4

(a)

h = –3

[1]

(b)

k = 4

[1]

(c)

Paksi simetri :

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x = 3

[1]

Halaman 2

Matematik Tambahan SPM Skema Jawapan Kertas 1 Set 1 7 . 2

6 ( x+3 ) x ½

– 3x

= 2

3(x+3) = –3x 8 x = – 11 x = 

11 8

8 . log 2 ( x – 1 ) + log 2 4 ( x – 1 ) 4x – 4 x2–4x + 4 (x–2)2 x

2 = = = = =

9.

. 2

–2(x+ 1)

[1]

+ [ –2x – 2 ]

[1] [1]

2 = log 2 22 = log 2 4

= 2 log 2 x log 2 x 2 x2 0 0 2

[1]

[1] [1]

 32 x 3  log 2  2   log 2 32  log 2 x 3  log 2 y 2  y  = log 2 2 5  3 log 2 x  2 log 2 y

[1] [1]

= 5+3p–2r 10 .

Sn = Tn = T 5 = = = =

[1]

5n –3 S n – S n–1 S5 – S4 [ 5(5) – 3 ] – [ 5(4) – 3 ] 22 – 17 5

11 . (a) r =

2 = 1

[1] [1]

– 2

[1]

(b) T 1 , T 2 , ……… T 5 , T 6 , ……….. T 15 Hasil tambah

 S 15 =

 S5 =

=

S

15

– S5

1 (2)15  1 2 = – 5461.5 (2)  1

1 (2) 5  1 2 = (2)  1

S 15 – S 5 =

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– 5456

– 5.5

;

[1]

[1] [1]

Halaman 3

Matematik Tambahan SPM Skema Jawapan Kertas 1 Set 1 12 . 47, 44, 41, ………… (a) d= 44 – 47 = –3 (b) Sn=

[1]

n [ 2 a  ( n  1) d ] 2

n [ 2( 47)  (n  1)( 3)] = 2 3n2 – 97n– 2650

–1325

[1]

= 0

(3n + 53)(n – 50) = 0 n = 50

[1] [1]

13 . y = p x q ( a ) log

10

y = log

log

10

p = 2

p + q log 10 x

p = 10 2 = 100

[1]

20 1  04 2

[1]

q = ( b ) log

10

10

y = 2 + ½ log

x = 10 , log 10 y log 10 y y y

= = = =

10

x

2 + ½ log 10 10 2.5 10 2.5 316 . 2

[1]

[1]

Kaedah Alternatif y= pxq = 100 x ( 10 ) ½ = 316 . 2 14 .

p q x 2 2 2 q 1 = x 3 3 q 1  3 2 = ( q  1) 3

y = y

p 2 p

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[1] [1]

Halaman 4

Matematik Tambahan SPM Skema Jawapan Kertas 1 Set 1



8 i  5 j

15 . ( a ) OB =

~

[1]

~



( b ) Vektor unit dalam arah OB



16 . ( a ) AB

= =



( b ) OR

=

 OR

=

 OR

17

=

 8i  5 j

[1]

89

  AO + OB   2    3 

– 

10    6

12     3

[1]

2(OB)  1(OA) 2 1

  2  1  10  2    1  21   6   3 

 6

=   5

[1]

[1]

 

1 . + sek 2 x = 3 kot 2 x tan 2 x + sek 2 x = 3

[1]

tan 2 x + 1 + tan 2 x = 3 2 tan 2 x = 2 tan 2 x = 1 tan x = ± 1 x = 45 o , 135 o , 225 o , 315 o

4 18 . ( a ) kos POR = = 0.8 5 POR = 0 . 6435 r

[1] [1,1] [1] [1]

( b ) s PQ = j  r = 5 x 0.6435 = 3 . 2175 RQ = 1 cm ,

PR = 3 cm ( Teorem Pythagoras )

Perimeter kawasan berlorek = 3 + 1 + 3. 2175 = 7 . 2175

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[1]

[1]

Halaman 5

Matematik Tambahan SPM Skema Jawapan Kertas 1 Set 1

19 . f ( x ) =

4 x (5  2 x ) 2

u = 4x

v =(5–2x)2

du 4 dx

dv  ( 2)(5  2 x )( 2) dx

dy dv du u v dx dx dx f ' ( x)  4 x[ 4(5  2 x)]  (5  2 x) 2 (4)

[1]

= 4 8 x2 – 160 x + 100

[1]

f ' ' ( x)  96 x  160

20 .

[ 1 ] salah satu

(a) x = t + 3 t 2

dx  1  6t dt

[1]

y = 2t – 1

dy 2 dt

[ 1 ] salah satu

dx dx dt   dy dt dy  (1  6t )  

1 2

[1]

1 (1  6t ) 2

(b) y  3.98  4 = -0.02

[1]

dy 2 dt dy y  dt t y t  dy dt

 0.02 2  0.01 

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[1]

Halaman 6

Matematik Tambahan SPM Skema Jawapan Kertas 1 Set 1 21 .

 dy   4 f ( x )dx

3 dy   3 f ( x)dx 4 2 2 3 1  2 x   3 f ( x)dx  4  3x 2   2 2

[1]

3 1  4 1  4   4  12 12  1 = 2

=

22 . ( a )

6

[1] [1]

C 3  4 C 2  20  6 = 120

[1]

( b ) Jumlah pemain lelaki dan perempuan = 6 + 4 = 10 Bilangan cara pilih 5 orang pemain =

10

C 5  252

[1]

Bilangan cara pilih 1 orang pemain lelaki dan 4 orang pemain perempuan =

6

C1  4 C 4  6  1  6

[1]

Bilangan pasukan yang mesti mengandungi sekurang – kurangnya 2 orang pemain lelaki =

23 .

xm ;

2 

x

2

x

2



 x

 720

m 2 = 144 – 9h2 =

246

[1]

;  2  9h 2

2

n 720 9h2=  m2 5

m

252 – 6 =

144  9h 2

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[1] [1] [1]

Halaman 7

Matematik Tambahan SPM Skema Jawapan Kertas 1 Set 1 24 . ( a ) Kebarangkalian bahawa susunan itu ialah TUPS =

1 1 1 1 1     6 5 4 3 360

[1]

( b ) Kebarangkalian bahawa susunan itu tidak mengandungi hiasan P 5

= =

6

C4 C4 1 3

[1] [1]

25 . varians = 2 = 9 sisihan piawai =  = 3 min =  = 10 P ( X < r ) = 0 . 975 Z =

X  

r  10   X  10 P    = 0 . 975 3   3

[1]

r  10   P Z   = 0 . 975 3   r  10   1 – P Z   = 0 . 975 3   r  10   P Z   3  

[1]

= 1 – 0. 975 = 0 . 025

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= 1.96

[1]

=

[1]

15.88

Halaman 8