Mate Tambahan > Skema Spm M3t Kertas 1 Set 2
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Matematik Tambahan SPM Skema Jawapan Kertas 1 set 2 1. a) f(4) = 2(4)+4 = 12 b)
[1][1] [1][1]
2x + 4 = 2 Î x = –1
2.
f(2) = 2
[1]
2k + 4(2) = 2 Î k = –3
[1]
40 + 3 x = x 5 + 2x
3.
[1]
x 2 + x − 20 = 0
⇒
(x + 5)(x − 4) = 0
atau
x=
−1 ±
(1) 2 − 4(1)(−20) 2(1)
[1]
x = –5 , 4 4.
k +5 = 1+ n 2k
Hasiltambah punca-punca
4k = n(1) 2k
Hasildarab punca-punca
[1]
⇒ n=2
k +5 =3 ⇒ k = 1 2k 5
a)
[1]
[1]
[1]
Persamaan paksi simetri
1 − 2 + − 2 x= 2 x+h=0 x = −h 5 ∴ −h = − 4 5 ∴h = 4
=
−5 4
[1]
[1]
1 ,0 atau ( 0,2 ) dalam 2
Gantikan mana2 titik, ( -2,0) atau − persamaan
y = − k + 2(x + h )
2
2
5 − k + 2 x + = 0 4 k = Learning To Score 2005 Jabatan Pelajaran Perak
9 8
[1] [1] Halaman
1
Matematik Tambahan SPM Skema Jawapan Kertas 1 set 2 6
( x + 1 ) (x – 5 ) ≥ 16.
+
X2 – 5x + x – 5 – 16 ≥ 0
+ [1]
X2 – 4x – 21 ≥ 0
-3
7 –
(x–7)(x+3)≥0
[1]
x ≤ –3 atau x ≥ 7 7.
log y 2 = x ⇒
[1]
log 8 2 =x log 8 y
[1]
1
log 8 (8)3 ⇒ =x log 8 y ⇒
8
log 8 y =
1 3x
[1]
y = 1 log 3 2y + 5
[1]
y = -3
[1]
3 x + 2 = 216 + 3 x
(3)x (3)2 − 3 x
= 216
8(3) = 216 x
3 x = 27 ⇒ 3 x = 33 10
[1]
log 3 y = 1 + log 3 (2 y + 5) log 3 y − log 3 (2 y + 5) = 1
9
[1]
T8 = S 8 − S 7 =
=
1 1 ( (32(7) − (7) 2 ) 32(8) − (8) 2 ) 8 8 2
1 8
Learning To Score 2005 Jabatan Pelajaran Perak
[1] [1] [1][1] [1] [1] [1] Halaman
2
Matematik Tambahan SPM Skema Jawapan Kertas 1 set 2 11
k +1 k + 4 = k − 3 k +1
[1]
(k + 1)2 = (k − 3)(k + 4) [1]
k = -13 12
13
a 1− r 1 S∞ = 2 1 1 − 4 2 = 3 S∞ =
m=
[1]
[1]
3+5 = −2 0−4
[1]
log10 y = log10 p − x log10 q pintasan -y = 3 = log10 p
p = 10 3 = 1000 kecerunan = − log10 q = −2
q = 10 2 = 100 14
[1] [1] [1]
2x + 3y = 6 y= -
M AD = −
x=0
2 x +2 3
2 3
3y = 6 y=2
[ 1, kedua-dua ]
A(0,2) M AD × M BA = −1 3 2 M BA = = 2 −n 4 n=− 3 Learning To Score 2005 Jabatan Pelajaran Perak
[1] [1] Halaman
3
Matematik Tambahan SPM Skema Jawapan Kertas 1 set 2 15
1 4−6 3− 2 , = −1 , 2 2 2
a)
Koordinat T =
b)
ST : TR = 2 : 3
1 2m + 3(−3) 2n + 3(5) , = 2 2+3 2+3
T −1 ,
−1 =
2m + 3(−3) 1 2n + 3(5) , = 2+3 2 2+3
m = 2
,
25 4
a)
[1]
[1] [1]
[2]
a) 2 a + 2 b ~
17
n = –
− 25 4
R = 2 ,
16
[1]
~
t = ku ~
~
3 i + m j = k (2 i + 3 j ) ~
~
2k=3
,
3k = m
3 2
,
3 3 = m 2
k=
[1] [1]
9 2
m =
18
~
~
[1]
4 sin ( x – 15o) kos ( x – 15o) =
3
2 sin ( x – 15o) kos ( x – 15o) =
3 2
sin 2 ( x – 15 )
=
3 2
2 x –30o = 60o , 120o , 420o , 480o x
Learning To Score 2005 Jabatan Pelajaran Perak
= 45 , 75 , 225 ,255
[1] [1] [1]
Halaman
4
Matematik Tambahan SPM Skema Jawapan Kertas 1 set 2 19
0.65 radian = 37o15’
12 = 15.0753 kos37 o15′
OQ =
[1]
Perimeter kawasan berlorek = PQ + QR
[1]
+ SPR
= 9.1225 + ( 15.0753 – 12 ) + 12 ( 0.65 )
[1]
= 19.9978cm
[1]
y = 5t 4 − 4t 3 dan
20
x = t2
dx dy = 2t , = 20 t3 – 12 t2 dt dt
[ 1, salah satu ]
dy dy dt = × dx dt dx =
(20t
=
10t 2 − 6t
3
)
1 − 12t 2 2t
[1] [1]
f ′( x) = 6(2 x 3 + 1) (6 x 2 ) = 36 x 2 (2 x 3 + 1)
21
5
[1]
5
f ′′( x) = (2 x 3 + 1) (72 x ) + 36 x 2 (5)(2 x 3 + 1) (6 x 2 )
[1]
f ′′(1) = (3) (72) + 36(5)(3) (6)
[1]
5
4
5
4
[1]
= 104976 22
8
a)
∫ g ( x)dx
4
=
2
2
= 8
b)
∫ g ( x)dx + 5
∫ g ( x)dx
[1]
8
+
14
x2 m 2 2
∫ mx dx
= 43
2
8
+
m
Learning To Score 2005 Jabatan Pelajaran Perak
[1]
4
+ 9 = 14
∫ g ( x)dx 2
8
= 43
=
29 30
[1]
[1] Halaman
5
Matematik Tambahan SPM Skema Jawapan Kertas 1 set 2 23
a)
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
[1]
b)
2x5x 4x3x2x1
[1]
c)
1x 4x 3x2x1+ 1x 4x 3x2x1
[1]
= 48
[1]
24
a)
= 240
P ( LLL ) = P(L) x P(L) x P(L) =
b)
3 7
x
2 5
1 2
=
3 35
[1]
P ( GLL ) = P(G) x P(L) x P(L)
4 2 x 7 5 4 = 35 =
25
x
x
[1]
1 2
P ( X ≥ 2) = 1 − [P( X = 0) + P( X = 1)]
6 100
= 1 − 10C0
=
0.061265
Learning To Score 2005 Jabatan Pelajaran Perak
0
10
1
94 10 6 + C1 100 100
94 100
[1]
9
[1] [1]
Halaman
6
[1][1] [1][1]
2x + 4 = 2 Î x = –1
2.
f(2) = 2
[1]
2k + 4(2) = 2 Î k = –3
[1]
40 + 3 x = x 5 + 2x
3.
[1]
x 2 + x − 20 = 0
⇒
(x + 5)(x − 4) = 0
atau
x=
−1 ±
(1) 2 − 4(1)(−20) 2(1)
[1]
x = –5 , 4 4.
k +5 = 1+ n 2k
Hasiltambah punca-punca
4k = n(1) 2k
Hasildarab punca-punca
[1]
⇒ n=2
k +5 =3 ⇒ k = 1 2k 5
a)
[1]
[1]
[1]
Persamaan paksi simetri
1 − 2 + − 2 x= 2 x+h=0 x = −h 5 ∴ −h = − 4 5 ∴h = 4
=
−5 4
[1]
[1]
1 ,0 atau ( 0,2 ) dalam 2
Gantikan mana2 titik, ( -2,0) atau − persamaan
y = − k + 2(x + h )
2
2
5 − k + 2 x + = 0 4 k = Learning To Score 2005 Jabatan Pelajaran Perak
9 8
[1] [1] Halaman
1
Matematik Tambahan SPM Skema Jawapan Kertas 1 set 2 6
( x + 1 ) (x – 5 ) ≥ 16.
+
X2 – 5x + x – 5 – 16 ≥ 0
+ [1]
X2 – 4x – 21 ≥ 0
-3
7 –
(x–7)(x+3)≥0
[1]
x ≤ –3 atau x ≥ 7 7.
log y 2 = x ⇒
[1]
log 8 2 =x log 8 y
[1]
1
log 8 (8)3 ⇒ =x log 8 y ⇒
8
log 8 y =
1 3x
[1]
y = 1 log 3 2y + 5
[1]
y = -3
[1]
3 x + 2 = 216 + 3 x
(3)x (3)2 − 3 x
= 216
8(3) = 216 x
3 x = 27 ⇒ 3 x = 33 10
[1]
log 3 y = 1 + log 3 (2 y + 5) log 3 y − log 3 (2 y + 5) = 1
9
[1]
T8 = S 8 − S 7 =
=
1 1 ( (32(7) − (7) 2 ) 32(8) − (8) 2 ) 8 8 2
1 8
Learning To Score 2005 Jabatan Pelajaran Perak
[1] [1] [1][1] [1] [1] [1] Halaman
2
Matematik Tambahan SPM Skema Jawapan Kertas 1 set 2 11
k +1 k + 4 = k − 3 k +1
[1]
(k + 1)2 = (k − 3)(k + 4) [1]
k = -13 12
13
a 1− r 1 S∞ = 2 1 1 − 4 2 = 3 S∞ =
m=
[1]
[1]
3+5 = −2 0−4
[1]
log10 y = log10 p − x log10 q pintasan -y = 3 = log10 p
p = 10 3 = 1000 kecerunan = − log10 q = −2
q = 10 2 = 100 14
[1] [1] [1]
2x + 3y = 6 y= -
M AD = −
x=0
2 x +2 3
2 3
3y = 6 y=2
[ 1, kedua-dua ]
A(0,2) M AD × M BA = −1 3 2 M BA = = 2 −n 4 n=− 3 Learning To Score 2005 Jabatan Pelajaran Perak
[1] [1] Halaman
3
Matematik Tambahan SPM Skema Jawapan Kertas 1 set 2 15
1 4−6 3− 2 , = −1 , 2 2 2
a)
Koordinat T =
b)
ST : TR = 2 : 3
1 2m + 3(−3) 2n + 3(5) , = 2 2+3 2+3
T −1 ,
−1 =
2m + 3(−3) 1 2n + 3(5) , = 2+3 2 2+3
m = 2
,
25 4
a)
[1]
[1] [1]
[2]
a) 2 a + 2 b ~
17
n = –
− 25 4
R = 2 ,
16
[1]
~
t = ku ~
~
3 i + m j = k (2 i + 3 j ) ~
~
2k=3
,
3k = m
3 2
,
3 3 = m 2
k=
[1] [1]
9 2
m =
18
~
~
[1]
4 sin ( x – 15o) kos ( x – 15o) =
3
2 sin ( x – 15o) kos ( x – 15o) =
3 2
sin 2 ( x – 15 )
=
3 2
2 x –30o = 60o , 120o , 420o , 480o x
Learning To Score 2005 Jabatan Pelajaran Perak
= 45 , 75 , 225 ,255
[1] [1] [1]
Halaman
4
Matematik Tambahan SPM Skema Jawapan Kertas 1 set 2 19
0.65 radian = 37o15’
12 = 15.0753 kos37 o15′
OQ =
[1]
Perimeter kawasan berlorek = PQ + QR
[1]
+ SPR
= 9.1225 + ( 15.0753 – 12 ) + 12 ( 0.65 )
[1]
= 19.9978cm
[1]
y = 5t 4 − 4t 3 dan
20
x = t2
dx dy = 2t , = 20 t3 – 12 t2 dt dt
[ 1, salah satu ]
dy dy dt = × dx dt dx =
(20t
=
10t 2 − 6t
3
)
1 − 12t 2 2t
[1] [1]
f ′( x) = 6(2 x 3 + 1) (6 x 2 ) = 36 x 2 (2 x 3 + 1)
21
5
[1]
5
f ′′( x) = (2 x 3 + 1) (72 x ) + 36 x 2 (5)(2 x 3 + 1) (6 x 2 )
[1]
f ′′(1) = (3) (72) + 36(5)(3) (6)
[1]
5
4
5
4
[1]
= 104976 22
8
a)
∫ g ( x)dx
4
=
2
2
= 8
b)
∫ g ( x)dx + 5
∫ g ( x)dx
[1]
8
+
14
x2 m 2 2
∫ mx dx
= 43
2
8
+
m
Learning To Score 2005 Jabatan Pelajaran Perak
[1]
4
+ 9 = 14
∫ g ( x)dx 2
8
= 43
=
29 30
[1]
[1] Halaman
5
Matematik Tambahan SPM Skema Jawapan Kertas 1 set 2 23
a)
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
[1]
b)
2x5x 4x3x2x1
[1]
c)
1x 4x 3x2x1+ 1x 4x 3x2x1
[1]
= 48
[1]
24
a)
= 240
P ( LLL ) = P(L) x P(L) x P(L) =
b)
3 7
x
2 5
1 2
=
3 35
[1]
P ( GLL ) = P(G) x P(L) x P(L)
4 2 x 7 5 4 = 35 =
25
x
x
[1]
1 2
P ( X ≥ 2) = 1 − [P( X = 0) + P( X = 1)]
6 100
= 1 − 10C0
=
0.061265
Learning To Score 2005 Jabatan Pelajaran Perak
0
10
1
94 10 6 + C1 100 100
94 100
[1]
9
[1] [1]
Halaman
6
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