1. Q2 +6, R +2 ad
Transcript of 1. Q2 +6, R +2 ad
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1. π¦ β€ 2π₯ + 6, π¦ β₯ π₯ + 2 ad π¦ < 6
Make sure π¦ = 6 is dotted line. / Pastikan π¦ = 6 adalah garis titik-titik.
2. _
(a) β πππ
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(b) First find the length ππ. / Mula, cari panjang WU.
ππ = β82 + 62
= β100
= 10 ππ
tan β πππ =ππ
ππ
=6
10
β πππ = tanβ16
10
= 30.963 @ 30Β°58β²
3. Solve. / Selesaikan.
π₯(3π₯ β 1) = 4(2 + π₯)
3π₯2 β π₯ = 8 + 4π₯
3π₯2 β π₯ β 4π₯ β 8 = 0
3π₯2 β 5π₯ β 8 = 0
(3π₯ β 8)(π₯ + 1) = 0
π₯ = β1,8
3
4. Simultaneous linear equations. / Persamaan linear serentak.
1
2π£ + 2π€ = 1 β¦ β¦ (1)
2π£ β 3π€ = β7 β¦ β¦ (2)
1
2π£ = 1 β 2π€
π£ = 2(1 β 2π€)
π£ = 2 β 4π€ β¦ β¦ (3)
Substitute (3) into (2). / Masukkan (3) kedalam (2).
2(2 β 4π€) β 3π€ = β7
4 β 8π€ β 3π€ = β7
4 β 11π€ = β7
4 + 7 = 11π€
11π€ = 11
π€ =11
11
π = π
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Substitute π€ = 1 into (3). / Masukkan π€ = 1 kedalam (3).
π£ = 2 β 4π€
= 2 β 4(1)
π = βπ
5. Trapezium π΄π΅πΉπΈ.
π΄π΅ = π·πΆ = 14 ππ
πΈπΉ = π»πΊ = 12 ππ
π΄πΈ = π·π» = 7 ππ
π΄π· = π΅πΆ = 8 ππ
Cone π = 3, β = 6
Volume of the trapezium prism
1
2Γ (πΈπΉ + π΄π΅) Γ πΈπ΄ Γ π΄π·
=1
2Γ (12 + 14) Γ 7 Γ 8
= 728 ππ3
Volume of the cone.
1
3Γ πππ π ππππ Γ βπππβπ‘
=1
3Γ (
22
7) (3)2 Γ 6
= 564
7 ππ3
Volume of remaining solid.
728 β 564
7= 671
3
7 ππ3
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6. _
(a) β5 > β6 (ππ ππΈ) and 33 = 27 (ππ ππΈ)
So, TRUE.
(b) Implication 1: πΌπ π3 = β8, π‘βππ π = β2
Implikasi 1 π½πππ π3 = β8, ππππ π = β2
Implication 2 : πΌπ π = β2, π‘βππ π3 = β8
Implikasi 2: π½πππ π = β2, ππππ π3 = β8.
(c) 3π2 + 1 , π = 1, 2, 3, 4, β¦ β¦
7. ππ β₯ ππ
πππ = πππ
πππ =12
4
= 3
(a) π¦ = ππ₯ + π
β6 = 3(3) + π
= 9 + π
π = β6 β 9
= β15
Equation QR π = ππ β ππ
(b) At π₯ β πππ‘ππππππ‘, π¦ = 0.
π¦ = 3π₯ β 15
0 = 3π₯ β 15
15 = 3π₯
π₯ =15
3
π = π
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8. π½ = {π΄, πΉ, π} , πΎ = {7, 6, 9, 4}
(a) Sample space. / Ruang sample
{(π¨, π), (π¨, π), (π¨, π), (π¨, π), (π, π), (π, π), (π, π), (π, π), (π·, π), (π·, π), (π·, π), (π·, π)}
(b) _
(i) {(π, 7), (π, 9)}
2
12=
π
π
(ii) {(πΉ, 7), (πΉ, 6), (πΉ, 9), (πΉ, 4), (π΄, 7), (π, 7)}
6
12=
π
π
9. _
(a) Perimeter shaded region. / Ukur lilit kawasan berlorek.
πΆπ’ππ£π πππ + ππ’ππ£π πππ + π π
=180Β°
360°à 2 à (
22
7) Γ 7 +
360Β° β 240Β°
360°à 2 Γ
22
7Γ 14 + 14
= 22 + 291
3+ 14
= πππ
π ππ
(b) Area of shaded region. / Luas kawasan yang berlorek.
π πππ‘ππ ππππ β π πππ‘ππ ππππ
=120Β°
360Β°Γ
22
7Γ 142 β
1
2Γ
22
7Γ 72
= 2051
3β 77
= ππππ
π ππ2
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10. _
(a) Length of time car is stationary. / Tempoh masa kereta itu berhenti.
1.5 β 0.5 = π. π βππ’π
(b)
(i) The vehicles meet after 1.2 hours. / Kenderaan itu bertemu selepas 1.2 jam.
1.2 βππ’ππ = 1βππ’π 12ππππ’π‘ππ
6.30 π. π. +1βππ’π 12ππππ’π‘ππ = π. ππ π. π
(ii) The distance from town B. / Jarak antara bandar B.
120 β 50 = ππ ππ
(c) Average speed. / Purata halaju.
120
2.5= ππ ππββ1
11. π (2 π
β2 3) & π (
2 13 β2
)
(a) _
(i) The matric has no inverse if ππ β ππ = 0. / Matrik tersebut tidak mempunyai
songsang sekiranya ππ β ππ = 0
(2)(3) β (π)(β2) = 0
6 + 2π = 0
2π = β6
π =β6
2
= βπ
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(ii) Inverse of Q. / Songsang bagi Q.
1
ππ β ππ(
π βπβπ π
)
πβ1 =1
(2)(β2) β (1)(3)(
β2 β1β3 2
)
= βπ
π(
βπ βπβπ π
)
(b) 2π₯ + π¦ = 4
3π₯ β 2π¦ = 13
(2 13 β2
) (π₯π¦) = (
413
)
(π₯π¦) = β
1
7(
β2 β1β3 2
) (4
13)
= β1
7(
(β2)(4) + (β1)(13)(β3)(4) + 2(13)
)
= β1
7(
β2114
)
= (3
β2)
π = π, π = βπ
12. _
(a) Complete the table below. / Selesaikan jadual dibawah.
π¦ = β48
π₯
π₯ β4 β3.2 β2 β1 1 1.2 2 3.2 4
π¦ 12 15 24 48 β48 βππ β24 β15 βππ
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(b) _
(c) _
(i) π₯ = 1.5, π¦ = βππ
(ii) π¦ = 22, π₯ = βπ. π
(d) Refer Graph / Rujuk graf.
13. Transformation. / Penjelmaan.
(a) Transformation V is a translation of (24
).
Penjelmaan V ialah translasi (24
).
Transformation W is a rotation of 180Β° about centre (β4, 0).
Penjelmaan W ialah putaran 180Β° pada pusat (β4, 0).
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(i) π2 = (0, 10)
(ii) ππ = (β2, 2)
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(b) πππ π is the image of πΎπΏππ under a combined transformation ππ.
πππ π ialah imej bagi πΎπΏππ di bawah gabungan penjelmaan ππ.
(i) Transformation π is a reflection about line ππ.
Penjelmaan π ialah pantulan pada garisan ππ.
(ii) _
Transformation π is an enlargement with scale factor 3 about point (3, 3).
Penjelmaan π ialah pembesaran dengan skala faktor 3 pada titik (3, 3).
(c) π΄πππ πππ π = π΄πππ πΎπΏππ Γ π2
π΄πππ πππ π = 25.4 Γ 32
= 228.6
Area of shaded region. / Luas kawasan yang berlorek.
228.6 β 25.4 = πππ. π π’πππ‘2
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14. _
(a) _
Age (Year) / Umur
(Tahun) Midpoint / Titik Tengah
Frequency / Kekerapan
11 β 15 13 6
16 β 20 18 9
21 β 25 23 5
26 β 30 28 5
31 β 35 33 2
36 β 40 38 6
41 β 45 43 2
(b) Modal class = 16 β 20
(c) Mean / Min
13(6) + 18(9) + 23(5) + 28(5) + 33(2) + 38(6) + 43(2)
6 + 9 + 5 + 5 + 2 + 6 + 2
=875
35
= 25
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(d) _
(e) 9 + 6 = ππ visitors / Pelawat
15. _
(a) _
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(b) _
(i) _
(ii) _
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16. π (45Β°π, 100Β°πΈ), π (45Β°π, 30Β°πΈ), ππ ππ π‘βπ ππππππ‘ππ ππ π‘βπ ππππ‘β. / PR adalah diameter
bumi.
(a) Latitute R = 45Β°π
Longitude π = 100Β°πΈ β 180Β° = β80Β°πΈ = 80Β°π
πΉ (ππΒ°π΅, ππΒ°πΎ)
(b) 5100 nautical miles.
5100
60β²= 85Β°
45Β°π β 85Β° = β40Β°π = 40Β°π
πΊ (ππΒ°π΅, πππΒ°π¬)
(c) Difference longitude angle between π and π.
Perbezaan sudut longitude di antara P dan Q.
100Β° β 30Β° = 70Β°
Distance = 70Β° Γ 60β² Γ cos 45Β° = ππππ. ππ ππ
(d) π (45Β°π, 80Β°π) π (40Β°π, 100Β°πΈ)
Difference angle / Perbezaan sudut.
180Β° β 45Β° β 40Β° = 95Β°
π·ππ π‘ππππ π½ππππ = 95Β° Γ 60β² = 5700 ππ
670 =5700
π‘πππ
π‘πππ =5700
670
= π. π πππππ