spm matematik tambahan gerak gempur pertama 2012
DESCRIPTION
soalan gerak gempur yang berformat SPMTRANSCRIPT
SMK KOTA MASAI 2
PEPERIKSAAN GERAK GEMPUR SPM 2012 3472/1PERTAMA
MATEMATIK TAMBAHANKertas 1Julai 20122 jam Dua jam
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1
NAMA:_________________________________
KELAS:_________________________________
1. Tulis nombor kad pengenalan dan angka giliran anda pada petak yang disediakan
2. Kertas soalan ini adalah dalam dwibahasa.
3. Soalan dalam bahasa Inggeris mendahului soalan yang sepadan dalam bahasa Melayu
4. Calon dibenarkan menjawan keseluruhan atau sebahagian soalan sama ada dalam bahasa Inggeris atau bahasa Melayu
5. Calon dikehendaki membaca maklumat di halaman belakang kertas soalan.
Untuk Kegunaan PemeriksaSoalan Markah Penuh Markah Diperolehi
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JUMLAH 80
Kertas soalan ini mengandungi 20 halaman bercetak
1. Given g : x |2x – 6|, findDiberi g : x |2x – 6|, cari
(a) g(8). [1 mark/1 markah]
(b) g(–3). [1 mark/1 markah]
(c) the object of 4.objek bagi 4.
[2 marks/2 markah]Answer/Jawapan:(a)
(b)
(c)
2. Given g : x → x – 6 and the composite function gf : x → 2x – 3, findDiberi g : x → x – 6 dan fungsi gubahan gf : x → 2x – 3, cari(a) f (x).(b) the value of x when fg(x) = –13.
nilai x apabila fg(x) = –13.[4 marks/4 markah]
Answer/Jawapan:(a)
(b)
3. Given the functions g : x → 5x + 2 and h : x → x2 – 4x + 3, findDiberi fungsi g : x → 5x + 2 dan h : x → x2 – 4x + 3, cari
(a) g–1(6).
(b) hg(x). [4 marks]
[4 markah]
Answer/Jawapan: (a) .....................................
2
(b) .....................................
4.Given –3 and are the roots of an equation 4x² + bx + c = 0.Find the values of b and c.
Diberi –3 dan ialah punca-punca persamaan 4x² + bx + c = 0.Cari nilai b dan nilai c.
[3 marks/3 markah]
Answer/Jawapan:
5. If one root of the quadratic equation x2 – 2x + 3k = 0 is the reciprocal of the other root, findJika satu punca persamaan kuadratik x2 – 2x + 3k = 0 ialah salingan punca yang satu lagi, cari
(a) the value of k. nilai k.
[2 marks/2 markah](b) the two roots of the equation.
kedua-dua punca persamaan itu. [2 marks/2 markah]
Answer/Jawapan:(a)
(b)
3
6. (a) Solve the following quadratic equation:Selesaikan persamaan kuadratik berikut:
3x2 + 5x – 2 = 0
(b) The quadratic equation hx2 + kx + 3 = 0, where h and k are constants, has two equal roots.Express h in terms of k.Persamaan kuadratik hx2 + kx + 3 = 0, dengan keadaan h dan k ialah pemalar, mempunyai dua punca sama.Ungkapkan h dalam sebutan k.
[4 marks][4 markah]
Answer/Jawapan: (a) ......................................(b) ......................................
7. Find the range of the values of x for which x(x – 3) < 4(3 – x).Cari julat nilai x supaya x(x – 3) < 4(3 – x).
[2 marks/2 markah]
Answer/Jawapan:
8. Diagram shows the graph of a quadratic function f(x) = (x + h)2 + k, where h and k are constants. The graph of f (x) has a minimum point (1, 2) and intersects the y-axis at y = 3.Rajah menunjukkan graf bagi suatu fungsi kuadratik f(x) = (x + h)2 + k, dengan h dan k ialah pemalar. Graf f(x) mempunyai titik minimum (1, 2) dan menyilang paksi-y di y = 3.
Diagram/Rajah
4
(a) Find the values of h and k.
Cari nilai h dan nilai k.(b) If the graph of f (x) is reflected in the y-axis, write the function represented by the resulting graph.
Jika graf f(x) itu dipantulkan pada paksi-y, tulis fungsi yang diwakili oleh graf yang terhasil.[4 marks/4 markah]
Answer/Jawapan:(a)
(b)
9. The quadratic function f (x) = –x2 + 4x + a2, where a is a constant, has maximum value 8.Find the values of a.
[3 marks]Fungsi kuadratik f(x) = –x2 + 4x + a2, dengan keadaan a ialah pemalar, mempunyai nilai maksimum 8.Cari nilai-nilai yang mungkin bagi a.
[3 markah]
Answer/Jawapan: a = ....................................
10. Given 4(23x) = 81 – x, find the value of x.Diberi 4(23x) = 81 – x, cari nilai x.
[3 marks/3 markah]
5
Answer/Jawapan:
11. Solve the equation 63x − 2 = 8x.Selesaikan persamaan 63x − 2 = 8x.
[4 marks/4 markah] Answer/Jawapan:
12. The points A(–1, p), B(2, –1) and C(4, 5) are collinear.Find the value of p.Titik-titik A(–1, p), B(2, –1) dan C(4, 5) adalah segaris.Cari nilai p.
[2 marks/2 markah]
Answer/Jawapan:
13. A set of data consists of six numbers. The sum of the numbers is 72 and the sum of the squares of the numbers is 944.Satu set data mempunyai enam nombor. Hasil tambah bagi nombor-nombor itu ialah 72 dan hasil tambah bagi kuasa dua nombor-nombor itu ialah 944.
Find, for the six numbers,Cari, bagi enam nombor itu, (a) the mean.
min. [1 mark/1 markah]
(b) the standard deviation.sisihan piawai.
6
[2 marks/2 markah]
Answer/Jawapan:(a)
(b)
14. A set of positive integers consists of 2, 6 and h. The variance for this set of integers is
Find the value of h.
Satu set integer positif terdiri daripada 2, 6, dan h. Varians bagi set integer ini ialah
Cari nilai h. [3 marks/markah]
Answer/Jawapan: h = .................................................
15. Diagram shows a sector OPQ with centre O and a radius of 4 cm.Rajah menunjukkan sektor OPQ berpusat O dan berjejari 4 cm.
Diagram/Rajah
Given the area of the sector is 12 cm2.Diberi luas sektor itu ialah 12 cm2.
Find
7
Cari (a) the value of θ, in radians.
nilai θ, dalam radian.(b) the perimeter, in cm, of the sector.
perimeter, dalam cm, sektor itu.[3 marks/markah]
Answer/Jawapan:(a)
(b)
16. The curve y = –x2 – 8x + 14 has a maximum point at x = q, where q is a constant.Find the value of q.Lengkung y = –x2 – 8x + 14 mempunyai titik minimum di x = q, dengan q ialah pemalar.Cari nilai q.
[3 marks/markah]
Answer/Jawapan: q = ..................................................
17. The curve y = ax2 + bx has a gradient of 2 at the point (1, −2).Find the values of a and b.Lengkung y = ax2 + bx mempunyai kecerunan 2 pada titik (1, −2).Cari nilai a dan nilai b.
[4 marks/4 markah]
8
Answer/Jawapan:
18. It is given that −14, −10, −6, …, is an arithmetic progression.Find the sum of the first ten terms.Diberi bahawa −14, −10, −6, …, ialah satu janjang aritmetik.Cari hasil tambah sepuluh sebutan pertama.
[2 marks/2 markah]
Answer/Jawapan:
19. The first three terms of a geometric progression are 81, 54, 36.Find the sum to infinity of the geometric progression.Tiga sebutan pertama suatu janjang geometri ialah 81, 54, 36.Cari hasil tambah hingga sebutan ketakterhinggaan bagi janjang geometri itu.
[3 marks/markah]
Answer/Jawapan: .......................................................
20. The first three terms of an arithmetic progression are 3h, k, h + 2.Tiga sebutan pertama suatu janjang aritmetik ialah 3h, k, h + 2.
(a) Express k in terms of h.Ungkapkan k dalam sebutan h.
(b) Find the 10th term of the progression in terms of h.Cari sebutan ke-10 bagi janjang itu dalam sebutan h.
[4 marks][4 markah]
Answer/Jawapan:(a)
9
(b)
21. Diagram shows a straight line graph of lg y against x.Rajah menunjukkan graf garis lurus lg y melawan x.
Diagram/Rajah
Express y in terms of x.Ungkapkan y dalam sebutan x.
[3 marks/3 markah]Answer/Jawapan:
22.
Given and , find the value of
Diberi dan , cari nilai [2 marks/2 markah]
10
Answer/Jawapan: ......................................
23.Diagram shows two vectors, and in a Cartesian plane.
Rajah menunjukkan dua vektor dan pada satu satah Cartesan.
Diagram/Rajah
(a)Express in the form .
Ungkapkan dalam bentuk .(b)
Given find the values of m and n.
Diberi cari nilai m dan nilai n.[2 marks/2 markah]
Answer/Jawapan:(a)
(b)
24. Diagram shows seven letter cards.Rajah menunjukkan tujuh keping kad huruf.
Diagram/Rajah
11
FindCari(a) the number of arrangements, in a row, of all the cards which begin with a vowel.
bilangan susunan yang disusun dalam sebaris dengan semua kad itu yang bermula dengan huruf vokal.(b) the number of these arrangements which end with the letter V.
bilangan susunan itu yang berakhir dengan huruf V.[4 marks/4 markah]
Answer/Jawapan:(a)
(b)
25.The probability that Shariman and Jeffry wake up early on a certain day are and respectively.Find the probability thatKebarangkalian bahawa Shariman dan Jeffry bangun awal pagi pada satu hari tertentu
masing-masing ialah dan .Cari kebarangkalian bahawa
(a) both of them wake up early.kedua-dua mereka bangun awal pagi.
(b) only one of them wakes up early.hanya seorang daripada mereka bangun awal pagi.
[3 marks/3 markah]Answer/Jawapan:(a)
12
(b)
SMK KOTA MASAI 2
Jawapan
SPM Matematik Tambahan Tingkatan 4,5 - GERAK GEMPUR 1 2012 Kertas 1
1. (a) 10(b) 12(c) 1 or 5
2. (a) gf (x) = 2x – 3 g[f(x)] = 2x – 3 f(x) – 6 = 2x – 3 f(x) = 2x + 3(b) fg(x) = –13 f(x – 6) = –13 2(x – 6) + 3 = –13 2x – 12 + 3 = –13 2x = –4 x = –2
3.g(x) = 5x + 2h(x) = x2 – 4x + 3
(a) Let y = g–1(6) g(y) = 6 5y + 2 = 6 5y = 4
y =
Thus, g–1(6) =
(b) hg(x) = h(5x + 2) = (5x + 2)2 – 4(5x + 2) + 3 = 25x2 + 20x + 4 – 20x – 8 + 3 = 25x2 – 1
4.
x² – x + 3x – = 0
13
4x2 – x + 12x – 3 = 0 4x2 + 11x – 3 = 0 …… ①
Compare ① with 4x2 + bx + c = 0.Thus, b = 11 and c = –3.
5. (a) k =
(b) 1, 1
6. (a) 3x2 + 5x – 2 = 0(3x – 1)(x + 2) = 0
x = or –2
(b) hx2 + kx + 3 = 0
For two equal roots, b2 – 4ac = 0. k2 – 4h(3) = 0 k2 – 12h = 0
h =
7.
8. (a) The graph of f(x) = (x + h)2 + k has a minimum point (1, 2). Thus, h = –1 and k = 2.
(b) If the graph of f(x) is reflected in the y-axis, the minimum point is (–1, 2). The function represented by the resulting graph is f(x) = (x + 1)2 + 2.
9. f (x) = –x2 + 4x + a2
= –(x2 – 4x) + a2
= –[x2 – 4x + (–2)2 – (–2)2] + a2
= –[(x – 2)2 – 4] + a2
= –(x – 2)2 + 4 + a2
Given the maximum value is 8.Thus, 4 + a2 = 8 a2 = 4 a = –2 or 2
10. 4(23x) = 81 – x
22(23x) = 23(1 – x)
22 + 3x = 23 – 3x
2 + 3x = 3 – 3x6x = 1
x =
14
11. x =1.087
12. Given A(–1, p), B(2, –1) and C(4, 5) are collinear.
13. (a) 12(b) 3.651
14. 7
15. (a)
(b) Perimeter of sector= 2r + rθ= 2(4) + 4(1.5)= 14 cm
16. –4
17. a = 4, b = –6
18. 40
19. 243
20. Arithmetic progression: 3h, k, h + 2, ...
(a) d = k – 3h = (h + 2) – k k – 3h = h + 2 – k 2k = 4h + 2 k = 2h + 1
(b) a = 3h and d = k – 3h = (2h + 1) – 3h = 1 – h
T10 = a + 9d = 3h + 9(1 – h) = 3h + 9 – 9h = 9 – 6h
The 10th term is 9 – 6h.
21.Gradient of the straight line = = 2
15
Y-intercept = 1The equation of the straight line isY = mX + clg y = 2x + 1Thus, y = 102x + 1
22. –21
23. (a)
(b)
Thus, m = 13 and n = –1.
24. (a) Number of arrangements = 4 × 6! = 2 880
(b) Number of arrangements = 1 × 6! = 720
25. (a) P(Both of them wake up early)
(b) P(Only one of them wakes up early)
16