Firasat SPM - Matematik Tambahan

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Format Peperiksaan SPM Matematik Tambahan (3472) Bil. Perkara Kertas 1 (3472/1) Kertas 2 (3472/2) 1 Jenis Instrumen Ujian Bertulis 2 Jenis Item • Subjektif Respons Terhad • Subjektif Respons Terhad Berstruktur 3 Bilangan Soalan Bahagian A 12 soalan (64 markah) (Jawab semua soalan) Bahagian B 3 soalan (16 markah) (Jawab dua soalan) Bahagian A 7 soalan (50 markah) (Jawab semua soalan) Bahagian B 4 soalan (30 markah) (Jawab tiga soalan) Bahagian C 4 soalan (20 markah) (Jawab dua soalan) 4 Jumlah Markah 80 100 5 Konstruk • Mengingat & Memahami • Mengaplikasi • Menganalisis • Menilai • Mencipta • Mengingat & Memahami • Mengaplikasi • Menganalisis • Menilai • Mencipta 6 Tempoh Ujian 2 jam 2 jam 30 minit 7 Cakupan Konstruk Standard kandungan dan standard pembelajaran dalam Dokumen Standard Kurikulum dan Pentaksiran (DSKP) KSSM (Tingkatan 4 dan Tingkatan 5) 8 Aras Kesukaran Rendah : Sederhana : Tinggi 5 : 3 : 2 9 Kaedah Penskoran Analitik 10 Alatan Tambahan Kalkulator saintifik yang tidak boleh diprogram Firasat Add Math(Format).indd 1 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 3472/2 KM1–1 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT Kertas peperiksaan ini mengandungi 9 halaman bercetak. JANGAN BUKA KERTAS PEPERIKSAAN INI SEHINGGA DIBERITAHU MAKLUMAT UNTUK CALON INFORMATION FOR CANDIDATES 1. Kertas peperiksaan ini mengandungi dua bahagian: Bahagian A dan Bahagian B. This question paper consists of two sections: Section A and Section B. 2. Jawab semua soalan dalam Bahagian A dan mana-mana dua soalan daripada Bahagian B. Answer all questions in Section A and any two questions from Section B. 3. Tulis jawapan anda dalam ruang yang disediakan dalam kertas peperiksaan. Write your answers in the spaces provided in the question paper. 4. Tunjukkan langkah-langkah dalam kerja mengira anda. Ini boleh membantu anda untuk mendapatkan markah. Show your working. It may help you to get marks. 5. Sekiranya anda hendak menukar jawapan, batalkan jawapan yang telah dibuat. Kemudian tulis jawapan yang baharu. If you wish to change your answer, cross out the answer that you have done. Then write down the new answer. 6. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. The diagrams in the questions provided are not drawn to scale unless stated. 7. Markah yang diperuntukkan bagi setiap soalan ditunjukkan dalam kurungan. The marks allocated for each question are shown in brackets. 8. Anda dibenarkan menggunakan kalkulator saintifik. You may use a scientific calculator. Untuk Kegunaan Pemeriksa Kod Pemeriksa: Bahagian Soalan Markah Penuh Markah Diperoleh A 1 2 2 4 3 5 4 4 5 8 6 6 7 6 8 7 9 6 10 5 11 5 12 6 B 13 8 14 8 15 8 Jumlah 80 2 jam Dua jam 3472/1 SIJIL PELAJARAN MALAYSIA 2023 Kertas Model SPM SULIT 2 3 4 5 6 7 8 1 MATEMATIK TAMBAHAN Kertas 1 NO. KAD PENGENALAN ANGKA GILIRAN Firasat Add Math(S1).indd 1 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

KM1–2 SULIT SULIT 3472/1 Firasat SPM 2023: Matematik Tambahan 3472/1 Untuk Kegunaan Pemeriksa Bahagian A Section A [64 markah/marks] Jawab semua soalan. Answer all questions.  1 Tentukan jenis punca bagi persamaan kuadratik 2x2 – 5x + 3 = 0 tanpa menyelesaikan persamaan itu. Determine the type of roots of the quadratic equation 2x2 – 5x + 3 = 0 without solving the equation. [2 markah/marks] Jawapan/Answer:  2 Rajah 1 menunjukkan sektor POQ dan sektor ROS berpusat O. Diberi bahawa OP = 18 cm dan POQ = 2 3 radian. Diagram 1 shows sectors POQ and ROS with the same centre O. It is given that OP = 18 cm and POQ = 2 3 radian. 18 cm 2 3 O rad P R S Q Rajah 1/Diagram 1 Jika nisbah luas sektor POQ kepada luas kawasan berlorek PQSR ialah 9 : 7, cari panjang OS, dalam cm. If the ratio of the area of sector OPQ to the area of the shaded region PQSR is 9 : 7, find the length of OS, in cm. [4 markah/marks] Jawapan/Answer: 1 2 2 4 Firasat Add Math(S1).indd 2 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

KM1–3 [Lihat halaman sebelah SULIT SULIT © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 3472/1 Untuk Kegunaan Pemeriksa  3 Diberi ln (x + 1) – ln x = 2, cari nilai x. Given that ln (x + 1) – ln x = 2, find the value of x. [Guna/Use e = 2.718] [5 markah/marks] Jawapan/Answer:  4 Rajah 2 menunjukkan sebahagian daripada graf garis lurus penyuaian terbaik log2 y melawan log2 x. Diagram 2 shows part of the line of best fit of log2 y against log2 x. Ungkapkan y dalam sebutan x. Express y in terms of x. [4 markah/marks] Jawapan/Answer:  5 Hasil tambah dua sebutan pertama bagi suatu janjang geometri ialah 150. Sebutan ketiga melebihi sebutan kedua sebanyak 45. The sum of the first two terms of a geometric progression is 150. The third term exceeds the second term by 45. (a) Cari dua nilai nisbah sepunya yang mungkin bagi janjang itu. Find two possible values of common ratio of the progression. [5 markah/marks] (b) Cari sebutan pertama bagi janjang geometri di 5(a) dengan keadaan hasil tambah ketakterhinggaan wujud. Seterusnya, cari hasil tambah ketakterhinggaan itu. Find the first term of the geometric progression in 5(a) such that the sum to infinity exists. Hence, calculate the sum to infinity. [3 markah/marks] Jawapan/Answer: (a) (b) 4 4 5 8 3 5 (–3, 4) (–2, 1) log2 y log2 x O Rajah 2/Diagram 2 Firasat Add Math(S1).indd 3 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

KM1–4 SULIT SULIT 3472/1 Firasat SPM 2023: Matematik Tambahan 3472/1 Untuk Kegunaan Pemeriksa  6 Diberi bahawa tangen kepada lengkung y = 3x2 + hx + k pada titik (–2, 9) adalah berserenjang dengan garis lurus x – 10y + 5 = 0. Cari nilai h dan nilai k. It is given that the tangent to the curve y = 3x2 + hx + k at the point (–2, 9) is perpendicular to the straight line x – 10y + 5 = 0. Find the value of h and of k. [6 markah/marks] Jawapan/Answer:  7 Diberi bahawa garis lurus y = x + k tidak bersilang dengan bulatan y2 + x2 = 2. Cari julat nilai k. It is given that the straight line y = x + k does not intersect the circle y2 + x2 = 2. Find the range of values of k. [6 markah/marks] Jawapan/Answer:  8 Diberi fungsi f(x) = x + 1 x – 1 , x ≠ 1, g(x) = x + m dan (gf) –1(x) = x – 1 x – 3 , x ≠ 3, cari nilai m. Given that the functions f(x) = x + 1 x – 1 , x ≠ 1, g(x) = x + m and (gf)–1(x) = x – 1 x – 3 , x ≠ 3, find the value of m. [7 markah/marks] Jawapan/Answer: 8 7 7 6 6 6 Firasat Add Math(S1).indd 4 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

KM1–5 [Lihat halaman sebelah SULIT SULIT © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 3472/1 Untuk Kegunaan Pemeriksa  9 Penyelesaian secara lukisan berskala tidak diterima. Solution by scale drawing is not accepted. Rajah 3 menunjukkan garis lurus CD berserenjang dengan garis lurus AB dan bertemu pada titik D. Titik C terletak pada paksi-y. Diagram 3 shows a straight line CD which is perpendicular to the straight line AB and meets at point D. Point C lies on the y-axis. y x O C D B(15, 0) A(0, 5) Rajah 3/Diagram 3 (a) Tulis persamaan garis lurus AB dalam bentuk pintasan. Write the equation of AB in the intercept form. [1 markah/mark] (b) Diberi 3AD = 2DB, cari koordinat titik D. Given that 3AD = 2DB, find the coordinates of point D. [2 markah/marks] (c) Cari pintasan-y bagi garis lurus CD. Find the y-intercept of the straight line CD. [3 markah/marks] Jawapan/Answer: (a) (b) (c) 10 Diberi p ~ =  x 7 dan q ~ =  6 –2, cari nilai-nilai x jika  p ~ + q ~  = 13. Given that p ~ =  x 7 and q ~ =  6 –2, find the values of x if  p ~ + q ~  = 13. [5 markah/marks] Jawapan/Answer: 9 6 10 5 Firasat Add Math(S1).indd 5 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

KM1–6 SULIT SULIT 3472/1 Firasat SPM 2023: Matematik Tambahan 3472/1 Untuk Kegunaan Pemeriksa 11 Rajah 4 menunjukkan lengkung y = x2 + 2x – 1 bersilang dengan lengkung y = –x2 + 2x + 7. Diagram 4 shows the curve y = x2 + 2x – 1 intersects the curve y = –x2 + 2x + 7. y x O 2 –2 y = x2 + 2x – 1 y = –x2 + 2x + 7 Rajah 4/Diagram 4 Hitung luas rantau berlorek. Calculate the area of the shaded region. [5 markah/marks] Jawapan/Answer: 12 Cari bilangan cara huruf-huruf daripada perkataan “TESSELLATIONS” dapat disusun jika Find the number of ways in which the letters from the word “TESSELLATIONS” can be arranged if (a) tiada syarat, there is no condition, [2 markah/marks] (b) huruf vokal sentiasa bersama. the vowels are always together. [4 markah/marks] Jawapan/Answer: (a) (b) 11 5 12 6 Firasat Add Math(S1).indd 6 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

KM1–7 [Lihat halaman sebelah SULIT SULIT © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 3472/1 Untuk Kegunaan Pemeriksa Bahagian B Section B [16 markah/marks] Bahagian ini mengandungi tiga soalan. Jawab dua soalan. This section contains three questions. Answer two questions. 13 Seutas dawai dipotong kepada n bahagian. Panjang setiap bahagian bertambah secara berturutan mengikut janjang geometri. Diberi bahawa panjang dawai bahagian ketujuh ialah lapan kali panjang dawai bahagian keempat. A wire is cut into n parts. The length of each part increases successively according to the geometric progression. It is given that the length of the seventh part is eight times the length of the fourth part. (a) Cari nisbah sepunya. Find the common ratio. [3 markah/marks] (b) Jika jumlah panjang dawai itu ialah 3 069 cm dan panjang dawai bahagian pertama ialah 3 cm, cari If the total length of the wire is 3 069 cm and the length of the first part is 3 cm, find (i) nilai n, the value of n, (ii) panjang, dalam cm, dawai bahagian terakhir. the length, in cm, of the last part of the wire. [5 markah/marks] Jawapan/Answer: (a) (b) (i) (ii) 13 8 Firasat Add Math(S1).indd 7 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

KM1–8 SULIT SULIT 3472/1 Firasat SPM 2023: Matematik Tambahan 3472/1 Untuk Kegunaan Pemeriksa 14 Buktikan bahawa/Prove that sin 2x 1 + kos 2x/cos 2x = tan x [2 markah/marks] Seterusnya, cari nilai bagi setiap yang berikut dalam bentuk surd. Hence, find the value of each of the following in the form of surd. (a) tan 15° (b) tan 67.5° [6 markah/marks] Jawapan/Answer: (a) (b) 14 8 Firasat Add Math(S1).indd 8 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

KM1–9 [Lihat halaman sebelah SULIT SULIT © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 3472/1 Untuk Kegunaan Pemeriksa 15 8 15 Dalam Rajah 5, AGHB ialah sebuah sukuan bulatan berpusat A dan AHD ialah lengkok sebuah bulatan berpusat B dengan jejari 5 cm. Diberi bahawa CD = 3 cm. In Diagram 5, AGHB is a quadrant with centre A and AHD is an arc of a circle with centre B and a radius of 5 cm. It is given that CD = 3 cm. G H D A B C Rajah 5/Diagram 5 [Guna/Use π = 3.142] (a) Cari CBD, dalam radian, betul kepada empat tempat perpuluhan. Find CBD, in radians, correct to four decimal places. [2 markah/marks] (b) Berdasarkan geometri deduktif, terangkan mengapa ABH = 60°. Based on deductive geometry, explain why ABH = 60°. [2 markah/marks] (c) Seterusnya, cari hasil tambah panjang lengkok GH dan panjang lengkok HD. KBAT Mengaplikasi Hence, find the sum of the arc length GH and the arc length HD. [4 markah/marks] Jawapan/Answer: (a) (b) (c) KERTAS PEPERIKSAAN TAMAT END OF QUESTION PAPER Firasat Add Math(S1).indd 9 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

KM1–10 SULIT 3472/2 3472/2 Firasat SPM 2023: Matematik Tambahan SULIT Kertas peperiksaan ini mengandungi 7 halaman bercetak. 2 1 2 jam Dua jam tiga puluh minit 3472/2 JANGAN BUKA KERTAS PEPERIKSAAN INI SEHINGGA DIBERITAHU MAKLUMAT UNTUK CALON INFORMATION FOR CANDIDATES 1. Kertas peperiksaan ini adalah dalam dwibahasa. 2. Soalan dalam bahasa Melayu mendahului soalan yang sepadan dalam bahasa Inggeris. 3. Calon dikehendaki membaca maklumat di halaman belakang kertas peperiksaan ini. SIJIL PELAJARAN MALAYSIA 2023 Kertas Model SPM SULIT MATEMATIK TAMBAHAN Kertas 2 2 3 4 5 6 7 8 1 Firasat Add Math(S1).indd 10 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 3472/2 KM1–11 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT Bahagian A Section A [50 markah/marks] Jawab semua soalan. Answer all questions.  1 Selesaikan persamaan serentak berikut: Solve the following simultaneous equations: r + s + t = 0 5r – 3s + 5t = 12 10r + s – 5t = 6 [5 markah/marks] 2 Diberi fungsi f : x → 4x + h dan fungsi songsangnya f –1 : x → x + 5 k , k ≠ 0, cari Given that the function f : x → 4x + h and its inverse function f –1 : x → x + 5 k , k ≠ 0, find (a) nilai h dan nilai k, the value of h and of k, [3 markah/marks] (b) nilai b jika f –1f(b) = b2 – 2. the value of b if f–1f(b) = b2 – 2 [3 markah/marks] 3 Fungsi kuadratik f(x) = –x2 + 6x – 8 boleh diungkapkan dalam bentuk f(x) = –(x – 3)2 + q, dengan keadaan q ialah pemalar. The quadratic function f(x) = –x2 + 6x – 8 can be expressed in the form f(x) = –(x – 3)2 + q, where q is a constant. (a) Cari nilai q. Find the value of q. [3 markah/marks] (b) Lakar graf y = f(x). Sketch the graph of y = f(x). [4 markah/marks] 4 Tunjukkan bahawa Show that 2 log4 x + 2 log4 y = log2 xy [4 markah/marks] Seterusnya, cari nilai x dan nilai y yang memuaskan persamaan serentak berikut. Hence, find the value of x and of y that satisfy the following simultaneous equation. log2 xy = 10 log4 x log4 y = 3 2 [5 markah/marks] 5 (a) Dalam Rajah 1, sudut-sudut A dan B adalah dengan keadaan A + B = 45°. Tinggi segi tiga itu ialah h cm. In Diagram 1, the angles A and B are such that A + B = 45°. The height of the triangle is h cm. 2 cm 3 cm h cm A B Rajah 1/Diagram 1 Dengan mengembangkan tan (A + B), cari nilai h. By expanding tan (A + B), find the value of h. [4 markah/marks] (b) Diberi kos x = – 4 5 dan sin y = – 12 13 , dengan keadaan x dan y masing-masing ialah sudutsudut dalam sukuan kedua dan ketiga, cari nilai sin (x + y). Given that cos x = – 4 5 and sin y = – 12 13 , such that x and y are angles in the second and third quadrants respectively, find the value of sin (x + y). [4 markah/marks]  6 Rajah 2 menunjukkan trapezium ABCD, dengan keadaan A →B = 2a ~, A →D = 3b ~ dan B →C = 3 5 A →D. R dan S masing-masing ialah titik tengah bagi BC dan CD. AD dan RS dipanjangkan sehingga bertemu pada titik T. Diagram 2 shows trapezium ABCD such that A →B = 2a ~, A →D = 3b ~ and B →C = 3 5 A →D. R and S are the midpoints of BC and CD respectively. AD and RS are extended until meet at point T. A B C D R S T Rajah 2/Diagram 2 Firasat Add Math(S1).indd 11 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

KM1–12 SULIT 3472/2 3472/2 Firasat SPM 2023: Matematik Tambahan SULIT (a) Cari dalam sebutan a ~ dan b ~, Find in terms of a ~ and b ~ , (i) R →S, (ii) R →A. [3 markah/marks] (b) Diberi R →T = kR→S dan A →T = mA→D, cari R →T dalam sebutan Given that R →T = kR→S and A →T = mA→D, find R →T in terms of (i) k, a ~ dan/and b ~, (ii) m, a ~ dan/and b ~. [2 markah/marks] (c) Seterusnya, cari nilai k dan nilai m. Hence, find the value of k and of m. [3 markah/marks] 7 (a) Diberi y = (x + 2)4 (2x – 3)3 , cari dy dx . Given that y = (x + 2)4 (2x – 3)3 , find dy dx . [4 markah/marks] (b) Hitung/Calculate  1 –1 1 (3 – x)3 dx. [3 markah/marks] Bahagian B Section B [30 markah/marks] Bahagian ini mengandungi empat soalan. Jawab tiga soalan. This section contains four questions. Answer three questions. 8 (a) Diberi bahawa tangen kepada lengkung y = px3 + kx pada titik (1, 1) mempunyai kecerunan –5. Cari nilai p dan nilai k. It is given that the tangent to the curve y = px3 + kx at the point (1, 1) has a gradient of –5. Find the value of p and of k. [4 markah/marks] (b) Rajah 3 menunjukkan rantau yang dibatasi oleh lengkung y = 12 x + 2, garis lurus x = 1 dan x = k serta paksi-x. Diagram 3 shows a region bounded by the curve y = 12 x + 2, the straight lines x = 1 and x = k and the x-axis. O 1 k x y y = 12 x + 2 Rajah 3/Diagram 3 Apabila rantau berlorek itu dikisarkan melalui 360° pada paksi-x, isi padu janaan ialah 24p unit3 . Cari nilai k. When the shaded region is revolved through 360° about the x-axis, the generated volume is 24p units3 . Find the value of k. [6 markah/marks] 9 Penyelesaian secara lukisan berskala tidak diterima. Solution by scale drawing is not accepted. Dalam Rajah 4, garis lurus AB adalah berserenjang dengan garis lurus BC. Persamaan garis lurus BC ialah 2x – y – 5 = 0. In Diagram 4, the straight line AB is perpendicular to the straight line BC. The equation of the straight line BC is 2x – y – 5 = 0. y x O C B A(–10, –5) 2x – y – 5 = 0 Rajah 4/Diagram 4 (a) Cari/Find (i) persamaan garis lurus AB, the equation of the straight line AB, (ii) koordinat titik B. the coordinates of point B. [4 markah/marks] (b) Tembereng garis AB dipanjangkan ke titik D dengan keadaan AB : BD = 2 : 3. Hitung luas segi tiga ADO. The line segment AB is extended to point D such that AB : BD = 2 : 3. Calculate the area of the triangle ADO. [4 markah/marks] (c) Titik P bergerak dengan keadaan APB sentiasa 90°. Cari persamaan lokus bagi titik P. A point P moves such that APB is always 90°. Find the equation of the locus of point P. [2 markah/marks] Firasat Add Math(S1).indd 12 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 3472/2 KM1–13 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT 10 Gunakan kertas graf untuk menjawab soalan ini. Use a graph paper to answer this question. Jadual 1 menunjukkan nilai-nilai bagi dua pemboleh ubah, T dan P, yang diperoleh daripada suatu eksperimen. Pemboleh ubah P dan T dihubungkan oleh persamaan P2 = qrT, dengan keadaan q dan r ialah pemalar. Table 1 shows the values of two variables, T and P, obtained from an experiment. The variables P and T are related by the equation P2 = qrT , where q and r are constants. T –5 5 15 20 25 P 2.05 3.55 6.18 8.13 10.71 Jadual 1/Table 1 (a) Plot graf log10 P melawan T, dengan menggunakan skala 2 cm kepada 5 unit pada paksi-T dan 2 cm kepada 0.1 unit pada paksilog10 P. Seterusnya, lukis garis lurus penyuaian terbaik. Plot the graph of log10 P against T, by using a scale of 2 cm to 5 units on the T-axis and 2 cm to 0.1 unit on the log10 P-axis. Hence, draw the line of best fit. [4 markah/marks] (b) Menggunakan graf anda di 10(a), Using your graph in 10(a), (i) cari nilai P apabila T = 10, find the value of P when T = 10, (ii) ungkapkan P2 = qrT dalam bentuk linear dan seterusnya cari nilai q dan nilai r. express P2 = qrT in linear form and hence, find the value of q and of r. [6 markah/marks] 11 (a) Didapati bahawa 70% daripada bilangan murid di sebuah bandar menyertai kelas tuisyen. Cari bilangan minimum murid yang perlu dipilih supaya kebarangkalian sekurang-kurangnya seorang murid menyertai kelas tuisyen adalah melebihi 0.95. It is found that 70% of the number of students in a city attend tuition classes. Find the minimum number of students to be chosen so that the probability of at least one student attending the tuition classes is more than 0.95. [5 markah/marks] (b) Markah Matematik Tambahan dalam suatu peperiksaan di sebuah sekolah bertaburan secara normal dengan min 70 dan sisihan piawai 8. The Additional Mathematics marks in an examination of a school are normally distributed with a mean of 70 and a standard deviation of 8. Hitung/Calculate (i) kebarangkalian bahawa seorang murid yang dipilih secara rawak memperoleh markah antara 65 dengan 73, the probability that a student chosen at random obtains marks between 65 and 73, (ii) markah minimum untuk mendapat gred A jika 10% daripada bilangan murid yang menduduki peperiksaan itu mendapat gred A. the minimum marks required for a grade of A if 10% of the number of students who sat for the examination scored a grade of A. [5 markah/marks] Bahagian C Section C [20 markah/marks] Bahagian ini mengandungi empat soalan. Jawab dua soalan. This section contains four questions. Answer two questions. 12 Gunakan kertas graf untuk menjawab soalan ini. Use a graph paper to answer this question. Seorang pengusaha bot pelancongan menyediakan x perjalanan ke Pulau P dan y perjalanan ke Pulau Q setiap hari berdasarkan kekangan berikut: A tour boat operator provides x trips to Island P and y trips to Island Q each day based on the following constraints: I Jumlah bilangan perjalanan yang disediakan tidak melebihi 14. The total number of trips provided is not more than 14. II Bilangan perjalanan ke Pulau P tidak melebihi tiga kali bilangan perjalanan ke Pulau Q. The number of trips to Island P is not more than three times the number of trips to Island Q. III Tambang perjalanan ke Pulau P dan Pulau Q masing-masing ialah RM60 dan RM30. Jumlah tambang yang diperoleh dalam sehari melebihi RM480. The fare per trip to Island P and Island Q are RM60 and RM30 respectively. The total fare collected per day is more than RM480. (a) Tulis tiga ketaksamaan, selain daripada x  0 dan y  0, yang memenuhi semua kekangan yang diberi. Write three inequalities, other than x  0 and y  0 which satisfy all the given constraints. [3 markah/marks] (b) Menggunakan skala 2 cm kepada 2 perjalanan pada kedua-dua paksi, bina dan lorek rantau yang memenuhi semua kekangan yang diberi. Using a scale of 2 cm to 2 trips on both axes, construct and shade the region R that satisfies all the given constraints. [3 markah/marks] Firasat Add Math(S1).indd 13 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

KM1–14 SULIT 3472/2 3472/2 Firasat SPM 2023: Matematik Tambahan SULIT (c) Menggunakan graf di 12(b), cari Using the graph in 12(b), find (i) julat bilangan perjalanan ke Pulau Q jika 8 perjalanan disediakan ke Pulau P pada suatu hari tertentu, the range of the number of trips to Island Q if 8 trips are provided to Island P on a certain day, (ii) keuntungan maksimum yang diperoleh dalam sehari jika keuntungan bagi setiap perjalanan ke Pulau P dan Pulau Q masingmasing ialah RM30 dan RM20. the maximum profit obtained per day if the profits from a trip to Island P and to Island Q are RM30 and RM20 respectively. [4 markah/marks] 13 Rajah 5 menunjukkan carta palang yang mewakili nilai jualan bulanan bagi empat barang keperluan, A, B, C dan D, yang dijual di sebuah kedai runcit. Diagram 5 shows a bar chart that represents the values of the monthly sales of four essential items, A, B, C and D, sold at a grocery shop. A B C D 1 600 1 200 800 400 0 Nilai jualan bulanan (RM) Value of monthly sales (RM) Barang keperluan Essential item Rajah 5/Diagram 5 Jadual 2 menunjukkan harga seunit bagi setiap barang keperluan pada tahun 2020 dan 2021 dan indeks harga pada tahun 2021 berasaskan tahun 2020. Table 2 shows the unit price of each essential item in the years 2020 and 2021 and the price indices in the year 2021 based on the year 2020. Barang keperluan Essential item Harga seunit (RM) Unit price (RM) Indeks harga pada tahun 2021 berasaskan tahun 2020 Price index in the year 2021 based on the year 2020 Tahun 2020 Year 2020 Tahun 2021 Year 2021 A 1.50 x 120 B 3.50 4.55 y C 2.60 3.25 125 D z 2.70 135 Jadual 2/Table 2 (a) Cari nilai x, y dan z. Find the value of x, of y and of z. [3 markah/marks] (b) Hitung indeks gubahan pada tahun 2021 berasaskan tahun 2020. Calculate the composite index in the year 2021 based on the year 2020. [3 markah/marks] (c) Jumlah jualan bulanan bagi barang keperluan pada tahun 2021 ialah RM4 000. Hitung jumlah jualan bulanan yang sepadan pada tahun 2020. The total monthly sales of the essential items in the year 2021 is RM4 000. Calculate the corresponding total monthly sales in the year 2020. [2 markah/marks] (d) Cari indeks harga bagi barang keperluan C pada tahun 2023 berasaskan tahun 2020 jika kadar peningkatan harga dari tahun 2021 ke tahun 2023 adalah sama dengan kadar peningkatan harga dari tahun 2020 ke tahun 2021. Find the price index of item C in the year 2023 based on the year 2020 if the rate of increase of the price from the year 2021 to the year 2023 is the same as the rate of increase from the year 2020 to the year 2021. [2 markah/marks] 14 Penyelesaian secara lukisan berskala tidak diterima. Solution by scale drawing is not accepted. Rajah 6 menunjukkan sebuah sisi empat ABCD. Diagram 6 shows a quadrilateral ABCD. A 6 cm 5.5 cm 8 cm 7.5 cm 143° B D C E Rajah 6/Diagram 6 Diberi bahawa luas segi tiga ADC adalah sama dengan luas segi tiga BCD, BE = 8 cm, DCB = 143° dan ADC ialah sudut cakah. It is given that the area of triangle ADC is equal to the area of triangle BCD, BE = 8 cm, DCB = 143° and ADC is an obtuse angle. (a) Hitung/Calculate (i) ADC, (ii) panjang, dalam cm, bagi DE, the length, in cm, of DE, (iii) panjang, dalam cm, bagi CE. the length, in cm, of CE. [7 markah/marks] (b) (i) Lakar sebuah segi tiga B’C’D’ yang mempunyai bentuk yang berlainan daripada segi tiga BCD, dengan keadaan B’D’ = BD, D’C’ = DC dan D’B’C’ = DCB. Sketch a triangle A’B’C’ that has a different shape from triangle BCD, such that B’D’ = BD, D’C’ = DC and D’B’C’ = DCB. (ii) Seterusnya, nyatakan D’C’B’. Hence, state D’C’B’. [3 markah/marks] Firasat Add Math(S1).indd 14 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 3472/2 KM1–15 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT 15 Penyelesaian secara lakaran graf tidak diterima. Solution by graph sketching is not accepted. Dalam Rajah 7, A dan B ialah dua titik tetap pada satu garis lurus dengan keadaan AB = 7.5 m. In Diagram 7, A and B are two fixed points on a straight line such that AB = 7.5 m. A B 7.5 m Zarah P Particle P Zarah Q Particle Q Rajah 7/Diagram 7 Pada suatu ketika, zarah P melalui titik A dengan halaju vP = 2t + 3, manakala zarah Q melalui titik B dengan halaju vQ = 5 – t, dengan keadaan t ialah masa, dalam saat, selepas zarah P dan zarah Q masing-masing melalui titik A dan titik B. At an instant, particle P passes through point A with a velocity of vP = 2t + 3, while particle Q passes through point B with a velocity of vQ = 5 – t, where t is the time, in seconds, after particles P and Q pass through points A and B respectively. [Anggapkan gerakan ke arah kanan sebagai arah positif.] [Consider the motion to the right as the positive direction.] (a) Cari/Find (i) sesaran zarah P dan sesaran zarah Q, dalam m, dari titik tetap A dalam sebutan t, the displacement of particle P and the displacement of particle Q, in m, from the fixed point A in terms of t, (ii) julat nilai t apabila kedua-dua zarah P dan Q bergerak ke kanan, the range of values of t when both particles P and Q moves to the right, (iii) jarak di antara zarah P dengan zarah Q apabila zarah Q berhenti seketika. the distance between particle P and particle Q when particle Q stops instantaneously. [7 markah/marks] (b) (i) Tentukan masa apabila kedua-dua zarah P dan Q bertemu. Determine the time when both particles P and Q meet. (ii) Seterusnya, hitung jarak, dalam m, dari titik A apabila kedua-dua zarah itu bertemu. Hence, calculate the distance, in m, from point A when both particles meet. [3 markah/marks] KERTAS PEPERIKSAAN TAMAT END OF QUESTION PAPER Firasat Add Math(S1).indd 15 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

KM1–16 SULIT 3472/2 3472/2 Firasat SPM 2023: Matematik Tambahan SULIT MAKLUMAT UNTUK CALON INFORMATION FOR CANDIDATES 1. Kertas peperiksaan ini mengandungi tiga bahagian: Bahagian A, Bahagian B dan Bahagian C. This question paper consists of three sections: Section A, Section B and Section C. 2. Jawab semua soalan dalam Bahagian A, mana-mana tiga soalan daripada Bahagian B dan mana-mana dua soalan daripada Bahagian C. Answer all questions in Section A, any three questions from Section B and any two questions from Section C. 3. Tunjukkan kerja mengira anda. Show your working. 4. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. The diagrams in the questions provided are not drawn to scale unless stated. 5. Markah yang diperuntukkan bagi setiap soalan atau ceraian soalan ditunjukkan dalam kurungan. The marks allocated for each question or sub-part of a question are shown in brackets. 6. Anda dibenarkan menggunakan kalkulator saintifik. You may use a scientific calculator. Firasat Add Math(S1).indd 16 12/06/2023 7:20 PM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 3472/2 KM2–1 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT Kertas peperiksaan ini mengandungi 9 halaman bercetak. JANGAN BUKA KERTAS PEPERIKSAAN INI SEHINGGA DIBERITAHU MAKLUMAT UNTUK CALON INFORMATION FOR CANDIDATES 1. Kertas peperiksaan ini mengandungi dua bahagian: Bahagian A dan Bahagian B. This question paper consists of two sections: Section A and Section B. 2. Jawab semua soalan dalam Bahagian A dan mana-mana dua soalan daripada Bahagian B. Answer all questions in Section A and any two questions from Section B. 3. Tulis jawapan anda dalam ruang yang disediakan dalam kertas peperiksaan. Write your answers in the spaces provided in the question paper. 4. Tunjukkan langkah-langkah dalam kerja mengira anda. Ini boleh membantu anda untuk mendapatkan markah. Show your working. It may help you to get marks. 5. Sekiranya anda hendak menukar jawapan, batalkan jawapan yang telah dibuat. Kemudian tulis jawapan yang baharu. If you wish to change your answer, cross out the answer that you have done. Then write down the new answer. 6. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. The diagrams in the questions provided are not drawn to scale unless stated. 7. Markah yang diperuntukkan bagi setiap soalan ditunjukkan dalam kurungan. The marks allocated for each question are shown in brackets. 8. Anda dibenarkan menggunakan kalkulator saintifik. You may use a scientific calculator. Untuk Kegunaan Pemeriksa Kod Pemeriksa: Bahagian Soalan Markah Penuh Markah Diperoleh A 1 6 2 6 3 6 4 5 5 5 6 5 7 5 8 6 9 5 10 5 11 5 12 5 B 13 8 14 8 15 8 Jumlah 80 2 jam Dua jam 3472/1 SIJIL PELAJARAN MALAYSIA 2023 Kertas Model SPM SULIT MATEMATIK TAMBAHAN Kertas 1 NO. KAD PENGENALAN ANGKA GILIRAN 2 3 4 5 6 7 8 1 FIRASAT Add Math(S2).indd 1 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

KM2–2 SULIT SULIT 3472/1 Firasat SPM 2023: Matematik Tambahan 3472/1 Untuk Kegunaan Pemeriksa Bahagian A Section A [64 markah/marks] Jawab semua soalan. Answer all questions.  1 Diberi bahawa f(x) = x – 3 dan g(x) = 2x + 1. It is given that f(x) = x – 3 and g(x) = 2x + 1. (a) Cari nilai p jika fg–1 p 2 = –1. Find the value of p if fg–1 p 2 = –1. [3 markah/marks] (b) Tentukan nilai k jika ff–1(k2 ) = 6k – 9. Determine the value of k if ff–1(k2 ) = 6k – 9. [3 markah/marks] Jawapan/Answer: (a) (b) 2 Dalam satu acara merejam lembing, jarak mengufuk dan jarak mencancang bagi pergerakan lembing oleh seorang peserta masing-masing diwakili oleh x m dan y m. Pergerakan lembing itu diwakili oleh y = – 1 360 x2 + 3 20x + 1. In a javelin throwing event, the horizontal distance and the vertical distance of the motion of the spear by a participant are represented by x m and y m respectively. The motion of the spear is represented by y = – 1 360 x2 + 3 20x + 1. Cari/Find (a) ketinggian lembing apabila ia dilepaskan oleh peserta itu, the height of the spear when it is released by the participant, [1 markah/mark] (b) jarak mengufuk maksimum, dalam m, yang dicapai oleh peserta itu. the maximum horizontal distance, in m, that the participant achieved. [5 markah/marks] Jawapan/Answer: (a) (b) 1 6 2 6 FIRASAT Add Math(S2).indd 2 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

KM2–3 [Lihat halaman sebelah SULIT SULIT © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 3472/1 Untuk Kegunaan Pemeriksa 3 Beza antara punca bagi persamaan kuadratik 4x2 + 4x = 7 + q, dengan keadaan q ialah pemalar, ialah 2. Cari nilai q. The difference between the roots of the quadratic equation 4x2 + 4x = 7 + q, where q is a constant, is 2. Find the value of q. [6 markah/marks] Jawapan/Answer: 4 Cari julat nilai k jika graf bagi fungsi kuadratik f(x) = (k – 6)x2 – 8x + k menyilang paksi-x pada dua titik yang berbeza. Seterusnya, nyatakan nilai-nilai k jika paksi-x ialah tangen kepada lengkung itu. Find the range of values of k if the graph of the quadratic function f(x) = (k – 6)x2 – 8x + k intersects the x-axis at two different points. Hence, state the values of k if the x-axis is a tangent to the curve. [5 markah/marks] Jawapan/Answer: 5 Selesaikan persamaan serentak: Solve the simultaneous equations: x + y 2 + 1 = 0 1 x + 2 y – 1 2 = 0 [5 markah/marks] Jawapan/Answer: 3 6 4 5 5 5 FIRASAT Add Math(S2).indd 3 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

KM2–4 SULIT SULIT 3472/1 Firasat SPM 2023: Matematik Tambahan 3472/1 Untuk Kegunaan Pemeriksa 6 Ungkapkan 8 – 3 2 4 + 3 2 dalam bentuk a + b 2 , dengan keadaan a dan b ialah pemalar. Express 8 – 3 2 4 + 3 2 in the form of a + b 2 , where a and b are constants. [5 markah/marks] Jawapan/Answer: 7 Diberi log3 T – log9 V = 2, ungkapkan T dalam sebutan V. Given that log3 T – log9 V = 2, express T in terms of V. [5 markah/marks] Jawapan/Answer: 8 Suatu janjang aritmetik mempunyai 10 sebutan. Hasil tambah semua sebutan ialah 155 dan hasil tambah sebutan-sebutan genap ialah 85. An arithmetic progression has 10 terms. The sum of all the terms is 155 and the sum of the even terms is 85. Cari KBAT Mengaplikasi Find (a) sebutan pertama dan beza sepunya, the first term and the common difference, [5 markah/marks] (b) sebutan terakhir janjang itu. the last term of the progression. [1 markah/mark] Jawapan/Answer: (a) (b) 8 6 7 5 6 5 FIRASAT Add Math(S2).indd 4 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

KM2–5 [Lihat halaman sebelah SULIT SULIT © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 3472/1 Untuk Kegunaan Pemeriksa 9 Rajah 1 menunjukkan sebuah segi empat selari ABCD. Koordinat titik A, B dan C masing-masing ialah (2, 0), (13, 3) dan (16, 12). Diagram 1 shows a parallelogram ABCD. The coordinates of the points A, B and C are (2, 0), (13, 3) and (16, 12) respectively. y x O D C(16, 12) B(13, 3) A(2, 0) Rajah 1/Diagram 1 Cari/Find (a) koordinat titik D, the coordinates of point D, [3 markah/marks] (b) persamaan pembahagi dua sama serenjang AC. the equation of the perpendicular bisector of AC. [2 markah/marks] Jawapan/Answer: (a) (b) 10 Diberi titik-titik O(0, 0), P(–2, 3), Q(13, 8) dan R terletak pada garis PQ dengan keadaan P →R = 3 5 P →Q, cari Given that the points O(0, 0), P(–2, 3), Q(13, 8) and R lies on line PQ such that P →R = 3 5 P →Q, find (a) O →R, [2 markah/marks] (b) vektor unit dalam arah P →R dengan menyatakan jawapan anda dalam bentuk surd. the unit vector in the direction of P →R by stating your answer in the form of surds. [3 markah/marks] Jawapan/Answer: (a) (b) 9 5 10 5 FIRASAT Add Math(S2).indd 5 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

KM2–6 SULIT SULIT 3472/1 Firasat SPM 2023: Matematik Tambahan 3472/1 Untuk Kegunaan Pemeriksa 11 Rajah 2 menunjukkan sektor AOB bagi sebuah bulatan berpusat O. Diagram 2 shows sector AOB of a circle with centre O. A B 265° 18' O q Rajah 2/Diagram 2 Diberi bahawa luas sektor minor AOB ialah 82.65 cm2 . Hitung It is given that the area of the minor sector AOB is 82.65 cm2 . Calculate (a) nilai θ, dalam radian, the value of θ, in radians, [2 markah/marks] (b) luas, dalam cm2 , tembereng berlorek. the area, in cm2 , of the shaded segment. [Guna/Use p = 3.142] [3 markah/marks] Jawapan/Answer: (a) (b) 12 Pemboleh ubah x dan y dihubungkan oleh persamaan ax2 + by2 – 3 = 0, dengan keadaan a dan b ialah pemalar. Satu graf garis lurus diperoleh dengan memplot y2 melawan x2 , seperti yang ditunjukkan dalam Rajah 3. The variables x and y are related by the equation ax2 + by2 – 3 = 0, where a and b are constants. A straight line graph is obtained by plotting y2 against x2 , as shown in Diagram 3. Cari nilai a dan nilai b. Find the value of a and of b. [5 markah/marks] Jawapan/Answer: 11 5 12 5 y2 x2 – 4 O 8 Rajah 3/Diagram 3 FIRASAT Add Math(S2).indd 6 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

KM2–7 [Lihat halaman sebelah SULIT SULIT © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 3472/1 Untuk Kegunaan Pemeriksa Bahagian B Section B [16 markah/marks] Bahagian ini mengandungi tiga soalan. Jawab dua soalan. This section contains three questions. Answer two questions. 13 Rajah 4 menunjukkan sebuah bongkah dengan jisim m kg digantung pada hujung spring yang berayun. Diagram 4 shows a block with a mass of m kg that is suspended from the bottom end of an oscillating spring. Rajah 4/Diagram 4 Tempoh ayunan, T saat, bagi spring itu diberi oleh T = 2p m 4 . The period of oscillation, T seconds, of the spring is given by T = 2p m 4 . Cari KBAT Mengaplikasi Find (a) dT dm , [4 markah/marks] (b) perubahan hampir dalam tempoh ayunan apabila jisim bongkah bertambah daripada 0.10 kg kepada 0.11 kg. the approximate change in the period of oscillation when the mass of the object increases from 0.10 kg to 0.11 kg. [Nyatakan jawapan anda dalam sebutan p.] [State your answer in terms of p.] [4 markah/marks] Jawapan/Answer: (a) (b) 13 8 FIRASAT Add Math(S2).indd 7 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

KM2–8 SULIT SULIT 3472/1 Firasat SPM 2023: Matematik Tambahan 3472/1 Untuk Kegunaan Pemeriksa 14 Rajah 5 menunjukkan sebuah segi tiga ABC dengan keadaan panjang sisi-sisinya mengikut janjang geometri. Diagram 5 shows a triangle ABC such that the lengths of its sides follow geometric progression. A B C 32 cm Rajah 5/Diagram 5 Diberi panjang sisi yang terpanjang ialah 32 cm dan perimeter segi tiga itu ialah 56 cm, cari Given that the longest side is 32 cm and the perimeter of the triangle is 56 cm, find KBAT Mengaplikasi (a) nisbah sepunya, the common ratio, [6 markah/marks] (b) panjang, dalam cm, sisi yang terpendek. the length, in cm, of the shortest side. [2 markah/marks] Jawapan/Answer: (a) (b) 14 8 FIRASAT Add Math(S2).indd 8 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

KM2–9 [Lihat halaman sebelah SULIT SULIT © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 3472/1 Untuk Kegunaan Pemeriksa 15 8 15 Jisim sekampit beras, dalam kg, bertaburan secara normal dengan min μ dan sisihan piawai σ. Jika 97.5% daripada kampit beras itu mempunyai jisim melebihi 30.5 kg dan 88.49% daripada kampit beras itu mempunyai jisim kurang daripada 46.3 kg, cari nilai μ dan nilai σ. The mass of a bag of rice, in kg, is normally distributed with mean μ and standard deviation σ. If 97.5% of the bags of rice have masses of more than 30.5 kg and 88.49% of the bags of rice have masses of less than 46.3 kg, find the value of μ and of σ. [8 markah/marks] Jawapan/Answer: KERTAS PEPERIKSAAN TAMAT END OF QUESTION PAPER FIRASAT Add Math(S2).indd 9 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

KM2–10 SULIT 3472/2 3472/2 Firasat SPM 2023: Matematik Tambahan SULIT Kertas peperiksaan ini mengandungi 7 halaman bercetak. 2 1 2 jam Dua jam tiga puluh minit 3472/2 JANGAN BUKA KERTAS PEPERIKSAAN INI SEHINGGA DIBERITAHU MAKLUMAT UNTUK CALON INFORMATION FOR CANDIDATES 1. Kertas peperiksaan ini adalah dalam dwibahasa. 2. Soalan dalam bahasa Melayu mendahului soalan yang sepadan dalam bahasa Inggeris. 3. Calon dikehendaki membaca maklumat di halaman belakang kertas peperiksaan ini. SIJIL PELAJARAN MALAYSIA 2023 Kertas Model SPM SULIT 2 3 4 5 6 7 8 1 MATEMATIK TAMBAHAN Kertas 2 FIRASAT Add Math(S2).indd 10 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 3472/2 KM2–11 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT Bahagian A Section A [50 markah/marks] Jawab semua soalan. Answer all questions. 1 Vektor kedudukan bagi bot P, t jam selepas meninggalkan pelabuhan A ialah t(15i ∼ + 30j ∼ ), manakala vektor kedudukan bagi bot Q ialah (10i ∼ + 40j ∼ ) + t(10i ∼ + 10j ∼ ). The position vector of boat P, t hours after leaving port A is t(15i∼ + 30j∼ ), while the position vector of boat Q is (10i∼ + 40j∼ ) + t(10i∼ + 10j∼ ). (a) Nyatakan vektor kedudukan awal bagi bot Q. State the initial position vector of boat Q. [1 markah/mark] (b) Hitung jarak di antara kedudukan awal bot Q dengan pelabuhan A. Calculate the distance between the initial position of boat Q and port A. [1 markah/mark] (c) Tentukan halaju Determine the velocity of (i) bot P/boat P, (ii) bot Q/boat Q. [4 markah/marks] (d) Hitung masa, dalam jam, apabila bot P dan bot Q bertembung. Calculate the time, in hours, when boat P and boat Q collide. [3 markah/marks] 2 Diberi bahawa f : x → 6 – 2x. It is given that f : x → 6 – 2x. (a) Cari/Find (i) f 2 (x), (ii) (f 2 )–1(x). [3 markah/marks] (b) Lakar graf y = |f 2 (x)| untuk 0  x  3. Seterusnya, cari julat bagi graf itu. Sketch the graph of y = |f 2 (x)| for 0  x  3. Hence, find the range of the graph. [3 markah/marks] 3 Jika α dan β ialah punca-punca bagi persamaan kuadratik 2x2 – 4x + 5 = 0, cari persamaan kuadratik yang mempunyai punca-punca  α – 1 β dan β – 1 a  . If α and β are the roots of the quadratic equation 2x2 – 4x + 5 = 0, find the quadratic equation that has the roots  α – 1 β  and  β – 1 a  . [7 markah/marks] 4 Sebuah kilang menghasilkan tiga jenis minuman. Jadual 1 menunjukkan campuran serbuk kopi, serbuk koko dan gula untuk menghasilkan tiga jenis minuman itu. A factory produces three types of drinks. Table 1 shows the mixture of coffee powder, cocoa powder and sugar to produce the three types of drinks. Jenis Type Serbuk kopi Coffee powder Serbuk koko Cocoa powder Gula Sugar Jumlah kos (RM) Total cost (RM) P 60% 30% 10% 6.25 Q 40% 30% 30% 4.21 R 30% 70% 0% 3.23 Jadual 1/Table 1 Kos bagi serbuk kopi, serbuk koko dan gula yang diperlukan masing-masing ialah RMx, RMy and RMz. Tentukan nilai x, nilai y dan nilai z. The costs of coffee powder, cocoa powder and sugar required are RMx, RMy and RMz respectively. Determine the value of x, of y and of z. [8 markah/marks] 5 Rajah 1 menunjukkan sebuah segi tiga bersudut tegak ABC. Diagram 1 shows a right-angled triangle ABC. A (2 + 4 3 ) cm (4 + 2 3 ) cm B C Rajah 1/Diagram 1 Cari panjang, dalam cm, bagi BC. Nyatakan jawapan anda dalam bentuk surd. Find the length, in cm, of BC. State your answer in surd form. [5 markah/marks] FIRASAT Add Math(S2).indd 11 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

KM2–12 SULIT 3472/2 3472/2 Firasat SPM 2023: Matematik Tambahan SULIT 6 Rajah 2 menunjukkan sektor AOB bagi sebuah bulatan berpusat O. Diagram 2 shows the sector AOB of a circle with centre O. O B C A 12 cm Rajah 2/Diagram 2 Diberi bahawa jejari ialah 12 cm dan BC = 8 cm. It is given that the radius is 12 cm and BC = 8 cm. Cari Find (a) BOC, dalam radian, BOC, in radians, [4 markah/marks] (b) perimeter, dalam cm, kawasan berlorek. the perimeter, in cm, of the shaded region. [5 markah/marks] [Guna/Use π = 3.142] 7 Hitung bilangan cara enam orang pekerja duduk mengelilingi sebuah meja bulat untuk bermesyuarat dengan keadaan dua orang pekerja tertentu mesti duduk bersama. Calculate the number of ways six workers can sit at a round table to have a meeting such that two particular workers must sit together. [6 markah/marks] Bahagian B Section B [30 markah/marks] Bahagian ini mengandungi empat soalan. Jawab tiga soalan. This section contains four questions. Answer three questions. 8 Sehelai kertas berbentuk segi empat sama dilipat separuh sebanyak 12 kali. Diberi bahawa ketebalan kertas itu ialah 0.0075 cm. A square piece of paper is folded in half 12 times. It is given that the thickness of the paper is 0.0075 cm. (a) Tunjukkan bahawa ketebalan lipatan kertas itu membentuk suatu janjang geometri. Seterusnya, nyatakan nisbah sepunya. Show that the thickness of the paper folding forms a geometric progression. Hence, state the common ratio. [4 markah/marks] (b) Cari ketebalan, dalam cm, lipatan kertas selepas kertas itu dilipat sebanyak 12 kali. Find the thickness, in cm, of the paper folding after it is folded 12 times. [2 markah/marks] (c) Tentukan bilangan kali minimum lipatan kertas itu supaya ketebalannya melebihi 15 cm. Determine the minimum number of times of the paper folding in order that its thickness exceeds 15 cm. [4 markah/marks] 9 Rajah 3 menunjukkan sebuah gabungan pepejal yang terdiri daripada sebuah kuboid dan sebuah piramid. Diagram 3 shows a composite solid which consists of a cuboid and a pyramid. A E H V G C B D F x cm 2 cm (10 – 2x) cm Rajah 3/Diagram 3 Tinggi piramid itu ialah separuh daripada tinggi kuboid. The height of the pyramid is half of the height of the cuboid. (a) Tunjukkan bahawa isi padu, V cm3 , pepejal gabungan itu diberi oleh V = 70x – 14x2 3 . KBAT Mengaplikasi Show that the volume, V cm3 , of the composite solid is given by V = 70x – 14x2 3 . [4 markah/marks] (b) Cari isi padu maksimum, dalam cm3 , pepejal gabungan itu. Find the maximum volume, in cm3 , of the composite solid. [6 markah/marks] FIRASAT Add Math(S2).indd 12 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 3472/2 KM2–13 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT 10 (a) Buktikan bahawa 2 sin (x – y) kos (x – y) – kos (x + y) = kot y – kot x. KBAT Mengaplikasi Prove that 2 sin (x – y) cos (x – y) – cos (x + y) = cot y – cot x. [5 markah/marks] (b) Selesaikan persamaan tan (45° + z) = 4 tan (45° – z) untuk 0°  x  360°. Solve the equation tan (45° + z) = 4 tan (45° – z) for 0°  x  360°. [5 markah/marks] 11 (a) Diketahui bahawa 1 daripada 3 orang murid di sebuah sekolah menonton berita di televisyen setiap hari. 10 orang murid dipilih secara rawak dari sekolah itu. Cari It is known that 1 out of 3 students of a school watches the news on television every day. 10 students are chosen at random from the school. Find (i) min dan sisihan piawai bagi bilangan murid yang menonton berita di televisyen setiap hari, the mean and the standard deviation of the number of students who watch the news on television every day, (ii) kebarangkalian bahawa kurang daripada 3 orang murid yang menonton berita di televisyen setiap hari. the probability that less than 3 students watch the news on television every day. [5 markah/marks] (b) Jisim bersih setin kerepek ubi kentang bertaburan secara normal dengan min μ g dan sisihan piawai 4.5 g. The net mass of a tin of potato chips is normally distributed with a mean of μ g and a standard deviation of 4.5 g. (i) Diberi 2.28% tin kerepek ubi kentang mempunyai jisim melebihi 210 g, cari nilai μ. Given that 2.28% of the tins of potato chips have masses of more than 210 g, find the value of μ. (ii) Cari peratusan tin kerepek ubi kentang yang mempunyai jisim antara 196 g dengan 198 g. Find the percentage of the tins of potato chips that have masses between 196 g and 198 g. [5 markah/marks] Bahagian C Section C [20 markah/marks] Bahagian ini mengandungi empat soalan. Jawab dua soalan. This section contains four questions. Answer two questions. 12 Penyelesaian secara lukisan berskala tidak diterima. Solution by scale drawing is not accepted. Rajah 4 menunjukkan sebuah segi tiga PQR. Diagram 4 shows a triangle PQR. P Q R 8 m 11 m 40° Rajah 4/Diagram 4 (a) Cari/Find (i) PQR, (ii) panjang, dalam m, bagi PR. the length, in m, of PR. [5 markah/marks] (b) Dengan mengambil kira kes berambiguiti, lakar segi tiga P’Q’R’. Seterusnya, cari Taking into consideration the ambiguous case, sketch triangle P’Q’R’. Hence, find (i) P’Q’R’, (ii) luas, dalam m2 , segi tiga P’Q’R’. the area, in m2 , of triangle P’Q’R’. [5 markah/marks] FIRASAT Add Math(S2).indd 13 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

KM2–14 SULIT 3472/2 3472/2 Firasat SPM 2023: Matematik Tambahan SULIT 13 Sejenis biskut dibuat dengan menggunakan empat bahan, A, B, C dan D. Jadual 2 menunjukkan harga bagi bahan-bahan itu pada tahun 2018 dan 2021. A type of biscuits is made using four ingredients, A, B, C and D. Table 2 shows the prices of the ingredients in the years 2018 and 2021. Bahan Ingredient Harga per kg (RM) Price per kg (RM) Tahun 2018 Year 2018 Tahun 2021 Year 2021 A a 8.40 B 3.00 3.60 C b c D 5.00 5.50 Jadual 2/Table 2 (a) Indeks harga bagi bahan A pada tahun 2021 berasaskan tahun 2018 ialah 105. Cari nilai a. The price index of ingredient A in the year 2021 based on the year 2018 is 105. Find the value of a. [2 markah/marks] (b) Indeks harga bagi bahan C pada tahun 2021 berasaskan tahun 2018 ialah 120. Harga per kg bagi bahan C pada tahun 2021 ialah RM1.50 lebih daripada harganya yang sepadan pada tahun 2018. Hitung nilai b dan nilai c. The price index of ingredient C in the year 2021 based on the year 2018 is 120. The price per kg of ingredient C in the year 2021 is RM1.50 more than its corresponding price in the year 2018. Calculate the value of b and of c. [3 markah/marks] (c) Indeks gubahan bagi kos membuat biskut itu pada tahun 2021 berasaskan tahun 2018 ialah 113.46. The composite index for the cost of making the biscuit in the year 2021 based on the year 2018 is RM113.46. Hitung/Calculate (i) harga bagi setin biskut pada tahun 2021 jika harganya yang sepadan pada tahun 2018 ialah RM12.40, the price of a tin of biscuits in the year 2021 if its corresponding price in the year 2018 is RM12.40, (ii) nilai k jika nisbah bagi bahan A, B, C dan D yang digunakan ialah 5 : 3 : k : 1. the value of k if the ratio of the ingredients A, B, C and D used is 5 : 3 : k : 1. [5 markah/marks] 14 Suatu zarah bergerak dari satu titik tetap O di sepanjang satu garis lurus. Halaju zarah itu, v m s–1, diberi oleh v = 3t2 – 10t – 8, dengan keadaan t ialah masa dalam saat, selepas meninggalkan O. A particle moves from a fixed point O along a straight line. The velocity of the particle, v m s–1, is given by v = 3t2 – 10t – 8, where t is the time in seconds, after leaving O. [Anggapkan gerakan ke arah kanan sebagai arah positif.] [Assume the motion to the right as the positive direction.] Cari/Find (a) pecutan, dalam m s–2, zarah itu apabila ia berehat seketika, the acceleration, in m s–2, of the particle when it is momentarily at rest, [4 markah/marks] (b) julat nilai t apabila zarah itu bergerak ke kanan, the range of values of t when the particle moves to the right, [2 markah/marks] (c) jumlah jarak, dalam m, yang dilalui oleh zarah itu dalam 7 saat yang pertama. the total distance, in m, travelled by the particle in the first 7 seconds. [4 markah/marks] 15 Gunakan kertas graf untuk menjawab soalan ini. Use a graph paper to answer this question. Sebuah syarikat perabot ingin menghabiskan stok lama dengan mengadakan jualan murah. Terdapat dua jenis sofa yang dijual seperti ditunjukkan dalam Jadual 3. A furniture company wants to clear its old stocks by having a sale. There are two types of sofas sold as shown in Table 3. Jenis sofa Type of sofa Harga (RM) Price (RM) Keuntungan (RM) Profit (RM) PVC 3 200 1 000 Berkulit penuh Full leather 8 000 1 500 Jadual 3/Table 3 FIRASAT Add Math(S2).indd 14 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

SULIT 3472/2 KM2–15 [Lihat halaman sebelah © Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 SULIT Jualan murah itu adalah berdasarkan kekangan berikut: The sale is subject to the following constraints: I Terdapat 80 set sofa yang tinggal dalam stok lama. There are 80 sets of sofas left in the old stocks. II Jumlah perolehan daripada jualan itu mestilah sekurang-kurangnya RM224 000. The total revenue from the sale must be at least RM224 000. III Bilangan set sofa PVC yang dijual melebihi bilangan set sofa berkulit penuh sebanyak 10 atau lebih. The number of sets of the PVC sofas sold exceed the number of sets of the full leather sofas by 10 or more. Syarikat itu menjual x set sofa PVC dan y set sofa berkulit penuh. The company sold x sets of PVC sofas and y sets of full leather sofas. (a) Tulis tiga ketaksamaan, selain daripada x  0 dan y  0 yang memenuhi semua kekangan di atas. Write three inequalities, other than x  0 and y  0 that satisfy all the above constraints. [3 markah/marks] (b) Menggunakan skala 2 cm kepada 10 set sofa pada kedua-dua paksi, bina dan lorek rantau R yang memenuhi semua kekangan yang diberi. Using a scale of 2 cm to 10 sets of sofas on both axes, construct and shade the region R which satisfies all the given constraints. [3 markah/marks] (c) Gunakan graf yang dibina di 15(b) untuk menjawab soalan-soalan berikut. Use the graph constructed in 15(b) to answer the following questions. (i) Jika 30 set sofa berkulit penuh yang dijual, cari bilangan minimum dan maksimum set sofa PVC yang dijual. If 30 sets of the full leather sofas are sold, find the minimum and maximum number of sets of the PVC sofas sold. (ii) Cari keuntungan maksimum yang diperoleh daripada jualan murah itu. Find the maximum profit obtained from the sale. [4 markah/marks] KERTAS PEPERIKSAAN TAMAT END OF QUESTION PAPER FIRASAT Add Math(S2).indd 15 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

KM2–16 SULIT 3472/2 3472/2 Firasat SPM 2023: Matematik Tambahan SULIT MAKLUMAT UNTUK CALON INFORMATION FOR CANDIDATES 1. Kertas peperiksaan ini mengandungi tiga bahagian: Bahagian A, Bahagian B dan Bahagian C. This question paper consists of three sections: Section A, Section B and Section C. 2. Jawab semua soalan dalam Bahagian A, mana-mana tiga soalan daripada Bahagian B dan mana-mana dua soalan daripada Bahagian C. Answer all questions in Section A, any three questions from Section B and any two questions from Section C. 3. Tunjukkan kerja mengira anda. Show your working. 4. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. The diagrams in the questions provided are not drawn to scale unless stated. 5. Markah yang diperuntukkan bagi setiap soalan atau ceraian soalan ditunjukkan dalam kurungan. The marks allocated for each question or sub-part of a question are shown in brackets. 6. Anda dibenarkan menggunakan kalkulator saintifik. You may use a scientific calculator. FIRASAT Add Math(S2).indd 16 12/06/2023 7:21 PM PENERBIT ILMU BAKTI SDN. BHD.

•J1• Firasat SPM 2023: Matematik Tambahan - Jawapan Kertas Model 1 Kertas 1 Bahagian A  1 2x2 – 5x + 3 = 0 a = 2, b = –5, c = 3 b2 – 4ac = (–5)2 – 4(2)(3) [K1] = 1 [Positif/Positive] Maka, persamaan kuadratik itu mempunyai dua punca nyata dan berbeza. [N1] Hence, the quadratic equation has two real and distinct roots.  2 Katakan/Let OS = r 1 2 (18)2  2 3 1 2 r2  2 3 – 1 2 (18)2  2 3 = 9 7 [K1] (18)2 (2) r2 (2) – (18)2 (2) = 9 7 648 2r2 – 648 = 9 7 [K1] 9(2r2 – 648) = 4 536 18r2 – 5 832 = 4 536 r2 = 576 [K1] r = 24 [N1]  3 ln (x + 1) – ln x = 2 ln  x + 1 x  = 2 [K1] loge  x + 1 x  = 2 x + 1 x = e2 [K1] x + 1 = e2 x e2 x – x = 1 x(e2 – 1) = 1 [K1] x = 1 e2 – 1 x = 1 2.7182 – 1 [K1] x = 0.1566 [N1]  4 log2 y = mlog2 x + c Bagi/For (–3, 4): log2 x = –3 dan/and log2 y = 4, 4 = –3m + c ...........① [K1] Bagi/For (–2, 1): log2 x = –2 dan/and log2 y = 1, 1 = –2m + c ...........② [K1] ① – ②: –m = 3 m = –3 Daripada/From ①: 4 = –3(–3) + c c = –5 Maka/Thus, log2 y = –3log2 x – 5 [K1] log2 y + 3log2 x = –5 log2 y + log2 x3 = –5 log2 (yx3 ) = –5 yx3 = 2–5 yx3 = 1 32 y = 1 32x3 [N1]  5 (a) S2 = 150 a + ar = 150 a(1 + r) = 150 .......① [K1] T3 – T2 = 45 ar2 – ar = 45 ar(r – 1) = 45 ........② [K1] ① ② : a(1 + r) ar(r – 1) = 150 45 [K1] 1 + r r2 – r = 10 3 3 + 3r = 10r2 – 10r 10r2 – 13r – 3 = 0 (2r – 3)(5r + 1) = 0 [K1] r = 3 2 atau/or r = – 1 5 [N1] (b) Untuk S∞ wujud, |r|  1. For S∞ to exist, |r|  1. Maka, r = – 1 5 akan digunakan. Thus, r = – 1 5 will be used. Daripada/From ①: a1 + – 1 5 = 150 [K1] 4 5 a = 150 a = 1871 2 [N1] Maka/Hence, S∞ = 187 1 2 1 – – 1 5 = 156 1 4 [N1]  6 y = 3x2 + hx + k x – 10y + 5 = 0 (–2, 9) Kecerunan garis 10y = x + 5 ⇒ y = 1 10x + 1 2 ialah 1 10. [K1] The gradient of the line 10y = x + 5 ⇒ y = 1 10x + 1 2 is 1 10. Oleh itu, kecerunan tangen pada titik (–2, 9) ialah –10. [K1] Thus, the gradient of the tangent at the point (–2, 9) is –10. y = 3x2 + hx + k dy dx = 6x + h [K1] –10 = 6(–2) + h h = 2 [N1] Lengkung melalui titik (–2, 9). The curve passes through the point (–2, 9). y = 3x2 + 2x + k 9 = 3(–2)2 + 2(–2) + k [K1] 9 = 12 – 4 + k k = 1 [N1]  7 y = x + k ................① y2 + x2 = 2 .............② Gantikan ① ke dalam ②: Substitute ① into ②: (x + k) 2 + x2 = 2 [K1] x2 + 2kx + k2 + x2 – 2 = 0 2x2 + 2kx + k2 – 2 = 0 [K1] a = 2, b = 2k, c = k2 – 2 b2 – 4ac  0 (2k) 2 – 4(2)(k2 – 2)  0 [K1] 4k2 – 8k2 + 16  0 –4k2 + 16  0 4k2 – 16  0 k2 – 4  0 (k + 2)(k – 2)  0 [K1] –2 2 k Julat nilai k yang dikehendaki ialah k  –2 atau k  2. [N2] The required range of values of k is k  –2 atau k  2.  8 gf(x) = g[f(x)] = f(x) + m [K1] = x + 1 x – 1 + m [K1] = x + 1 + m(x – 1) x – 1 = x + 1 + mx – m x – 1 [K1] Katakan/Let Tukar perkara rumus kepada x. Change the subject of the formula to x. y = x + 1 + mx – m x – 1 y(x – 1) = x + 1 + mx – m [K1] yx – y = x + 1 + mx – m yx – x – mx = y + 1 – m x(y – 1 – m) = y + 1 – m x = y + 1 – m y – 1 – m (gf) –1(y) = y + 1 – m y – 1 – m [K1] Diberi/Given that (gf) –1(x) = x – 1 x – 3, x ≠ 3. Dengan perbandingan, By comparison, 1 – m = –1 m = 2 [N1] atau/or –1 – m = –3 m = 3 – 1 m = 2 [N1]  9 (a) Persamaan AB ialah/The equation of AB is x 15 + y 5 = 1. [P1] Jawapan P = Pengetahuan/Knowledge, K = Kaedah/Method, N = Nilai/Value FIRASAT Add Math(Jaw).indd 1 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J2• (b) 3AD = 2DB AD DB = 2 3 AD : DB = 2 : 3 A(0, 5) B(15, 0) D 2 3 D =  3(0) + 2(15) 2 + 3 , 3(5) + 2(0) 2 + 3  [K1] D = (6, 3) [N1] (c) mAB = 0 – 5 15 – 0 = – 1 3 \ mCD = 3 [K1] Persamaan CD ialah Equation of CD is y – 3 = 3(x – 6) y – 3 = 3x – 18 y = 3x – 15 [K1] Maka, pintasan-y ialah –15. [N1] Hence, the y-intercept is –15. 10 p ~ + q ~ =  x 7 +  6 –2 =  x + 6 5  [K1] Kedua-dua belah persamaan dikuasa duakan untuk memansuhkan punca kuasa dua. Both sides of the equation are squared to eliminate the square roots.  p ~ + q ~  = 13  (x + 6)2 + 52 = 13 [K1] (x + 6)2 + 25 = 132 [K1] x2 + 12x + 36 + 25 – 169 = 0 x2 + 12x – 108 = 0 (x – 6)(x + 18) = 0 [K1} x = 6 atau/or x = –18 [N1] 11 Luas rantau berlorek Area of the shaded region = 2 –2 (–x2 + 2x + 7) – (x2 + 2x – 1) dx [K1] = 2 –2 (–2x2 + 8) dx [K1] = – 2x3 3 + 8x 2 –2 [K1] = – 2(2)3 3 + 8(2) – – 2 3 (–2)3 + 8(–2) [K1] = 32 3 – – 32 3  = 211 3 unit2 /units2 [N1] 12 (a) Bilangan susunan tanpa syarat Number of arrangements without restrictions = 13! 2!3!2!2! [K1] = 129 729 600 [N1] (b) Jika 5 huruf vokal diletakkan bersama, mereka akan dikira sebagai 1 unit. Bersama-sama dengan 8 huruf konsonan, terdapat 9 unit. If the 5 vowels have to be grouped together, they will be counted as 1 unit. Together with the 8 consonants, there are 9 units. EEAIO T S S S L L T N 3 3 3 3 3 3 3 3 3 Ini menghasilkan/This gives = 9! 3!2!2! [K1] = 15 120 [K1] Tetapi 5 huruf vokal juga boleh disusun sesama mereka dalam kumpulan mereka. But the 5 vowels also can be arranged among themselves. E E A I O 3 3 3 3 3 Ini menghasilkan/This gives 5! 2! = 60 [K1] Dengan menggunakan petua pendaraban, bilangan susunan Using the multiplication rule, the number of arrangements = 15 120 × 60 = 907 200 [N1] Bahagian B 13 (a) T7 = 8T4 ar6 = 8ar3 [K1] r6 r3 = 8 [K1] r3 = 8 r = 2 [N1] (b) (i) Sn = 3 069 3(2n – 1) 2 – 1 = 3 069 [K1] 2n – 1 = 1 023 2n = 1 024 [K1] 2n = 210 n = 10 [N1] (ii) T10 = ar9 = 3(2)9 [K1] = 1 536 cm [N1] 14 Sebelah kiri/LHS = sin 2x 1 + kos/cos 2x = 2 sin x kos/cos x 1 + (2 kos2 /cos2 x – 1) [K1] = 2 sin x kos/cos x 2 kos2 /cos2 x = sin x kos/cos x = tan x [N1] = Sebelah kanan/RHS (a) tan 15° = sin 30° 1 + kos/cos 30° = 1 2 1 +  3 2 [K1] = 1 2 2 +  3 2 = 1 2 +  3 [K1] = 2 –  3 (2 +  3 )(2 –  3 ) = 2 –  3 1 = 2 –  3 [N1] (b) tan 67.5° = sin 135° 1 + kos/cos 135° = sin (180° – 135°) 1 – kos/cos (180° – 135°) = sin 45° 1 – kos/cos 45° = 1  2 1 – 1  2 [K1] = 1  2 – 1 =  2 + 1 ( 2 – 1)( 2 + 1) [K1] =  2 + 1 2 – 1 =  2 + 1 [N1] 15 (a) sin CBD = 3 5 CBD = 36.8699° [K1] = 36.8699° × 3.142 180° = 0.6436 rad [N1] (b) 5 cm 5 cm 5 cm 5 cm 5 cm 4 cm 3 cm A B C D H G Bagi lengkok/For arc AHD, BD = BH = AB = 5 cm Bagi sukuan bulatan AGHB, For quadrant AGHB, AG = AH = AB = 5 cm [K1] Oleh itu/Thus, AH = BH = AB = 5 cm Maka, ∆ABH ialah segi tiga sama sisi. [N1] Therefore, ∆ABH is an equilateral triangle. Maka/Hence, ABH = 60° FIRASAT Add Math(Jaw).indd 2 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

•J3• Firasat SPM 2023: Matematik Tambahan - Jawapan (c) ABH = 60° = p 3 rad Seterusnya/Hence, HBD = p – p 3 – 0.6436 rad = 1.4511 rad [K1] GAH = p 2 – p 3 = p 6 rad Panjang lengkok GH Arc length GH = 5 × p 6 = 5p 6 cm [K1] Panjang lengkok HD Arc length HD = 5 × 1.4511 = 7.2555 cm [K1] Maka, hasil tambah panjang lengkok GH dan panjang lengkok HD Hence, the sum of the arc length GH and the arc length HD = 5(3.142) 6 + 7.2555 = 9.874 cm [N1] Kertas 2 Bahagian A 1 r + s + t = 0 .....................① 5r – 3r + 5t = 12 .............② 10r + s – 5t = 6 ...............③ r + s + t = 0 ..........① (–) 10r + s – 5t = 6 ..........③ –9r + 6t = –6 –3r + 2t = –2 ........④ [K1] 3r + 3s + 3t = 0 ..........① × 3 (+) 5r – 3s + 5t = 12 ........② 8r + 8t = 12 2r + 2t = 3...........⑤ [K1] –3r + 2t = –2 ........④ (–) 2r + 2t = 3 ..........⑤ –5r = –5 r = 1 [N1] Daripada ④/From ④: –3(1) + 2t = –2 2t = 1 t = 1 2 [N1] Daripada ①/From ①: 1 + s + 1 2 = 0 s = –11 2 [N1] 2 (a) Katakan/Let f –1(x) = y f(y) = x 4y + h = x [K1] y = x – h 4 f –1(x) = x – h 4 [K1] Tetapi diberi bahawa f –1(x) = x + 5 k . But it is given that f –1(x) = x + 5 k . Maka, dengan perbandingan, h = –5 dan k = 4. [N1] Hence, by comparison, h = –5 and k = 4. (b) f –1f(b) = b2 – 2 b = b2 – 2 [K1] b2 – b – 2 = 0 (b – 2)(b + 1) = 0 [K1] b = 2 atau/or b = –1 [N1] 3 (a) h(x) = –x2 + 6x – 8 = –(x2 – 6x + 8) = – x2 – 6x +  –6 2  2 –  –6 2  2 + 8 [K1] = –(x2 – 6x + 9 – 9 + 8) = –[(x – 3)2 – 1] = –(x – 3)2 + 1 [K1] Maka/Thus, q = 1 [N1] (b) Titik maksimum/Maximum point = (3, 1) [K1] Pintasan-y/y-intercept = –8 Pada paksi-x/At the x-axis, y = 0. –x2 + 6x – 8 = 0 x2 – 6x + 8 = 0 (x – 2)(x – 4) = 0 x = 2 atau/or x = 4 [K1] Maka, lengkung akan menyilang paksi-x pada (2, 0) dan (4, 0). Thus, the curve will intersect the x-axis at (2, 0) and (4, 0). [K1] [N1] y x (3, 1) 2 4 –8 O 4 Sebelah kiri/LHS = 2log4 x + 2log4 y = 2log2 x log2 4 + 2log2 y log2 4 [K1] = 2log2 x log2 22 + 2log2 y log2 22 [K1] = 2log2 x 2 + 2log2 y 2 [K1] = log2 x + log2 y = log2 xy [N1] = Sebelah kanan/RHS Katakan/Let log4 x = f dan/and log4 y = g. log2 xy = 2log4 x + 2log4 y = 2f + 2g [K1] log2 xy = 10 2f + 2g = 10 f + g = 5 ............① [K1] log4 x log4 y = 3 2 f g = 3 2 f = 3 2 g ........② [K1] Gantikan ② ke dalam ①. Substitute ② into ①. 3 2 g + g = 5 5 2 g = 5 g = 2 log4 y = 2 y = 44 y = 16 [N1] Daripada/From ②, f = 3 2 g = 3 2 (2) = 3 log4 x = 3 x = 43 x = 64 [N1] 5 (a) tan (A + B) = tan 45° = 1 tan A + tan B 1 – tan A tan B = 1 [K1] tan A + tan B = 1 – tan A tan B 2 h + 3 h = 1 –  2 h  3 h [K1] 5 h = 1 – 6 h2 5h = h2 – 6 h2 – 5h – 6 = 0 [K1] (h – 6)(h + 1) = 0 h = 6 atau/or h = –1 [N1] (b) 3 –4 5 x [K1] –5 –12 13 y [K1] sin (x + y) = sin x kos y + kos x sin y =  3 5– 5 13  + – 4 5– 12 13  [K1] = 33 65 [N1] FIRASAT Add Math(Jaw).indd 3 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J4• 6 A B C D R S T 3b ~ 2a ~ (a) (i) R →S = R→C + C→S R →S = 1 2 B →C + 1 2 C →D R →S = 1 2  3 5 A →D + 1 2 C →D R →S = 1 2  3 5 (3b ~)  + 1 2 C →B + B→A + A→D R →S = 9 10 b ~ + 1 2 – 9 5 b ~ – 2a ~ + 3b ~ [K1] R →S = 9 10 b ~ + 1 2  6 5 b ~ – 2a ~ R →S = 9 10 b ~ +  3 5 b ~ – a ~ R →S = – a ~ + 3 2 b ~ [N1] (ii) R →A = R→B + B→A = C →R + B→A = – 9 10 b ~ – 2a ~ [P1] (b) (i) R →T = kR→S R →T = k–a ~ + 3 2 b ~ R →T = –ka~ + 3 2 kb~ [P1] (ii) R →T = R →A + A→T R →T = – 9 10 b ~ – 2a ~ + mA→D R →T = – 9 10 b ~ – 2a ~ + m(3b ~) R →T = –2a ~ + 3m – 9 10  b ~ [N1] (c) Menyamakan pekali a ~, Equating the coefficients of a ~, –k = –2 k = 2 [P1] Menyamakan pekali b ~, Equating the coefficients of b ~, 3m – 9 10 = 3 2 k [K1] 3m – 9 10 = 3 2 (2) 3m = 39 10 m = 13 10 [N1] 7 (a) y = (x + 2)4 (2x – 3)3 dy dx = (x + 2)4 (3)(2x – 3)2 (2) + (2x – 3)3 (4)(x + 2)3 [K2] dy dx = 6(x + 2)4 (2x – 3)2 + 4(2x – 3)3 (x + 2)3 dy dx = 2(x + 2)3 (2x – 3)2 × [3(x + 2) + 2(2x – 3)] [K1] dy dx = 2(x + 2)3 (2x – 3)2 (7x) dy dx = 14x(x + 2)3 (2x – 3)2 [N1] (b) 1 –1 1 (3 – x)3 dx = 1 –1 (3 – x)–3 dx =  (3 – x)–2 –2(–1)  1 –1 [K1] =  1 2(3 – x)2  1 –1 = 1 2(3 – 1)2 – 1 2(3 + 1)2 [K1] = 1 8 – 1 32 = 3 32 [N1] Bahagian B 8 (a) y = px3 + kx dy dx = 3px2 + k [K1] Pada titik (1, 1), kecerunan ialah –5. At the point (1, 1), the gradient is –5. dy dx = –5 3p(1)2 + k = –5 3p + k = –5 ..........① [K1] Lengkung melalui titik (1, 1). The curve passes through the point (1, 1). 1 = p(1)3 + k(1) p + k = 1 ...............② [K1] ① – ②: 2p = –6 p = –3 Daripada ②/From ②, –3 + k = 1 k = 4 [N1] (b) Isi padu janaan = 24p Generated volume = 24p k 1 py2 dx = 24p k 1 y2 dx = 24 k 1  12 x + 2 2 dx = 24 [K1] 144k 1 (x + 2)–2 dx = 24 [K1]  (x + 2)–1 –1  k 1 = 1 6 [K1] – 1 x + 2  k 1 = 1 6 [K1] – 1 k + 2 + 1 1 + 2 = 1 6 [K1] – 1 k + 2 = – 1 6 k = 4 [N1] 9 (a) (i) Kecerunan garis lurus BC: Gradient of the straight line BC: 2x – y – 5 = 0 ⇒ y = 2x – 5 \ mBC = 2 \ mAB = – 1 mBC = – 1 2 Maka, persamaan garis lurus AB ialah Hence, the equation of the straight line AB is y – y1 = m(x – x1 ) y – (–5) = – 1 2 [x – (–10)] [K1] Pada titik/At point A(–10, –5), x1 = –10, y1 = –5. 2(y + 5) = –(x + 10) 2y + 10 = –x – 10 2y = –x – 20 [N1] (ii) Persamaan BC: Equation of BC: 2x – y – 5 = 0 ........① Persamaan AB: Equation of AB: x + 2y + 20 = 0 ......② 4x – 2y – 10 = 0 ..① × 2 (+) x + 2y + 20 = 0 ..② 5x + 10 = 0 x = –2 [K1] Daripada ①/From ①: 2(–2) – y – 5 = 0 y = –9 Maka, titik B ialah (–2, –9). [N1] Hence, point B is (–2, –9). (b) A(–10, –5) B(–2, –9) D(h, k) 2 3 B = (–2, –9)  3(–10) + 2h 2 + 3 , 3(–5) + 2k 2 + 3  = (–2, –9) [K1]  –30 + 2h 5 , –15 + 2k 5  = (–2, –9) Menyamakan koordinat-x: Equating the x-coordinates: –30 + 2h 5 = –2 2h = 20 [K1] h = 10 Menyamakan koordinat-y: Equating the y-coordinates: –15 + 2k 5 = –9 2k = –30 k = –15 Maka, titik D ialah (10, –15). Hence, point D is (10, –15).[K1] FIRASAT Add Math(Jaw).indd 4 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

•J5• Firasat SPM 2023: Matematik Tambahan - Jawapan Luas ∆ADO/Area of ∆ADO = 1 2 –10 10 0 –10 –5 –15 0 –5  = 1 2 150 – (–50) = 1 2200 = 100 unit2 /units2 [N1] (c) Katakan P ialah titik (x, y). Let P be point (x, y). Oleh sebab APB = 90°, maka AP berserenjang dengan PB. Since APB = 90°, thus AP is perpendicular to PB. (mAP)(mPB) = –1  y – (–5) x – (–10) y – (–9) x – (–2)  = –1 (y + 5)(y + 9) (x + 10)(x + 2) = –1 [K1] (y + 5)(y + 9) = –(x + 10)(x + 2) y2 + 14y + 45 = –(x2 + 12x + 20) x2 + y2 + 12x + 14y + 65 = 0 Maka, persamaan lokus bagi P ialah Hence, the equation of the locus of P is x2 + y2 + 12x + 14y + 65 = 0 [N1] 10 (a) T –5 5 15 20 25 P 2.05 3.55 6.18 8.13 10.71 log10 P 0.31 0.55 0.79 0.91 1.03 [K1] Graf log10 P melawan T Graph of log10 P against T 5 10 15 20 25 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 25 – 5 = 20 1.03 – 0.55 = 0.48 0.43 [K2] [N1] 0.67 –5 T log10 P (b) (i) Apabila/When T = 10, log10 P = 0.67 P = 4.68 [N1] (ii) P2 = qrT 2log10 P = log10 q + Tlog10 r 2log10 P = Tlog10 r + log10 q log10 P = 1 2 Tlog10 r + 1 2 log10 q [N1] 1 2 log10 r = Kecerunan/Gradient 1 2 log10 r = 0.48 20 [K1] log10 r = 0.048 r = 1.12 [N1] 1 2 log10 q = Pintasan-Y/Y-intercept 1 2 log10 q = 0.43 [K1] log10 q = 0.86 q = 7.24 [N1] 11 (a) X – Bilangan murid yang menyertai kelas tuisyen Number of students attend tuition classes X ~ B(n, 0.7) P(X  1)  0.95 1 – P(X = 0)  0.95 1 – n C0 (0.7)0 (0.3)n  0.95 [K1] 1 – 0.3n  0.95 [K1] 1 – 0.95  0.3n 0.3n  0.05 nlg 0.3  lg 0.05 [K1] n  lg 0.05 lg 0.3 n  2.49 [K1] Maka, nilai integer bagi n ialah 3. [N1] Hence, the integer value of n is 3. (b) X – Markah Matematik Tambahan/Marks of Additional Mathematics X ~ N(70, 82 ) (i) P(65  X  73) = P 65 – 70 8  Z  73 – 70 8  [K1] = P(–0.625  Z  0.375) = 1 – Q(0.625) – Q(0.375) = 1 – 0.2660 – 0.3538 = 0.3802 [N1] (ii) P(X  m) = 0.1 PZ  m – 70 8  = 0.1 [K1] 0.1 1.282 m – 70 8 = 1.282 [K1] m = 80.256 Maka, markah minimum untuk mendapat gred A ialah 81. [N1] Hence, the minimum marks to obtain grade A is 81. Bahagian C 12 (a) I x + y  14 [P1] x 0 14 y 14 0 II x  3y ⇒ 3y  x [P1] x 0 12 y 0 4 III 60x + 30y  480 2x + y  16 [P1] x 0 8 y 16 0 (b) 2 4 6 8 10 12 14 16 18 16 14 12 10 8 6 4 2 0 x y [K2] [N1] Maksimum/Maximum (10, 4) x + y = 14 2x + y = 16 30x + 20y = 60 x = 3y R (c) (i) Jika terdapat tepat 8 perjalanan ke Pulau P, lukis garis lurus x = 8. If exactly 8 trips to Island P are provided, then draw the straight line x = 8. [K1] Maka, julat bilangan perjalanan ke Pulau Q ialah 3  y  6. [N1] Hence, the range of the number of trips to Island Q is 3  y  6. (ii) Keuntungan/Profit = 30x + 20y Lukis garis lurus/Draw the straight line 30x + 20y = 60. x 0 2 y 3 0 Titik optimum ialah (10, 4). [K1] The optimal point is (10, 4). Keuntungan maksimum Maximum profit = 30(10) + 20(4) = RM380 [N1] 13 (a) x 1.50 × 100 = 120 x = 120 × 1.50 100 x = 1.80 [N1] y = 4.55 3.50 × 100 = 130 [P1] 2.70 z × 100 = 135 z = 2.70 × 100 135 z = 2.00 [N1] FIRASAT Add Math(Jaw).indd 5 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J6• (b) – I2021/2020 = (120 × 1 600) + (130 × 400) + (125 × 1 200) + (135 × 800) 1 600 + 400 + 1 200 + 800 [K2] = 502 000 4 000 = 125.5 [N1] (c) – I2021/2020 = 125.5 [K1] RM4 000 Q2020 × 100 = 125.5 Q2020 = RM4 000 × 100 125.5 Q2020 = RM3 187.25 [N1] (d) +25.5% +25.5% I2020 I2021 I2023 (100) (125.5) (?) I2023/2020 = 100 + 25.5 100 × 125.5[K1] = 157.5 [N1] 14 (a) (i) Luas/Area of ∆ADC = Luas/Area of ∆BCD 1 2 (7.5)(6)sin ADC = 1 2 (6)(5.5)sin 143° 22.5 sin ADC = 9.92994 [K1] sin ADC = 0.44133 asas/Basic  = 26.19° ADC = 180° – 26.19° = 153.81° [N1] (ii) Dalam/In ∆BCD, DB2 = 62 + 5.52 – 2(6)(5.5) kos/cos 143° DB2 = 118.96 [K1] DB = 10.907 cm DE = DB – EB = 10.907 – 8 = 2.907 cm [N1] (iii) Dalam/In ∆BCD, sin CDB 5.5 = sin 143° 10.907 [K1] sin CDB = sin 143° 10.907 × 5.5 sin CDB = 0.30347 CDB = 17.67° Dalam/In ∆CDE, CE2 = 2.9072 + 62 – 2(2.907)(6) kos/cos 17.67° [K1] CE2 = 11.21246 CE = 3.349 cm [N1] (b) (i) [K1] [N1] C' 6 cm 7.5 cm 2.907 cm 8 cm 143° 5.5 cm 6 cm 37° 37° D' A B' C E (ii) D'C'B' = 37° [P1] 15 (a) (i) vP = 2t + 3 sP = (2t + 3) dt sP = t2 + 3t + c Apabila/When t = 0, sP = 0 . Maka/Thus, c = 0. sP = t2 + 3t [N1] vQ = 5 – t sQ = (5 – t) dt sQ = 5t – t2 2 + c [K1] Apabila/When t = 0, sQ = 7.5. Maka/Thus, c = 7.5. sQ = 5t – 0.5t2 + 7.5 [N1] (ii) Apabila/When vP  0, 2t + 3  0 t  – 3 2 ........① Apabila/When vQ  0, 5 – t  0 t  5 .............② [K1] Menggabungkan ① dan ②: Combining ① and ②: 0  t  5 [N1] (iii) Apabila/When vQ = 0, t = 5 Apabila/When t = 5, sP = 52 + 3(5) = 40 sQ = 5(5) – 0.5(5)2 + 7.5 = 20 [K1] Jarak di antara zarah P dengan zarah Q Distance between particles P and Q = 40 – 20 = 20 m [N1] (b) (i) Apabila zarah P dan zarah Q bertemu, When particles P and Q meet, sP = sQ t 2 + 3t = 5t – 0.5t2 + 7.5 [K1] 1.5t2 – 2t – 7.5 = 0 3t2 – 4 – 15 = 0 (t – 3)(3t + 5) = 0 t = 3 atau/or t = – 5 3 t = – 5 3 tidak diterima/is not accepted. \ t = 3 [N1] (ii) Apabila/When t = 3, sP = 32 + 3(3) = 18 m [N1] Kertas Model 2 Kertas 1 Bahagian A 1 (a) Katakan/Let g–1(x) = y g(y) = x 2y + 1 = x y = x – 1 2 g–1(x) = x – 1 2 [K1] fg–1  p 2  = –1 f  p 2 – 1 2  = –1 f  p – 2 4  = –1 p – 2 4 – 3 = –1 [K1] p – 2 4 = 2 p – 2 = 8 p = 10 [N1] (b) ff–1(k2 ) = 6k – 9 k2 = 6k – 9 [K1] ff–1(x) = x k2 – 6k + 9 = 0 [K1] (k – 3)2 = 0 k = 3 [N1] 2 (a) y = – 1 360 x2 + 3 20 x + 1 Apabila lembing itu dilepaskan, x = 0. [N1] When the spear is released, x = 0. y = – 1 360 (0)2 + 3 20 (0) + 1 = 1 m (b) Apabila/When y = 0, – 1 360 x2 + 3 20 x + 1 = 0 [N1] x2 – 54x – 360 = 0 x = –b ±  b2 – 4ac 2a x = –(–54) ±  (–54)2 – 4(1)(–360) 2(1) x = 54 ±  4 356 2 [K1] x = 54 ± 66 2 [K1] x = 60 atau/or x = –6 [K1] x = –6 tidak diterima. x = –6 is not accepted. Oleh itu/Thus, x = 60 Maka, jarak mengufuk maksimum yang dicapai ialah 60 m. [N1] Hence, the maximum horizontal distance set is 60 m. 3 4x2 + 4x = 7 + q 4x2 + 4x – 7 – q = 0 [K1] a = 4, b = 4, c = –7 – q Katakan satu punca ialah α. Maka, punca yang satu lagi ialah α – 2. [K1] Let one root be α. Then, the other root is α – 2. Hasil tambah punca-punca/Sum of roots = – b a α + (α – 2) = – 4 4 [K1] 2α – 2 = –1 2α = 1 α = 1 2 [K1] Hasil darab punca-punca/Product of roots = – c a α(α – 2) = –7 – q 4 [K1] 4 1 21 2 – 2 = –7 – q –3 = –7 – q q = –4 [N1] FIRASAT Add Math(Jaw).indd 6 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

•J7• Firasat SPM 2023: Matematik Tambahan - Jawapan 4 f(x) = (k – 6)x2 – 8x + k a = k – 6, b = –8, c = k b2 – 4ac  0 (–8)2 – 4(k – 6)(k)  0 [K1] 64 – 4k2 + 24k  0 16 – k2 + 6k  0 k2 – 6k – 16  0 (k + 2)(k – 8)  0 [K1] –2 8 k Julat nilai k yang dikehendaki ialah –2  k  8. The required range of values of k is –2  k  8. Jika lengkung menyentuh paksi-x (iaitu paksi-x ialah tangen kepada lengkung), b2 – 4ac = 0. [K1] If the curve touches the x-axis (which is the x-axis is a tangent to the curve), b2 – 4ac = 0. Maka/Hence, k = –2 atau/or k = 8. [N2] 5 x + y 2 + 1 = 0 2x + y + 2 = 0 y = –2x – 2 .........① [K1] 1 x + 2 y – 1 2 = 0 2y + 4x – xy = 0 ................② [K1] Gantikan ① ke dalam ②: Substitute ① into ②: 2(–2x – 2) + 4x – x(–2x – 2) = 0 –4x – 4 + 4x + 2x2 + 2x = 0 2x2 + 2x – 4 = 0 x2 + x – 2 = 0 (x + 2)(x – 1) = 0 [K1] x = –2 atau/or x = 1 Apabila/When x = –2, y = –2(–2) – 2 = 2 Apabila/When x = 1, y = –2(1) – 2 = –4 [K1] Penyelesaian ialah/The solutions are x = –2, y = 2 atau/or x = 1, y = – 4 [N1] 6 8 – 3 2 4 + 3 2 = (8 – 3 2 )(4 – 3 2 ) (4 + 3 2 )(4 – 3 2 ) [K2] = 32 – 24 2 – 12 2 + 18 16 – 18 [K1] = 50 – 36 2 –2 [K1] = 18 2 – 25 [N1] 7 log3 T – log9 V = 2 log3 T – log3 V log3 9 = 2 [K1] log3 T – log3 V 2 = 2 [K1] log3 9 = log3 32 = 2 log3 3 = 2(1) = 2 2 log3 T – log3 V = 4 log3 T2 – log3 V = 4 [K1] log3 T2 V = 4 [K1] T2 V = 34 T2 = 34 V T = 9V [N1] 8 (a) S10 = 155 10 2 [2a + (10 – 1)d] = 155 5(2a + 9d) = 155 2a + 9d = 31.........① [K1] T2 + T4 + T6 + T8 + T10 = 85 (a + d) + (a + 3d) + (a + 5d) + (a + 7d) + (a + 9d) = 85 5a + 25d = 85 ..........② [K1] 10a + 45d = 155 ........① × 5 (−) 10a + 50d = 170 ........② × 2 –5d = –15 [K1] d = 3 [N1] Daripada ①/From ①: 2a + 9(3) = 31 2a = 4 a = 2 [N1] (b) T10 = a + 9d = 2 + 9(3) = 29 [N1] 9 Pembahagi dua sama serenjang AC./Perpendicular bisector of AC. C(16, 12) B(13, 3) A(2, 0) D(h, k) x y O (a) Katakan koordinat titik D ialah (h, k). Let the coordinates of point D be (h, k). Titik tengah BD = Titik tengah AC Midpoint of BD = Midpoint of AC  h + 13 2 , k + 3 2  =  2 + 16 2 , 0 + 12 2   h + 13 2 , k + 3 2  = (9, 6) [K1] Menyamakan koordinat-x: Equating the x-coordinate: h + 13 2 = 6 h + 13 = 18 h = 5 Menyamakan koordinat-y: Equating the y-coordinate: k + 3 2 = 6 k + 3 = 12 k = 9 [K1] Maka, koordinat titik D ialah (5, 9). [N1] Hence, the coordinates of point D are (5, 9). (b) Titik tengah AC/ Midpoint of AC = (9, 6) Kecerunan AC/ Gradient of AC = 12 – 0 16 – 2 = 6 7 Kecerunan pembahagi dua sama serenjang Gradient of the perpendicular bisector = – 1  6 7 = – 7 6 [K1] Persamaan pembahagi dua sama serenjang ialah Equation of the perpendicular bisector is y – 6 x – 9 = – 7 6 6(y – 6) = –7(x – 9) 6y – 36 = –7x + 63 6y = –7x + 99 [N1] 10 (a) P → R = 3 5 P → Q O → R – O→ P = 3 5 (O → Q – O→ P) O → R –  –2 3  = 3 5 13 8  –  –2 3  O → R –  –2 3  = 3 5  15 5  [K1] O → R =  9 3 +  –2 3  O → R =  7 6 [N1] (b) P → R = 3 5 P → Q = 3 5  15 5  =  9 3 |P → R| =  92 + 32 =  90 =  9 × 10 = 3 10 [K1] Vektor unit dalam arah P → R Unit vector in the direction of P→ R = 1 3 10  9 3 [K1] = 1 3 10   10  10 9 3 =  10 30  9 3 =  3 10 10  10 10  [N1] FIRASAT Add Math(Jaw).indd 7 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J8• 11 (a) 360° – 265°18’ = 94.7° 94.7° = 94.7° × 3.142 180° [K1] = 1.653 rad [N1] (b) Luas sektor minor AOB = 82.65 cm2 Area of minor sector AOB = 82.65 cm2 1 2 × r 2 × 1.653 = 82.65 r 2 = 100 r = 10 cm [K1] Area of the shaded segment Luas tembereng berlorek = 1 2 × 102 × (1.653 – sin 94.7°) [K1] = 32.82 cm2 [N1] 12 ax2 + by2 – 3 = 0 by2 = –ax2 + 3 y2 = –  a b(x2 ) + 3 b [K1] Pintasan-Y/ Y-intercept = – 4 3 b = – 4 [K1] b = – 3 4 [N1] Kecerunan/Gradient = – a b – a – 3 4 = 4 8 [K1] a = 4 8 × 3 4 a = 3 8 [N1] Bahagian B 13 (a) T = 2p m 4  1 2 T = 2p m 1 2 2  [K1] T = pm 1 2 [K1] dT dm = 1 2 pm 1 2 – 1 [K1] dT dm = 1 2 pm– 1 2 dT dm = p 2 m [N1] (b) δT δm ≈ dT dm δT ≈ dT dm × δm δT ≈ 1 2p 1  m × δm [K1] δT ≈ 1 2p 1  0.10 × (0.11 – 0.10) δT ≈ 1 2p 1  0.10 × (0.01) [K1] δT ≈ 0.01581p [N1] 14 Katakan/Let AB = a, BC = ar dan/ and AC = ar2 = 32 cm. ar2 = 32 .....① [K1] Perimeter = 56 cm a + ar + ar2 = 56 [K1] a + ar + 32 = 56 a + ar = 24 a(1 + r) = 24 ........② [K1] ① ② : ar2 a(1 + r) = 32 24 [K1] r2 1 + r = 4 3 3r2 = 4 + 4r 3r2 – 4r – 4 = 0 (r – 2)(3r + 2) = 0 [K1] r = 2 atau/or r = – 2 3 r = – 2 3 tidak diterima. r = – 2 3 is not accepted. \ r = 2 [N1] (b) Daripada ①/ From ①: ar2 = 32 a(2)2 = 32 a = 8 [K1] Maka, panjang sisi yang terpendek ialah 8 cm. [N1] Hence, the length of the shortest side is 8 cm. 15 X – Jisim sekampit beras, dalam kg Mass of a bag of rice, in kg X ~ N(µ, σ2 ) Diberi/Given P(X  30.5) = 97.5% = 0.975 P Z  30.5 – µ σ  = 0.975 [K1] 0.025 0.975 –1.96 30.5 – µ σ = –1.96 [K1] 30.5 – µ = –1.96σ .......① [K1] Diberi/Given P(X  46.3) = 88.49% = 0.8849 PZ  46.3 – µ σ  = 0.8849 0.8849 0.1151 1.2 46.3 – µ σ = 1.2 [K1] 46.3 – µ = 1.2σ ....................② [K1] 30.5 – µ = –1.96σ ....... ① (–) 46.3 – µ = 1.2σ ........... ② [K1] –15.8 = –3.16σ σ = 5 [N1] Daripada ①/From ①, 30.5 – µ = –1.96(5) μ = 40.3 [N1] Kertas 2 Bahagian A 1 (a) Apabila/ When t = 0, sQ ~ = (10i ~ + 40j ~ ) + (0)(10i ~ + 10j ~ ) = 10i ~ + 40j ~ [N1] (b) |sQ ~ |=  102 + 402 =  1 700 = 41.23 km [N1] (c) (i) s ~ = tv ~ sP ~ = t(15i ~ + 30j ~ ) [K1] Dengan perbandingan/By comparison, vP ~ = 15i ~ + 30j ~ [N1] (ii) sQ ~ = (10i ~ + 40j ~ ) + t(10i ~ + 10j ~ ) [K1] Sesaran awal Initial position s ~ = tv ~ Dengan perbandingan By comparison, vQ ~ = 10i ~ + 10j ~ [N1] (d) Apabila bot P dan bot Q bertembung, When boat P and boat Q collide, sP ~ = sQ ~ t(15i ~ + 30j ~ ) = (10i ~ + 40j ~ ) + t(10i ~ + 10j ~ ) [K1] 15ti ~ + 30tj ~ = (10 + 10t)i ~ + (40 + 10t)j ~ Dengan perbandingan, By comparison, 15t = 10 + 10t [K1] 5t = 10 t = 2 Maka, kedua-dua bot akan bertembung selepas 2 jam. Hence, both boats will collide after 2 hours. [N1] 2 (a) (i) f 2 (x) = f(6 – 2x) = 6 – 2(6 – 2x) = 6 – 12 + 4x = 4x – 6 [N1] (ii) Katakan/Let (f 2 )–1(x) = w f 2 (w) = x 4w – 6 = x [K1] w = x + 6 4 (f 2 )–1(x) = x + 6 4 [N1] (b) y = | f 2 (x)|= |4x – 6| Pada paksi-x/On the x-axis, y = 0 |4x – 6| = 0 x = 3 2 = 11 2 [K1] x 0 11 2 3 y 6 0 6 f(0) = |4(0) – 6| = |–6| = 6 [N1] y O x 6 (3, 6) 1 1 2 FIRASAT Add Math(Jaw).indd 8 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

•J9• Firasat SPM 2023: Matematik Tambahan - Jawapan Julat ialah/ The range is 0  f(x)  6. [N1] 3 2x2 – 4x + 5 = 0 a = 2, b = –4, c = 5 Hasil tambah punca-punca/Sum of roots = – b a α + β = –  –4 2  α + β = 2 [K1] Hasil darab punca-punca/Product of roots = c a αβ = 5 2 [K1] Hasil tambah punca-punca baharu/Sum of new roots = α – 1 β  + β – 1 α  = α + β –  1 α + 1 β  = 2 –  β + α αβ  [K1] = 2 –  2 5 2  = 2 – 4 5 = 6 5 [K1] Hasil darab punca-punca baharu Product of new roots = α – 1 β β – 1 α  = αβ – 1 – 1 + 1 αβ [K1] = 5 2 – 2 + 2 5 = 9 10 [K1] Persamaan kuadratik yang diperlukan ialah The required quadratic equation is x2 – 6 5 x + 9 10 = 0 10x2 – 12x + 9 = 0 [N1] 4 0.6x + 0.3y + 0.1z = 6.25 .........① 0.4x + 0.3y + 0.3z = 4.21 .........② 0.3x + 0.7y = 3.23 .........③ 1.8x + 0.9y + 0.3z = 18.75 .. ① × 3 (−) 0.4x + 0.3y + 0.3z = 4.21 ............② 1.4x + 0.6y = 14.54 ............④ ③ × 1.4: 0.42x + 0.98y = 4.522 .... ⑤ ④ × 0.3: 0.42x + 0.18y = 4.362 .... ⑥ ⑤ − ⑥: 0.8y = 0.16 [K1] y = 0.20 [N1] Daripada ④/From ④: 1.4x + 0.6(0.20) = 14.54 1.4x = 14.42 x = 10.30 [N1] Daripada ①/From ①: 0.6(10.30) + 0.3(0.20) + 0.1z = 6.25 [K1] 0.1z = 0.01 z = 0.10 [N1] 5 BC =  AC2 – AB2 BC =  (2 + 4 3 ) 2 – (4 + 2 3 ) 2 [K1] BC =  4 + 16 3 + 48 – (16 + 16 3 + 12) [K1] BC =  24 [K1] BC =  4 × 6 [K1] BC = 2 6 unit/units [N1] [P1] [P1] [P1] 6 (a) sin BOC = 8 12 [K1] BOC = 41.81° [K1] BOC = 41.81° × 3.142 180° [K1] BOC = 0.7298 rad [N1] (b) OC2 = 122 – 82 = 80 OC =  80 = 8.9443 [K1] Perimeter kawasan berlorek Perimeter of the shaded region = BC + AC + Panjang lengkok AB/Arc length AB = 8 + (12 – 8.9443) + 12(0.7298) [K3] = 8 + 3.0557 + 8.7576 = 19.81 cm [N1] 7 P1 P2 P3 P4 P5 P6 3 3 3 3 3 Bilangan cara ialah/The number of ways is = (5 – 1)! [K2] = 4! [K1] = 24 [K1] Tetapi P1 dan P2 boleh saling tukar kedudukan. But P1 and P2 can interchange positions. Maka, jumlah cara Hence, the number ways = 24 × 2 [K1] = 48 [N1] Bahagian B 8 (a) T1 = 0.0075 × 2 = 0.015 cm T2 = 0.015 × 2 = 0.03 cm T3 = 0.03 × 2 = 0.06 cm [K1] T2 T1 = 0.03 0.015 = 2 T3 T2 = 0.06 0.03 = 2 [K1] Oleh sebab T2 T1 = T3 T2 = 2 (pemalar), maka ketebalan lipatan kertas itu membentuk suatu janjang geometri dengan keadaan a = 0.015 dan r = 2. [K1] [N1] Since T2 T1 = T3 T2 = 2 (constant), then the thickness of the paper folding form a geometric progression such that a = 0.015 and r = 2. (b) T12 = arn – 1 = 0.015(2)12 – 1 [K1] = 30.72 cm [N1] (c) Sn = a(rn – 1) r – 1  15 0.015(2n – 1) 2 – 1  15 [K1] 2n – 1  15 0.015 2n  1 001 n log10 2  log10 1 001 [K1] n(0.3010)  3 n  3 0.3010 n  9.967 [K1] Bilangan kali minimum lipatan kertas/The minimum number of folding = 10 [N1] 9 (a) Isi padu/Volume V = 2x(10 – 2x) + 1 3 × 2(10 – 2x) × 1 2 x [K2] = 20x – 4x2 + 1 3 x(10 – 2x) = 20x – 4x2 + 10 3 x – 2 3 x2 = 60x – 12x2 + 10x – 2x2 3 [K1] = 70x – 14x2 3 [N1] [Tertunjuk/Shown] (b) dV dx = 1 3 (70 – 28x) [K1] Apabila V adalah maksimum, When V is maximum, dV dx = 0 1 3 (70 – 28x) = 0 [K1] 28x = 70 x = 5 2 [K1] d2 V dx2 = – 28 3 (< 0) [K1] Maka, Vmaksimum /Hence, Vmaximum = 70 5 2  – 14  5 2  2 3 [K1] = 291 6 cm3 [N1] 10 (a) Sebelah kiri/LHS = 2 sin (x – y) kos/cos (x – y) – kos/cos (x + y) = 2(sin x kos/cos y – kos/cos x sin y) kos/cos x kos/cos y + sin x sin y – (kos/cos x kos/cos y – sin x sin y) [K1] FIRASAT Add Math(Jaw).indd 9 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J10• = 2(sin x kos/cos y – kos/cos x sin y) 2 sin x sin y [K1] = (sin x kos/cos y – kos/cos x sin y) sin x sin y [K1] = sin x kos/cos y sin x sin y – kos/cos x sin y sin x sin y [K1] = kos/cos y sin y – kos/cos x sin x = kot/cot y – kot/cot x [N1] = Sebelah kanan/RHS (b) tan (45° + z) = 4 tan (45° – z) tan 45° + tan z 1 – tan 45° tan z = 4 tan 45° – tan z 1 + tan 45° tan z [K1] 1 + tan z 1 – tan z = 4 1 – tan z 1 + tan z [K1] (1 + tan z)2 = 4(1 – tan z)2 1 + 2 tan z + tan2 z = 4(1 – 2 tan z + tan2 z) 3 – 10 tan z + 3 tan2 z = 0 3 tan2 z – 10 tan z + 3 = 0 (3 tan z – 1)(tan z – 3) = 0 [K1] tan z = 1 3 atau/or tan z = 3 [K1] z = 18.43°, 71.57°, 198.43°, 251.57° [N1] 11 (a) X – Bilangan murid yang menonton berita setiap hari X – Number of students who watch the news every day X ~ B10, 1 3  (i) Min/Mean = np = 10 × 1 3 = 3 1 3 [N1] Sisihan piawai/Standard deviation =  npq =  10 × 1 3 × 2 3 = 1.491 [N1] (ii) P(X  3) = P(X = 0) + P(X = 1) + P(X = 2) = 10C0  1 3  0  2 3  10 + 10C1  1 3  1  2 3  9 + 10C2  1 3  2  2 3  8 [K2] = 0.2291 [N1] (b) X – Jisim bersih setin kerepek ubi kentang X – Net mass of a tin of potato chips X ~ N(µ, 4.52 ) (i) P(X  210) = 0.0228 PZ  210 – µ 4.5  = 0.0228 [K1] 0.0228 2 z 210 – µ 4.5 = 2 [K1] 210 – µ = 9 µ = 201 [N1] (ii) P(196  X  198) = P 196 – 201 4.5  Z  198 – 201 4.5  = P(–1.111  Z  –0.667) [K1] = Q(0.667) – Q(1.111) = 0.2523 – 0.1333 = 0.119 = 11.9% [N1] Bahagian C 12 (a) (i) Dalam segi tiga PQR, menggunakan petua sinus, In triangle PQR, using the sine rule, sin QPR 11 = sin 40° 8 [K1] sin QPR = sin 40° 8 × 11 sin QPR = 0.88383 QPR = 62.11° [K1] PQR = 180° – 40° – 62.11° = 77.89° [N1] (ii) Menggunakan petua sinus/Using the sine rule, PR sin PQR = 8 sin 40° [K1] PR sin 77.89° = 8 sin 40° PR = 8 sin 40° × sin 77.89° PR = 12.1688 = 12.17 m [N1] (b) [N1] P P' R' Q' 8 m 62.11° 62.11° 22.11° 117.89° 40° 11 m (i) Q’P’R’ = 180° – Q’P’P = 180° – Q’PP’ = 180° – 62.11° = 117.89° [K1] P’Q’R’ = 180° – 117.89° – 40° = 22.11° [N1] (ii) Luas segi tiga P’Q’R’ Area of triangle P’Q’R’ = 1 2 × 8 × 11 × sin 22.11° [K1] = 16.56 m2 [N1] 13 (a) Bahan A/Ingredient A: I2021/2018 = 105 Q2021 Q2018 × 100 = 105 8.40 a × 100 = 105 a = 8.40 × 100 105 [K1] = 8.00 [N1] (b) Bahan C/Ingredient C: I2021/2018 = 120 c b × 100 = 120 [K1] c = 120b 100 c = 1.20b ........... ① c – b = 1.50 .............② Gantikan ① ke dalam ②: Substitute ① into ②: 1.20b – b = 1.50 0.20b = 1.50 b = 1.50 0.20 = 7.50 [N1] Daripada ①/From ①: c = 1.20(7.50) = 9.00 [N1] (c) (i) I – 2021/2018 = 113.46 Q2021 Q2018 × 100 = 113.46 Q2021 12.40 × 100 = 113.46 [K1] Q2021 = 113.46 × 12.40 100 = 14.07 [N1] (ii) Bahan Ingredient Harga per kg (RM) Price per kg (RM) I2021/2018 Tahun 2018 Year 2018 Tahun 2021 Year 2021 A 8.00 8.40 105 B 3.00 3.60 120 C 7.50 9.00 120 D 5.00 5.50 110 I – 2021/2018 = 113.46 ΣIw Σw = 113.46 (105 × 5) + (120 × 3) + (120k) + (110 × 1) 5 + 3 + k + 1 = 113.46 120k + 995 k + 9 = 113.46 [K1] 120k + 995 = 113.46k + 1 021.14 [K1] 6.54k = 26.14 k = 4 [N1] FIRASAT Add Math(Jaw).indd 10 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

•J11• Firasat SPM 2023: Matematik Tambahan - Jawapan 14 (a) Apabila/When v = 0, 3t2 – 10t – 8 = 0 (3t + 2)(t – 4) = 0 [K1] t = – 2 3 atau/or t = 4 t = – 2 3 tidak diterima/is not accepted \ t = 4 [K1] a = dv dt = 6t – 10 [K1] Apabila/When t = 4, a = 6(4) – 10 = 14 m s–2 [N1] (b) Apabila zarah itu bergerak ke kanan, When the particle moves to the right, v  0 3t2 – 10t – 8  0 (3t + 2)(t – 4)  0 [K1] 4 t – 2 3 t  – 2 3 tidak diterima/ is not accepted. \ t  4 [N1] (c) s =  v dt s =  (3t2 – 10t – 8) dt [K1] s = 3t3 3 – 10t 2 2 – 8t + c Apabila/When t = 0, s = 0. Oleh itu/Thus, c = 0. \ s = t 3 – 5t2 – 8t [K1] t 0 4 7 s 0 –48 42 Zarah berpatah balik. The particle reverses its direction. –48 0 42 t = 4 t = 7 t = 0 Maka, jumlah jarak yang dilalui dalam 7 saat yang pertama ialah Hence, the total distance travelled in the first 7 seconds is = 48 + 48 + 42 [K1] = 138 m [N1] 15 (a) I x + y  80 [P1] II 3 200x + 8 000y  224 000 2x + 5y  140[P1] III x – y  10 x – 10  y y  x – 10 [P1] (b) x + y = 80 x 0 80 y 80 0 2x + 5y = 140 x 0 70 y 28 0 y = x – 10 x 60 10 y 50 0 (b) [K2] [N1] 10 20 30 40 50 60 70 80 80 70 60 50 40 30 20 10 0 y x 1 000x + 1 500y = 15 000 28 15 Maksimum/Maximum (45, 35) x + y = 80 x – y = 10 R 2x + 5y = 140 (c) (i) Jika/If y = 30, xminimum = 40 dan/and x maksimum/maximum = 50 [N1] (ii) Keuntungan/Profit = 1 000x + 1 500y Lukis garis lurus/ Draw the straight line 1 000x + 1 500y = 15 000 [K1] x 0 15 y 10 0 Titik optimum ialah (45, 35). The optimal point is (45, 35). [K1] Maka, keuntungan maksimum Hence, the maximum profit = 1 000(45) + 1 500(35) = RM97 500 [N1] Kertas Model 3 Kertas 1 Bahagian A 1 y = 2x x – 1 1 y = x – 1 2x [K1] 1 y = 1 2 –  1 21 x 1 y = –  1 21 x + 1 2 [K1] Bagi titik/For point (n, 24), 1 x = n dan/and 1 y = 24. 24 = – 1 2n + 1 2 [K1] 48 = –n + 1 n = – 47 [N1] Bagi titik/For point (2, m), 1 x = 2 dan/and 1 y = m. m = – 1 2(2) + 1 2 m = – 1 2 [N1] 2 (a) g(5) = −14 1 – h(5) = –14 5h = 15 h = 3 [N1] gf(x) = 10 – 6x g(2x – k) = 10 – 6x 1 – 3(2x – k) = 10 – 6x [K1] 1 – 6x + 3k = 10 – 6x 3k = 9 k = 3 [N1] (b) Jika/If ff–1(m – 2) = 3 m m – 2 = [K1] 3 m ff–1(x) = x sentiasa benar. ff–1(x) = x is always true. m2 – 2m = 3 m2 – 2m – 3 = 0 [K1] (m + 1)(m – 3) = 0 m = –1 atau/or m = 3 [N1] FIRASAT Add Math(Jaw).indd 11 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J12• 3 kx2 – 5x + k + 1 = 0 a = k, b = –5, c = k + 1 Katakan punca-punca ialah 2α dan 3α. Let the roots be 2α and 3α. Hasil tambah punca-punca Sum of roots: 2α + 3α = 5 k [K1] 5α = 5 k α = 1 k [K1] Hasil darab punca-punca/Product of roots: (2α)(3α) = k + 1 k [K1] 6α2 = k + 1 k 6 1 k  2 = k + 1 k 6 k 2 = k + 1 k [K1] 6 k = k + 1 1 k2 + k = 6 k2 + k – 6 = 0 [K1] (k – 2)(k + 3) = 0 k = 2 atau/or k = –3 [N1] 4 y = t – 3x ................ ① y = 3 x ...................... ② Gantikan ① ke dalam ②: Substitute ① into ②: t – 3x = 3 x [K1] tx – 3x2 = 3 3x2 – tx + 3 = 0 [K1] a = 3, b = –t , c = 3 Jika garis lurus tidak bersilang dengan lengkung, maka If a straight line does not intersect the curve, then b2 – 4ac  0 (–t)2 – 4(3)(3)  0 [K1] t2 – 36  0 (t + 6)(t – 6)  0 [K1] –6 6 t Julat nilai t yang dikehendaki ialah –6  t  6. [N1] The required range of values of t is –6  t  6. 5 Luas JKMN/ Area of JKMN = 80p xy = 80p y = 80p x ............① [K1] Panjang MN melebihi panjang lengkok LM sebanyak 6p cm. The length of MN exceeds the arc length LM by 6p cm. x – y p 2 = 6p ..........②: [K1] s = jθ Gantikan ① ke dalam ②: Substitute ① into ②: x –  80p x p 2 = 6p [K1] 2x2 – 80p2 = 12px x2 – 40p2 – 6px = 0 x2 – 6px – 40p2 = 0 [K1] (x – 10p)(x + 4p) = 0 x = 10p atau/or x = –4p x = –4p tidak diterima. x = –4p is not accepted. \ x = 10p [N1] Daripada ①/ From ①: y = 80p 10p = 8 [N1] 6 32x = 9 272y 32x = 32 (33 )2y 32x = 32 – 6y [K1] 2x = 2 – 6y x = 1 – 3y .........① [K1] log4 2 + log4 (2x – 3y) = log4 (x + 5) log4 2(2x – 3y) = log4 (x + 5) [K1] 4x – 6y = x + 5 3x – 6y = 5 ..........② [K1] Gantikan ① ke dalam ②: Substitute ① into ②: 3(1 – 3y) – 6y = 5 3 – 15y = 5 15y = –2 y = – 2 15 [N1] Daripada ①/From ①: x = 1 – 3– 2 15 = 7 5 [N1] 7 RA = 2RC  (x – 6)2 + (y – 10)2 = [K1] 2 (x – 9)2 + (y – 6)2 (x – 6)2 + (y – 10)2 = [K1] 22 [(x – 9)2 + (y – 6)2 ] x2 – 12x + 36 + y2 – 20y + 100 [K1] = 4(x2 – 18x + 81 + y2 – 12y + 36) x2 – 12x + y2 – 20y + 136 = 4x2 + 4y2 – 72x – 48y + 468 3x2 + 3y2 – 60x – 28y + 332 = 0 [N1] 8 (a) S →R = 2 3 P →Q = 2 3 (9a ~) = 6a ~ [K1] P →R = P →S + S →R = 2b ~ + 6a ~ [N1] (b) (i) P →A = 1 3 P →Q = 1 3 (9a ~) = 3a ~ A →B = h(P →S) = h(2b ~) = 2hb~ P →B = P →A + A →B = 3a ~ + 2hb~ [N1] (ii) Oleh sebab P, B dan R adalah segaris, maka Since P, B and R are collinear, then P →B = kP→R 3a ~ + 2hb~ = k(6a ~ + 2b ~) [K1] 3a ~ + 2hb~ = 6ka ~ + 2kb ~ Menyamakan pekali a ~, Equating the coefficients of a ~, 6k = 3 k = 1 2 [N1] Menyamakan pekali b ~, Equating the coefficients of b ~, 2k = 2h h = k h = 1 2 [N1] 9 f(x) = 3x 2x – 4 u v u = 3x v = 2x – 4 du dx = 3 dv dx = 2 f’(x) = v du dx – u dv dx v2 = (2x – 4)(3) – 3x(2) (2x – 4)2 [K1] = 6x – 12 – 6x (2x – 4)2 = –12 (2x – 4)2 = –12(2x – 4)–2 [K1] f"(x) = 24(2x – 4)–3(2) = 48 (2x – 4)3 [K1] f"(3) = 48 (2 × 3 – 4)3 [K1] = 48 8 = 6 [N1] 10 4 2 f(x) dx + 9 4 f(x) dx + 12 9 f(x) dx – 3 8 g(x) dx = 12 2 f(x) dx + 8 3 g(x) dx [K2] = 20 + 12 [K1] = 32 [N1] 11 dy dx = 36 (2x + 1)3 y =  36 (2x + 1)3 dx [K1] y =  36(2x + 1)–3 dx FIRASAT Add Math(Jaw).indd 12 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

•J13• Firasat SPM 2023: Matematik Tambahan - Jawapan y = 36 (2x + 1)–2 –2(2)  + c [K1] y = –9 1 (2x + 1)2  + c [K1] x = 1 dan/and y = 4 4 = –9 1 (2(1) + 1)2  + c 4 = –1 + c c = 5 [K1] Maka, persamaan lengkung ialah y = – 9 (2x + 1)2 + 5. [N1] Hence, the equation of the curve is y = – 9 (2x + 1)2 + 5. 12 (a) Bilangan cara untuk memilih 2 orang murid lelaki dan 2 orang murid perempuan Number of ways to select 2 boys and 2 girls = 3 C2 × 7 C2 = 63 [K1][N1] 3 orang murid lelaki, pilih 2. 3 boys, choose 2. 7 orang murid perempuan, pilih 2. 7 girls, choose 2. (b) Bilangan yang ada. Numbers that are available. 3 7 Bilangan cara memilih wakil murid Number of ways to select student representatives Bilangan murid lelaki Number of boys Bilangan murid perempuan Number of girls 1 3 3 C1 × 7 C3 2 2 3 C2 × 7 C2 3 1 3 C3 × 7 C1 4 0 Tidak mungkin/ Impossible Bilangan yang diperlukan. Numbers that are required. Maka, bilangan cara memilih dengan keadaan sekurangkurangnya seorang murid lelaki dipilih Hence, the number of selections such that at least 1 boy selected = 3 C1 × 7 C3 + 3 C2 × 7 C2 + 3 C3 × 7 C1 = 105 + 63 + 7 [K3] = 175 [N1] Bahagian B 13 4x + 2y + 2z = 190 ............① [P1] 3x + 4y + 3z = 295 ............② [P1] 2x + 4y + 2z = 250 ............③ [P1] ① − ③: 2x – 2y = –60 ....................④ ① × 3 − ② × 2: 6x – 2y = –20 .......⑤ [K1] ⑤ − ④: 4x = 40 x = 10 Daripada ④/From ④: 2(10) – 2y = –60 [K1] 2y = 20 + 60 y = 40 Daripada ①/From ①: 4(10) + 2(40) + 2z = 190 2z = 70 z = 35 Harga sehelai baju-T, sepasang seluar sukan dan sepasang kasut masing-masing ialah RM10, RM40 dan RM35. [N3] The price of a T-shirt, a pair of sport pants and a pair of shoes are RM10, RM40 and RM35 respectively. 14 (a) mBC = mAO = 3 Katakan A ialah titik (h, k). Let A be point (h, k). Diberi/Given OA = 5 10 h2 + k2 = 25(10) h2 + k2 = 250 ......... ① Titik A(h, k) terletak pada garis lurus y = 3x. Point A(h, k) lies on the straight line y = 3x. Oleh itu/Thus, k = 3h ...... ② [K1] Gantikan ② ke dalam ①: Substitute ② into ①: h2 + (3h)2 = 250 10h2 = 250 h2 = 25 h = 5 Daripada ②/From ②: k = 3(5) = 15 Maka, koordinat titik A ialah (5, 15). [N1] Hence, the coordinates of point A are (5, 15). (b) mOA = 3 Oleh itu/Thus, mAB = – 1 3 Persamaan AB ialah The equation of AB is y – 15 = – 1 3 (x – 5) 3y – 45 = –x + 5 3y = –x + 50 [K1] Pada titik B (paksi-y)/On point B (y-axis), x = 0. 3y = –0 + 50 [K1] y = 50 3 = 162 3 Maka, koordinat titik B ialah 0, 16 2 3 . [N1] Hence, the coordinates of point B are 0, 16 2 3 . (c) mBC = mAO = 3 Persamaan BC ialah The equation of BC is y – 50 3 = 3(x – 0) 3y – 50 = 9x 3y = 9x + 50 ..... ③ [K1] Persamaan OC ialah The equation of OC is y + 2x = 0 y = –2x .....④ [K1] Gantikan ④ ke dalam ③: Substitute ④ into ③: 3(–2x) = 9x + 50 15x = –50 x = – 10 3 = – 31 3 Daripada ④/From ④: y = –2– 10 3  = 20 3 = 62 3 Maka, koordinat titik C ialah –31 3 , 6 2 3. [N1] Hence, the coordinates of point C are –31 3 , 6 2 3. 15 (a) Sudut bagi setiap sektor Angle of each sector = 360° 8 = 45° = 45° × 3.142 180° = 0.7855 rad [K1] x x 45° 8 cm sin 22.5° = x 8 x = 3.0615 [K1] Panjang sisi/Length of side = 6x = 6 × 3.0615 = 18.37 cm [N1] (b) Luas kawasan berlorek Area of the shaded region = 18.3692 – 1 2 (8)2 (0.7855) × 4 [K1] = 337.42 – 100.544 = 236.9 cm2 [N1] (c) Panjang setiap lengkok Length of each arc = 8 × 0.7855 = 6.284 cm [K1] FIRASAT Add Math(Jaw).indd 13 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J14• Perimeter kawasan tidak berlorek Perimeter of the unshaded region = 4(6.284 + 8 + 8) [K1] = 89.14 cm [N1] Kertas 2 Bahagian A 1 (a) Katakan/Let f–1(x) = y f(y) = x hy – k y – 5 = x [K1] hy – k = xy – 5x hy – xy = k – 5x y(h – x) = k – 5x y = k – 5x h – x [K1] f –1(x) = k – 5x h – x = 5x – k x – h Tetapi diberi bahawa But it is given that f –1 : x → 5x – 6 x – 2 , x ≠ 2 Dengan perbandingan, By comparison, k = 6 dan/and h = 2. [N2] (b) f : x → 2x – 6 x – 5 , x ≠ 5 f(x) = x Pemetaan sendiri/Self mapping 2x – 6 x – 5 = x [K1] 2x – 6 = x2 – 5x x2 – 7x + 6 = 0 (x – 1)(x – 6) = 0 [K1] x = 1 atau/or x = 6 [N1] 2 (a) –x2 + 4x  x – 4 –x2 + 3x + 4  0 x2 − 3x − 4  0 [K1] (x + 1)(x – 4)  0 [K1] –1 4 x Julat nilai x yang dikehendaki ialah x  –1 atau x  4. [N1] The required range of values of x is x  –1 or x  4. (b) f(x) = (k + 2)x2 – 12x + 2(k – 1) a = k + 2, b = –12, c = 2(k – 1) Jika graf fungsi kuadratik f(x) sentiasa berada di bawah paksi-x, maka If a quadratic function graph f(x) is always below the x-axis, then b2 – 4ac  0 (–12)2 – 4(k + 2)(2)(k – 1)  0 [K1] 144 – 8(k2 + k – 2)  0 18 – (k2 + k – 2)  0 18 – k2 – k + 2  0 k2 + k – 20  0 (k – 4)(k + 5)  0 [K1] [K1] –5 4 k Julat nilai k yang dikehendaki ialah k  – 5 atau k  4. [N1] The required range of values of k is k  – 5 or k  4. 3 loga x2 y = p loga x2 + loga y = p [K1] 2 loga x + loga y = p ............... ① [K1] loga y x3 = q loga y – loga x3 = q [K1] loga y – 3 loga x = q .............. ② [K1] ① − ②: 5 loga x = p – q loga x = p – q 5 [K1] Daripada ①/From ①: 2 p – q 5  + loga y = p [K1] 2p – 2q + 5 loga y = 5p loga y = 3p + 2q 5 loga x y = loga x – loga y = p – q 5 –  3p + 2q 5  = p – q – 3p – 2q 5 = –2p – 3q 5 [N1] 4 (a) A dan B berada dalam sukuan kedua kerana sin A adalah positif dan kos B adalah negatif dalam sukuan kedua. A and B lie in the second quadrant because sin A is positive and cos B is negative in the second quadrant. y x 17 A –15 O 8 (–15, 8) tan A = 8 –15 y x B O 25 –24 7 tan B = 7 –24 [K1] tan (A + B) = tan A + tan B 1 – tan A tan B = – 8 15 + – 7 24  1 – – 8 15 – 7 24  [K2] = – 33 40 38 45 = – 297 304 [N1] (b) kos/cos B = – 24 25 2 kos2 /cos2 B 2 – 1 = – 24 25 [K1] 2 kos2 /cos2 B 2 = 1 – 24 25 2 kos2 /cos2 B 2 = 1 25 kos2 /cos2 B 2 = 1 50 [K1] kos/cos B 2 = 1  50 kos/cos B 2 = 1  25 × 2 kos/cos B 2 = 1 5 2 kos/cos B 2 =  2 5(2) kos/cos B 2 =  2 10 [N1] 5 ‘TUESDAY’ mempunyai 4 huruf konsonan dan 3 huruf vokal. ‘TUESDAY’ has 4 consonants and 3 vowels. Bilangan kod yang dapat dibentuk jika tiada syarat yang dikenakan The number of codes that can be formed if no condition is applied = 7 P5 = 2 520 [K1] Katakan K mewakili huruf konsonan dan V mewakili huruf vokal. Let K represents a consonant and V represents a vowel. Bilangan V Number of V Bilangan K Number of K 2 3 1 4 4 K 3 V Kes 1/Case 1: 2V dan/and 3K K1 K2 K3 4 P3 × 3 P2 [K1] 4 konsonan, susun 3. 4 consonants, arrange 3. K1 K2 K3 4 P3 × 3 P2 K1 K2 K3 4 P3 × 3 P2 3 kad tinggal, susun 2 kad. 3 cards left, arrange 2 cards. FIRASAT Add Math(Jaw).indd 14 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

•J15• Firasat SPM 2023: Matematik Tambahan - Jawapan Jumlah bilangan kod (bagi kes 1) Total number of codes (for case 1) = 3(4 P3 × 3 P2 ) [K1] = 3(24 × 6) = 3(144) = 432 ............① [K1] Kes 2/Case 2: 1V dan/and 4K K1 K2 K3 K4 V1 4! × 3 P1 4 konsonan, susun semua. 4 consonants, arrange all. V1 K1 K2 K3 K4 4! × 3 P1 Jumlah bilangan kod (bagi kes 2) Total number of codes (for case 2) = 2(4! × 3 P1 ) [K1] = 2(24 × 3) = 2 × 72 = 144 .............② [K1] ① + ②: 432 + 144 = 576 Maka, kebarangkalian bahawa kod yang dibentuk itu mengandungi sekurang-kurangnya 3 huruf konsonan yang disusun sebelahmenyebelah ialah Hence, the probability that the code formed consists of at least 3 consonants which are arranged side by side is = 576 2 520 = 8 35 [N1] 6 X – Jisim durian, dalam kg X – Mass of the durians, in kg X ~ N(3, 0.52 ) (a) P(X  3.6) = PZ  3.6 – 3 0.5  [K1] = P(Z  1.2) [K1] = 1 – 0.1151 = 0.8849 [N1] (b) P(X  m) = 65% PZ  m – 3 0.5  = 0.65 [K1] [K1] 0.35 0.65 –0.385 z m – 3 0.5 = – 0.385 [K1] m = 2.808 [N1] 7 S P O Q T R C 18 cm 12 cm 6 cm 6 cm 6 cm (a) Dalam ∆OCQ/ In ∆OCQ, kos/cos OCQ = 6 12 = 1 2 OCQ = 60° COQ = 180° − 90° − 60° = 30° POQ = 2 × COQ = 2 × 30° = 60° = 60° × 3.142 180° = 1.0473 rad [K1] OP = OQ =  122 – 62 =  108 = 10.3923 cm [K1] PS = QT = 18 – OQ = 18 – 10.3923 = 7.6077 cm [K1] Panjang lengkok SRT Arc length SRT = 18(1.0473) = 18.8514 cm [K1] Panjang lengkok PRQ Arc length PRQ = 6 × refleks PCQ/ reflex PCQ = 6 × 240° × 3.142 180°  = 6 × 4.1893 = 25.136 cm [K1] Perimeter kawasan berlorek Perimeter of the shaded region = Panjang lengkok SRT + Panjang lengkok PRQ + PS + QT Arc length SRT + Arc length PRQ + PS + QT = 18.8514 + 25.136 + 7.6077 + 7.6077 = 59.20 cm [N1] (b) Luas kawasan berlorek Area of the shaded region = Luas sektor OST – Luas sisi empat – Luas sektor PRQ Area of sector OST – Area of quadrilateral – Area of sector PRQ = 1 2 (18)2 (1.0473) – 6(10.3923) – 1 2 (6)2 (4.1893) [K1] CQ × OQ = 169.6626 – 62.3538 – 75.4074 = 31.90 cm2 [N1] 3 kad tinggal, susun 1 vokal. 3 cards left, arrange 1 vowel. Bahagian B 8 (a) V (m) 10 20 30 40 50 P (%) 6.7 9.1 12.0 14.8 18.1 P 2.59 3.02 3.46 3.85 4.25 [K1] [K2] [N1] 10 20 30 40 50 60 70 80 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0 V P 50 2.2 4.25 – 2.2 = 2.05 (b) P = 1 a2 (V + b) 2 P = 1 a (V + b) P = 1 a V + b a [K1] (i) 1 a = Kecerunan/Gradient 1 a = 4.25 – 2.2 50 [K1] 1 a = 2.05 50 1 a = 0.041 a = 24.39 [N1] b a = Pintasan-Y/Y-intercept b 24.39 = 2.2 [K1] b = 53.66 [N1] FIRASAT Add Math(Jaw).indd 15 12/06/2023 7:23 PM PENERBIT ILMU BAKTI SDN. BHD.

© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J16• (ii) Apabila/When V = 32, Daripada graf,/From the graph, P = 3.5 P = 12.25% [N1] 9 (a) mAB = 9 – 0 8 – (–4) = 3 4 mPQ = – 1 mAB = – 4 3 [K1] Persamaan garis lurus PQ ialah The equation of the straight line PQ is y – 10 x – 1 = – 4 3 [K1] 3y – 30 = –4(x – 1) 3y – 30 = –4x + 4 3y = –4x + 34 [K1] Pada paksi-x, y = 0, On the x-axis, y = 0, 3(0) = –4x + 34 0 = –4x + 34 x = 34 4 = 17 2 Maka, koordinat titik Q ialah 8 1 2 , 0. [N1] Hence, the coordinates of point Q are 8 1 2 , 0. Persamaan AB ialah The equation of AB is y – 0 x + 4 = 9 – 0 8 – (–4) y x + 4 = 3 4 4y = 3x + 12 [K1] Persamaan PQ/Equation of PQ: 3y = –4x + 34 ..................... ① Persamaan AB/ Equation of AB: 4y = 3x + 12 ........................ ② ② × 3: 12y = 9x + 36 (−) ① × 4: 12y = –16x + 136 0 = 25x – 100 25x = 100 x = 4 [K1] Gantikan x = 4 ke dalam ②. Substitute x = 4 into ②. 4y = 3(4) + 12 4y = 24 y = 6 Koordinat titik C ialah (4, 6). The coordinates of point C are (4, 6). [N1] (b) D(x, y), P(1, 10), B(8, 9) (mPD)(mDB) = –1  y – 10 x – 1 y – 9 x – 8 = –1 [K1] (y – 10)(y – 9) = –(x – 1)(x – 8) [K1] y2 – 19y + 90 = –(x2 – 9x + 8) y2 – 19y + 90 = –x2 + 9x – 8 x2 + y2 – 9x – 19y + 98 = 0 [N1] 10 (a) Sebuah telaga minyak menghasilkan 100 000 tong minyak mentah pada tahun pertama. Oleh itu, a = 100 000. An oil well produces 100 000 barrels of oil in the first year. Thus, a = 100 000. [K1] Bilangan tong minyak mentah yang dihasilkan berkurang sebanyak 5% setiap tahun. Oleh itu, r = 0.95. [K1] The number of barrels of oil produced decreases by 5% each year. Thus, r = 0.95. T5 = ar4 = 100 000(0.95)4 [K2] = 81 450 [N1] (b) S12 = 100 000(1 – 0.9512) 1 – 0.95 [K2] = 919 279 [N1] (c) S = a 1 – r = 100 000 1 – 0.95 [K1] = 2 000 000 [N1] 11 (a) X – Bilangan murid yang berbasikal ke sekolah X – Number of students who cycle to school X ~ B(10, 0.1) (i) P(X = 3) = 10C3 (0.1)3 (0.9)7 [K1] = 0.0574 [N1] (ii) P(X  2) = 1 – P(X = 0) – P(X = 1) – P(X = 2) = 1 – 10C0 (0.1)0 (0.9)10 – 10C1 (0.1)1 (0.9)9 – 10C2 (0.1)2 (0.9)8 [K2] = 0.07019 [N1] (b) X – Masa yang diambil, dalam minit X – Time taken, in minutes X ~ N(24, 122 ) (i) P(X  36) = PZ  36 – 24 12  [K1] = P(Z  1) = 0.1587 [N1] (ii) P(X  t) = 100 800 PZ  t – 24 12  = 0.125 [K1] 0.125 –1.151 z t – 24 12 = –1.151 [K1] t = 10.19 [N1] Bahagian C 12 (a) Dalam ∆PRQ/ In ∆PRQ, sin PRQ 10 = sin 40° 7 [K1] sin PRQ = sin 40° 7 × 10 sin PRQ = 0.918268 Sudut asas/Basic angle = 66.67° [K1] PRQ = 180° − 66.67° = 113.33° [N1] (b) Dalam ∆PRS/In ∆PRS, PS2 = 72 + 82 – 2(7)(8) kos/cos 66.67° [K1] PS2 = 68.64505 PS = 8.285 cm [N1] (c) Luas segi tiga PST = 30 cm2 Area of triangle PST = 30 cm2 1 2 (8.285)(12) sin SPT = 30[K1] sin SPT = 0.6035 SPT = 37.12° [N1] (d) Dalam ∆PRQ/In ∆PRQ, RPQ = 180° – 40° – 113.33° = 26.67° Luas PSRQ/Area of PSRQ = Luas ∆PRS + Luas ∆PRQ Area of ∆PRS + Area of ∆PRQ = 1 2 (7)(8) sin 66.67° + [K1] 1 2 (7)(10) sin 26.67° [K1] = 25.7107 + 15.7098 = 41.42 cm2 [N1] 13 (a) x = 9.00 8.00 × 100 = 112.5 [N1] y 3.00 × 100 = 105 y = 105 × 3.00 100 y = 3.15 [N1] 4.80 z × 100 = 120 z = 4.80 × 100 120 z = 4.00 [N1] (b) I – 2020/2018 = ΣIw Σw = (112.5 × 20) + (105 × 30) + (108 × 10) + (120 × 40) 20 + 30 + 10 + 40 [K2] = 11 280 100 = 112.8 [N1] FIRASAT Add Math(Jaw).indd 16 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.