•J17• Firasat SPM 2023: Matematik Tambahan - Jawapan (c) P2020 P2018 × 100 = I – 2020/2018 P2020 RM43.00 × 100 = 112.8 [K1] P2020 = 112.8 × RM43.00 100 P2020 = RM48.50 [N1] (d) I – 2018 +12.8% I – 2020 +10% I – 2023 (100) (112.8) (?) I – 2023/2018 = 100 + 10 100 × 112.8 [K1] = 124.08 [N1] 14 (a) Pada titik O, pecutan ialah 12 m s–2. Oleh itu, apabila t = 0, a = 12. At point O, the acceleration is 12 m s–2. Thus, when t = 0, a = 12. a = m + kt 12 = m + (0)k m = 12 [N1] a = 12 + kt v = a dt v = (12 + kt) dt v = 12t + kt2 2 + c [K1] Zarah itu melalui titik O dengan halaju 15 m s–1, iaitu apabila t = 0, v = 15. Oleh itu, c = 15. The particle passes through point O with a velocity of 15 m s–1, which is when t = 0, v = 15. Thus, c = 15. \ v = 12t + kt2 2 + 15 [K1] Zarah itu berehat seketika apabila t = 5. Oleh itu, apabila t = 5, v = 0. The particle is instantaneously at rest when t = 5. Thus, when t = 5, v = 0. v = 12t + kt2 2 + 15 0 = 12(5) + k(5)2 2 + 15 0 = 60 + 25 2 k + 15 k = –6 [N1] (b) Apabila/When k = –6, v = 12t – 6 t2 2 + 15 = 12t – 3t2 + 15 Pada halaju maksimum At maximum velocity, dv dt = 0 12 – 6t = 0 t = 2 [K1] Apabila/ When t = 2, v = 12(2) – 3(2)2 + 15 = 27 d2 v dt2 = –6 (< 0) Oleh itu, v adalah maksimum. Maka, halaju maksimum ialah 27 m s–1. Therefore, v is a maximum. Hence, the maximum velocity is 27 m s–1. [N1] (c) Cari s dalam sebutan t dahulu. Find s in terms of t first. s = v dt s = (12t – 3t2 + 15) dt [K1] s = 12t 2 2 – 3t3 3 + 15t + c s = 6t2 – t3 + 15t + c Apabila/When t = 0, s = 0. Maka/Thus c = 0. \ s = 6t 2 – t 3 + 15t [K1] t (s) 0 5 6 s (m) 0 100 90 Rajah bagi gerakan zarah itu adalah seperti yang ditunjukkan di bawah. The diagram of the motion of the particle is as shown below. 0 90 100 t = 0 t = 6 t = 5 Maka, jumlah jarak yang dilalui oleh zarah Hence, the total distance travelled by the particle = 100 + 10 [K1] = 110 m [N1] 15 (a) I x + y 10 [P1] II 2 400x + 1 600y 19 200 3x + 2y 24 [P1] III y 2x [P1] (b) x + y = 10 x 0 10 y 10 0 3x + 2y = 24 x 0 8 y 12 0 y = 2x x 0 6 y 0 12 [K2] [N1] 2 4 6 8 10 12 10 8 6 4 2 0 x y 280x + 200y = 560 5 2.8 R 3x + 2y = 24 x + y = 10 y = 2x Maksimum/Maximum (4, 6) (c) (i) Jika luas tanah yang ditanam dengan tomato ialah 4 km2 , julat luas tanah yang ditanam dengan kubis ialah 2 x 5. If the area planted with tomatoes is 4 km2 , the range of values of the area planted with cabbages is 2 x 5. [N1] (ii) Keuntungan/Profit = 280x + 200y Lukis garis lurus/Draw the straight line 280x + 200y = 560 [K1] x 0 2 y 2.8 0 Titik optimum ialah/The optimal point is (4, 6). [K1] Maka, keuntungan maksimum/Hence, the maximum profit = 280(4) + 200(6) = RM2 320 [N1] Kertas Model 4 Kertas 1 Bahagian A 1 (a) fg(x) = f x + 2 x – 2 = 1 – 2 x + 2 x – 2 [K1] = x – 2 – 2(x + 2) x – 2 = x – 2 – 2x – 4 x – 2 = – x – 6 x – 2 fg(x) = x + 6 2 – x , x ≠ 2 [N1] FIRASAT Add Math(Jaw).indd 17 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J18• (b) Katakan/Let y = x + 6 2 – x y(2 – x) = x + 6 [K1] 2y – xy = x + 6 x + xy = 2y – 6 x(1 + y) = 2y – 6 x = 2y – 6 1 + y Maka/Hence, (fg)–1(x) = 2x – 6 1 + x , x ≠ –1 [N1] 2 (a) Lengkung kuadratik y1 = x2 – 2x + hk + k menyentuh paksi-x pada satu titik sahaja. The quadratic curve y1 = x2 – 2x + hk + k touches the x-axis at only one point. Maka/Thus, b2 – 4ac = 0. (–2)2 – 4(1)(hk + k) = 0 4 – 4hk – 4k = 0 1 – hk – k = 0 ......... ① [K1] y2 = 3(x – 1)2 + k – 1 = 3(x2 – 2x + 1) + k – 1 = 3x2 – 6x + 3 + k – 1 = 3x2 – 6x + k + 2 [K1] Lengkung kuadratik y2 = 3x2 – 6x + k + 2 menyentuh paksi-x pada satu titik sahaja. The quadratic curve y2 = 3x2 – 6x + k + 2 touches the x-axis at only one point. Maka/Thus, b2 – 4ac = 0 (–6)2 – 4(3)(k + 2) = 0 [K1] 36 – 12k – 24 = 0 –12k + 12 = 0 12k = 12 k = 1 [N1] Gantikan k = 1 ke dalam ①. Substitute k = 1 into ①. 1 – h(1) – (1) = 0 h = 0 [N1] (b) Apabila h = 0 dan k = 1, When h = 0 and k = 1, y1 = x2 – 2x + hk + k = x2 – 2x + 0 + 1 y1 = x2 – 2x + 1 y2 = 3x2 – 6x + k + 2 = 3x2 – 6x + 1 + 2 = 3x2 – 6x + 3 [K1] Apabila/When y1 = 0, x2 – 2x + 1 = 0 (x – 1)2 = 0 x = 1 Apabila/When y2 = 0, 3x2 – 6x + 3 = 0 x2 – 2x + 1 = 0 (x – 1)2 = 0 x = 1 [N1] 3 f(x) = x2 – x – 6 = (x + 2)(x – 3) Pintasan-y/y-intercept = –6 [K1] Apabila/When f(x) = 0, (x + 2)(x – 3) = 0 x = –2 atau/or x = 3 [K1] f(x) = x2 – x – 6 = x2 – x + – 1 2 2 – – 1 2 2 – 6 = x2 – x + 1 4 – 1 4 – 6 [K1] = x – 1 2 2 – 25 4 [K1] x –3 5 f(x) 6 14 [N2] y (–3, 6) (5, 14) –2 –6 0 3 1 2 , – 25 4 x 4 (a) 4 + 2 2 – 2 = 4 + 2 2 – 2 × 2 + 2 2 + 2 [K1] = 8 + 6 2 + 2 4 – 2 = 10 + 6 2 2 = 5 + 3 2 [N1] (b) 1 16 log5 x = logx 5 1 16 log5 x = 1 log5 x [K1] (log5 x)2 = 16 log5 x = ±4 x = 54 atau/or x = 5–4 x = 625 atau/or x = 1 625 [N2] 5 mAB = 6 – 0 0 – (–3) = 2 mBC = – 1 mAB = – 1 2 [K1] Persamaan garis lurus BC ialah The equation of the straight line BC is y – 6 = – 1 2 (x – 0) 2y – 12 = – x 2y = – x + 12 [K1] Pada titik C, y = 0. At point C, y = 0. 2(0) = – x + 12 x = 12 Maka, titik C ialah (12, 0). [K1] Thus, point C is (12, 0). Oleh sebab ABC = BCD = 90o , garis lurus DC adalah selari dengan garis lurus AB. Since ABC = BCD = 90o , the straight line DC is parallel to the straight line AB. Maka/Hence, mDC = mAB = 2. [K1] Persamaan garis lurus CD ialah Equation of the straight line CD is y – 0 = 2(x – 12) y = 2x – 24 2x – y = 24 [K1] 2x 24 – y 24 = 24 24 x 12 + y (–24) = 1 [N1] 6 T1 = r + r + 1 4 (2πr) = 2r + 1 2 πr T2 = 3r + 3r + 1 4 (2π × 3r) = 6r + 3 2 πr T3 = 5r + 5r + 1 4 (2π × 5r) = 10r + 5 2 πr [K1] T2 – T1 = 6r + 3 2 πr – 2r + 1 2 πr = 4r + πr T3 – T2 = 10r + 5 2 πr – 6r + 3 2 πr = 4r + πr [K1] Oleh sebab T2 – T1 = T3 – T2 = 4r + πr (pemalar), maka perimeter sukuan bulatan membentuk janjang aritmetik. [N1] Since T2 – T1 = T3 – T2 = 4r + πr (a constant), thus the perimeters of the quadrants form an arithmetic progression. Beza sepunya/Common difference = 4r + πr [N1] 7 Jika titik-titik A, B dan C adalah segaris, maka If the points A, B and C are collinear, then A →B = hB →C O →B – O →A = h(O →C – O →B) 4i ∼ + 2j ∼ – (9i ∼ – 10i ∼) = h[(ki ∼ – 2j ∼ ) – (4i ∼ + 2j ∼ )] [K1] –5i ∼ + 12j ∼ = h[(k – 4)i ∼ – 4j ∼ ] –5i + 12j = h(k – 4)i ∼ – 4hj ∼ [K1] Bandingkan pekali i ∼, Compare the coefficients of i ∼, h(k – 4) = –5 .............. ① [K1] Bandingkan pekali j ∼ , Compare the coefficients of j ∼ , –4h = 12 ................... ② [K1] h = –3 Gantikan h = –3 ke dalam ①. Substitute h = –3 into ①. –3(k – 4) = –5 –3k + 12 = –5 k = 17 3 [N1] 8 y = a bx log10 y = log10 a – x log10 b log10 y = – x log10 b + log10 a [K1] Bagi titik (–1, 8), x = –1 dan log10 y = 8. [K1] For the point (–1, 8), x = –1 and log10 y = 8. 8 = –(–1) log10 b + log10 a 8 = log10 b + log10 a ................① Bagi titik (3, 0), x = 3 dan log10 y = 0. [K1] For the point (3, 0), x = 3 and log10 y = 0. 0 = –3 log10 b + log10 a ...........② ① – ② : 8 = 4 log10 b 2 = log10 b b = 102 b = 100 [N1] FIRASAT Add Math(Jaw).indd 18 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
•J19• Firasat SPM 2023: Matematik Tambahan - Jawapan Daripada/From ①: 8 = log10 100 + log10 a 8 = 2 + log10 a log10 a = 6 a = 106 a = 1 000 000 [N1] 9 P Q R r r π – θ r θ Panjang lengkok PR = Panjang lengkok RQ + Diameter PQ Arc length PR = Arc length RQ + Diameter PQ r(π – θ) = rθ + 2r [N1] rπ – rθ = rθ + 2r π – θ = θ + 2 2θ = π – 2 [K1] θ = π – 2 2 θ = 3.142 – 2 2 θ = 0.571 [K1] θ = 0.571 × 180° 3.142 [K1] θ = 32o 43' [N1] 10 2 kosek2 /cosec2 θ + kot/cot θ – 3 = 0 2(kot2 /cot2 θ + 1) + kot/cot θ – 3 = 0 [K1] 2 kot2 /cot2 θ + 2 + kot/cot θ – 3 = 0 2 kot2 /cot2 θ + kot/cot θ – 1 = 0 (2 kot/cot θ – 1)(kot/cot θ + 1) = 0 [K1] kot/cot θ = 1 2 atau/or kot/cot θ = – 1 Apabila/When kot/cot θ = 1 2 , 1 tan θ = 1 2 tan θ = 2 asas/Basic = 63.43o [K1] θ = 63.43o atau/or 243.43o Apabila/When kot/cot θ = – 1, 1 tan θ = – 1 tan θ = – 1 [K1] asas/Basic = 45o θ = 135o atau/or 315o ∴θ = 63.43o , 135o , 243.43°, 315o [N2] 11 y = 6x – x2 Apabila/When x = k, y = 6k – k2 Maka, titik A ialah (k, 6k – k2 ).[K1] Thus, point A is (k, 6k – k2 ). dy dx = 6 – 2x [K1] Apabila/When x = k, mAB = dy dx = 6 – 2k [K1] 6k – k2 – 10 k – 3 = 6 – 2k [K1] 6k – k2 – 10 = (6 – 2k)(k – 3) 6k – k2 – 10 = 6k – 18 – 2k2 + 6k k2 – 6k + 8 = 0 (k – 2)(k – 4) = 0 [K1] k = 2 atau/or k = 4 k = 4 tidak diterima/is not accepted. ∴ k = 2 [N1] 12 dy dx = (2x – 8)3 y = ∫ (2x – 8)3 dx y = (2x – 8)4 4(2) + c y = (2x – 8)4 8 + c [K1] Apabila lengkung mempunyai nilai minimum, When the curve has a minimum value, dy dx = 0 (2x – 8)3 = 0 2x – 8 = 0 x = 4 [K1] Oleh sebab nilai minimum ialah –2, maka y = –2. Since the minimum value is –2, then y = – 2. Maka, titik minimum ialah (4, –2). Hence, the minimum point is (4, –2). y = (2x – 8)4 8 + c [K1] –2 = [2(4) – 8]4 8 + c –2 = 0 + c c = –2 [K1] Maka, persamaan lengkung ialah y = (2x – 8)4 8 – 2. [N1] Hence, the equation of the curve is y = (2x – 8)4 8 – 2. Bahagian B 13 x + y + z = 150 ..............① [P1] Bagi setiap dua unit rumah mewah yang dibina, pemaju perumahan itu mesti membina satu unit rumah mampu milik. Cuba bayangkan bahawa 10 unit rumah mewah dibina, maka 5 unit rumah mampu milik mesti dibina. Maka, y + z = 2x. For every two units luxury houses built, the housing developer must build one unit affordable house. Imagine that 10 luxury houses are built, then 5 affordable houses must be built. Hence, y + z = 2x. y + z = 2x 2x – y – z = 0 .....................② [P1] 350x + 400y + 470z = 60 300 35x + 40y + 47z = 6 030......③ [P1] ① + ②: x + y + z = 150 (+) 2x – y – z = 0 3x = 150 x = 50 Gantikan x = 50 ke dalam persamaan ①. Substitute x = 50 into equation ①. 50 + y + z = 150 y + z = 100 ................. ④ [K1] Gantikan x = 50 ke dalam persamaan ③. Substitute x = 50 into equation ③. 35(50) + 40y + 47z = 6 030 40y + 47z = 4 280 ............. ⑤ [K1] 40y + 40z = 4 000 .......④ × 40 (–) 40y + 47z = 4 280 .......⑤ –7z = –280 z = 40 Gantikan z = 40 ke dalam persamaan ④. Substitute z = 40 into equation ④. y + 40 = 100 y = 60 Maka, bilangan rumah berkembar satu tingkat, rumah teres dua tingkat dan rumah berkembar dua tingkat yang akan dibina masing- masing ialah 50 unit, 60 unit dan 40 unit. [N3] Hence, the number of single storey semidetached house, double storey terrace house and double storey semi-detached house that are to be built are 50 units, 60 units and 40 units respectively. 14 (a) (i) y = – 4 3 x mOA = – 4 3 \ mAB = – 1 – 4 3 = 3 4 [K1] 4y = kx + 25 y = k 4 x + 25 4 \ mAB = k 4 k 4 = 3 4 ⇒ k = 3 [N1] (ii) 4y = 3x + 25....................① y = – 4 3 x ...................... ② Gantikan ② ke dalam ①. Substitute ② into ①. 4– 4 3 x = 3x + 25 [K1] – 16x = 9x + 75 25x = – 75 x = – 3 Daripada/ From ②: y = – 4 3 (–3) = 4 Maka, koordinat titik A ialah (–3, 4). [N1] Hence, the coordinates of point A are (–3, 4). (b) (i) A(–3, 4) B(h, k) 3 1 Q0, 25 4 Menyamakan koordinat-x, Equating the x-coordinate, (–3) + 3h 3 + 1 = 0 [K1] –3 + 3h = 0 h = 1 FIRASAT Add Math(Jaw).indd 19 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J20• Menyamakan koordinat-y, Equating the y-coordinate, 1(4) + 3k 3 + 1 = 25 4 [K1] 4 + 3k = 25 k = 7 Maka, koordinat titik B ialah (1, 7). [N1] Hence, the coordinates of point B are (1, 7). (ii) mCB = mOA = – 4 3 Persamaan BC ialah Equation of BC is y – 7 = – 4 3 (x – 1) 3y – 21 = – 4x + 4 3y = – 4x + 25 [N1] 15 (a) (i) OM = OC + CM = c ∼ + 1 2 CB = c ∼ + 1 2 (CO + OB ) = c ∼ + 1 2 (– c ∼ + b ∼) [K1] = 1 2 c ∼ + 1 2 b ∼ = 1 2 (c ∼ + b ∼) ........... ① [N1] (ii) NB DB = 3 4 DB = 4 3 NB = 4 3 (NO + OB ) = 4 3 – 1 2 c ∼ + b ∼ [K1] = – 2 3 c ∼ + 4 3 b ∼ [N1] (b) DC = DB + BC = – 2 3 c ∼ + 4 3 b ∼ + (–b ∼ + c ∼) [K1] = 1 3 c ∼ + 1 3 b ∼ = 1 3 (c ∼ + b ∼) [K1] = 1 3 (2OM) Daripada/From ①: c ~ + b ~ = 2OM = 2 3 OM Oleh sebab DC = 2 3 OM, maka DC dapat diungkapkan sebagai gandaan skalar OM. Oleh itu, DC adalah selari dengan OM. [N2] Since DC = 2 3 OM, then DC can be expressed as a scalar multiple of OM. Hence, DC is parallel to OM. Kertas 2 Bahagian A 1 (a) × 0.9 × 0.9 After 1st bounce \a = 90 After 2nd bounce 100, 90, 81, … Tn 25 90(0.9)n – 1 25 [K1] (0.9)n – 1 25 90 log (0.9)n – 1 log 25 90 [K1] (n – 1) log 0.9 log 25 90 (n – 1)(–0.0458) –0.5563[K1] n – 1 – 0.5563 – 0.0458 n – 1 12.15 n 13.15 [K1] Maka, bilangan lantunan minimum ialah 14. [N1] Hence, the minimum number of bounces is 14. (b) Jumlah jarak yang dilalui Total distance travelled = 100 + 2S∞ = 100 + 2 90 1 – 0.9 [K1] = 1 900 cm [N1] 2 (a) Bilangan cara kesemua 10 orang murid itu dapat duduk The number of ways all the 10 students can be seated Tiga orang murid di sebelah kiri meja. Three students on the left of the table. = 5 P3 × 7! + 5 P3 × 7! [K2] = 604 800 [N1] Tiga orang murid di sebelah kanan meja. Three students on the right of the table. (b) Tiga titik pada satu garis lurus tidak boleh membentuk segi tiga. Three points on one straight line cannot form triangle. Maka, bilangan segi tiga yang dapat dibentuk Hence, the number of triangles that can be formed = 10 C3 – 4C3 – 6C3 [K2] = 120 – 4 – 20 = 96 [N1] 3 (a) dy dx = – 2 x3 + 3 x4 [K1] d2 y dx2 = 6 x4 – 12 x5 [K1] x4 dy dx + d2 y dx2 + x2 y + 5 = 0 x4 – 2 x3 + 3 x4 + 6 x4 – 12 x5 + x2 1 x2 – 1 x3 + 5 = 0 –2x + 9 – 12 x + 1 – 1 x + 5 = 0 [K1] –2x + 15 – 13 x = 0 –2x2 + 15x – 13 = 0 2x2 – 15x + 13 = 0 (2x – 13)(x – 1) = 0 x = 13 2 atau/or x = 1 [N1] (b) (i) 8 cm r cm h cm (8 – h) cm Katakan jejari permukaan membulat air ialah r cm. Let the radius of the circular surface of the water be r cm. r2 = 82 – (8 – h)2 [K1] r2 = 64 – (64 – 16h + h2 ) r2 = 16h – h2 Luas permukaan membulat air: The area of the circular surface of the water: A = πr 2 A = π(16h – h2 ) [N1] [Tertunjuk/Shown] (ii) dA dh = π(16 – 2h) Dengan menggunakan petua rantai, By using the chain rule, dA dt = dA dh × dh dt = π(16 – 2h) × 0.5 [K1] = π[16 – 2(5)] × 0.5 = 3π cm2 s–1 [N1] 4 (a) x = – y2 + 10 … ① x = (y – 2)2 … ② Gantikan ② ke dalam ①: Substitute ② into ①: (y – 2)2 = – y2 + 10 [K1] y2 – 4y + 4 = – y2 + 10 2y2 – 4y – 6 = 0 y2 – 2y – 3 = 0 (y – 3)(y + 1) = 0 [K1] y = 3 atau/or y = – 1 Daripada/From ②: Apabila/When y = 3, x = (3 – 2)2 = 1 Maka, koordinat titik A ialah (1, 3). [N1] Hence, the coordinates of point A are (1, 3). Apabila/When y = –1, x = (–1 – 2)2 = 9 Maka, koordinat titik B ialah (9, –1). [N1] Hence, the coordinates of point B are (9, –1). FIRASAT Add Math(Jaw).indd 20 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
•J21• Firasat SPM 2023: Matematik Tambahan - Jawapan (b) Luas rantau berlorek Area of the shaded region = ∫ 3 –1 (–y2 + 10) dy [K1] – ∫ 3 –1 (y – 2)2 dy = ∫ 3 –1 (–y2 + 10) dy – ∫ 3 –1 (y2 – 4y + 4) dy = ∫ 3 –1 (–2y2 + 4y + 6) dy = – 2y3 3 + 2y2 + 6y 3 –1 [K1] = – 2 3 (3)3 + 2(3)2 + 6(3) – – 2 3 (–1)3 + 2(–1)2 + 6(–1) [K1] = 18 – – 10 3 = 21 1 3 unit2 /units2 [N1] 5 y x O 1 y = x2 + 2 3 2 x = 1 Isi padu janaan Generated volume = π∫0 3 12 dy – π∫2 3 (y – 2) dy [K1] y = x2 + 2 x2 = y – 2 = π[y]0 3 – π y2 2 – 2y 3 2 [K1] = π(3 – 0) – π 32 2 – 2(3) – 22 2 – 2(2) [K2] = 3π – 1 2 π [K1] = 5 2 π unit3 /units3 [N1] 6 (a) Sebelah kiri/LHS = 2 kot/ cot x 2 – kosek2 / cosec2 x = 2 kos/ cos x sin x 2 – 1 sin2 x [K1] = 2 kos/cos x sin x 2 sin2 x – 1 sin2 x = 2 kos/ cos x sin x × sin2 x 2 sin2 x – 1 [K1] = 2 sin x kos/ cos x 2 sin2 x – 1 = sin 2x –(1 – 2 sin2 x) [K1] = sin 2x – cos 2x [K1] = – tan 2x (Sebelah kanan/RHS) (b) (i) 1 4p 1 2p 3 4p y = – 5x p x y O (p, –5) y = –tan 2x p [N2] (ii) 2 kot/cot x 2 – kosek2 /cosec2 x – 5x π = 0 – tan 2x = 5x π tan 2x = – 5x π [K1] Lakar garis lurus y = – 5x π . Sketch the straight line y = – 5x π . Bilangan penyelesaian = 2[N1] Number of solutions = 2 7 (a) 3x + 4 – 2x + 1 = 1 3x + 4 – 2x + 1 2 = 12 [K1] 3x + 4 + (2x + 1) – 2 3x + 4 2x + 1 = 1 2 3x + 4 2x + 1 = 5x + 4 2 3x + 4 2x + 1 2 = (5x + 4)2 [K1] 4(3x + 4)(2x + 1) = 25x2 + 40x + 16 4(6x2 + 11x + 4) = 25x2 + 40x + 16 24x2 + 44x + 16 = 25x2 + 40x + 16 [K1] x2 – 4x = 0 x(x – 4) = 0 x = 0 atau/or x = 4 Apabila x = 0, panjang bahagian yang lebih pendek ialah 1 m dan panjang bahagian yang lebih panjang ialah 2 m. [N1] When x = 0, the length of the shorter part ialah 1 m and the length of the longer part is 2 m. Apabila x = 4, panjang bahagian yang lebih pendek ialah 3 m dan panjang bahagian yang lebih panjang ialah 4 m. [N1] When x = 4, the length of the shorter part ialah 3 m and the length of the longer part is 4 m. (b) Oleh sebab/Since MQN = 90o , mMQ × mQN = –1 y – 4 x – 1 y – 0 x – 3 = –1 [K1] y2 – 4y x2 – 4x + 3 = –1 y2 – 4y = –x2 + 4x – 3 x2 + y2 – 4x – 4y + 3 = 0 Maka, persamaan lokus Q ialah Hence, the equation of the locus of Q is x2 + y2 – 4x – 4y + 3 = 0. [N1] Bahagian B 8 (a) [K2] x 1.00 2.25 4.00 6.25 9.00 y 100 005 9 995 998 101 10 x 1.0 1.5 2.0 2.5 3.0 log10 y 5 4 3 2 1 0.5 1.0 1.5 2.0 2.5 3.0 7 6 5 4 3 2 1 0 x log10 y 4.6 fi [K2][N1] (b) (i) y = ab x log10 y = log10 a + log10 b x log10 y = log10 a + x log10 b log10 y = x log10 b + log10 a log10 a = Pintasan-Y/Y-intercept log10 a = 7 [K1] a = 10 000 000 [N1] log10 b = Kecerunan/Gradient log10 b = 5 – 1 1 – 3 [K1] log10 b = –2 b = 1 100 [N1] (ii) Apabila/When x = 1.44, x = 1.44 = 1.2 Daripada graf/From the graph, log10 y = 4.6 y = 39 811 [N1] FIRASAT Add Math(Jaw).indd 21 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J22• 9 A 1 K B M O 3 2 5 14y ∼ 2x ∼ L (a) (i) O →M = 5 7 O →B = 5 7 (14y ∼ ) = 10y ∼ [N1] (ii) A →K = 1 4 A →B = 1 4 (–2x ∼ + 14y ∼ ) = – 1 2 x ∼ + 7 2 y ∼ [N1] (b) (i) A →L = pA →M = p(A →O + O →M = p(–2x ∼ + 10y ∼ ) = –2px∼ + 10py ∼ [N1] (ii) K →L = qK →O = qK →A + A →O = q 1 4 B →A – 2x ∼ = q 1 4 (–14y ∼ + 2x ∼) – 2x ∼ [K1] = q– 7 2 y ∼ + 1 2 x ∼ – 2x ∼ = q– 7 2 y ∼ – 3 2 x ∼ = – 7 2 qy ∼ – 3 2 qx∼ [N1] (c) A →K = A →L + L →K 7 2 y ∼ – 1 2 x ∼ = –2px ∼ + 10py ∼ + 7 2 qy ∼ + 3 2 qx∼ [K1] 7 2 y ∼ – 1 2 x ∼ = 10p + 7 2 qy ∼ + –2p + 3 2 qx ∼ Bandingkan pekali x ∼: Compare the coefficients of x∼: – 1 2 = – 2p + 3 2 q – 4p + 3q = –1 … ① [K1] Bandingkan pekali y ∼ : Compare the coefficients of y ∼ : 10p + 7 2 q = 7 2 20p + 7q = 7 … ② [K1] –20p + 15q = – 5 … ① × 5 (+) 20p + 7q = 7 … ② 22q = 2 q = 1 11 [N1] Daripada/From ①: –4p + 3 1 11 = –1 –4p = – 14 11 p = 7 22 [N1] 10 (a) X – Bilangan guru yang mempunyai tekanan darah tinggi X – Number of teachers who have high blood pressure X~B9, 1 5 (i) P(X = 3) = 9 C3 1 5 3 4 5 6 [K1] = 0.1762 [N1] (ii) P(X 2) = 1 – P(X = 0) – P(X = 1) – P(X = 2) = 1 – 9 C0 1 5 0 4 5 9 – 9 C1 1 5 1 4 5 8 – 9 C2 1 5 2 4 5 7 [K1] = 1 – 0.1342 – 0.3020 – 0.3020 = 0.2618 [N1] (b) X – Aras glukosa, dalam mg/dL X – Glucose level, in mg/dL X ~ N(115,152 ) (i) P(100 < X < 136) = P 100 – 115 15 < Z < 136 – 115 15 [K1] = P(–1 < Z < 1.4) = 1 – 0.1587 – 0.0808 [K1] = 0.7605 [N1] 0.1587 0.0808 –1 1.4 (ii) P(X 140) = PZ 140 – 115 15 = P(Z 1.667) [K1] = 0.0478 [K1] Katakan jumlah pekerja = N Let the total number of employees = N 0.0478N = 9 N = 9 0.0478 N = 188 (betul kepada integer terdekat/correct to the nearest integer) [N1] 11 A M B R P Q O 30° 30° 60° 60° 6 cm 6 cm 6 cm (a) Dalam/In ΔPQO, sin 30° = 6 PO 1 2 = 6 PO PO = 12 cm [K1] PM = PO + OM = 12 + 6 = 18 cm PA = PB = PM = 18 cm [K1] Panjang lengkok AMB Arc length AMB = 18 × 60° × 3.142 180° [K1] = 18.85 cm [N1] (b) Dalam/ In ΔPQO, PQ = 122 – 62 = 10.39 cm [K1] Luas ΔPOQ = Luas ΔPOR Area of ΔPOQ = Area of ΔPOR = 1 2 × 6 × 10.39 = 31.17 cm2 [K1] Luas sektor QMR Area of sector QMR = 1 2 × 62 × 240° × 3.142 180° [K1] = 75.41 cm2 Luas sektor APB Area of sector APB = 1 2 × 182 × 60° × 3.142 180° [K1] = 169.67 cm2 FIRASAT Add Math(Jaw).indd 22 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
•J23• Firasat SPM 2023: Matematik Tambahan - Jawapan Maka, luas kawasan berlorek Hence, area of the shaded region = Luas sektor APB – Luas ΔPOQ – Luas ΔPOR – Luas sektor QMR Area of sector APB – Area of ΔPOQ – Area of ΔPOR – Area of sector QMR = 169.67 – 31.17 – 31.17 – 75.41 [K1] = 31.92 cm2 [N1] Bahagian C 12 (a) I x + y 500 [P1] II x 3y [P1] III 5x + 3y 1 500 [P1] (b) Bagi/ For x + y = 500 x 0 400 y 500 100 Bagi/ For x = 3y x 0 300 y 0 100 Bagi/ For 5x + 3y = 1 500 x 0 300 y 500 0 50 100 150 200 250 300 350 400 450 500 450 400 350 300 250 200 150 100 50 0 x y R 5x + 3y = 1 500 x + y = 500 x = 3y Maksimum/Maximum (375, 125) (c) (i) Apabila/When x = 150, yminimum = 250 \ Bilangan rim minimum kertas 70 g yang dihasilkan = 250 [N1] The minimum number of reams of the 70 g papers = 250 (ii) Keuntungan/ Profit = 5x + 3y Lukis garis lurus Draw the straight line 5x + 3y = 1 500. [K1] Sudah dilukis di (b). Already drawn in (b). Daripada graf, titik optimum ialah (375, 125). From the graph, the optimal point is (375, 125). [K1] Maka, jumlah keuntungan maksimum Hence, the maximum total profit = 5(375) + 3(125) = RM2 250 [N1] 13 (a) v = –10 + 7t – t2 Apabila/When t = 0, v = –10 + 7(0) – (0)2 = –10 Maka, halaju awal zarah ialah –10 m s-1. [N1] Hence, the initial velocity of the particle is –10 m s-1. (b) Pada halaju maksimum, At maximum velocity, dv dt = 0 7 – 2t = 0 [K1] 2t = 7 t = 3.5 Apabila/When t = 3.5, v = –10 + 7(3.5) – (3.5)2 = 2.25 d2 v dt2 = – 2 (< 0) Maka, halaju maksimum ialah 2.25 m s-1. [N1] Hence, the maximum velocity is 2.25 m s-1. (c) Apabila zarah itu bergerak ke kanan, When the particle moves to the right, v 0 –10 + 7t – t2 0 [K1] t2 – 7t + 10 0 (t – 2)(t – 5) 0 2 5 t Maka, julat nilai t ialah 2 < t < 5. [N1] Hence, the range of values of t is 2 < t < 5. (d) Apabila/When v = 0, –10 + 7t – t2 = 0 [K1] t2 – 7t + 10 = 0 (t – 2)(t – 5) = 0 t = 2 atau/or t = 5 Oleh itu, zarah berhenti seketika untuk kali kedua apabila t = 5. [K1] Therefore, the particle stops instantaneously for the second time when t = 5. s = ∫ v dt s = ∫ (–10 + 7t – t2 ) dt s = – 10t + 7t2 2 – t3 3 + c Apabila/When t = 0, s = 0. Oleh itu/Thus, c = 0. \s = –10t + 7t2 2 – t3 3 [K1] t (s) 0 2 5 s (m) 0 –82 3 –41 6 [K2] [N1] FIRASAT Add Math(Jaw).indd 23 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J24• –82 3 –41 6 0 s (m) t = 2 t = 5 t = 0 Maka, jumlah jarak yang dilalui Hence, the total distance travelled = 82 3 + 8 2 3 – 41 6 [K1] = 82 3 + 41 2 = 13 1 6 m [N1] 14 (a) x = 1.75 2.50 × 100 = 70 [N1] y 4.50 × 100 = 130 y = 130 × 4.50 100 y = 5.85 [N1] 5.70 z × 100 = 114 z = 5.70 × 100 114 z = 5.00 [N1] Kertas Model 5 Kertas 1 Bahagian A 1 mAB = 0 – 2 6 – 3 = – 2 3 Bagi/For x 6 + y p = 1, mJK = – Pintasan-y/ y-intercept Pintasan-x/ x-intercept = – p 6 [K1] mAB × mJK = –1 – 2 3 × – p 6 = –1 [K1] p 9 = –1 p = –9 [N1] 2 y = hx k log10 y = log10 hx k 1 2 log10 y = log10 hx – log10 k 1 2 log10 y = xlog10 h – 1 2 log10 k Pada titik 2, 9 2, x = 2 dan log10 y = 9 2 . At the point 2, 9 2, x = 2 and log10 y = 9 2 . 9 2 = 2 log10 h – 1 2 log10 k ......① [K1] Pada titik 4, 17 2 , x = 4 dan log10 y = 17 2 . At the point 4, 17 2 , x = 4 and log10 y = 17 2 . 17 2 = 4 log10 h – 1 2 log10 k .....② [K1] ② – ①: 4 = 2 log10 h log10 h = 2 h = 102 h = 100 [N1] Daripada/From ①: 9 2 = 2(2) – 1 2 log10 k 1 2 log10 k = – 1 2 log10 k = –1 k = 10–1 k = 1 10 [N1] 3 S∞ = 6 a 1 – r = 6…….① [K1] T1 + T2 = 16 3 a + ar = 16 3 3a(1 + r) = 16…….② [K1] (b) – I = 114.6 ∑Iw ∑w = 114.6 (70 × 1) + (120 × 3) + (130 × 2) + 114h 1 + 3 + 2 + h = 114.6 [K1] 690 + 114h 6 + h = 114.6 690 + 114h = 687.6 + 114.6h 0.6h = 2.4 h = 4 [N1] (c) Q2019 Q2018 × 100 = 114.6 Q2019 1 500 × 100 = 114.6 [K1] Q2019 = 114.6 × 1 500 100 Q2019 = RM1 719 [N1] (d) Alat tulis B/Stationery B I 2023/2018 = 150 Q2023 Q2018 × 100 = 150 Q2023 Q2018 = 150 100 Daripada Jadual 2, From Table 2, I 2019/2018 = 120 Q2019 Q2018 × 100 = 120 Q2018 Q2019 = 100 120 [K1] I2023/2019 = Q2023 Q2019 × 100 = Q2023 Q2018 × Q2018 Q2019 × 100 [K1] = 150 100 × 100 120 × 100 = 125 [N1] 15 (a) (i) Dalam ΔABC, menggunakan petua sinus, In ΔABC, using the sine rule, BC sin 35° = 7 sin 40° [K1] BC = 7 sin 40° × sin 35° = 6.246 cm [N1] (ii) ECD = ACB = 180° – 35° – 40° = 105° Dalam ΔECD, mengggunakan petua kosinus, In ΔECD, using the cosine rule, DE2 = 112 + 42 – 2(11)(4) kos/cos 105° [K1] DE2 = 159.7761 DE = 12.64 cm [N1] (b) (i) [N2] 30° C' C B' A' 35° 40° 105° 75° 75° 7 cm 7 cm (ii) A'C'B' = A'C'C' = 180° – 105° = 75° [N1] (iii) Dalam ∆A'B'C', menggunakan petua sinus, In ∆A'B'C', using the sine rule, C'B' sin 65° = 7 sin 40° [K1] C’B’ = 7 sin 40° × sin 65° = 9.8698 cm Luas ∆A'B'C' Area of ∆A'B'C' = 1 2 × 7 × 9.8698 × sin 75° [K1] = 33.37 cm2 [N1] FIRASAT Add Math(Jaw).indd 24 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
•J25• Firasat SPM 2023: Matematik Tambahan - Jawapan ① ②: a 1 – r 3a(1 + r) = 6 16 [K1] 1 3(1 – r)(1 + r) = 3 8 9(1 – r 2 ) = 8 9 – 9r 2 = 8 9r 2 = 1 r2 = 1 9 r = ± 1 3 [N1] Daripada/From ②: Apabila/When r = 1 3 , 3a1 + 1 3 = 16 4a = 16 a = 4 [N1] Apabila/When r = – 1 3 , 3a1 – 1 3 = 16 2a = 16 a = 8 [N1] 4 y = p 2 + qx y = p(2 + qx) –1 dy dx = –p(2 + qx) –2(q) = –pq (2 + qx)2 [K1] Pada titik (1, 1), kecerunan lengkung ialah 3 5 . At the point (1, 1), the gradient of the curve is 3 5 . – pq (2 + q(1))2 = 3 5 –5pq = 3(4 + 4q + q2 )…….① [K1] Lengkung itu melalui titik (1, 1). The curve passes through the point (1, 1). 1 = p 2 + q(1) p = q + 2….… ② [K1] Gantikan ② ke dalam ①: Substitute ② into ①: –5q(2 + q) = 3(4 + 4q + q2 ) –10q – 5q2 = 12 + 12q + 3q2 8q2 + 22q + 12 = 0 4q2 + 11q + 6 = 0 (q + 2)(4q + 3) = 0 q = –2 atau/or q = – 3 4 [N1] Daripada/From ②: Apabila/When q = –2, p = –2 + 2 p = 0 [N1] Apabila/When q = – 3 4 , p = – 3 4 + 2 p = 11 4 [N1] 5 2x2 + 7x + 19 = 0 Punca-punca ialah 2a – 3 dan 2β – 3. The roots are 2a – 3 and 2β – 3. Hasil tambah punca/Sum of roots = – b a (2α – 3) + (2β – 3) = – 7 2 [K1] 2(α + β) – 6 = – 7 2 α + β = 5 4 [K1] Hasil darab punca/Product of roots = c a (2α – 3)(2β – 3) = 19 2 [K1] 4αβ – 6α – 6β + 9 = 19 2 4αβ – 6(α + β) = 1 2 4αβ – 6 5 4 = 1 2 4αβ = 8 αβ = 2 [K1] Maka, persamaan yang dikehendaki ialah Hence, the required equation is x2 – 5 4 x + 2 = 0 4x2 – 5x + 8 = 0 [N1] 6 j = j o e –0.2t j o 2 = j o e –0.2t [K1] 1 2 = e –0.2t –0.2t = loge 1 2 [K1] –0.2t = ln 1 2 t = ln 1 2 –0.2 [K1] t = 3.466 [K1] Maka, bilangan tahun minimum ialah 4. [N1] Hence, the minimum number of years is 4. 7 Luas segi empat selari/Area of the parallelogram = 7 – 1 ( 7 – 2)t = 7 – 1 [K1] t = 7 – 1 7 – 2 [K1] = ( 7 – 1)( 7 + 2) ( 7 – 2)( 7 + 2) [K1] = 7 + 7 – 2 7 – 4 [K1] = 5 + 7 3 = 5 3 + 7 3 [N1] 8 (a) Jika titik-titik F, G dan H adalah segaris, maka FG→ = kGH→ , dengan keadaan k ialah pemalar. If the points F, G and H are collinear, then FG → = kGH→ . 6p ~ – 4q ~ = k4p ~ – (2u – 1)q ~ 6p ~ – 4q ~ = 4kp ~ – k(2u – 1)q ~ [K1] Menyamakan pekali p ~ , Equating the coefficients of p ~ , 4k = 6 k = 3 2 [K1] Menyamakan pekali q ~ , Equating the coefficients of q ~ , –4 = –k(2u – 1) [K1] –4 = – 3 2 (2u – 1) –8 = –3(2u –1) –8 = –6u + 3 6u = 11 u = 11 6 [N1] (b) FG→ = 3 2 GH→ FG → GH → = 3 2 FG→ : GH→ = 3 : 2 [N1] 9 dy dx = (2x + 1)3 y = (2x + 1)3 dx [K1] y = (2x + 1)4 4(2) + c y = (2x + 1)4 8 + c [K1] Lengkung melalui titik The curve passes through point 1 2 , –3. –3 = 2 1 2 + 1 4 8 + c [K1] –3 = 2 + c c = –5 Maka, persamaan lengkung ialah Hence, the equation of the curve is y = (2x + 1)4 8 – 5. [N1] 10 l = T72 = a + 71d = a + 71(20) [K1] = a + 1 420 [K1] S72 = 55 440 n 2 (a + l) = 55 440 72 2 (a + a + 1 420) = 55 440 [K1] 36(2a + 1 420) = 55 440 72a + 51 120 = 55 440 [K1] 72a = 4 320 a = 60 [K1] …, T61, T62, T63, …, T72 FIRASAT Add Math(Jaw).indd 25 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J26• Hasil tambah 12 sebutan terakhir Sum of the last 12 terms = S72 – S60 = 72 2 [2(60) + (72 – 1)(20)] – 60 2 [2(60) + (60 – 1)(20) [K2] = 55 440 – 39 000 = 16 440 [N1] 11 (a) X – Bilangan mentol lampu yang mempunyai jangka hayat kurang daripada setahun X – Number of light bulbs with lifespan of less than a year X ~ B(3, p) P(X = 0) = 1 125 3 C0p0q3 = 1 125 [K1] (1)(1)q3 = 1 125 q = 1 5 [K1] p = 1 – 1 5 = 4 5 [N1] (b) Bilangan mentol lampu yang masih dapat digunakan selepas setahun Number of light bulbs that are still functioning after a year = q × 50 = 1 5 × 50 [K1] = 10 [N1] 12 n Cr – 1 = 36 n! (r – 1)!(n – r + 1)! = 36… ① [K1] n Cr = 84 n! r!(n – r)! = 84… ② [K1] n Cr + 1 = 126 n! (r – 1)!(n – r – 1)! = 126… ③ [K1] ② ① : n! r!(n – r)! n! (r – 1)!(n – r + 1)! = 84 36 (r – 1)!(n – r + 1)! r!(n – r)! = 7 3 (r – 1)!(n – r + 1) (n – r)! r(r – 1)! (n – r)! = 7 3 n – r + 1 r = 7 3 … ④ [K1] ③ ② : n! (r + 1)!(n – r – 1)! n! r!(n – r)! = 126 84 [K1] r!(n – r)! (r + 1)!(n – r – 1)! = 126 84 r!(n – r) (n – r –1)! (r + 1)(r)!(n – r – 1)! = 3 2 n – r r + 1 = 3 2 2(n – r) = 3(r + 1) 2n – 2r = 3r + 3 2n = 5r + 3 n = 5r + 3 2 … ⑤ [K1] Gantikan ⑤ ke dalam ④: Substitute ⑤ into ④: 5r + 3 2 – r + 1 r = 7 3 5r + 3 – 2r + 2 2 = 7r 3 3r + 5 2 = 7r 3 9r + 15 = 14r 5r = 15 r = 3 [K1] Maka/Hence, r C2 = 3 C2 = 3 [N1] Bahagian B 13 (a) Kecerunan garis AMC Gradient of the line AMC = 3 – 9 –5 – (–9) = – 3 2 Kecerunan pembahagi dua sama serenjang Gradient of the perpendicular bisector = – – 2 3 = 2 3 [K1] Persamaan pembahagi dua sama serenjang ialah Equation of the perpendicular bisector y – 3 = 2 3 [x – (–5)] [K1] 3y – 9 = 2x + 10 3y = 2x + 19 [N1] (b) Bagi titik B(–8, k), For point B(–8, k), 3k = 2(–8) + 19 3k = 3 k = 1 Maka, titik B ialah (–8, 1). [K1] Thus, point B is (–8, 1). Katakan C ialah titik (p, q). Let C be point (p, q). Menyamakan koordinat-x, Equating the x-coordinates, –9 + p 2 = –5 –9 + p = –10 p = –1 Menyamakan koordinat-y, Equating the y-coordinates, 9 + q 2 = 3 9 + q = 6 q = –3 Maka, titik C ialah (–1, –3). [K1] Hence, point C is (–1, –3). Luas/Area of ABCD = 1 2 –9 –8 –1 1 –9 9 1 –3 9 9 = 1 2 –9 + 24 – 9 + 9 – (–72 – 1 – 3 – 81) = 1 2 (172) = 86 unit2 /units2 [N1] (c) AD = 9 + 1 = 10 unit/units [x – (–9)]2 + (y – 9)2 = 10 [K1] x2 + 18x + 81 + y2 – 18y + 81 = 100 Maka, persamaan lokus titik P ialah x2 + y2 + 18x – 18y + 62 = 0. [N1] Hence, the equation of locus of point P is x2 + y2 + 18x – 18y + 62 = 0. 14 (a) f(x) = –x2 – x + 2 = – (x2 + x – 2) = –x2 + x + 1 2 2 – 1 2 2 – 2 = –x2 + x + 1 4 – 1 4 – 2 = –x + 1 2 2 – 9 4 = –x + 1 2 2 + 9 4 [K1] Titik maksimum ialah Maximum point is – 1 2 , 21 4. [K1] Pintasan-y ialah/The y-intercept is 2. Pada paksi-x/At the x-axis, y = 0. –x2 – x + 2 = 0 x2 + x – 2 = 0 (x + 2)(x – 1) = 0 x = –2 atau/or x = 1 Pintasan-x ialah –2 dan 1./The x-intercepts are –2 and 1. [K1] Graf y = f(x) = –x2 – x + 2 adalah seperti berikut: The graph of y = f(x) = –x2 – x + 2 is as follows: 1 2 – 1 2 , 2 1 4 –2 0 x y [N1] FIRASAT Add Math(Jaw).indd 26 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
•J27• Firasat SPM 2023: Matematik Tambahan - Jawapan (b) – x2 – x + 2 < 0 x2 + x – 2 > 0 (x + 2)(x – 1) > 0 –2 1 x [K1] Julat nilai x yang dikehendaki ialah x < –2 atau x > 1. [N1] The required range of values of x is x < –2 or x > 1. (c) 1 2 Ujian garis mengufuk Horizontal line test – 1 2 , 2 1 4 –2 0 x y [K1] Dengan menggunakan ujian garis mengufuk, terdapat lebih daripada satu titik persilangan. Oleh itu, f(x) bukan fungsi satu dengan satu. Maka, f(x) tidak mempunyai fungsi songsang. [N1] By using the horizontal line test, there are more than one point of intersection. Thus, f(x) is not a one-to-one function. Hence, f(x) has no inverse function. 15 (a) kos/ cos 30° = 1 – sin2 30° = 1 – h2 [K1] sin 40° = 1 – kos2 / cos2 40° = 1 – k2 [K1] sin 70° = sin (30° + 40°) = sin 30° kos/ cos 40° + kos/ cos 30° sin 40° [K1] = hk + 1 – h2 × 1 – k2 [N1] (b) kos/ cos 80° = 2 kos2 / cos2 40° – 1 [K1] = 2k2 – 1 [N1] (c) kos/ cos 40° = 2 kos2 / cos2 20° – 1 k = 2 kos2 / cos2 20° – 1 [K1] k + 1 2 = kos2 / cos2 20° kos/ cos 20° = k + 1 2 [N1] Kertas 2 Bahagian A 1 Jumlah panjang semua sisi/Sum of all sides = 32 cm 8x + 4y = 32 2x + y = 8 y = 8 – 2x…….① [K1] Jumlah luas permukaan/Total surface area = 40 cm2 2x2 + 4xy = 40 x2 + 2xy = 20…….② [K1] Gantikan ① ke dalam ②: Substitute ① into ②: x2 + 2x(8 – 2x) = 20 [K1] x2 + 16x – 4x2 – 20 = 0 –3x2 + 16x – 20 = 0 3x2 – 16x + 20 = 0 (3x – 10)(x – 2) = 0 x = 10 3 atau/or x = 2 [K1] Daripada/From ①: Apabila/When x = 10 3 , y = 8 – 2 10 3 = 4 3 [K1] Tidak diterima kerana ia tidak memuaskan syarat y > x. It is not accepted because the condition y > x is not satisfied. Apabila/When x = 2, y = 8 – 2(2) = 4 [K1] Maka, isi padu kuboid Hence, the volume of the cuboid = 2 × 2 × 4 = 16 cm3 [N1] 2 Jejari Radius Lilitan Circumference Sebutan Term j 2pj T1 j + 1 2pj(j + 1) T2 j + 2 2p(j + 2) T3 … … … (a) Diberi panjang dawai ialah 78p cm, maka Given that the length of the wire is 78p cm, thus S6 = 78p n 2 [2a + (n – 1)d] = 78p 6 2 [2(2pj) + 5(2p)] = 78p [K2] 3(4pj + 10p) = 78p 4pj + 10p = 26p [K1] 4pj = 16p j = 16p 4p j = 4 Maka, jejari bagi bulatan terkecil ialah 4 cm. [N1] Hence, the radius of the smallest circle is 4 cm. (b) a = 2pj = 2p(4) = 8p [K1] Diberi/Given Sn = 200p, n 2 [2(8p) + (n – 1)(2p)] = 200p [K1] 8np + n(n – 1)p = 200p 8np + n2 p – np = 200p 7np + n2 p = 200p 7n + n2 = 200 n2 + 7n – 200 = 0 n = –7 ± 72 – 4(1)(–200) 2(1) [K1] = –7 ± 849 2(1) = –18.07 atau/ or 11.07 n = –18.07 tidak diterima/ is not accepted. ∴ n = 11.07 Maka, bilangan bulatan lengkap yang dapat dibentuk ialah 11. [N1] Hence, the number of complete circles that can be formed is 11. 3 y = 2(x –2)2 + 3q y = 2(x2 – 4x + 4) + 3q y = 2x2 – 8x + 8 + 3q [K1] H.T.P/S.O.R. = – –8 2 = 4…….① H.D.P./P.O.R. = 8 + 3q 2 …….②[K1] y = x2 + x – px – 5 y = x2 + (1 – p)x – 5 H.T.P/S.O.R. = – (1 – p) 2 = p – 1…③ H.D.P./P.O.R. = –5 …④ [K1] Menyamakan ① dan ③: Equating ① and ③: p – 1 = 4 p = 5 [N1] Menyamakan ② dan ④: Equating ② and ④: 8 + 3q 2 = –5 [K1] 3q + 8 = –10 3q = –18 q = –6 [N1] 4 y = 4 – x2 – 16 x2 dy dx = –2x + 32 x3 d2 y dx2 = –2 – 96 x4 [K1] Pada titik pusingan/At turning points, dy dx = 0 –2x + 32 x3 = 0 32 x3 = 2x x4 = 16 x = ±2 [K1] Apabila/When x = 2, y = 4 – 22 – 16 22 y = –4 Maka, (2, –4) ialah titik pusingan. [N1] Thus, (2, –4) is a turning point. d2 y dx2 = –2 – 96 24 = –8 (Negatif/Negative) [K1] Maka, (2, –4) ialah titik maksimum. [N1] Hence, (2, –4) is a maximum point. Apabila/When x = –2, y = 4 – (–2)2 – 16 (–2)2 = –4 FIRASAT Add Math(Jaw).indd 27 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J28• Maka, (–2, –4) ialah titik pusingan. [N1] Thus, (–2, –4) is a turning point. d2 y dx2 = –2 – 96 (–2)4 = –8 (Negatif/Negative) Maka, (–2, –4) ialah titik maksimum. [N1] Hence, (–2, –4) is a maximum point. 5 (a) A 2a ~ 6a ~ 2b ~ 4b X ~ Y P O B (i) OX → = OA → + AX → = 8a ~ + 2b ~ [P1] (ii) BY → = BA → + AY → = –6b ~ – 2a ~ [P1] (b) OP → = λOX → OP → = λ8a ~ + 2b ~ = 8λa ~ + 2λb ~… ① [K1] Katakan/Let BP → = kBY → . OP → – OB → = kOY → – OB → OP → – OB → = kOY → – kOB → OP → – 8a ~ + 6b ~ = 6ka ~ – k(8a ~ + 6b ~) OP → = 6ka ~ – 8ka ~ – 6kb ~ + 8a ~ + 6b ~ OP → = –2ka ~ + 8a ~ + 6b ~ – 6kb ~ OP → = (8 – 2k)a ~ + (6 – 6k)b ~… ② [K1] Dengan menyamakan ① dan ②: Equating ① and ② : (8 – 2k)a ~ + (6 – 6k)b ~ = 8λa ~ + 2λb ~ Bandingkan pekali a ~, Compare the coefficients of a ~, 8 – 2k = 8λ 4 – k = 4λ … ③ [K1] Bandingkan pekali b ~, Compare the coefficients of b ~, 6 – 6k = 2λ 12 – 12k = 4λ … ④ [K1] ④ – ③: 8 – 11k = 0 k = 8 1 1 Oleh itu/Thus, BP → = 8 1 1 BY → . Maka/Hence, BP : BY = 8 : 11 [N1] 6 (a) L = Luas segi empat tepat PQRS – Luas segi tiga PMN Area of rectangle PQRS – Area of triangle PMN L = 60 × 40 – 1 2 (3x)(40 – x) [K1] L = 2 400 – 60x + 3 2 x2 [Tertunjuk/Shown] [N1] (b) dL dx = –60 + 3x [K1] dL dt = dL dx × dx dt 36 = (–60 + 3x) × dx dt [K1] 36 = [–60 + 3(25)] × dx dt 36 = 15 × dx dt dx dt = 2.4 cm s–1 [K1] x = ∫ 2.4 dt x = 2.4t + c Apabila/When t = 0, x = 25 25 = 2.4(0) + c c = 25 ∴ x = 2.4t + 25 [K1] Apabila/When t = 5, x = 2.4(5) + 25 = 37 Maka/Hence, SM = 37 cm [N1] 7 Apabila/When sin 3 2 x = 0, 3 2 x = 0°, 180°, 360°, 540° x = 0°, 120°, 240°, 360° x = 0, 2p 3 , 4p 3 , 2p [K1] Langkah/Step 1: Lakar graf/Sketch the graph of y = 3 sin 3 2 x Langkah/Step 2: Lakar graf/Sketch the graph of y = 3 sin 3 2 x Langkah/Step 3: Lakar graf/Sketch the graph of y = 3 sin 3 2 x + 1 3 sin 3 2 x = 3 2p x – 1 3 sin 3 2 x + 1 = 3 2p x Lukis garis lurus/Sketch the straight line y = 3x 2p . [N1] x 0 2p y 0 3 4 3 2 1 0 y = 3 sin 3 2 x + 1 (2π, 3) y = 3x 2π x y p 3 2p 3 4p 3 5p 3 p 2p [K4] [N1] Bilangan penyelesaian/Number of solutions = Bilangan titik persilangan/ Number of points of intersection = 4 [N1] Bahagian B 8 (a) x 1 2 3 4 5 6 y 2.667 0.583 0.222 0.104 0.053 0.028 x2 y 2.67 2.33 2.00 1.66 1.33 1.00 [K1] 3.0 2.5 2.0 1.5 1.0 0.5 0 x x2 y Graf x2 y melawan x Graph of x2 y against x 4.5 1 2 3 4 5 6 6 – 0 = 6 3 – 1 = 2 [K1][N1] (b) (i) y = h kx + 1 kx2 x2 y = h kx + 1 k [P1] 1 k = Pintasan-Y/Y-intercept 1 k = 3.0 [K1] k = 1 3 [N1] h k = Kecerunan/Gradient h 1 3 = – 2 6 [K1] h = – 2 6 × 1 3 h = – 1 9 [N1] (ii) Apabila x = 4.5, daripada graf, When x = 4.5 , from the graph, x2 y = 1.5 (4.5)2 y = 1.5 y = 1.5 (4.5)2 [K1] y = 0.074 [N1] 9 (a) y = 3 x – 1 y2 = 9(x – 1) Pada paksi-x/At the x-axis, y = 0 02 = 9(x – 1) x = 1 [K1] Isi padu janaan = 9 2 π unit3 Generated volume = 9 2 π units3 π∫ k 1 y2 dx = 9 2 π ∫ k 1 9(x – 1) dx = 9 2 [K1] ∫ k 1 (x – 1) dx = 1 2 x2 2 – x k 1 = 1 2 [K1] k2 2 – k – 1 2 – 1 = 1 2 [K1] FIRASAT Add Math(Jaw).indd 28 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
•J29• Firasat SPM 2023: Matematik Tambahan - Jawapan k2 2 – k + 1 2 = 1 2 k2 2 – k = 0 k2 – 2k = 0 k = 2 [N1] (b) Luas rantau berlorek Area of the shaded region = ∫ 2 1 3 x – 1 dx [K1] = ∫ 2 1 3(x – 1) 1 2 dx [K1] = 3 (x – 1) 3 2 3 2 2 1 [K1] = 2(x – 1) 3 2 2 1 = 2(2 – 1) 3 2 – 0 [K1] = 2 unit2 /units2 [N1] 10 X ~ N(μ, 1002 ) [K1] P(X > 1 169) < 0.117 P Z > 1 169 – μ 100 < 0.117 [K1] 0.117 1.19 z [K1] Z adalah lebih besar atau sama dengan 1.19 kerana luas rantau berlorek adalah kurang atau sama dengan 0.117. Z is greater or equal to 1.19 because the area of the shaded region is less than or equal to 0.117. 1 169 – μ 100 1.19 [K1] 1 169 – μ 119 μ 1 050 [K1] P(X > 879) > 0.877 Z adalah lebih kecil atau sama dengan –1.16 kerana luas rantau berlorek adalah lebih besar atau sama dengan 0.877. Z is less than or equal to –1.16 because the area of the shaded region is greater than or equal to 0.877. PZ > 879 – μ 100 > 0.877 [K1] 0.877 0.123 –1.16 z [K1] 879 – μ 100 < –1.16 [K1] 879 – μ < –116 μ > 995 [K1] Maka, julat nilai μ ialah 995 < μ < 1 050. [N1] Hence, the range of values of μ is 995 < μ < 1 050. 11 (a) Oleh sebab jejari sebuah bulatan sentiasa sama, maka OP = OQ = OR = 12 cm. Oleh sebab sisi bertentangan sebuah rombus adalah sama, maka PQ = OR dan RQ = OP. [P2] Since the radius of a circle is always the same, then OP = OQ = OR = 12 cm. Since the opposite sides of a rhombus are always the same, then PQ = OR and RQ = OP. Oleh itu, ΔOPQ dan ΔOQR ialah segi tiga sama sisi dengan keadaan Therefore, ΔOPQ and ΔOQR are equilateral triangles such that ∠POR = ∠POQ + ∠QOR = 60° + 60° = 120° [P1] Maka/Hence, ∠POR = 120° × π 180° = 2 3 π rad ∠POR = 2 3 × 3.142 = 2.09467 rad [K1] (b) (i) Q K O 12 cm 60° kos/ cos 60° = 12 OK 1 2 = 12 OK OK = 12 × 2 OK = 24 cm [K1] Maka, panjang lengkok JLK Hence, the arc length JLK = 24 × 2.09467 [K1] = 50.27 cm [N1] (ii) Luas tembereng berlorek Area of the shaded segment = 1 2 (24)2 (2.09467 – sin 120°) [K2] = 353.8 cm2 [N1] Bahagian C 12 O L P Q 22 m (a) sQ = 6t3 – 2t vQ = dsQ dt = 18t2 – 2 [K1] Apabila/When t = 0, vQ = 18(0)2 – 2 = –2 Maka, halaju awal zarah Q ialah –2 m s–1. [N1] Hence, the initial velocity of particle Q is –2 m s–1. (b) Apabila zarah Q bertukar arah, When particle Q changes direction, vQ = 0 18t2 – 2 = 0 [K1] t 2 = 1 9 t = 1 3 t(s) 0 1 3 2 sQ(m) 0 – 4 9 44 – 4 9 0 44 s(m) t = 1 3 t = 0 t = 2 [K1] Maka, jumlah jarak yang dilalui oleh zarah Q dalam 2 saat yang pertama Hence, the total distance travelled by particle Q in the first 2 seconds = 4 9 + 4 9 + 44 = 44 8 9 m [N1] (c) vP = 18t2 + 20 sP = ∫ vP dt sP = ∫ (18t 2 + 20) dt [K1] sP = 6t3 + 20t + c Apabila/When t = 0, sP (dari titik/from point O) = 0. Oleh itu/Thus, sP = 6t3 + 20t sQ = 6t3 – 2r + 22 [K1] Apabila t = 0, jarak zarah Q dari titik O ialah 22 m. When t = 0, the distance of particle Q from point O is 22 m. Apabila zarah P dan zarah Q bertemu, When the particles P and Q met, sP = sQ 6t3 + 20t = 6t3 – 2t + 22 [K1] 22t = 22 t = 1 [K1] Maka, apabila zarah P dan zarah Q bertemu, jarak zarahzarah dari titik O Hence, when particles P and Q met, the distance of the particles from point O = 6(1)3 + 20(1) = 26 m [N1] 13 (a) z 50 × 100 = 120 z = 120 × 50 100 z = 60 [N1] FIRASAT Add Math(Jaw).indd 29 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 •J30• (b) y x × 100 = 125 y = 125x 100 y = 1.25x … ① [K1] y – x = 15 y = x + 15 … ② [K1] Gantikan ② ke dalam ①: Substitute ② into ①: x + 15 = 1.25x 0.25x = 15 x = 60 [N1] Gantikan x = 60 ke dalam ②: Substitute x = 60 into ②: y = 60 + 15 = 75 [N1] (ii) Bahan Material Harga (RM) pada tahun Price (RM) in the year I2022/2020 w 2020 2022 K 40 44 110 h L 25 40 160 3 M 60 75 125 4 N 50 60 120 1 [K1] (c) (i) I – 2022/2020 = 130 RM260 Q2020 × 100 = 130 [K1] Q2020 = RM260 × 100 130 = RM200 [K1] I – 2022/2020 = 130 110h + (160 × 3) + (125 × 4) + (120 × 1) h + 3 + 4 + 1 = 130 [K1] 110h + 1 100 h + 8 = 130 110h + 1 100 = 130h + 1 040 20h = 60 h = 3 [N1] 14 (a) (i) ∠SPQ = 180° – 78° – 32° = 70° Dalam ΔPSQ, menggunakan petua sinus, In ΔPSQ, using the sine rule, QS sin 70° = 16 sin 32° [K1] QS = 16 sin 32° × sin 70° QS = 28.37 cm [N1] (ii) Dalam ∆SRQ, menggunakan petua kosinus, In ∆SRQ, using the cosine rule, 28.372 = 122 + 182 – 2(12)(18) kos/ cos ∠QRS [K1] kos/ cos ∠QRS = 122 + 182 – 28.372 2(12)(18) kos/ cos ∠QRS = –0.7798 ∠QRS = 141.24° [N1] (iii)Luas/Area of PQRS = Luas/Area of ΔPSQ + Luas/Area of ΔSRQ = 1 2 × 16 × 28.37 × sin 78° + 1 2 × 12 × 18 × sin 141.24° [K2] = 222.0 + 67.61 = 289.61 cm2 [N1] (b) (i) [K1] [N1] Sʹ Qʹ Rʹ 12 cm R 18 cm 141.24° 38.76° 38.76° (ii) ∠SʹRʹQʹ = ∠SʹRRʹ = 180° – 141.24° = 38.76° [N1] 15 (a) I x + y 40 [P1] x 0 40 y 40 0 II y – x < 10 [P1] x 0 30 y 10 40 III y > 15 [P1] (b) [K2] [N1] 5 10 15 20 25 30 35 40 x y 0.6 8 Minimum (5, 15) 8x + 10y = 80 40 35 30 25 20 15 10 5 0 x + y = 40 y = 15 Maksimum/ Maximum (15, 25) R y – x = 10 (c) (i) Bilangan maksimum guru dari Sekolah A ialah 25 orang. [P1] The maximum number of teachers from School A is 25. (ii) Tuntutan/Claim = 8x + 10y Lukis garis lurus/Draw the straight line 8x + 10y = 80. x 0 10 y 8 0 Titik optimum ialah (5, 15) bagi tuntutan minimum dan (15, 25) bagi tuntutan maksimum. The optimal points are (5, 15) for minimum claim and (15, 25) for maximum claim. Tuntutan minimum/ Minimum claim = 8(5) + 10(15) [K1] = RM190 Tuntutan maksimum/ Maximum claim = 8(15) + 10(25) [K1] = RM370 Maka, julat tuntutan ialah Hence, the range of claim is RM190 < Tuntutan/ Claim < RM370 [N1] FIRASAT Add Math(Jaw).indd 30 12/06/2023 7:24 PM PENERBIT ILMU BAKTI SDN. BHD.