JAWAPAN DAN ULASAN Pustaka Yakin Pelajar Sdn. Bhd. FORMAT TERKINI SPM TERKINI KERTAS MODEL FORMAT INSTRUMEN MATEMATIK TAMBAHAN SIJIL PELAJARAN MALAYSIA 2021 KERTAS 1 Bahagian A No. Skema Pemarkahan Marking Scheme Markah Marks Markah Total Total Marks 1 ∠AOB major = 2π – π 3 = 5 3 π Luas / Area = 1 2 (2)2 5 3 π = 10 3 π cm2 1 1 2 2 m = 7 – 1 5 – 2 = 2 c – 1 0 – 2 = 2 c – 1 = –4 c = –3 ln y = 2 ln x − 3 ln y − ln x2 = −3 ln y x2 = –3 y x2 = e–3 y = x2 e3 1 1 1 1 4 PA N EL GU R U PA K A R & G U R U C E M E R LA NG © Pustaka Yakin Pelajar Sdn. Bhd. J1 3472 • MATEMATIK TAMBAHAN
3472 • MATEMATIK TAMBAHAN J2 © Pustaka Yakin Pelajar Sdn. Bhd. No. Skema Pemarkahan Marking Scheme Markah Marks Markah Total Total Marks 3 (a) Sn < 650 217 1 – 23 n 1 – 23 < 650 1 – 23 n < 650 651 1 – 650 651 < 23 n 1 651 < 23 n 23 n > 1 651 n < lg 1 651 lg 23 n < 15.97 n = 15 (b) S∞ − T5 = 217 1 – 23 – 217 23 4 = 651 – 3 472 81 = 608 11 81 1121 5 4 (a) dydx = 1 – 4x3 (b) y = –2x m1 = –2 m2 = 12 1 – 4x3 = 12 12 = 4x3 x = 2 Gantikan x = 2 ke dalam Substitute x = 2 into y = x + 2x2 = 2 + 2x2 = 2 + 24 = 52 P 2, 52 1111 4
© Pustaka Yakin Pelajar Sdn. Bhd. J3 3472 • MATEMATIK TAMBAHAN No. Skema Pemarkahan Marking Scheme Markah Marks Markah Total Total Marks 5 (a) f(x) > 0 x2 + 3x – 4 > 0 (x + 4)(x – 1) > 0 –4 1 x < –4, x > 1 (b) (i) px + q – (x2 + 3x – 4) = 0 –x2 + (p – 3)x + q + 4 = 0 x2 – (p – 3)x – q – 4 = 0 α + β = p – 3 αβ = –q – 4 (ii) x2 + rx + r – 10 = 0 2 α + 2 β = –r dan / and 2 α 2 β = r – 10 2 α + 2 β = –r 2 β + α αβ = –r 2(p – 3) –q – 4 = –r r = 2(p – 3) q + 4 2 α 2 β = r – 10 4 αβ = r – 10 4 –q – 4 = r – 10 r = –4 q + 4 + 10 2(p – 3) q + 4 = –4 q + 4 + 10 2(p – 3) = –4 + 10(q + 4) 2(p – 3) = 10q + 36 p – 3 = 5q + 18 p = 5q + 21 1 1 1 1 1 1 1 7
3472 • MATEMATIK TAMBAHAN J4 © Pustaka Yakin Pelajar Sdn. Bhd. No. Skema Pemarkahan Marking Scheme Markah Marks Markah Total Total Marks 6 (a) D B C 8 6 θ tan θ = 6 8 θ = tan–1 6 8 = 0.6436 radian (b) BD = √ 62 + 82 = 10 cm ∠ABH = π 3 rad ∠HBD = π – π 3 – 0.6436 = 1.451 rad ∠GAH = π 2 – π 3 = π 6 rad EF = √ 122 + 182 = 21.63 cm SGH = 10 π 6 = 5π 3 cm atau / or SHD = 10(1.451) = 14.51 Perimeter = 10 + 5π 3 + 14.51 + 2 + 21.63 = 53.38 cm 1 1 1 1 1 1 6 7 (a) ED + DB + BE = 0 (b) EC = 11i + 4i + 10j = 15i + 10j (c) 1 1 + 3 4 t = 5 p 1 + 3t = 5 t = 4 3 1 + 4t = p 1 + 4 4 3 = p 3 + 16 3 = p p = 19 3 1 1 1 1 1 1 6
© Pustaka Yakin Pelajar Sdn. Bhd. J5 3472 • MATEMATIK TAMBAHAN No. Skema Pemarkahan Marking Scheme Markah Marks Markah Total Total Marks 8 (a) A(p, q) B(x, y) C(r, s) m n q – y y – s = m n nq – ny = my – ms ms + nq = (m + n)y y = ms + nq m + n x – p r – x = m n nx – np = mr – mx (m + n)x = mr + np x = mr + np m + n (x, y) = mr + np m + n , ms + nq m + n (b) (i) A(3, 2) B(6, 0) C(r, s) m n m n = 6 – 3 r – 6 m n = 3 r – 6 m n r – 6 3 r 9 m n = 2 – 0 0 – s m n = 2 –s m n 2 –s s –2 1 1 1 1 1 1 6 (ii) mAB = 2 – 0 3 – 6 = – 2 3 m = – p 6 – p 6 = – 2 3 p = 4
3472 • MATEMATIK TAMBAHAN J6 © Pustaka Yakin Pelajar Sdn. Bhd. No. Skema Pemarkahan Marking Scheme Markah Marks Markah Total Total Marks 9 (a) g(x) = –3x2 + 8x – 4 1 5 (b) Kawasan berlorek / Shaded area = ∫ 2 1(x3 – 6x2 + 12x – 6) dx – [–x3 + 4x2 – 4x] 2 1 = x4 4 – 6x3 3 + 12x2 2 – 6x + x3 – 4x2 + 4x 2 1 = x4 4 – x3 + 2x2 – 2x 2 1 = 24 4 – 23 + 2(2)2 – 2(2) – 1 4 – 13 + 2(1)2 – 2(1) = 16 4 – 8 + 8 – 4 – 1 4 – 1 + 2 – 2 = 0 – – 3 4 = 3 4 unit2 2 1 1 10 (a) (x, x + 20, x + 40) mm 1 (b) a = x T72 = x + (72 – 1)(20) = x + 1 420 S72 = 55 440 72 2 (x + x + 1 420) = 55 440 2x + 1 420 = 55 440 36 x = 60 18 kepingan terakhir / Last 18 terms: T55, T56, ..., T72 S18 = 18 2 [2(60 + 30(20)) + (18 – 1)(20)] = 14 940 mm (c) Cara 1 / Method 1 60 + (72 – 1)(20) = 1 480 1 480 500 2πr = 1 480 r = 1 480 2 = 740 π 60 + (72 – 1)(20) = 1 480 dan / and r = 740 π 1 1 1 1 1
© Pustaka Yakin Pelajar Sdn. Bhd. J7 3472 • MATEMATIK TAMBAHAN No. Skema Pemarkahan Marking Scheme Markah Marks Markah Total Total Marks V1 = πr2 h = π 740 π 2 (500) = 273 800 000 π mm3 = 2.738 × 108 π mm3 Cara 2 / Method 2 1 480 500 2πr = 500 r = 500 2π = 250 π V2 = πr2 h = π 250 π 2 (1 480) = 92 500 000 π mm3 = 9.25 × 108 π mm3 V1 = π 740 π 2 (500) dan / and V2 = π 250 π 2 (1 480) Cara 1 membentuk silinder dengan isi padu maksimum: Method 1 forms a cylinder with maximum volume: V1 = 2.738 × 108 π mm3 dan / and r = 740 π 1 1 8 11 (a) (i) 6! = 720 (ii) 1 kad / card 3 kad / cards atau / or 2 kad / cards 2 kad / cards 1 kad / card 3 kad / cards = 4 × 2! × 2! × 3 = 48 cara / ways 2 kad / cards 2 kad / cards : 15 atau / or 17 atau / or 19 26 atau / or 62 = 6 cara / ways : 51 atau / or 57 atau / or 59 62 = 3 cara / ways 1
3472 • MATEMATIK TAMBAHAN J8 © Pustaka Yakin Pelajar Sdn. Bhd. No. Skema Pemarkahan Marking Scheme Markah Marks Markah Total Total Marks : 26 51 atau / or 57 atau / or 59 + 26 71 atau / or 75 atau / or 79 + 26 91 atau / or 95 atau / or 97 = 9 cara / ways : 62 71 atau / or 75 atau / or 79 = 3 cara / ways : 62 91 atau / or 95 atau / or 97 = 3 cara / ways ATAU / OR 1 P1 × 1 P1 × 3 P1 × 3 P1 atau / or 1 P1 × 1 P1 × 2 P1 × 3 P1 atau / or 1 P1 × 3 P1 × 1 P1 × 1 P1 atau / or 2 P1 × 3 P1 × 1 P1 × 1 P1 atau / or 4 P2 × 1 × 1 Jumlah bilangan cara / Total number of ways: = 48 + 6 + 3 +9 + 3 + 3 = 72 2 1 7 (b) (i) n = 0 (ii) p C3 = 26(p – 2) p! (p – 3)!3! = 26(p – 2) p(p – 1)(p – 2)(p – 3)! (p – 3)!(6) = 26(p – 2) p(p – 1) = 26(6) p(p – 1) = 13(12) p = 13 1 1 1 12 (a) P(X = 2) + P(X = 3) = 1 – 81 128 – 1 64 = 45 128 1 4 (b) 5 C2 (p) 2 (q) 3 + 5 C3 (p) 3 (q) 2 = 45 128 10p2 q3 + 10p3 q2 = 45 128 10p2 q2 (q + p) = 45 128 p2 q2 (q + p) = 9 256 p2 q2 (1) = 9 256 pq = 3 16 p = 3 16q Tertunjuk / Shown 1 1 1
© Pustaka Yakin Pelajar Sdn. Bhd. J9 3472 • MATEMATIK TAMBAHAN No. Skema Pemarkahan Marking Scheme Markah Marks Markah Total Total Marks 13 (a) 3 4x – 1 = 1 3 x = 5 2 1 1 8 (b) f(m + 1) = 3fg(p) 3 4(m + 1) – 1 = 3 3 4p2 + 3 3(4m + 4 – 1) = 4p2 + 3 12m + 9 = 4p2 + 3 m = 4p2 – 6 12 m = 2p2 – 3 6 1 1 (c) (i) 3 4g(x) – 1 = 3 4x2 + 3 4g(x) – 1 = 4x2 + 3 g(x) = x2 + 1 (ii) Ujian garis mengufuk Horizontal line test y x x1 x2 O 1 g(x) Bukan hubungan satu kepada satu Not one to one relationship 1 1 1 1 14 (a) (2x + 1)2 + √ 2x 2 = 3√ 2 2 4(22x ) + 2x = 18 4(22x )2 + 2x – 18 = 0 [4(2x ) + 9][2x – 2] = 0 4(2x ) + 9 = 0 2x = – 9 4 2x > 0 2x = – 9 4 (tidak mungkin / not possible) 2x – 2 = 0 2x = 2 x = 1 1 1 1 1 Bahagian B
3472 • MATEMATIK TAMBAHAN J10 © Pustaka Yakin Pelajar Sdn. Bhd. No. Skema Pemarkahan Marking Scheme Markah Marks Markah Total Total Marks (b) 5 + √ 7 t 2 = 9 √ 7 – 1 5 + √ 7 t = 18√ 7 – 18 t = 18√ 7 – 18 5 + √ 7 × 5 – √ 7 5 – √ 7 = 90√ 7 – 126 – 90 + 18√ 7 25 – 7 = 108√ 7 – 216 18 = 6√ 7 – 12 1 1 1 1 8 15 (a) sin x = t kos / cos x = ± √ 1 – t 2 0 < x < 90° kos / cos x = √ 1 – t 2 sin 2x = 2t√ 1 – t 2 1 1 8 (b) 2 kos2 / cos2 θ 2 – 1 = k kos / cos θ 2 = ± 1 + k 2 270° < θ < 360° 135° < θ 2 < 180° kos / cos θ 2 < 0 kos / cos θ 2 = – 1 + k 2 1 1 (c) tan 45° = 2 tan 22.5° 1 – tan2 22.5° Biarkan / Let tan 22.5° = t 1 = 2t 1 – t 2 1 – t 2 = 2t t 2 + 2t – 1 = 0 t = –2 ± √ 22 – 4(1)(–1) 2(1) = –2 ± √ 8 2 t = –2 + √ 8 2 dan / and t = –2 – √ 8 2 tan 22.5° > 0 tan 22.5° = –2 + 2√ 2 2 = √ 2 – 1 1 1 1 1