49 Kertas 1 / Paper 1 Bahagian A / Section A 7.1 Pembahagi Tembereng Garis/ Divisor of a Line Segment 1 Diberi bahawa P(−5, u), Q(3, −2) dan R(7, 9) adalah segaris dengan keadaan PQ : QR = m : n. Tentukan It is given that P(−5, u), Q(3, −2) and R(7, 9) are collinear such that PQ : QR = m : n. Determine (a) m : n, (b) nilai u. / the value of u. BT ms.178–180 Aras R [5 markah/marks] Jawapan/Answer: (a) (b) 7.2 Garis Lurus Selari dan Garis Lurus Serenjang/ Parallel Lines and Perpendicular Lines 2 Rajah 1 menunjukkan tiga garis lurus, PQ, QS dan SR dengan keadaan e, f, g dan h ialah pemalar. BT ms.186–187 Aras S Diagram 1 shows three straight lines, PQ, QS and SR where e, f, g and h are constants. Ungkapkan / Express (a) e dalam sebutan g, e in terms of g, (b) h dalam sebutan f dan g. h in terms of f and g. [5 markah/marks] Jawapan/Answer: (a) (b) Bidang Pembelajaran: Geometri BAB 7 Geometri Koordinat Coordinate Geometry Rajah 1/ Diagram 1 x y O h P Q ey = 2x fi 8 + = 1 S R y g x f PT SPM Add Math Tkt 4-Bab 7-8.indd 49 15/01/2025 11:08 AM PENERBIT ILMU BAKTI SDN. BHD.
50 7.3 Luas Poligon/ Areas of Polygons 3 Rajah 2 menunjukkan garis lurus BPC dengan keadaan BP : PC = 3 : 2. Aras R Diagram 2 shows a straight line BPC such that BP : PC = 3 : 2. Cari / Find (a) koordinat C, BT ms.178–180 the coordinates of C, (b) luas, dalam unit2 , bagi ΔABC. BT ms.193–194 the area, in units2 , of ΔABC. [4 markah/marks] Jawapan/Answer: (a) (b) 4 Diberi koordinat bagi dua titik, G(10, 0) dan H(0, −4). Pembahagi dua sama serenjang bagi GH menyentuh titik Q dan R. Cari Aras S Given the coordinates of two points, G(10, 0) and H(0, −4). The perpendicular bisector of GH touches points Q and R. Find (a) persamaan QR, BT ms.186–187 the equation of QR, (b) luas ΔQOR dengan keadaan O ialah asalan. BT ms.193–194 the area of ΔQOR such that O is the origin. [5 markah/marks] Jawapan/Answer: (a) (b) Rajah 2/ Diagram 2 x y 9 C A O P(3, 4) B(–3, 1) PT SPM Add Math Tkt 4-Bab 7-8.indd 50 15/01/2025 11:08 AM PENERBIT ILMU BAKTI SDN. BHD.
51 5 Tentukan sama ada titik-titik (–2, 7), (1, 4) dan (7, –2) adalah segaris atau tidak. BT ms.193–194 Aras R Determine whether the points (–2, 7), (1, 4) and (7, –2) are collinear or not. [3 markah/marks] Jawapan/Answer: 7.4 Persamaan Lokus/ Equations of Loci 6 Persamaan lokus bagi titik bergerak L diberi oleh x2 + y2 + 3x – 3y – 8 = 0. Tunjukkan bahawa The equation of locus of a moving point L is given by x2 + y2 + 3x – 3y – 8 = 0. Show that BT ms.203–204 Aras S (a) lokus L menyilang paksi-y pada dua titik, the locus of L intersects the y-axis at two points, (b) garis lurus y = x – 2 ialah tangen kepada lokus L. the straight line y = x – 2 is a tangent to the locus of L. [4 markah/marks] Jawapan/Answer: (a) (b) PT SPM Add Math Tkt 4-Bab 7-8.indd 51 15/01/2025 11:08 AM PENERBIT ILMU BAKTI SDN. BHD.
52 Bahagian B / Section B 7.1 Pembahagi Tembereng Garis/ Divisor of a Line Segment 7.2 Garis Lurus Selari dan Garis Lurus Serenjang/ Parallel Lines and Perpendicular Lines 7 (a) Rajah 3 menunjukkan garis lurus FH pada suatu satah Cartes. Diagram 3 shows a straight line FH on a Cartesian plane. Titik G terletak pada garis lurus FH dengan keadaan FG : GH = 4 : 1. Cari nilai p dan nilai q. BT ms.179–180 Aras S Point G lies on the straight line FH such that FG : GH = 4 : 1. Find the value of p and of q. [4 markah/marks] (b) Sebuah lampu isyarat akan dibina di persimpangan jalan lurus y = −4x + 12 dan jalan lurus berserenjang yang melalui titik P(6, 5) seperti yang ditunjukkan dalam Rajah 4. BT ms.186–187 Aras S A traffic light is to be constructed at the junction of a straight road y = −4x + 12 and a perpendicular straight road that passes through point P(6, 5) as shown in Diagram 4. Cari kedudukan lampu isyarat yang akan dibina itu. Find the position of the traffic light to be constructed. [4 markah/marks] Jawapan/Answer: (a) (b) Rajah 3/ Diagram 3 y x F(p, 5) H(8, 0) G(6, q) O Rajah 4/ Diagram 4 x y O P(6, 5) y = –4x + 12 PT SPM Add Math Tkt 4-Bab 7-8.indd 52 15/01/2025 11:08 AM PENERBIT ILMU BAKTI SDN. BHD.
53 Kertas 2 / Paper 2 Bahagian A / Section A 1 Rajah 1 menunjukkan dua garis lurus, PQ dan QR yang berserenjang antara satu sama lain. Aras S Diagram 1 shows two straight lines, PQ and QR which are perpendicular to each other. y x 2x + y = 9 Q(h, 1) R(0, –1) P(2, 5) O Rajah 1/ Diagram 1 Diberi bahawa persamaan garis lurus PQ ialah 2x + y = 9. It is given that the equation of the straight line PQ is 2x + y = 9. (a) Cari / Find (i) nilai h, BT ms.188–189 the value of h, (ii) persamaan garis lurus QR, BT ms.188–189 the equation of the straight line QR, (iii) luas, dalam unit2 , segi tiga PQR. BT ms.193–194 the area, in units2 , of triangle PQR. [6 markah/marks] (b) Garis lurus QR dipanjangkan ke titik S dengan keadaan SR : RQ = 3 : 2. Cari koordinat S. BT ms.178–180 Th e straight line QR is extended to point S such that SR : RQ = 3 : 2. Find the coordinates of S. [2 markah/marks] Jawapan/Answer: (a) (i) (ii) (iii) (b) PT SPM Add Math Tkt 4-Bab 7-8.indd 53 15/01/2025 11:08 AM PENERBIT ILMU BAKTI SDN. BHD.
54 Bahagian B / Section B 2 Rajah 2 menunjukkan sebuah segi empat tepat EFGH. Persamaan bagi garis lurus EF ialah y = 2x + 7. BT ms.197–198 Aras S Diagram 2 shows a rectangle EFGH. The equation of the straight line EF is y = 2x + 7. Cari / Find (a) persamaan garis lurus HG, the equation of the straight line HG, [2 markah/marks] (b) persamaan garis lurus EH, the equation of the straight line EH, [3 markah/marks] (c) koordinat H, the coordinates of H, [2 markah/marks] (d) luas, dalam unit2 , segi empat tepat EFGH. the area, in units2 , of rectangle EFGH. [3 markah/marks] Jawapan/Answer: (a) (b) (c) (d) y x E(–2, 3) G(8, 13) y = 2x + 7 H F O Rajah 2/ Diagram 2 PT SPM Add Math Tkt 4-Bab 7-8.indd 54 15/01/2025 11:08 AM PENERBIT ILMU BAKTI SDN. BHD.
55 3 Rajah 3 menunjukkan sebuah segi tiga EFG pada suatu satah Cartes. Aras T Diagram 3 shows a triangle EFG on a Cartesian plane. (a) Cari / Find BT ms.197–198 (i) koordinat G, / the coordinates of G, (ii) luas segi tiga EFG. / the area of triangle EFG. [5 markah/marks] (b) Garis lurus EF dipanjangkan ke titik H dengan keadaan EF : FH = 2 : 1. Cari koordinat H. BT ms.179–180 KBAT Menganalisis The straight line EF is extended to point H such that EF : FH = 2 : 1. Find the coordinates of H. [2 markah/marks] (c) Diberi titik L(x, y) bergerak dengan keadaan ∠GLH = 90°. Cari persamaan lokus bagi titik L. Given the point L(x, y) moves such that ∠GLH = 90°. Find the equation of locus of point L. BT ms.203–204 [3 markah/marks] Jawapan/Answer: (a) (i) (ii) (b) (c) y x y = 2x + 8 5y = x + 22 E(8, 6) F O G Rajah 3/ Diagram 3 PT SPM Add Math Tkt 4-Bab 7-8.indd 55 15/01/2025 11:08 AM PENERBIT ILMU BAKTI SDN. BHD.
56 4 Rajah 4 menunjukkan sebuah sisi empat EFGH. Aras S Diagram 4 shows a quadrilateral EFGH. Diberi bahawa persamaan garis lurus EH dan HG masingmasing ialah 2y = 3x + 32 dan y = –5. It is given that the equation of the straight line EH and HG are 2y = 3x + 32 and y = –5 respectively. (a) Cari / Find BT ms.188–189 (i) persamaan garis lurus EF, the equation of the straight line EF, [3 markah/marks] (ii) koordinat E, the coordinates of E, [2 markah/marks] (iii) persamaan garis lurus FG. the equation of the straight line FG. [2 markah/marks] (b) Titik P bergerak dengan keadaan jaraknya sentiasa 6 unit dari H. Cari persamaan lokus bagi titik P. Point P moves such that its distance from H is always 6 units. Find the equation of locus of point P. BT ms.200–201 [3 markah/marks] Jawapan/Answer: (a) (i) (ii) (iii) (b) Zon KBAT 1 Cari semua titik persilangan antara garis lurus y = 2x – 3 dengan sebuah bulatan berpusat (1, 1) dan berjejari 5 unit. BT ms.203–204 KBAT Mengaplikasi Find all the points of intersection between the straight line y = 2x – 3 and a circle with centre (1, 1) and a radius of 5 units. Jawapan/Answer: y O G H E F(3, 1) x –3 Rajah 4/ Diagram 4 PT SPM Add Math Tkt 4-Bab 7-8.indd 56 15/01/2025 11:08 AM PENERBIT ILMU BAKTI SDN. BHD.
57 Kertas 1 / Paper 1 Bahagian A / Section A 8.1 Vektor/ Vectors 1 Titik-titik F, G dan H adalah segaris. Diberi bahawa FG → = 6p ~ – 4q ~ dan GH → = 4p ~ + (2u – 1)q ~ , dengan keadaan u ialah pemalar. Cari BT ms.219 Aras R Th e points F, G and H are collinear. It is given that FG → = 6p ~ – 4q ~ and GH→ = 4p ~ + (2u – 1)q ~ , where u is a constant. Find (a) nilai u, / the value of u, (b) nisbah FG : GH. / the ratio of FG : GH. [5 markah/marks] Jawapan/Answer: (a) (b) 2 Vektor u ~ dan vektor w~ adalah bukan sifar dan tidak selari. Diberi bahawa (α + 5)u ~ = (6β − 2)w~, dengan keadaan α dan β ialah pemalar. Cari nilai α dan nilai β. BT ms.219 Aras R Th e vectors u~ and w~ are non-zero and not parallel. It is given that (α + 5)u~ = (6β − 2)w~, where α and β are constants. Find the value of α and of β. [2 markah/marks] Jawapan/Answer: Bidang Pembelajaran: Geometri BAB 8 Vektor Vectors PT SPM Add Math Tkt 4-Bab 7-8.indd 57 15/01/2025 11:08 AM PENERBIT ILMU BAKTI SDN. BHD.
58 8.2 Penambahan dan Penolakan Vektor/ Addition and Subtraction of Vectors 3 Rajah 1 menunjukkan sebuah segi tiga EFH dengan keadaan FG : GH = 2 : 3. Diagram 1 shows a triangle EFH where FG : GH = 2 : 3. Diberi bahawa EF → = 6h ~ dan EH → = 5k ~. Cari BT ms.222–223 Aras S It is given that EF→ = 6h~ and EH→ = 5k~. Find (a) FH → , (b) EG → . [4 markah/marks] Jawapan/Answer: (a) (b) 8.3 Vektor dalam Satah Cartes/ Vectors in a Cartesian Plane 4 Rajah 2 menunjukkan sebuah segi empat selari OKLM yang dilukis pada satu satah Cartes. Diagram 2 shows a parallelogram OKLM drawn on a Cartesian plane. Diberi bahawa O →M = 7i ~ + 3j ~ dan M →L = −3i ~ + 4j ~ . Cari M →K. BT ms.228 Aras S It is given that O→ M = 7i~ + 3j~ and M → L = −3i~ + 4j~ . Find M → K. [3 markah/marks] Jawapan/Answer: M L K y x O Rajah 2/ Diagram 2 y L K M x O Rajah 1/ Diagram 1 E H F G F E H G PT SPM Add Math Tkt 4-Bab 7-8.indd 58 15/01/2025 11:08 AM PENERBIT ILMU BAKTI SDN. BHD.
Praktis Topikal SPM Matematik Tambahan Tingkatan 4 – Jawapan J1 Bab 1 Fungsi Kertas 1 Bahagian A 1 y x O Graf ini bukan suatu fungsi kerana apabila diuji dengan garis mencancang, terdapat lebih daripada satu titik yang memotong graf itu. The graph is not a function because when tested with the vertical line, there is more than one point that cuts the graph. 2 f(x) = 3x – 2 f(p) = 3p – 2 10 = 3p – 2 3p = 12 p = 4 3 (a) f(x) = 1 |x + 2| = 1 x + 2 = –1 atau/or x + 2 = 1 x = –3 atau/or x = –1 (b) f(x) fi 2 |x + 2| fi 2 –2 fi x + 2 fi 2 –4 fi x fi 0 4 gf(x) = g[f(x)] = g(x + 4) = 2(x + 4)2 – 9 = 2(x2 + 8x + 16) – 9 = 2x2 + 16x + 32 – 9 = 2x2 + 16x + 23 5 (a) f(x) = 3x – 5 f 2 (x) = 3(3x – 5) – 5 = 9x – 15 – 5 = 9x – 20 (b) f 4 (x) = f 2 [f 2 (x)] = 9(9x – 20) – 20 = 81x – 180 – 20 = 81x – 200 (c) f 4 ( p 3 ) = 16 81( p 3 ) – 200 = 16 27p = 216 p = 8 6 (a) gf(x) = g[f(x)] = g(3 – x) = p(3 – x) 2 + q = p(9 – 6x + x2 ) + q = 9p – 6px + px2 + q = px2 – 6px + 9p + q Diberi/Given gf(x) = 3x2 – 18x + 15 Bandingkan pekali x2 dan pemalar, Compare the coeffi cient of x2 and constant, p = 3 , 9p + q = 15 9(3) + q = 15 q = –12 (b) g(x) = 3x2 – 12 g2 (x) = 3(3x2 – 12)2 – 12 g2 (0) = 3[3(0)2 – 12]2 – 12 = 3(144) – 12 = 420 7 Katakan/Let y = 3x – 5 3x = y + 5 x = y + 5 3 ∴ f –1(x) = x + 5 3 f –1(4) = 4 + 5 3 = 9 3 = 3 8 Katakan/Let y = 2x – 14 2x = y + 14 x = y + 14 2 ∴ f –1(x) = x + 14 2 = x 2 + 7 Bandingkan dengan f –1(x) = px + q, Compare with f –1(x) = px + q, p = 1 2, q = 7 9 (a) f(x) = 6 3x – 4 Katakan/Let y = 6 3x – 4 3xy – 4y = 6 3xy = 6 + 4y x = 6 + 4y 3y ∴ f –1(x) = 6 + 4x 3x , x ≠ 0 f –1g(x) = 6 + 4(6x – 3) 3(6x – 3) = 6 + 24x – 12 3(6x – 3) = 24x – 6 3(6x – 3) = 8x – 2 6x – 3 (b) f –1g(k) = 1 8k – 2 6k – 3 = 1 8k – 2 = 6k – 3 2k = –1 k = – 1 2 Bahagian B 10 (a) f(–2) = |2(–2) – 5| = |–9| = 9 (b) f(x) = 3 |2x – 5| = 3 2x – 5 = –3 atau/or 2x – 5 = 3 x = 1 atau/or x = 4 (c) f(x) = x |2x – 5| = x 2x – 5 = –x atau/or 2x – 5 = x x = 5 3 atau/or x = 5 (d) f(x) > 4 2x – 5 < –4 atau/or 2x – 5 > 4 2x < 1 atau/or 2x > 9 x < 1 2 atau/or x > 9 2 11 (a) g[h(x)] = 6 3x2 + 4 6 3h(x) – 2 = 6 3x2 + 4 3h(x) – 2 = 3x² + 4 3h(x) = 3x² + 6 h(x) = x² + 2 (b) hg(2) = h[g(2)] = h[ 6 3(2) – 2] = h( 6 4) = h( 3 2) = ( 3 2) 2 + 2 = 17 4 (c) g(p + 2) = 1 2 gh(q) 6 3(p + 2) – 2 = 1 2( 6 3q2 + 4 ) 2 3p + 4 = 1 3q2 + 4 2(3q² + 4) = 3p + 4 6q² + 8 = 3p + 4 3p = 6q² + 4 p = 6q2 + 4 3 p = 2q2 + 4 3 12 (a) Katakan/Let y = 3x – 18 2 2y = 3x – 18 3x = 2y + 18 x = 2y + 18 3 x = 2y 3 + 6 ∴ g(x) = 2x 3 + 6 (b) Katakan/Let y = 4x – 6 4x = y + 6 x = y + 6 4 ∴ f –1(x) = x + 6 4 f –1(4) = 4 + 6 4 = 10 4 = 5 2 Jawapan PT SPM Add Math Tkt 4-Jawapan.indd 1 07/02/2025 11:24 AM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2025 J2 (c) f –1g(x) = f –1[g(x)] = f –1( 2x 3 + 6) = ( 2x 3 + 6) + 6 4 = 2x 3 + 12 4 = 2x + 36 3 4 = 2x + 36 12 = x 6 + 3 13 (a) Katakan/Let y = 6x + q 6x = y – q x = y – q 6 x = y 6 – q 6 ∴ f –1(x) = x 6 – q 6 Bandingkan dengan f –1(x) = px – 12, Compare with f –1(x) = px – 12, p = 1 6 , q 6 = 12 q = 72 (b) f –1(12) = 1 6 (12) – 12 = –10 (c) f(x) = f –1(x) 6x + 72 = 1 6 x – 12 35 6 x = –84 x = – 72 5 (d) f –1(x) = 1 6 1 6 x – 12 = 1 6 1 6 x = 73 6 x = 73 Kertas 2 Bahagian A 1 (a) (i) Fungsi yang memetakan set B kepada set A ialah f –1. The function which maps set B to set A is f −1. Katakan/Let y = 4x + 3 4x = y – 3 x = y – 3 4 f –1(x) = x – 3 4 (ii) gf(x) = 24x + 8 gf[f –1(x)] = 24[f –1(x)] + 8 g(x) = 24 ( x – 3 4 ) + 8 = 6(x – 3) + 8 = 6x – 18 + 8 = 6x – 10 (b) gf(x) = 21x – 7 6[f(x)] – 10 = 21x – 7 6(4x + 3) – 10 = 21x – 7 24x + 18 – 10 = 21x – 7 3x = –15 x = –5 2 (a) 5 + 3x = 0 3x = –5 x = – 5 3 ∴u = – 5 3 (b) (i) f(–3) = –3 w – (–3) 5 + 3(–3) = –3 w + 3 –4 = –3 w + 3 = 12 w = 9 (ii) f(x) = x (9) – x 5 + 3x = x 9 – x = 5x + 3x² 3x² + 6x – 9 = 0 x² + 2x – 3 = 0 (x + 3)(x – 1) = 0 x = –3 atau/or x = 1 ∴ x = 1 (iii) f(x) = 9 – x 5 + 3x f( 1 3) = 9 – ( 1 3) 5 + 3( 1 3) = 13 9 f²( 1 3) = f[f( 1 3)] = f( 13 9 ) = 9 – ( 13 9 ) 5 + 3( 13 9 ) = 17 21 Bahagian B 3 (a) f(2) = 2 (2) – m (2) – 6 = 2 2 – m = –8 m = 10 (b) (i) u(2) = –4 p(2) (2) – 2q + 1 = –4 2p 3 – 2q = –4 2p = –12 + 8q p = –6 + 4q 1 u(3) = –9 p(3) (3) – 2q + 1 = –9 3p 4 – 2q = –9 3p = –36 + 18q p = –12 + 6q 2 Gantikan 1 ke dalam 2 , Substitute 1 into 2 , –6 + 4q = –12 + 6q –2q = –6 q = 3 Daripada/From 1 , p = – 6 + 4(3) = 6 (ii) u(x) = 6x x – 5 uv(x) = 6( 5x x – 6 ) ( 5x x – 6 ) – 5 = 30x x – 6 5x – 5x + 30 x – 6 = 30x 30 = x vu(x) = 5( 6x x – 5 ) ( 6x x – 5 ) – 6 = 30x x – 5 6x – 6x + 30 x – 5 = 30x 30 = x Kesimpulan: v(x) ialah fungsi songsang bagi u(x) dan sebaliknya. Conclusion: v(x) is the inverse function of u(x) and vice versa. (c) g(x) = 2x – 5 ⇒ g–1(x) = x + 5 2 fg[g–1(x)] = 6[g–1(x)] – 17 f(x) = 6( x + 5 2 ) – 17 = 3(x + 5) – 17 = 3x + 15 – 17 = 3x – 2 f(x) = 3x – 2 Zon KBAT 1 (a) Katakan/Let y = 4x – 3 x = y + 3 4 ∴ g–1(x) = x + 3 4 Katakan/Let y = 2x – 4 x = y + 4 2 ∴ h–1(x) = x + 4 2 (b) gh(x) = g(2x – 4) = 4(2x – 4) – 3 = 8x – 19 Katakan/Let y = 8x – 19 x = y + 19 8 ∴ (gh) –1(x) = x + 19 8 h–1g–1(x) = h–1( x + 3 4 ) = ( x + 3 4 ) + 4 2 = x + 3 + 16 4 2 = x + 19 8 Maka/Thus, (gh)–1(x) = h–1g–1(x) PT SPM Add Math Tkt 4-Jawapan.indd 2 07/02/2025 11:24 AM PENERBIT ILMU BAKTI SDN. BHD.
Praktis Topikal SPM Matematik Tambahan Tingkatan 4 – Jawapan J3 (c) –3 –3 2 5 O 2 5 (2,5) (5,2) y = x y = g–1(x) = x + 3 4 y = g(x) = 4x – 3 x y Graf y = g–1(x) ialah pantulan graf g(x) pada garis lurus y = x. Graph of y = g–1(x) is the reflection of the graph y = g(x) in the straight line y = x. Bab 2 Fungsi Kuadratik Kertas 1 Bahagian A 1 6x² – 4y – 30 = 0 y = 6x2 – 30 4 y < 6 6x2 – 30 4 < 6 6x² – 30 – 24 < 0 6x² < 54 x² < 9 x² – 9 < 0 (x + 3)(x – 3) < 0 x –3 3 Daripada graf/From the graph, –3 < x < 3. 2 (a) 2x² + ux – 12 = 0 x² + u 2 x – 6 = 0 HTP/SOR: 3α + 3β = – u 2 3(α + β) = – u 2 3(– 5 6) = – u 2 u = 5 (b) HDP/POR: 3α × 3β = –6 9αβ = –6 αβ = – 2 3 Persamaan kuadratik dengan puncapunca α dan β ialah Quadratic equation with the roots α and β is x² – (HTP/SOR)x + (HDP/POR) = 0 x² – (– 5 6)x + (– 2 3) = 0 x² + 5 6 x – 2 3 = 0 6x² + 5x – 4 = 0 3 3x² – 8x = 4x + 5 3x² – 12x – 5 = 0 x² – 4x – 5 3 = 0 HTP/SOR: α + β = – (–4) = 4 HDP/POR: αβ = – 5 3 (a) HTP/SOR = 1 α – 1 + 1 β – 1 = (β – 1) + (α – 1) (α – 1)(β – 1) = α + β – 2 αβ – (α + β) + 1 = (4) – 2 (– 5 3) – (4) + 1 = – 3 7 HDP/POR: 1 α – 1 × 1 β – 1 = 1 αβ – (α + β) + 1 = 1 (– 5 3) – (4) + 1 = – 3 14 Persamaan kuadratik: Quadratic equation: x² – (– 3 7)x + (– 3 14) = 0 14x² + 6x – 3 = 0 (b) HTP/SOR = β α + α β = β2 + α2 αβ = (α + β) 2 – 2αβ αβ = (4)2 – 2(– 5 3) (– 5 3) = – 58 5 HDP/POR = β α × α β = 1 Persamaan kuadratik: Quadratic equation: x² – (– 58 5 )x + 1 = 0 5x2 + 58x + 5 = 0 4 (a) hx2 + 6x + k = 4 hx2 + 6x + k – 4 = 0 a = h, b = 6, c = k – 4 b2 – 4ac = 0 (6)2 – 4(h)(k – 4) = 0 36 – 4h(k – 4) = 0 9 – h(k – 4) = 0 –h(k – 4) = –9 h = 9 k – 4 (b) x2 + x = 3kx – k2 x2 + x – 3kx + k2 = 0 x2 + (1 – 3k)x + k2 = 0 a = 1, b = 1 – 3k, c = k2 b2 – 4ac > 0 (1 – 3k) 2 – 4(1)(k2 ) > 0 1 – 6k + 9k2 – 4k2 > 0 1 – 6k + 5k2 > 0 (5k – 1)(k – 1) > 0 k 1 1 5 Daripada graf/From the graph, k < 1 5 atau/or k > 1 5 (a) 3x2 – 6px + p = 0 a = 3, b = –6p, c = p b2 – 4ac = 0 (–6p)2 – 4(3)(p) = 0 36p2 – 12p = 0 p(36p – 12) = 0 p = 0 atau/or 36p – 12 = 0 p = 12 36 = 1 3 (b) x2 – kx + 4 – 3x = 0 x2 – (k + 3)x + 4 = 0 a = 1, b = –(k + 3), c = 4 b2 – 4ac = 0 [–(k + 3)]2 – 4(1)(4) = 0 (k + 3)2 – 16 = 0 (k + 3 + 4)(k + 3 – 4) = 0 (k + 7)(k – 1) = 0 k = –7 atau/or k = 1 6 (a) f(x) = 3(x + 2)(x – 5) = 3[x² – 3x – 10] = 3[x² – 3x + (– 3 2) ² – (– 3 2) ² – 10] = 3[(x – 3 2) ² – 49 4 ] = 3(x – 3 2) ² – 147 4 ∴ h = – 3 2 , k = – 147 4 (b) (i) Titik pusingan/Turning point = ( 3 2 , – 147 4 ) (ii) x – 3 2 = 0 x = 3 2 Bahagian B 7 (a) 5x² – 2x + p = px – 2 5x² – 2x + p – px + 2 = 0 5x² – 2x – px + p + 2 = 0 5x² – (2 + p)x + p + 2 = 0 a = 5, b = –(2 + p), c = p + 2 b2 – 4ac = 0 [–(2 + p)]² – 4(5)(p + 2) = 0 4 + 4p + p² – 20p – 40 = 0 p² – 16p – 36 = 0 (p + 2)(p – 18) = 0 p + 2 = 0 , p – 18 = 0 p = –2 p = 18 (b) (i) f(x) = a(x + b)² + c Titik minimum/Minimum point (3, –5) ⇒ b = –3, c = –5 f(x) = a(x – 3)² – 5 Pada titik/At point (0, –1.5), –1.5 = a[(0) – 3]² – 5 9a = 3.5 a = 7 18 Maka/Thus, a = 7 18 , b = –3, c = –5 (ii) f(x) = 7 18 (x – 3)² – 5 ∴ g(x) = – 7 18 (x – 3)² + 5 Titik pusingan = (3, 5) Turning point = (3, 5) PT SPM Add Math Tkt 4-Jawapan.indd 3 07/02/2025 11:24 AM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2025 J4 8 (a) f(x) = ax² + nx Pada titik/At point (16, 0), 0 = a(16)² + n(16) n = –16a 1 Pada titik/At point (2, 21), 21 = a(2)² + n(2) 4a + 2n = 21 2 Gantikan 1 ke dalam 2 , Substitute 1 into 2 , 4a + 2(–16a) = 21 –28a = 21 a = – 3 4 Daripada/From 1 , n = – 16(– 3 4) = 12 ∴ a = – 3 4 , n = 12 (b) f(x) = – 3 4 x2 + 12x = – 3 4 (x² – 16x) = – 3 4[x² – 16x + (– 16 2 ) ² – (– 16 2 ) ² ] = – 3 4 [(x – 8)² – 64] = – 3 4 (x – 8)² + 48 Tinggi maksimum/Maximum height = 48 unit/units (c) Nilai tengah julat x: Middle value of the range of x: 0 + 16 2 = 8 Apabila/When x = 8, f(8) = – 3 4 (8)² + 12(8) = 48 ∴ Tinggi maksimum = 48 unit Maximum height = 48 units 9 (a) Pada titik/At point (0, –12), –12 = (0 – p)² + q p² + q = –12 1 Pada titik/At point (5, –7), –7 = (5 – p)² + q –7 = 25 – 10p + p² + q 2 Gantikan 1 ke dalam 2 , Substitute 1 into 2 , –7 = 25 – 10p + (–12) 10p = 20 p = 2 Daripada/From 1 , (2)² + q = –12 q = –16 (b) f(x) = (x – 2)² – 16 Titik minimum = (2, –16) Minimum point = (2, –16) Persamaan paksi simetri, x = 2 Equation of the axis of symmetry, x = 2 Apabila/When f(x) = 0, (x – 2)² – 16 = 0 x2 – 4x + 4 – 16 = 0 x2 – 4x – 12 = 0 (x + 2)(x – 6) = 0 x = –2 atau/or x = 6 f(x) f(x) = (x – 2)2 – 16 –2 0 6 x (2, –16) (0, –12) (c) f(x) = (x – 2)² – 16 h(x) = (x + 2)² – 16 Titik pusingan/Turning point = (–2, –16) Kertas 2 Bahagian A 1 (a) f(x) = –5x² – 5x + 3 = –5(x² + x – 3 5) = –5[x² + x + ( 1 2) ² – ( 1 2) ² – 3 5] = –5[(x + 1 2) ² – 17 20 ] = –5(x + 1 2) ² + 17 4 ∴ m = 1 2 , n = 17 4 (b) Titik maksimum/Maximum point = (– 1 2 , 17 4 ) Persamaan paksi simetri x = – 1 2 Equation of the axis of symmetry, x = – 1 2 Apabila/When x = 0, f(0) = –5(0)² – 5(0) + 3 = 3 f(x) x (– 1 2 , 17 4 ) (0, 3) O f(x) = –5x2 – 5x + 3 (c) px + 8 = –5x² – 5x + 3 5x² + px + 5x + 8 – 3 = 0 5x² + (p + 5)x + 5 = 0 a = 5, b = p + 5, c = 5 b² – 4ac = 0 (p + 5)² – 4(5)(5) = 0 p² + 10p + 25 – 100 = 0 p² + 10p – 75 = 0 (p + 15)(p – 5) = 0 p = –15 atau/or p = 5 p > 0, maka/thus p = 5 ⇒ y = 5x + 8 Selesaikan/Solve y = 5x + 8 dan/and y = –5x² – 5x + 3. 5x + 8 = –5x² – 5x + 3 5x² + 10x + 5 = 0 x² + 2x + 1 = 0 (x + 1)² = 0 x = –1 Apabila/When x = –1, y = 5(–1) + 8 = 3 Maka, titik persilangan = (–1, 3). Thus, the point of intersection = (–1, 3). 2 (a) f(x) = 4(x + 5)(2 – x) = 4(–x² – 3x + 10) = –4(x² + 3x – 10) = –4[x² + 3x + ( 3 2) ² – ( 3 2) ² – 10] = –4[(x + 3 2) ² – 49 4 ] = –4(x + 3 2) ² + 49 ∴ a = –4, p = 3 2 , q = 49 Nilai maksimum bagi f(x) ialah 49. The maximum value of f(x) is 49. (b) f(x) g(x) 4(x + 5)(2 – x) 4x + 20 4(x + 5)(2 – x) 4(x + 5) 4(x + 5)(2 – x) – 4(x + 5) 0 (x + 5)(2 – x) – (x + 5) 0 (–x² – 3x + 10) – x – 5 0 –x² – 4x + 5 0 x² + 4x – 5 0 (x + 5)(x – 1) 0 x –5 1 Daripada graf/From the graph, –5 x 1 3 (a) (i) –3 < k < 0 (ii) w(x) = –3(x – 2 + 6)² – 1 w(x) = –3(x + 4)² – 1 (b) Gantikan/Substitute y = x + k ke dalam/into x² + y² – 2x – 2y = 0. x² + (x + k)² – 2x – 2(x + k) = 0 x² + (x² + 2kx + k²) – 2x – 2x – 2k = 0 2x² + 2kx – 4x + k² – 2k = 0 2x² + (2k – 4)x + k² – 2k = 0 a = 2, b = 2k – 4, c = k2 – 2k b² – 4ac < 0 (2k – 4)² – 4(2)(k² – 2k) < 0 4k² – 16k + 16 – 8k² + 16k < 0 –4k² + 16 < 0 4k² – 16 > 0 k² – 4 > 0 (k – 2)(k + 2) > 0 k –2 2 Daripada graf/From the graph, k < –2 atau/or k > 2 Bahagian B 4 (a) f(x) = hx – 3x² – (h – 11) f(x) = –3x² + hx – (h – 11) = –3[x² – h 3 x + h – 11 3 ] = –3[x² – h 3 x + (– h 6 ) ² – (– h 6 ) ² + h – 11 3 ] = –3[(x – h 6 ) ² – h2 36 + h – 11 3 ] = –3(x – h 6 ) ² + h2 12 – (h – 11) = –3(x – h 6 ) ² + h2 12 – h + 11 (b) Nilai maksimum/Maximum value = 8 h2 12 – h + 11 = 8 h2 12 – h + 3 = 0 h² – 12h + 36 = 0 (h – 6)(h – 6) = 0 h = 6 f(x) = (6)x – 3x² – [(6) – 11] = –3x² + 6x + 5 Apabila f(x) menyilang paksi-x, f(x) = 0 When f(x) intersects the x-axis, f(x) = 0 b2 – 4ac = 62 – 4(–3)(5) = 96 (>0) Maka, graf fungsi f(x) menyilang paksi-x pada dua titik yang berbeza. Thus, the graph of function f(x) intersects the x-axis at two different points. PT SPM Add Math Tkt 4-Jawapan.indd 4 07/02/2025 11:24 AM PENERBIT ILMU BAKTI SDN. BHD.