49 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 BT ms.178–180 (a) m : n, (b) nilai u / the value of u. [5 markah/marks] Jawapan/Answer: (a) (b) 7.1 Pembahagi Tembereng Garis / Divisor of a Line Segment 7.3 Luas Poligon / Areas of Polygons 2 Rajah 1 menunjukkan garis lurus BPC dengan keadaan BP : PC = 3 : 2. Diagram 1 shows a straight line BPC such that BP : PC = 3 : 2. x y 9 C A O P(3, 4) B(–3, 1) Rajah 1/ Diagram 1 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] Bidang Pembelajaran: Geometri Geometri Koordinat Coordinate Geometry BAB 7 PT SPM Add Math Tkt 4-Bab 7-8.indd 49 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
50 Jawapan/Answer: (a) (b) 7.2 Garis Lurus Selari dan Garis Lurus Serenjang / Parallel Lines and Perpendicular Lines 3 Diberi dua titik G(10, 0) dan H(0, −4). Pembahagi dua sama serenjang bagi GH menyilang di titik Q dan R. Cari Given two points G(10, 0) and H(0, −4). e perpendicular bisector of GH intersects at 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) 4 Rajah 2 menunjukkan tiga garis lurus, PQ, QS dan SR, dengan keadaan e, f, g dan h ialah pemalar. Diagram 2 shows three straight lines, PQ, QS and SR, where e, f, g and h are constants. x y O h P Q ey = 2x 8 + = 1 S R y g x f Rajah 2/ Diagram 2 PT SPM Add Math Tkt 4-Bab 7-8.indd 50 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
51 Ungkapkan / Express BT ms.186–187 (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) 7.4 Persamaan Lokus / Equations of Loci 5 Persamaan lokus bagi titik bergerak L diberi oleh x2 + y2 + 3x – 3y – 8 = 0. Tunjukkan bahawa e equation of locus of a moving point L is given by x2 + y2 + 3x – 3y – 8 = 0. Show that BT ms.203–204 (a) lokus L bersilang dengan 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 28/02/2023 5:39 PM 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 6 (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 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 pada persimpangan jalan lurus y = −4x + 12 dan jalan lurus berserenjang yang melalui titik P(6, 5) seperti ditunjukkan dalam Rajah 4. A tra c 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. BT ms.186–187 Find the position of the tra c 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 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
53 Bahagian A / Section A 1 Rajah 1 menunjukkan dua garis lurus, PQ dan QR yang berserenjang antara satu sama lain. 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 persamaan garis lurus PQ ialah 2x + y = 9. Given 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. e straight line QR is extended to point S such that SR : RQ = 3 : 2. Find the coordinates of S. BT ms.178–180 [2 markah/marks] Jawapan/Answer: (a) (i) (ii) (iii) (b) PT SPM Add Math Tkt 4-Bab 7-8.indd 53 28/02/2023 5:39 PM 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. Diagram 2 shows a rectangle EFGH. e equation of the straight line EF is y = 2x + 7. Cari / Find BT ms.197–198 (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) 3 Rajah 3 menunjukkan sebuah segi tiga EFG pada suatu satah Cartes. 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 e 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. BT ms.203–204 Given the point L(x, y) moves such that ∠GLH = 90°. Find the equation of locus of point L. [3 markah/marks] y x E(–2, 3) G(8, 13) y = 2x + 7 H F O Rajah 2/ Diagram 2 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 54 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
55 Jawapan/Answer: (a) (i) (ii) (b) (c) 4 Rajah 4 menunjukkan sebuah sisi empat EFGH. Diagram 4 shows a quadrilateral EFGH. Diberi persamaan garis lurus EH dan HG masing-masing ialah 2y = 3x + 32 dan y = –5. Given 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. BT ms.200–201 Point P moves such that its distance from H is always 6 units. Find the equation of locus of point P. [3 markah/marks] y O G H E F(3, 1) x –3 Rajah 4/ Diagram 4 PT SPM Add Math Tkt 4-Bab 7-8.indd 55 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
56 Jawapan/Answer: (a) (i) (ii) (iii) (b) ZON KBAT 1 Tentukan titik-titik persilangan antara garis lurus y = 2x – 3 dengan sebuah bulatan yang berpusat di (1, 1) dan berjejari 5 unit. BT ms.203–204 KBAT Mengaplikasi Determine the points of intersection between the straight line y = 2x – 3 and a circle with centre (1, 1) and radius 5 units. Jawapan/Answer: PT SPM Add Math Tkt 4-Bab 7-8.indd 56 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
57 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 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) 8.2 Penambahan dan Penolakan Vektor / Addition and Subtraction of Vectors 2 Rajah 1 menunjukkan sebuah segi tiga EFH, dengan keadaan FG : GH = 2 : 3. Diagram 1 shows a triangle EFH, where FG : GH = 2 : 3. E H F G Rajah 1/ Diagram 1 Diberi bahawa EF → = 6h ~ dan EH → = 5k ~. Cari BT ms.222–223 It is given that EF→ = 6h~ and EH→ = 5k~. Find (a) FH → , (b) EG → . [4 markah/marks] Bidang Pembelajaran: Geometri Vektor Vectors BAB 8 PT SPM Add Math Tkt 4-Bab 7-8.indd 57 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
58 Jawapan/Answer: (a) (b) 8.3 Vektor dalam Satah Cartes / Vectors in a Cartesian Plane 3 Diberi bahawa u ~ = (m + 1 –9 ) dan w~ = ( m 1 ) , dengan keadaan m ialah pemalar. BT ms.232 It is given that u~ = ( m + 1 –9 ) and w~ = ( m 1 ), where m is a constant. (a) Ungkapkan vektor u ~ + w~ dalam sebutan m. Express the vector u~ + w~ in terms of m. (b) Diberi |u ~ + w~ | = 10 unit, cari nilai-nilai m. Given |u ~ + w~ | = 10 units, nd the values of m. [5 markah/marks] Jawapan/Answer: (a) (b) 8.1 Vektor / Vectors 8.3 Vektor dalam Satah Cartes / Vectors in a Cartesian Plane 4 (a) 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 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] (b) 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 It is given that O→ M = 7i~ + 3j~ and M → L = −3i~ + 4j~ . Find M → K. [3 markah/marks] M L K y x O Rajah 2/ Diagram 2 y L K M x O PT SPM Add Math Tkt 4-Bab 7-8.indd 58 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
59 Jawapan/Answer: (a) (b) 8.3 Vektor dalam Satah Cartes / Vectors in a Cartesian Plane 5 Dalam Rajah 3, OP → = −6i ~ + 2j ~ dan O →Q = 3i ~ + 7j ~ . In Diagram 3, OP→ = −6i~ + 2j~ and OQ→ = 3i~ + 7j~ . Tentukan / Determine (a) PQ → , BT ms.228 (b) vektor unit dalam arah PQ → . BT ms.230–231 the unit vector in the direction of PQ→ . [4 markah/marks] Jawapan/Answer: (a) (b) 6 Diberi bahawa p ~ = 9i ~ + j ~ dan q ~ = 6i ~ – nj~ , dengan keadaan n ialah pemalar. Cari BT ms.228–229 It is given that p~ = 9i~ + j ~ and q ~ = 6i~ – nj~ , where n is a constant. Find (a) p ~ – q ~ dalam bentuk xi ~ + yj ~ , / p ~ – q~ in the form of xi~ + yj~ , (b) nilai n jika |p ~ – q ~ | = √34 . / the value of n if |p~ – q~ | = √ 34 . [5 markah/marks] Rajah 3/ Diagram 3 y Q x P O PT SPM Add Math Tkt 4-Bab 7-8.indd 59 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
60 Jawapan/Answer: (a) (b) Bahagian B / Section B 8.2 Penambahan dan Penolakan Vektor / Addition and Subtraction of Vectors 8.3 Vektor dalam Satah Cartes / Vectors in a Cartesian Plane 7 (a) Dalam Rajah 4, PQS ialah sebuah segi tiga dengan keadaan PQ → = −3a ~ + 5b ~ dan PS → = 4a ~ + 9b ~. BT ms.221–223 In Diagram 4, PQS is a triangle such that PQ→ = −3a~ + 5b~ and PS → = 4a~ + 9b~. Diberi R terletak pada QS dengan keadaan QR : RS = 1 : 3, cari PR →. Given R lies on QS such that QR : RS = 1 : 3, nd PR →. [4 markah/marks] (b) Diberi vektor OA → = ( 4 p), OB→ = ( p 2 ) dan OC → = ( q 8 ) ialah tiga vektor pada suatu satah Cartes. Cari nilai p dan nilai q dengan keadaan AB → − BC→ = ( 6 q). BT ms.232–233 Given vector OA→ = ( 4 p ), OB→ = ( p 2 ) and OC → = ( q 8 ) are three vectors on a Cartesian plane. Find the value of p and of q such that AB→ − BC → = ( 6 q ). [4 markah/marks] Jawapan/Answer: (a) (b) S P Q R –3a + 5b ~~ 4a + 9b ~~ Rajah 4/ Diagram 4 PT SPM Add Math Tkt 4-Bab 7-8.indd 60 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
61 Bahagian A / Section A 1 Rajah 1 menunjukkan sebuah trapezium EFGH. Diberi bahawa EF → = 4d ~, EH → = 12e ~, EP→ = 2 3 EH → dan FG → = 3 4 EH →. BT ms.224–225 Diagram 1 shows a trapezium EFGH. It is given that EF→ = 4d~, EH→ = 12e~, EP → = 2 3 EH → and FG→ = 3 4 EH → . (a) Ungkapkan EG → dalam sebutan d ~ dan e ~. Express EG→ in terms of d~ and e~. [2 markah/marks] (b) Titik Q terletak di dalam trapezium EFGH dengan keadaan PQ → = (h − 1)EF → dan h ialah pemalar. Point Q lies inside the trapezium EFGH such that PQ→ = (h − 1)EF→ and h is a constant. (i) Ungkapkan EQ → dalam sebutan h, d ~ dan e ~. Express EQ→ in terms of h, d~ and e~. (ii) Jika titik-titik E, Q dan G adalah segaris, cari nilai h. If the points E, Q and G are collinear, nd the value of h. [6 markah/marks] Jawapan/Answer: (a) (b) (i) (ii) 2 Dalam Rajah 2, OP → = 15x ~ dan OQ→ = 10y ~ . Titik R terletak pada garis lurus OP dengan keadaan OR : RP = 3 : 2. Titik S terletak pada garis lurus OQ dengan keadaan OS : SQ = 2 : 3. Garis lurus PS dan QR bersilang pada titik T dengan keadaan PT → = αPS → dan QT → = βQR → , dengan keadaan α dan β ialah pemalar. BT ms.224–225 In Diagram 2, OP→ = 15x~ and OQ→ = 10y ~ . Point R lies on the straight line OP such that OR : RP = 3 : 2. Point S lies on the straight line OQ such that OS : SQ = 2 : 3. e straight lines PS and QR intersect at point T such that PT → = αPS → and QT→ = βQR →, where α and β are constants. (a) Ungkapkan OT → / Express OT→ (i) dalam sebutan α, x ~ dan y ~ , / in terms of α, x ~ and y~ , (ii) dalam sebutan β, x ~ dan y ~ . / in terms of β, x ~ and y~ . [4 markah/marks] (b) Cari nilai α dan nilai β. / Find the value of α and of β. [3 markah/marks] F G Q E H P Rajah 1/ Diagram 1 Rajah 2/ Diagram 2 O R P Q T S PT SPM Add Math Tkt 4-Bab 7-8.indd 61 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
62 Jawapan/Answer: (a) (i) (ii) (b) Bahagian B / Section B 3 Dalam Rajah 3, DEFH ialah sebuah sisi empat. DHG dan HJF ialah garis lurus. Diberi bahawa DE→ = 40m~, D→H = 16n ~, GF→ = 50m~ – 48n ~, DH → = 1 4 DG → dan HJ → = 3 5 HF → . BT ms.234 In Diagram 3, DEFH is a quadrilateral. DHG and HJF are straight lines. It is given that DE→ = 40m~, DH→ = 16n~, GF → = 50m~ – 48n~, DH→ = 1 4 DG → and HJ → = 3 5 HF →. (a) Ungkapkan dalam sebutan m~ dan n ~, Express in terms of m~ and n~, (i) EG →, (ii) HF →. [4 markah/marks] (b) Tunjukkan bahawa titik-titik E, J dan G adalah segaris. Show that the points E, J and G are collinear. [4 markah/marks] (c) Jika |m| = 4 dan |n| = 3, cari |EG →|. If |m| = 4 and |n| = 3, nd |EG →|. [2 markah/marks] Jawapan/Answer: (a) (i) (ii) (b) (c) G D E H F J Rajah 3/ Diagram 3 PT SPM Add Math Tkt 4-Bab 7-8.indd 62 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
63 4 Dalam Rajah 4, ABCD ialah sebuah sisi empat dengan keadaan garis lurus BD menyilang garis lurus CF di T. In Diagram 4, ABCD is a quadrilateral such that the straight line BD intersects the straight line CF at T. A B D C F T 12q ~ Rajah 4/ Diagram 4 Diberi bahawa DF → = 1 3 D →A, A →B = 1 2 F →C, D →A = 12p ~ , D→C = 12q ~ , D→T = h D→B dan F →T = kF→C. BT ms.224–225 It is given that DF→ = 1 3 D →A, A→B = 1 2 F →C, D→A = 12p~ , D→C = 12q~ , D→T = h D→B and F→T = kF→C. (a) Ungkapkan D →T / Express D→T (i) dalam sebutan h, p ~ dan q ~ , / in terms of h, p~ and q~ , (ii) dalam sebutan k, p ~ dan q ~ , / in terms of k, p~ and q~ , [4 markah/marks] (b) Cari nilai h dan nilai k. / Find the value of h and of k. [4 markah/marks] (c) Seterusnya, tentukan FT : AB. / Hence, determine FT : AB. [2 markah/marks] Jawapan/Answer: (a) (i) (ii) (b) (c) PT SPM Add Math Tkt 4-Bab 7-8.indd 63 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
64 5 Dalam Rajah 5, D, E, F, G, H dan J ialah enam buah khemah yang didirikan pada suatu tanah mengufuk. D, E dan F adalah segaris dengan keadaan DF = 4DE. Khemah G terletak pada garis JE dengan keadaan JE = 3GE. Khemah H terletak pada garis JF dengan keadaan JF = 3JH. In Diagram 5, D, E, F, G, H and J are six tents built on a horizontal ground. D, E and F are collinear such that DF = 4DE. Tent G is situated on the line JE such that JE = 3GE. Tent H is situated on the line JF such that JF = 3JH. G ED F H J Rajah 5/ Diagram 5 Diberi bahawa D →E = 12p ~ dan D →J = 36q ~ . BT ms.234 It is given that D→E = 12p ~ and D→J = 36q ~ . (a) Ungkapkan D →G dalam sebutan p ~ dan q ~ . Express D→G in terms of p~ and q~ . [4 markah/marks] (b) Tentukan sama ada kedudukan khemah D, G dan H adalah segaris. Determine whether the positions of tents D, G and H are collinear. [4 markah/marks] (c) Diberi | p ~ | = 2 m dan | q ~ | = 3 m, tentukan jarak, dalam m, di antara khemah D dengan khemah G. Given | p~ | = 2 m and | q~ | = 3 m, determine the distance, in m, between tent D and tent G. [2 markah/marks] Jawapan/Answer: (a) (b) (c) PT SPM Add Math Tkt 4-Bab 7-8.indd 64 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
65 ZON KBAT 1 Rajah 1 menunjukkan A, B, C dan D ialah empat buah menara komunikasi. B, C dan D adalah segaris dengan keadaan BC : CD = m : n. Diberi bahawa AB → = 4i ~ + 8j ~ , AC→ = 10i ~ + 5j ~ dan AD → = 12i ~ + hj~ , dengan keadaan h ialah pemalar. Diagram 1 shows A, B, C and D are four communication towers. B, C and D are collinears such that BC : CD = m : n. It is given that AB→ = 4i~ + 8j~ , AC → = 10i~ + 5j~ and AD→ = 12i~ + hj~ , where h is a constant. B A D C Rajah 1/ Diagram 1 Tentukan BT ms.234 KBAT Menganalisis Determine (a) nisbah m : n, the ratio of m : n, (b) nilai h. the value of h. Jawapan/Answer: (a) (b) PT SPM Add Math Tkt 4-Bab 7-8.indd 65 28/02/2023 5:39 PM PENERBIT ILMU BAKTI SDN. BHD.
66 Bahagian C / Section C 1 Rajah 1 menunjukkan sebuah segi tiga DEF. Diagram 1 shows a triangle DEF. (a) Hitung panjang, dalam cm, bagi DF. BT ms.252–253 Calculate the length, in cm, of DF. [2 markah/marks] (b) Sebuah sisi empat DEFG dibentuk dengan keadaan DF ialah pepenjuru, ∠DFG = 43° dan DG = 26 cm. Hitung dua nilai yang mungkin bagi ∠DGF. A quadrilateral DEFG is formed such that DF is a diagonal, ∠DFG = 43° and DG = 26 cm. Calculate two possible values of ∠DGF. BT ms.248 [2 markah/marks] (c) Dengan mempertimbangkan kes berambiguiti di (b), hitung BT ms.248 By considering the ambiguous case in (b), calculate (i) panjang, dalam cm, bagi FG, the length, in cm, of FG, (ii) luas, dalam cm2 , bagi sisi empat DEFG. the area, in cm2 , of quadrilateral DEFG. [6 markah/marks] Jawapan/Answer: (a) (b) (c) (i) (ii) D E F 24 cm 28 cm 68º Rajah 1/ Diagram 1 Pakej Elektif: Aplikasi Sains dan Teknologi Penyelesaian Segi Tiga Solution of Triangles BAB 9 PT SPM Add Math Tkt 4-Bab 9-10.indd 66 27/02/2023 12:29 PM PENERBIT ILMU BAKTI SDN. BHD.
67 2 Rajah 2 menunjukkan sebuah sisi empat EFGH. Diagram 2 shows a quadrilateral EFGH. E H F G 13 cm 9 cm 7 cm 50º Rajah 2/ Diagram 2 Diberi bahawa luas segi tiga EFG ialah 29.5 cm2 dan ∠EFG ialah sudut tirus. Hitung It is given that the area of triangle EFG is 29.5 cm2 and ∠EFG is an acute angle. Calculate (a) ∠EFG, BT ms.258 [2 markah/marks] (b) panjang, dalam cm, bagi EG, BT ms.252–253 the length, in cm, of EG, [2 markah/marks] (c) ∠GEH, BT ms.244–245 [3 markah/marks] (d) luas, dalam cm2 , bagi sisi empat EFGH. BT ms.258 the area, in cm2 , of quadrilateral EFGH. [3 markah/marks] Jawapan/Answer: (a) (b) (c) (d) PT SPM Add Math Tkt 4-Bab 9-10.indd 67 27/02/2023 12:29 PM PENERBIT ILMU BAKTI SDN. BHD.
68 3 Rajah 3 menunjukkan sebuah sisi empat PQRS. Diagram 3 shows a quadrilateral PQRS. 9 cm 5 cm 19.5 cm 48º Q R P 116º S Rajah 3/ Diagram 3 (a) Hitung / Calculate (i) panjang, dalam cm, bagi PR, BT ms.252–253 the length, in cm, of PR, (ii) ∠PRS. BT ms.244–245 [4 markah/marks] (b) Titik P’ terletak pada PR dengan keadaan P’S = PS. Point P’ lies on PR such that P’S = PS. (i) Lakar ΔP’SR. / Sketch ∆P’SR. BT ms.248 (ii) Hitung luas, dalam cm2 , bagi ΔP’SR. BT ms.258 Calculate the area, in cm2 , of ∆P’SR. [6 markah/marks] Jawapan/Answer: (a) (i) (ii) (b) (i) (ii) 4 (a) Rajah 4 menunjukkan dua buah segi tiga, EFG dan EGH. Diagram 4 shows two triangles, EFG and EGH. Diberi bahawa FGH ialah garis lurus dan ∠EGF ialah sudut tirus. Hitung BT ms.244–245 It is given that FGH is a straight line and ∠EGF is an acute angle. Calculate (i) ∠EGH, (ii) panjang, dalam cm, bagi EH. the length, in cm, of EH. [4 markah/marks] 12 cm 8 cm 54º F G H E 17º Rajah 4/ Diagram 4 PT SPM Add Math Tkt 4-Bab 9-10.indd 68 27/02/2023 12:29 PM PENERBIT ILMU BAKTI SDN. BHD.
69 (b) Rajah 5 menunjukkan sebuah kuboid bertapak segi empat tepat ABCD. Diagram 5 shows a cuboid with a rectangular base ABCD. Hitung / Calculate (i) ∠NAC, BT ms.252–253 (ii) luas, dalam cm2 , bagi ∆NAC, BT ms.258 the area, in cm2 , of ∆NAC, (iii) jarak terdekat, dalam cm, dari A ke garis lurus NC. the shortest distance, in cm, from A to the straight line NC. BT ms.258 [6 markah/marks] Jawapan/Answer: (a) (i) (ii) (b) (i) (ii) (iii) ZON KBAT 1 Rajah 1 menunjukkan sebuah sisi empat EFGH. Diagram 1 shows a quadrilateral EFGH. (a) Hitung / Calculate (i) panjang, dalam cm, bagi EG, / the length, in cm, of EG, BT ms.252–253 (ii) ∠EGF. BT ms.244–245 (b) Garis lurus GF dipanjangkan ke titik F' dengan keadaan EF' = EF. e straight line GF is extended to point F' such that EF' = EF. (i) Lakar ΔEF'G. / Sketch ΔEF'G. BT ms.248 (ii) Cari luas, dalam cm2 , bagi ΔEF'G. BT ms.259–260 KBAT Mengaplikasi Find the area, in cm2 , of ΔEF'G. 5 cm 6 cm 13 cm P C Q M B N A D Rajah 5/ Diagram 5 8 cm 17.4 cm 48º H 108º 6.4 cm G F E Rajah 1/ Diagram 1 PT SPM Add Math Tkt 4-Bab 9-10.indd 69 27/02/2023 12:29 PM PENERBIT ILMU BAKTI SDN. BHD.
70 Jawapan/Answer: (a) (i) (ii) (b) (i) (ii) 2 Rajah 2 menunjukkan sebuah segi tiga EFG. Diagram 2 shows a triangle EFG. (a) Hitung sudut cakah EGF. BT ms.244–245 Calculate the obtuse angle EGF. (b) Lakar dan labelkan sebuah lagi segi tiga selain segi tiga EFG, dengan keadaan panjang EF dan EG serta ∠EFG tidak berubah. Sketch and label another triangle other than EFG, such that the lengths of EF and EG and ∠EFG remain unchanged. BT ms.248 KBAT Mengaplikasi Jawapan/Answer: (a) (b) F E G 12.8 cm 10.4 cm 28° Rajah 2/ Diagram 2 PT SPM Add Math Tkt 4-Bab 9-10.indd 70 27/02/2023 12:29 PM PENERBIT ILMU BAKTI SDN. BHD.
71 Bahagian C / Section C 1 Sebiji kek diperbuat daripada lima jenis bahan, A, B, C, D dan E. Jadual 1 menunjukkan harga bahanbahan tersebut pada tahun 2019 dan tahun 2022. BT ms.281–282 A cake is baked by using ve ingredients, A, B, C, D and E. Table 1 shows the price of the ingredients for the year 2019 and 2022. Bahan Ingredient Harga sekilogram (RM) Price per kilogram (RM) Tahun 2019 Year 2019 Tahun 2022 Year 2022 A 6.00 7.20 B 3.00 q C 6.00 7.50 D r s E 2.50 3.25 Jadual 1/ Table 1 (a) Indeks harga bagi bahan B pada tahun 2022 berasaskan tahun 2019 ialah 115. Hitung nilai q. e price index of ingredient B in the year 2022 based on the year 2019 is 115. Calculate the value of q. [2 markah/marks] (b) Diberi bahawa indeks harga bagi bahan D pada tahun 2022 berasaskan tahun 2019 ialah 140. Harga sekilogram bagi bahan D pada tahun 2022 ialah RM1.60 lebih daripada harga sepadannya pada tahun 2019. Hitung nilai r dan nilai s. It is given that the price index of ingredient D in the year 2022 based on the year 2019 is 140. e price per kilogram of ingredient D in the year 2022 is RM1.60 more than its corresponding price in the year 2019. Calculate the value of r and of s. [3 markah/marks] (c) Indeks gubahan bagi kos pembuatan kek pada tahun 2022 berasaskan tahun 2019 ialah 123. Hitung e composite index for the cost of baking the cake in the year 2022 based on the year 2019 is 123. Calculate (i) kos pembuatan kek pada tahun 2019 jika kos sepadannya pada tahun 2022 ialah RM61.50, the cost of baking the cake in the year 2019 if its corresponding cost in the year 2022 is RM61.50, (ii) nilai n jika kuantiti bahan A, B, C, D dan E yang digunakan adalah dalam nisbah 3 : 5 : n : 2 : 1. the value of n if the quantities of ingredients A, B, C, D and E used are in the ratio of 3 : 5 : n : 2 : 1. [5 markah/marks] Pakej Elektif: Aplikasi Sains Sosial Nombor Indeks Index Number BAB 10 PT SPM Add Math Tkt 4-Bab 9-10.indd 71 27/02/2023 12:29 PM PENERBIT ILMU BAKTI SDN. BHD.
72 Jawapan/Answer: (a) (b) (c) (i) (ii) 2 Jadual 2 menunjukkan indeks harga bagi tiga jenama telefon pintar pada tahun 2018 berasaskan tahun 2016. Carta pai dalam Rajah 1 mewakili kuantiti relatif telefon pintar yang dijual di sebuah kedai. Table 2 shows the price indices of three brands of smartphones in the year 2018 based on the year 2016. e pie chart in Diagram 1 represents the relative quantity of the smartphones sold in a shop. BT ms.281–282 Telefon pintar Smartphone Indeks harga pada tahun 2018 berasaskan tahun 2016 Price index in the year 2018 based on the year 2016 P 160 Q 130 R 120 Jadual 2/ Table 2 (a) Jika harga telefon pintar jenama P pada tahun 2018 ialah RM1 800, cari harganya pada tahun 2016. If the price of a smartphone brand P in the year 2018 was RM1 800, nd its price in the year 2016. [2 markah/marks] (b) Hitung indeks gubahan bagi harga telefon pintar pada tahun 2018 berasaskan tahun 2016. Calculate the composite index for the price of the smartphones in the year 2018 based on the year 2016. [3 markah/marks] (c) Jumlah jualan telefon pintar pada tahun 2016 ialah RM540 000. Hitung jumlah jualan pada tahun 2018. e total sales of the smartphones in the year 2016 was RM540 000. Calculate the total sales in the year 2018. [2 markah/marks] (d) Harga telefon pintar jenama P bertambah sebanyak 15%, harga telefon pintar jenama Q bertambah sebanyak 30% dan harga telefon pintar jenama R kekal tidak berubah dari tahun 2018 ke tahun 2023. Hitung indeks gubahan bagi harga telefon pintar pada tahun 2023 berasaskan tahun 2016. e price of the smartphone brand P increases by 15%, the price of smartphone brand Q increases by 30% and the price of smartphone brand R remains unchanged from the year 2018 to the year 2023. Calculate the composite index for the price of the smartphones in the year 2023 based on the year 2016. [3 markah/marks] Rajah 1/ Diagram 1 60° 200° P R Q PT SPM Add Math Tkt 4-Bab 9-10.indd 72 27/02/2023 12:29 PM PENERBIT ILMU BAKTI SDN. BHD.
73 Jawapan/Answer: (a) (b) (c) (d) 3 Jadual 3 menunjukkan harga bagi lima jenis barang yang dijual di sebuah kedai pada tahun 2018 dan 2021 serta indeks harga barang tersebut pada tahun 2021 berasaskan tahun 2018. Carta palang dalam Rajah 2 mewakili jumlah jualan barang-barang itu. BT ms.281–282 Table 3 shows the price of ve items sold in a shop in the year 2018 and 2021 and the price indices in the year 2021 based on the year 2018. e bar chart in Diagram 2 represents the total sales of the items. Barang Item Harga (RM) Price (RM) Indeks harga pada tahun 2021 berasaskan tahun 2018 Price index in the year 2021 based on the year 2018 Tahun 2018 Year 2018 Tahun 2021 Year 2021 Beras Rice 1.80 2.70 150 Tepung gandum Wheat our 1.40 1.75 x Minyak masak Cooking oil y 4.50 180 Gula Sugar 2.00 z 160 Garam Salt 0.50 0.60 120 Jadual 3/ Table 3 PT SPM Add Math Tkt 4-Bab 9-10.indd 73 27/02/2023 12:29 PM PENERBIT ILMU BAKTI SDN. BHD.
74 0 10 20 30 40 50 60 70 80 90 100 Pemberat Weightage Beras Rice Tepung gandum Wheat our Minyak masak Cooking oil Gula Sugar Garam Salt Barang/Item Rajah 2/ Diagram 2 (a) Cari nilai-nilai x, y dan z. Find the values of x, y and z. [3 markah/marks] (b) Hitung indeks gubahan jualan barang pada tahun 2021 berasaskan tahun 2018. Calculate the composite index of the sales of the items in the year 2021 based on the year 2018. [4 markah/marks] (c) Jumlah jualan barang-barang itu pada tahun 2021 ialah RM637 200, hitung jumlah jualan pada tahun 2018. e total sales of the items in the year 2021 is RM637 200, calculate the the total sales in the year 2018. [3 markah/marks] Jawapan/Answer: (a) (b) (c) PT SPM Add Math Tkt 4-Bab 9-10.indd 74 27/02/2023 12:29 PM PENERBIT ILMU BAKTI SDN. BHD.
75 4 Jadual 4 menunjukkan indeks harga bagi tiga bahan, F, G dan H, yang digunakan dalam pembuatan sejenis kasut. BT ms.281–282 Table 4 shows the price indices of three materials, F, G and H, used in the production of a type of shoes. Bahan Material Indeks harga pada tahun 2018 berasaskan tahun 2014 Price index in the year 2018 based on the year 2014 Indeks harga pada tahun 2021 berasaskan tahun 2014 Price index in the year 2021 based on the year 2014 F 125 160 G 130 m H n 200 Jadual 4/ Table 4 (a) Cari indeks harga bagi bahan F pada tahun 2021 berasaskan tahun 2018. Find the price index for material F in the year 2021 based on the year 2018. [2 markah/marks] (b) Harga bagi bahan G pada tahun 2014 dan 2021 masing-masing ialah RM16 dan RM24. Cari e price of material G in the year 2014 and 2021 was RM16 and RM24 respectively. Find (i) nilai m, the value of m, (ii) harga bagi bahan G pada tahun 2018. the price of material G in the year 2018. [3 markah/marks] (c) Indeks gubahan bagi kos pembuatan kasut pada tahun 2018 berasaskan tahun 2014 ialah 123. Harga bagi bahan-bahan F, G dan H adalah dalam nisbah 3 : 2 : 4. Cari nilai n. e composite index for the cost of the production of the shoes in the year 2018 based on the year 2014 was 123. e price of the materials F, G and H are in the ratio 3 : 2 : 4. Find the value of n. [3 markah/marks] (d) Diberi harga kasut pada tahun 2018 ialah RM177. Cari harga sepadan kasut itu pada tahun 2014. Given the price of the shoes in the year 2018 was RM177. Find the corresponding price of the shoes in the year 2014. [2 markah/marks] Jawapan/Answer: (a) (b) (i) (ii) (c) (d) PT SPM Add Math Tkt 4-Bab 9-10.indd 75 27/02/2023 12:29 PM PENERBIT ILMU BAKTI SDN. BHD.
76 5 Jadual 5 menunjukkan perbelanjaan bulanan pada tahun 2019 dan 2021 serta indeks perbelanjaan pada tahun 2021 berasaskan tahun 2019 bagi Encik Hassan. Carta pai dalam Rajah 3 mewakili kuantiti relatif bagi perbelanjaan dalam sebulan. BT ms.281–282 Table 5 shows the monthly expenditure in the year 2019 and 2021 and the expenditure index in the year 2021 based on the year 2019 of Encik Hassan. e pie chart in Diagram 3 represents the relative quantity of the expenditure in a month. Perbelanjaan Expenditure Perbelanjaan bulanan (RM) Monthly expenditure (RM) Indeks perbelanjaan pada tahun 2021 berasaskan tahun 2019 Expenditure index in the year 2021 based on the year 2019 Tahun 2019 Year 2019 Tahun 2021 Year 2021 Sewa Rental 750 1 500 200 Pengangkutan Transport p 175 140 Makanan Food 320 576 180 Hiburan Entertainment 180 q 165 Bil utiliti Utility bills 80 140 r Jadual 5/ Table 5 Bil utiliti Utility bills Hiburan Entertainment Pengangkutan Transport Makanan Food Sewa Rental 40° 60° 150° Rajah 3/ Diagram 3 (a) Cari nilai-nilai p, q dan r. Find the values of p, q and r. [3 markah/marks] (b) Hitung indeks gubahan perbelanjaan pada tahun 2021 berasaskan tahun 2019. Calculate the composite index of the expenditure in the year 2021 based on the year 2019. [4 markah/marks] (c) Jumlah perbelanjaan pada tahun 2021 ialah RM3 400, hitung jumlah perbelanjaan pada tahun 2019. e total expenditure in the year 2021 was RM3 400, calculate the total expenditure in the year 2019. [3 markah/marks] PT SPM Add Math Tkt 4-Bab 9-10.indd 76 27/02/2023 12:29 PM PENERBIT ILMU BAKTI SDN. BHD.
77 Jawapan/Answer: (a) (b) (c) ZON KBAT 1 Diberi indeks gubahan bagi kos pembuatan sebiji kek pada tahun 2019 berasaskan tahun 2017 ialah 108 dan indeks gubahan bagi kos pembuatan kek itu pada tahun 2023 berasaskan tahun 2019 ialah 115. Hitung indeks gubahan bagi kos pembuatan kek pada tahun 2023 berasaskan tahun 2017. KBAT Menilai Given the composite index for the cost of baking a cake in the year 2019 based on the year 2017 was 108 and the composite index for the cost of baking the cake in the year 2023 based on the year 2019 is 115. Calculate the composite index for the cost of baking the cake in the year 2023 based on the year 2017. BT ms.280 Jawapan/Answer: PT SPM Add Math Tkt 4-Bab 9-10.indd 77 27/02/2023 12:29 PM PENERBIT ILMU BAKTI SDN. BHD.
78 [80 markah/marks] Masa/Time: 2 jam/2 hours Bahagian A / Section A [64 markah/marks] Jawab semua soalan. Answer all questions. 1 (a) Rajah 1 menunjukkan hubungan pasangan bertertib. Diagram 1 shows an ordered pairs relation. {(p, 2), (q, 4), (r, 4), (s, 6)} Rajah 1/ Diagram 1 Nyatakan objek bagi 4. State the object of 4. [2 markah/marks] (b) Diberi h : x → 2x + 10, cari nilai k dengan keadaan h–1(k) = 8. Given h : x → 2x + 10, fi nd the value of k such that h–1(k) = 8. [3 markah/marks] Jawapan/Answer: (a) (b) 2 Diberi fungsi g : x → 2x – 3 dan fg : x → 6x – 13. Cari Given the function g : x → 2x – 3 and fg : x → 6x – 13. Find (a) g−1(x), (b) f(x). [5 markah/marks] Jawapan/Answer: (a) (b) PEPERIKSAAN AKHIR TAHUN PT SPM Add Math Tkt 4-PAT.indd 78 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
79 3 Diberi fungsi f : x → 3x – 4 6 dan g : x → 3x + 8, cari Given the function f : x → 3x – 4 6 and g : x → 3x + 8, fi nd (a) f −1(x), (b) f −1g(x). [4 markah/marks] Jawapan/Answer: (a) (b) 4 (a) Diberi (−1, 12) ialah titik maksimum bagi fungsi kuadratik f(x) = 4h – (2k + x)2 , dengan keadaan h dan k ialah pemalar. Cari nilai h dan nilai k. Given (−1, 12) is the maximum point for a quadratic function f(x) = 4h – (2k + x)2 , where h and k are constants. Find the value of h and of k. [3 markah/marks] (b) Diberi persamaan kuadratik x2 + 3px = 4x – 9 mempunyai dua punca nyata yang sama. Cari nilainilai p. Given a quadratic equation x2 + 3px = 4x – 9 has two equal and real roots. Find the values of p. [3 markah/marks] Jawapan/Answer: (a) (b) 5 (a) Cari julat nilai x dengan keadaan (x − 3)2 ⩾ x + 17. Find the range of values of x such that (x − 3)2 ⩾ x + 17. [3 markah/marks] (b) Diberi hasil tambah punca dan hasil darab punca bagi persamaan kuadratik 4x2 − hx + k = 0 masingmasing ialah 5 2 dan 3 2 . Cari nilai h dan nilai k. Given the sum of roots and product of roots for the quadratic equation 4x2 − hx + k = 0 are 5 2 and 3 2 respectively. Find the value of h and of k. [3 markah/marks] Jawapan/Answer: (a) (b) PT SPM Add Math Tkt 4-PAT.indd 79 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
80 6 Rajah 2 menunjukkan graf bagi fungsi kuadratik f(x) = 25 – (x + w)2 . Diagram 2 shows the graph of a quadratic function f(x) = 25 – (x + w)2 . x –8 O 2 f(x) Rajah 2/ Diagram 2 Cari / Find (a) nilai w, the value of w, (b) koordinat titik maksimum, the coordinates of the maximum point, (c) persamaan paksi simetri. the equation of the axis of symmetry. [4 markah/marks] Jawapan/Answer: (a) (b) (c) 7 (a) Diberi bahawa persamaan kuadratik x(4x + p) = 3x – 1 tidak mempunyai punca nyata. Cari julat nilai p. It is given that the quadratic equation x(4x + p) = 3x − 1 has no real roots. Find the range of values of p. [3 markah/marks] (b) Selesaikan / Solve 3x + 2 = 648 + 3x [3 markah/marks] PT SPM Add Math Tkt 4-PAT.indd 80 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
81 Jawapan/Answer: (a) (b) 8 (a) Diberi log3 m = x dan log3 p = y. Ungkapkan log9 m2 √p dalam sebutan x dan y. Given log3 m = x and log3 p = y. Express log9 m2 √p in terms of x and y. [3 markah/marks] (b) Selesaikan / Solve log4 (x + 6) = 2 + log4 x [3 markah/marks] Jawapan/Answer: (a) (b) 9 (a) Selesaikan persamaan / Solve the equation 34x = 18 + 34x – 1 [3 markah/marks] (b) Diberi 6p, 12 p dan 4q ialah tiga sebutan berturutan bagi suatu janjang geometri. Ungkapkan q dalam sebutan p. Given 6p, 12 p and 4q are three consecutive terms of a geometric progression. Express q in terms of p. [3 markah/marks] Jawapan/Answer: (a) (b) PT SPM Add Math Tkt 4-PAT.indd 81 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
82 10 Diberi hasil tambah n sebutan pertama bagi suatu janjang aritmetik, Sn, diberi oleh Sn = 3n2 + 5n 2 . Cari Given the sum of the fi rst n terms of an arithmetic progression, Sn, is given by Sn = 3n2 + 5n 2 . Find (a) sebutan pertama, the fi rst term, (b) beza sepunya, the common diff erence, (c) sebutan ke-n. the nth term. [6 markah/marks] Jawapan/Answer: (a) (b) (c) 11 Rajah 3 menunjukkan dua garis lurus, PR dan QS. Diberi persamaan bagi QS ialah x 6 + y 4 = 1. Diagram 3 shows two straight lines, PR and QS. Given the equation of QS is x 6 + y 4 = 1. x S P(–4, 2) O y R Q Rajah 3/ Diagram 3 (a) Cari koordinat Q. Find the coordinates of Q. (b) Diberi bahawa PQ = 2QR, cari koordinat R. It is given that PQ = 2QR, fi nd the coordinates of R. [5 markah/marks] Jawapan/Answer: (a) (b) PT SPM Add Math Tkt 4-PAT.indd 82 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
83 12 (a) Diberi titik K ialah (–2, 3) dan titik L ialah (1, 4). Titik W bergerak dengan keadaan KW : WL = 1 : 2. Cari persamaan lokus bagi W. Given point K is (–2, 3) and point L is (1, 4). Point W moves such that KW : WL = 1 : 2. Find the equation of locus of W. [2 markah/marks] (b) Diberi u ~ = ( 4 5 ) dan v ~ = ( 6 h + 3 ). Jika vektor u ~ dan v ~ adalah selari, cari nilai h. Given u ~ = ( 4 5 ) and v~ = ( 6 h + 3 ). If the vectors u ~ and v ~ are parallel, fi nd the value of h. [3 markah/marks] Jawapan/Answer: (a) (b) Bahagian B / Section B [16 markah/marks] Jawab mana-mana dua soalan daripada bahagian ini. Answer any two questions from this section. 13 (a) Diberi bahawa lengkung bagi fungi kuadratik f(x) = x2 + 3x + 5 tidak bersilang dengan garis lurus y = 1 – kx, dengan keadaan k ialah pemalar. Cari julat nilai k. It is given that the curve of a quadratic function f(x) = x2 + 3x + 5 does not intersect with the straight line y = 1 – kx, where k is a constant. Find the range of values of k. [4 markah/marks] (b) Diberi sebutan ke-8 bagi suatu janjang aritmetik ialah 23 dan hasil tambah enam sebutan pertama ialah 3. Cari sebutan pertama bagi janjang tersebut. Given the 8th term of an arithmetic progression is 23 and the sum of the fi rst six terms is 3. Find the fi rst term of the progression. [4 markah/marks] Jawapan/Answer: (a) (b) PT SPM Add Math Tkt 4-PAT.indd 83 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
84 14 (a) Rajah 4 menunjukkan dua titik, E dan F pada satu garis lurus. Diagram 4 shows two points, E and F on a straight line. 1 x E(2, 3) F(6, 5) O 1 y Rajah 4/ Diagram 4 Ungkapkan y dalam sebutan x. Express y in terms of x. [4 markah/marks] (b) Rajah 5 menunjukkan garis lurus PQ pada suatu satah Cartes. Diagram 5 shows a straight line PQ on a Cartesian plane. Q(0.9, 1.2) P 3 log10 x log10 y O Rajah 5/ Diagram 5 Ungkapkan y dalam sebutan x. Express y in terms of x. [4 markah/marks] Jawapan/Answer: (a) (b) PT SPM Add Math Tkt 4-PAT.indd 84 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
85 15 (a) Diberi E(0, −3), F(2, 5), G(6, 3) dan H(7, p) ialah empat bucu bagi sebuah sisi empat EFGH. Jika luas bagi sisi empat itu ialah 7 unit2 , cari nilai-nilai yang mungkin bagi p. Given E(0, −3), F(2, 5), G(6, 3) and H(7, p) are four vertices of a quadrilateral EFGH. If the area of the quadrilateral is 7 units2 , fi nd the possible values of p. [4 markah/marks] (b) Maklumat berikut merujuk vektor a ~, b ~ dan w~. Th e following information refers to the vectors a~, b ~ and w~. a ~ = 4p ~ – 6q ~ b ~ = −3p ~ + 2q ~ w~ = (m – 2)a ~ + (m + n)b ~ dengan keadaan m dan n ialah pemalar. where m and n are constants. Cari nilai m dan nilai n jika w~ = 3p ~ – 2q ~ . Find the value of m and of n if w~ = 3p~ – 2q~ . [4 markah/marks] Jawapan/Answer: (a) (b) KERTAS PEPERIKSAAN TAMAT END OF QUESTION PAPER PT SPM Add Math Tkt 4-PAT.indd 85 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
86 [100 markah/marks] Masa: 2 jam 30 minit Time: 2 hours 30 minutes Bahagian A / Section A [50 markah/marks] Jawab semua soalan. Answer all questions. 1 Selesaikan persamaan serentak berikut. Solve the following simultaneous equations. 5x2 – 2y2 = 2 2x – y = 1 [6 markah/marks] 2 Selesaikan persamaan serentak 2x – y = 1 dan x2 + 3xy – y2 = 4. Beri jawapan anda betul kepada tiga tempat perpuluhan. Solve the simultaneous equations 2x – y = 1 and x2 + 3xy – y2 = 4. Give your answers correct to three decimal places. [7 markah/marks] 3 Rajah 1 menunjukkan satu pelan bagi persimpangan dua jalan raya lurus, PQ dan RS. Jalan raya RS diwakili oleh persamaan 2y = x + 14. Jalan raya PQ melalui titik Q(6, 0) dan berserenjang dengan jalan raya RS. Sebuah air pancut dibina di persimpangan jalan raya PQ dan RS. Diagram 1 shows a plan for the intersection of two straight roads, PQ and RS. Th e road RS is represented by the equation 2y = x + 14. Th e road PQ passes through the point Q(6, 0) and is perpendicular to the road RS. A fountain is built at the intersection of the roads PQ and RS. 2y = x + 14 Fountain Air pancut Q(6, 0) R P S Rajah 1/ Diagram 1 Cari / Find (a) persamaan yang mewakili jalan raya PQ, the equation that represents the road PQ, [4 markah/marks] (b) koordinat air pancut itu. the coordinates of the fountain. [3 markah/marks] PT SPM Add Math Tkt 4-PAT.indd 86 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
87 4 Rajah 2 menunjukkan sebuah segi tiga PQR. Titik M terletak pada garis lurus PQ dan titik N terletak pada garis lurus PR. Diagram 2 shows a triangle PQR. Point M lies on the straight line PQ and point N lies on the straight line PR. N R P Q M Rajah 2/ Diagram 2 Diberi bahawa 2PM = MQ, PN = 2NR, QR→ = 3u ~ dan PQ → = 4v ~. It is given that 2PM = MQ, PN = 2NR, QR→ = 3u~ and PQ→ = 4v~. (a) Ungkapkan dalam sebutan u ~ dan/atau v ~ Express in terms of u~ and/or v~ (i) PR →, (ii) M ⎯N →. [3 markah/marks] (b) Garis lurus MN dipanjangkan ke titik L dengan keadaan ML→ = ku~ + 2v ~, dengan keadaan k ialah pemalar. Cari nilai k. Th e straight line MN is extended to the point L such that ML→ = ku~ + 2v~, where k is a constant. Find the value of k. [4 markah/marks] 5 (a) Diberi px + p−x = √11 , tunjukkan bahawa (px + 3)(px − 3) = −p−2x Given px + p−x = √11 , show that (px + 3)(px − 3) = −p−2x [2 markah/marks] (b) (i) Permudahkan / Simplify log3 (3x + 4) – 7 log9 x2 + 6 log3 x [4 markah/marks] (ii) Seterusnya, selesaikan persamaan Hence, solve the equation log3 (3x + 4) – 7 log9 x2 + 6 log3 x = 2 [2 markah/marks] 6 (a) Rajah 3 menunjukkan graf bagi fungsi y = m(x + p)2 + q, dengan keadaan m, p dan q ialah pemalar. Diagram 3 shows a graph of the function y = m(x + p)2 + q, where m, p and q are constants. (3, –6) 12 x y O Rajah 3/ Diagram 3 Cari nilai-nilai m, p dan q. Find the values of m, p and q. [4 markah/marks] PT SPM Add Math Tkt 4-PAT.indd 87 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
88 (b) Jika α dan β ialah punca-punca bagi persamaan kuadratik 2x2 – 5x – 14 = 0, bentukkan persamaan kuadratik dengan punca-punca α 4 dan β 4 . If α and β are the roots of the quadratic equation 2x2 – 5x – 14 = 0, form a quadratic equation with the roots α 4 and β 4 . [4 markah/marks] 7 Permudahkan / Simplify log2 (7x − 3) − 6 log4 x2 + 4 log2 x. Seterusnya, selesaikan persamaan log2 (7x − 3) − 6 log4 x2 + 4 log2 x = 1. Hence, solve the equation log2 (7x – 3) – 6 log4 x2 + 4 log2 x = 1. [7 markah/marks] Bahagian B / Section B [30 markah/marks] Jawab mana-mana tiga soalan daripada bahagian ini. Answer any three questions from this section. 8 Ronald membeli 2 biji nanas, sebiji tembikai, dan 5 biji oren dengan harga RM20, manakala Jessie membeli 3 biji nanas, 3 biji tembikai, dan 4 biji oren dengan harga RM37. Harga sebiji nanas ialah RM2 lebih mahal daripada harga 2 biji oren. Ronald buys 2 pineapples, 1 watermelon, and 5 oranges for RM20, while Jessie buys 3 pineapples, 3 watermelons and 4 oranges for RM37. Th e price of a pineapple is RM2 more than the price of 2 oranges. (a) Tulis tiga persamaan daripada situasi di atas. Write down three equations from the above situation. [3 markah/marks] (b) Seterusnya, selesaikan sistem persamaan linear itu. Hence, solve the system of linear equations. [7 markah/marks] 9 Rajah 4 menunjukkan sebuah segi tiga XYZ. Diagram 4 shows a triangle XYZ. B Z A X C Y Rajah 4/ Diagram 4 Diberi A ialah titik tengah XY, XC : CZ = 2 : 3, XA → = 5u ~ dan YZ → = 5v ~. Given A is the midpoint of XY, XC : CZ = 2 : 3, XA→ = 5u~ and YZ→ = 5v~. (a) Ungkapkan dalam sebutan u ~ dan v ~ Express in terms of u~ and v ~ (i) AZ →, (ii) CZ →. [3 markah/marks] (b) Diberi u ~ = 2i ~ dan v = −i~ + 4j ~ , cari |CZ →|. Given u~ = 2i ~ and v ~ = −i ~ + 4j ~ , fi nd |CZ →|. [2 markah/marks] (c) Diberi BZ → = hAZ→ dan CB → = kCY→, dengan keadaan h dan k ialah pemalar. Cari nilai h dan nilai k. Given BZ → = hAZ → and CB → = kCY →, where h and k are constants. Find the value of h and of k. [5 markah/marks] PT SPM Add Math Tkt 4-PAT.indd 88 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
89 10 Rajah 5 menunjukkan garis lurus PQ. Diagram 5 shows a straight line PQ. y O x P(–3, 5) Q(6, –1) Rajah 5/ Diagram 5 (a) Hitung luas segi tiga POQ. Calculate the area of triangle POQ. [3 markah/marks] (b) Titik R(1, 2) terletak pada garis lurus PQ. Cari persamaan garis lurus yang melalui R dan berserenjang dengan PQ. Point R(1, 2) lies on the straight line PQ. Find the equation of the straight line that passes through R and is perpendicular to PQ. [4 markah/marks] (c) Titik L bergerak dengan keadaan jaraknya dari P adalah sentiasa dua kali jaraknya dari Q. Cari persamaan lokus bagi L. Point L moves such that its distance from P is always twice its distance from Q. Find the equation of locus of L. [3 markah/marks] 11 Gunakan kertas graf untuk menjawab soalan ini. Use a graph paper to answer this question. Jadual 1 menunjukkan nilai-nilai bagi dua pemboleh ubah, x dan y, yang diperoleh daripada suatu eksperimen. Satu garis lurus diperoleh apabila graf xy melawan x2 diplot. Table 1 shows the values of two variables, x and y, obtained from an experiment. A straight line is obtained when a graph of xy against x2 is plotted. x 1 2 3 4 5 6 y 7.4 7.3 8.87 10.85 13 15.2 Jadual 1/ Table 1 (a) Berdasarkan Jadual 1, bina satu jadual bagi nilai-nilai x2 dan xy. Based on Table 1, construct a table for the values of x2 and xy. [2 markah/marks] (b) Plot xy melawan x2 , dengan menggunakan skala 2 cm kepada 5 unit pada paksi-x2 dan 2 cm kepada 10 unit pada paksi-xy. Seterusnya, lukis garis lurus penyuaian terbaik. Plot xy against x2 , by using a scale of 2 cm to 5 units on the x2 -axis and 2 cm to 10 units on the xy-axis. Hence, draw the line of best fi t. [3 markah/marks] (c) Menggunakan graf di 11(b), Using the graph in 11(b), (i) cari nilai y apabila x = 5.5, fi nd the value of y when x = 5.5, (ii) ungkapkan y dalam sebutan x. express y in terms of x. [5 markah/marks] PT SPM Add Math Tkt 4-PAT.indd 89 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
90 Bahagian C / Section C [20 markah/marks] Jawab mana-mana dua soalan daripada bahagian ini. Answer any two questions from this section. 12 Jadual 2 menunjukkan indeks harga bagi empat jenis bahan, K, L, M dan N yang digunakan dalam pembuatan sejenis pen. Table 2 shows the price indices of four materials, K, L, M and N, used in the production of a type of pen. Bahan Material Indeks harga pada tahun 2021 berasaskan tahun 2016 Price index in the year 2021 based on the year 2016 Peratusan penggunaan (%) Percentage of usage (%) K 90 45 L 140 10 M 180 15 N h 30 Jadual 2/ Table 2 (a) Cari / Find (i) harga bahan M pada tahun 2021 jika harganya pada tahun 2016 ialah RM4.50, the price of material M in the year 2021 if its price in the year 2016 was RM4.50, (ii) harga bahan K pada tahun 2016 jika harganya pada tahun 2021 ialah RM6.75. the price of material K in the year 2016 if its price in the year 2021 was RM6.75. [4 markah/marks] (b) Indeks gubahan bagi kos pembuatan pen pada tahun 2021 berasaskan tahun 2016 ialah 122. Hitung Th e composite index for the cost of the production of pen in the year 2021 based on the year 2016 was 122. Calculate (i) nilai h, the value of h, (ii) harga sepadan bagi pen itu pada tahun 2021 jika harga pen itu pada tahun 2016 ialah RM64. the corresponding price of the pen in the year 2021 if the price of the pen in the year 2016 was RM64. [6 markah/marks] 13 Jadual 3 menunjukkan indeks harga satu kilogram bagi lima jenis buah yang dijual di sebuah kedai. Table 3 shows the price indices of one kilogram of fi ve types of fruits sold in a shop. Buah Fruit Indeks harga pada tahun 2019 berasaskan tahun 2018 Price index in the year 2019 based on the year 2018 Pemberat Weightage Betik/Papaya 110 4 Tembikai/Watermelon 125 3 Pisang/Banana p q Oren/Orange 102 6 Durian/Durian 135 5 Jadual 3/ Table 3 PT SPM Add Math Tkt 4-PAT.indd 90 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
91 Diberi indeks gubahan bagi harga buah-buahan pada tahun 2019 berasaskan tahun 2018 ialah 116.61. Given the composite index of the price of the fruits in the year 2019 based on the year 2018 was 116.61. (a) Harga pisang pada tahun 2018 ialah RM6.25 dan harganya pada tahun 2019 ialah RM7.25. Cari nilai Th e price of the banana in the year 2018 was RM6.25 and its price in the year 2019 was RM7.25. Find the value of (i) p, (ii) q. [4 markah/marks] (b) Cari harga oren pada tahun 2019 jika harganya pada tahun 2018 ialah RM12. Find the price of the orange in the year 2019 if its price in the year 2018 was RM12. [2 markah/marks] (c) Cari jumlah harga buah-buahan pada tahun 2019 jika jumlah harga pada tahun 2018 ialah RM495. Find the total price of the fruits in the year 2019 if the total price in the year 2018 was RM495. [2 markah/marks] (d) Harga durian pada tahun 2018 ialah RM29. Cari harga durian pada tahun 2023 jika harganya dijangka meningkat sebanyak 18% dari tahun 2019 ke tahun 2023. Th e price of the durian in the year 2018 was RM29. Find the price of the durian in the year 2023 if its price is expected to increase by 18% from the year 2019 to the year 2023. [2 markah/marks] 14 Rajah 6 menunjukkan sebuah sisi empat EFGH. Diagram 6 shows a quadrilateral EFGH. 32º H F 10.2 cm 14 cm 8 cm E 6 cm G Rajah 6/ Diagram 6 (a) Cari / Find (i) ∠EGF, (ii) panjang, dalam cm, bagi EG. the length, in cm, of EG. [4 markah/marks] (b) (i) Dengan menggunakan rumus Heron, hitung luas, dalam cm2 , bagi segi tiga EGH. By using Heron’s formula, calculate the area, in cm2 , of the triangle EGH. (ii) Cari jarak terdekat, dalam cm, dari titik H ke garis lurus EG. Find the shortest distance, in cm, from point H to the straight line EG. [4 markah/marks] (c) Lukis dan labelkan sebuah segi tiga selain daripada segi tiga EFG dengan keadaan FG’ = FG dan ∠FEG’ = ∠FEG. Draw and label another triangle other than triangle EFG such that FG’ = FG and ∠FEG’ = ∠FEG. [2 markah/marks] PT SPM Add Math Tkt 4-PAT.indd 91 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
92 15 Dalam Rajah 7, P, Q, R dan S ialah titik-titik pada satah mengufuk. Diberi ∠PRS dan ∠QRS ialah sudut cakah. In Diagram 7, P, Q, R and S are points on a horizontal plane. Given ∠PRS and ∠QRS are obtuse angles. 4 cm 11 cm 6 cm S R P 38º Q Rajah 7/ Diagram 7 Diberi bahawa PR = 11 cm, QS = 6 cm, RS = 4 cm dan ∠RQS = 38°. Jika luas segi tiga PRS ialah 8.95 cm2 , hitung It is given that PR = 11 cm, QS = 6 cm, RS = 4 cm and ∠RQS = 38°. If the area of the triangle PRS is 8.95 cm2 , calculate (a) ∠PRS, [2 markah/marks] (b) panjang, dalam cm, bagi PS, the length, in cm, of PS, [2 markah/marks] (c) ∠QSP, [3 markah/marks] (d) luas, dalam cm2 , bagi segi tiga PQS. the area, in cm2 , of the triangle PQS. [3 markah/marks] KERTAS PEPERIKSAAN TAMAT END OF QUESTION PAPER PT SPM Add Math Tkt 4-PAT.indd 92 28/02/2023 5:47 PM PENERBIT ILMU BAKTI SDN. BHD.
J1 Praktis Topikal SPM: Matematik Tambahan Tingkatan 4 – Jawapan BAB 1 Fungsi KERTAS 1 BAHAGIAN A 1 (a) f(x) = |3 – 2x| 0 = |3 – 2k| 0 = 3 – 2k k = 3 2 (b) Apabila/When x = 6, f(x) =|3 – 2(6)| = |–9| = 9 Julat/Range: 0 ⩽ f(x) ⩽ 9 2 (a) f(–2) = –13 –2m + n = –13 ––––❶ f(5) = 8 5m + n = 8 ––––❷ ❷ – ❶, 7m = 21 m = 3 Dari/From ❷, 5(3) + n = 8 n = – 7 (b) f(x) = x 3x – 7 = x 2x = 7 x = 7 2 3 (a) f(h) = 16 h2 + h + 4 = 16 h2 + h – 12 = 0 (h + 4)(h – 3) = 0 h = –4 atau/or 3 ∴ h = –4 (b) Apabila/When x = 0, g(x) = |3(0) – 6| = |–6| = 6 Maka/Thus, a = 6 Apabila/When g(x) = 0, |3x – 6| = 0 3x – 6 = 0 x = 2 Maka/Thus, b = 2 Apabila/When g(x) = 15, |3x – 6| = 15 3x – 6 = –15 atau/or 3x – 6 = 15 3x = –9 atau/or 3x = 21 x = –3 atau/or x = 7 Oleh sebab/Since x > 0, x = 7. Maka/Thus, c = 7 Maka/Hence, a = 6, b = 2, c = 7. 4 (a) f(x) =|4x + 3| f(–5) =|4(–5) + 3| =|–20 + 3| =|–17| = 17 (b) f(x) = 5 |4x + 3| = 5 4x + 3 = 5 atau/or 4x + 3 = –5 4x = 5 – 3 4x = –5 – 3 4x = 2 4x = –8 x = 1 2 x = –2 ∴ x = 1 2 atau/or x = –2 5 (a) f(x) = 3x + 7 fg(x) = 3g(x) + 7 3g(x) + 7 = 6x2 – 3x – 5 3g(x) = 6x2 – 3x – 12 g(x) = 2x2 – x – 4 (b) f(x) = mx + n f 2 (x) = m(mx + n) + n = m2 x + mn + n Diberi/Given f 2 (x) = 36x – 21 Bandingkan pekali x, Compare the coeffi cients of x, m2 = 36 m = 6 oleh sebab/since m > 0 Bandingkan pemalar, Compare the constants, mn + n = –21 (6)n + n = –21 7n = –21 n = –3 6 (a) gf(x) = 6(2x – 5) = 12x – 30 (b) f(x) = 2x + 7 maka/then fg(x) = 2g(x) + 7 Diberi/Given fg(x) = 2x – 1 Maka/Then 2g(x) + 7 = 2x – 1 2g(x) = 2x – 1 – 7 2g(x) = 2x – 8 g(x) = x – 4 7 (a) Diberi/Given f(x) = 2x – 6 dan/and fg(x) = 14 – 2x. fg(x) = 2g(x) – 6 14 – 2x = 2g(x) – 6 2g(x) = 20 – 2x g(x) = 10 – x gf(x) = 10 – f(x) = 10 – (2x – 6) = 16 – 2x (b) gf(p2 + 2) = 6p – 8 16 – 2(p2 + 2) = 6p – 8 16 – 2p2 – 4 = 6p – 8 2p2 + 6p – 20 = 0 p2 + 3p – 10 = 0 (p + 5)(p – 2) = 0 ∴ p = –5, p = 2 8 (a) Katakan/Let y = x + 4 2x – 5 2xy – 5y = x + 4 2xy – x = 5y + 4 x(2y – 1) = 5y + 4 x = 5y + 4 2y – 1 ∴ g–1(x) = 5x + 4 2x – 1 , x ≠ 1 2 g–1(–3) = 5(–3) + 4 2(–3) – 1 = –15 + 4 –6 – 1 = 11 7 (b) f(x) = ax + b f –1(–7) = 4 f(4) = –7 ∴ –7 = a(4) + b 4a + b = –7 1 f –1(3) = 9 f(9) = 3 ∴ 3 = a(9) + b 9a + b = 3 2 2 – 1 ; 5a = 10 a = 2 Daripada/From 1 , 4(2) +b = –7 b = –15 ∴ a = 2, b = –15 9 (a) Katakan/Let y = 6 – 4x 4x = 6 – y x = 6 – y 4 g–1(x) = 6 – x 4 (b) fg(x) = 7 – 8x fg(g–1(x)) = 7 – 8(g–1(x)) f(x) = 7 – 8( 6 – x 4 ) = 7 – 2(6 – x) = 7 – 12 + 2x = 2x – 5 f(7) = 2(7) – 5 = 14 – 5 = 9 BAHAGIAN B 10 (a) (i) x = 48 h(48) = 3 4 [272 – (48)] = 168 Kadar denyutan jantung Encik Ghazali ialah 168. Encik Ghazali’s heart rate is 168. (ii) h(x) = 3 4 (272 – x) 4 3 h(x) = 272 – x x = 272 – 4 3 h(x) ∴ h–1(x) = 272 – 4x 3 (b) (i) f(x) = mx + n f²(x) = m(mx + n) + n = m2 x + mn + n f 3 (x) = f²(x)f(x) = m2 (mx + n) + mn + n = m3 x + m2 n + mn + n Diberi/Given f 3 (x) = 27x – 65. Dengan perbandingan, By comparison, m3 = 27 ⇒ m = 3 m2 n + mn + n = –65 (3)2 n + (3)n + n = –65 13n = –65 n = –5 (ii) f²(x) = (3)2 x + (3)(–5) + (–5) = 9x – 20 f 4 (x) = f²(f²(x)) = 9(9x – 20) – 20 = 81x – 180 – 20 = 81x – 200 KERTAS 2 BAHAGIAN A 1 (a) h(y) = 16 3y – p h(3) = –2 JAWAPAN PT SPM Add Math Tkt 4-Jawapan.indd 1 01/03/2023 9:17 AM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J2 16 3(3) – p = –2 16 = –18 + 2p 2p = 34 p = 17 g(y) = p + qy g(y) = 17 + qy g(3) = –7 17 + q(3) = –7 3q = –24 q = –8 (b) g(y) = 17 – 8y y = 17 – g(y) 8 g–1(x) = 17 – x 8 (c) h(y) = 16 3y – 17 hg–1(x) = 16 3(g–1(x)) – 17 = 16 3( 17 – x 8 ) – 17 = 16 51 – 3x – 136 8 = 128 –3x – 85 = – 128 3x + 85 2 (a) Katakan/Let y = 2x – 3 2x = y + 3 x = y + 3 2 g–1(x) = x + 3 2 (b) g–1h(x) = ( x 3 + 2) + 3 2 = x 3 + 5 2 = x + 15 6 (c) h(x) = x 3 + 2 3h(x) = x + 6 x = 3h(x) – 6 h–1(x) = 3x – 6 Diberi/Given kh(x) = 2x + 8 kh(h–1(x)) = 2(3x – 6) + 8 k(x) = 6x – 12 + 8 k(x) = 6x – 4 3 (a) Katakan/Let y = 3x + 4 3x = y – 4 x = y – 4 3 g–1(x) = x – 4 3 Diberi/Given fg(x) = 6x + 1 fg(g–1(x)) = 6(g–1(x)) + 1 f(x) = 6 ( x – 4 3 ) + 1 = 2(x – 4) + 1 = 2x – 8 + 1 = 2x – 7 (b) gf(x) = 3(2x – 7) + 4 = 6x – 21 + 4 = 6x – 17 (c) 3fg(x – 2) = gf(x) 3[6(x – 2) + 1] = 6x – 17 3(6x – 12 + 1) = 6x – 17 3(6x – 11) = 6x – 17 18x – 33 = 6x – 17 12x = 16 x = 4 3 BAHAGIAN B 4 (a) (i) h(8) = 2(8) – 12 = 16 – 12 = 4 (ii) g(p – 1) = 1 2 h(8) + 1 3(p – 1) + 15 = 1 2 (4) + 1 3p – 3 + 15 = 2 + 1 3p = –9 p = –3 (iii) hg(x) = 2(3x + 15) – 12 = 6x + 30 – 12 = 6x + 18 (b) (i) y = |6x + 18| Apabila/When y = 0, x = –3 Apabila/When x = –6 , y = |6(–6) + 18| = 18 x = 0, y = |6(0) + 18| = 18 y 18 –3–6 x 0 y = |6x + 18| (ii) hg(q) = 2gh(q) 6q + 18 = 2[3(2q – 12) + 15] 6q + 18 = 2(6q – 36 + 15) 6q + 18 = 2(6q – 21) 6q + 18 = 12q – 42 60 = 6q q = 10 ZON KBAT 1 (a) h(x) = 2x + 7 x + 4 y = 2x + 7 x + 4 yx + 4y = 2x + 7 yx – 2x = 7 – 4y x(y – 2) = 7 – 4y x = 7 – 4y y – 2 h–1(x) = 7 – 4x x – 2 , x ≠ 2 y x –. 0 h(x) = 2x + 7 x + 4 7 4 h–1(x) = 7 – 4x x – 2 2 Domain bagi h–1/Domain of h–1: 7 4 ⩽ x < 2 BAB 2 Fungsi Kuadratik KERTAS 1 BAHAGIAN A 1 Hasil tambah punca/Sum of roots = p + 5 Hasil darab punca/Product of roots = 5p 2x2 + 4x + q = 0 x2 + 2x + q 2 = 0 Hasil tambah punca/Sum of roots: p + 5 = –2 p = –7 Hasil darab punca/Product of roots: 5p = q 2 5(–7) = q 2 –35 = q 2 q = –70 2 3x(x – 1) = (3 – x)(x + 1) 3x2 – 3x = 3x + 3 – x2 – x 4x2 – 5x – 3 = 0 a = 4, b = –5, c = –3 x = –(–5) ± √(–5)2 – 4(4)(–3) 2(4) x = –0.443 atau/or x = 1.693 3 (a) x = –4 3[(–4) + h] 2 = 75 Bahagikan kedua-dua belah persamaan dengan 3, Divide both sides by 3, (h – 4)2 = 25 (h – 4)2 – 25 = 0 (h – 4 – 5)(h – 4 + 5) = 0 (h – 9)(h + 1) = 0 h – 9 = 0 atau/or h + 1 = 0 h = 9 h = –1 (b) Hasil tambah punca/Sum of roots, α + β = 3 Hasil darab punca/Product of roots, αβ = – 5 Bagi persamaan dengan punca 1 α dan 1 β . For equation with roots 1 α and 1 β . HTP = 1 α + 1 β = β + α αβ = 3 –5 HOP = 1 α × 1 β = 1 αβ = 1 –5 Maka, persamaan kuadratik ialah Thus, the quadratic equation is x2 – ( 3 –5 )x + ( 1 –5 ) = 0 x2 + 3 5 x – 1 5 = 0 5x2 + 3x – 1 = 0 4 (a) HTP = 2 3 + (–11 2) = – 5 6 HDP = 2 3 × (–11 2) = –1 x2 – (– 5 6)x + (–1) = 0 x2 + 5 6 x – 1 = 0 6x2 + 5x – 6 = 0 a = 6, b = 5, c = –6 (b) Diberi/Given 2x + y = 3 y = 3 – 2x x2 > y2 x2 > (3 – 2x) 2 x2 > 9 – 12x + 4x2 3x2 – 12x + 9 < 0 x2 – 4x + 3 < 0 (x – 3)(x – 1) < 0 1 3 x ∴ 1 < x < 3 5 (a) hx2 + 6x + k = 4 hx2 + 6x + k – 4 = 0 PT SPM Add Math Tkt 4-Jawapan.indd 2 01/03/2023 9:17 AM PENERBIT ILMU BAKTI SDN. BHD.
J3 Praktis Topikal SPM: Matematik Tambahan Tingkatan 4 – Jawapan 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 x 1 1 5 Daripada graf/From the graph, k < 1 5 atau/or k > 1 6 (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 7 (a) p = –2 + 5 2 = 3 2 f(x) = (x – 3 2) 2 + q Pada titik/At point (5, 15), 15 = (5 – 3 2) 2 + q 15 = 49 4 + q q = 11 4 (b) T = ( 3 2 , 11 4 ) (c) x = 3 2 8 (a) h(x) = –5[(x – 4)2 – 2p] h(x) = –5(x – 4)2 + 10p Diberi titik maksimum (4, 6p – 12). Given the maximum point (4, 6p – 12). Maka/Then, 10p = 6p – 12 4p = –12 p = –3 (b) h(x) = 0 –5(x – 4)2 + 10p = 0 –5(x – 4)2 = –10p –5(x – 4)2 = –10(–3) (x – 4)2 = –6 Fungsi itu tidak mempunyai punca nyata kerana x – 4 = √–6 The function has no real roots because x – 4 = √–6 9 (a) x + 4 = 0 x = –4 (b) 3m – 12 = 9 3m = 21 m = 7 BAHAGIAN B 10 (a) 6x2 – 18x + q = 0 x² – 3x + q 6 = 0 Hasil tambah punca/Sum of roots: 3p + (7 – p) = 3 2p = –4 p = –2 Hasil darab punca/Product of roots: (3p)(7 – p) = q 6 q 6 = 3(–2)[7 – (–2)] q 6 = –6(9) q = –324 (b) 6x2 – 18x > –q 6x2 – 18x > –(–324) 6x2 – 18x – 324 > 0 x2 – 3x – 54 > 0 (x + 6)(x – 9) > 0 x –6 9 Daripada graf/From the graph, x < – 6 atau/or x > 9 (c) Punca-punca/Roots: p = –2 5 – 2p = 5 – 2(–2) = 9 Hasil tambah punca/Sum of roots: –2 + 9 = 7 Hasil darab punca/Product of roots: –2(9) = –18 Persamaan kuadratik: Quadratic equation: x² – 7x – 18 = 0 11 (a) x2 + 3x – 12 = 0 x2 + 3x = 12 x2 + 3x + ( 3 2) 2 = 12 + ( 3 2) 2 (x + 3 2) 2 = 57 4 x + 3 2 = ± 57 4 x = – 3 2 ± 57 2 x = – 3 2 + 57 2 atau/or – 3 2 – 57 2 ∴ x = 2.2749 atau/or –5.2749 (b) a = 3, b = –4, c = –5 x = – (–4)± (–4) 2 – 4(3)(–5) 2(3) = 4 ± 76 6 x = 4 + 76 6 atau/or x = 4 – 76 6 = 2.1196 = –0.7863 ∴ x = –0.7863, 2.1196 12 (a) Hasil tambah punca/Sum of roots, α + β = –4 Hasil darab punca/Product of roots, αβ = 7 Bagi persamaan kuadratik dengan punca-punca 2α + 3 dan 2β + 3. For a quadratic equation with the roots 2α + 3 and 2β + 3. Hasil tambah punca/Sum of roots = (2α + 3) + (2β + 3) = 2(α + β) + 6 = 2(–4) + 6 = –2 Hasil darab punca/Product of roots = (2α + 3)(2β + 3) = 4αβ + 6α + 6β + 9 = 4(αβ) + 6(α + β) + 9 = 4(7) + 6(–4) + 9 = 13 Maka, persamaan kuadratik ialah Thus, the quadratic equation is x2 – (–2)x + 13 = 0 x2 + 2x + 13 = 0 (b) (4 – λ)x2 – 2(λ – 3)x + λ = 3 (4 – λ)x2 – 2(λ – 3)x + λ – 3 = 0 a = 4 – λ, b = – 2(λ – 3), c = λ – 3 b2 – 4ac = 0 [– 2(λ – 3)]2 – 4(4 – λ)(λ – 3) = 0 4(λ – 3)2 – 4(4 – λ)(λ – 3) = 0 4(λ – 3)[(λ – 3) – (4 – λ)] = 0 4(λ – 3)[λ – 3 – 4 + λ] = 0 4(λ – 3)(2λ – 7) = 0 λ = 3 atau/or λ = 7 2 13 (a) 6x2 = (3h – 4)x + 5 Apabila/When x = 2, 6(2)² = (3h – 4)(2) + 5 24 = 6h – 8 + 5 6h = 27 h = 9 2 6x² = [3( 9 2) – 4]x + 5 6x² = 19 2 x + 5 12x² – 19x – 10 = 0 (x – 2)(12x + 5) = 0 x = 2 atau/or x = – 5 12 Maka/Thus, k 6 – 1 = – 5 12 k 6 = 7 12 k = 7 2 (b) A = 8p(60 – p) = 480p – 8p2 = –8[p2 – 60p] = –8[p2 – 60p + (–30)2 – (–30)2 ] = –8[(p – 30)2 – 900] = –8(p – 30)2 + 7 200 ∴ a = –8, m = –30, n = 7 200 Bagi nilai maksimum A, For maximum value of A, p – 30 = 0 p = 30 Jumlah panjang Total length = 2[8p + (60 – p)] = 2(7p + 60) = 2[7(30) + 60] = 540 m KERTAS 2 BAHAGIAN A 1 (a) g(x) = –6x² – 6x + 3 = –6(x2 + x – 1 2) = –6[x2 + x + ( 1 2) 2 – ( 1 2) 2 – 1 2] = –6[(x + 1 2) 2 – 3 4] PT SPM Add Math Tkt 4-Jawapan.indd 3 01/03/2023 9:17 AM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J4 = –6(x + 1 2) 2 + 9 2 ∴ m = 1 2 , n = 9 2 (b) a = –6 < 0 (Maksimum/Maximum). Titik maksimum/Maximum point = (– 1 2 , 9 2) Lengkung menyilang paksi-x pada dua titik yang berbeza. The curve intersects the x-axis at two different points. g(0) = –6(0)² – 6(0) + 3 = 3 g(x) 3 x 0 (– 1 2 , 9 2 ) • x = – 1 2 (c) g(x) = –6(x + 1 2) 2 + 9 2 1.56 = –6(x + 1 2) 2 + 9 2 6(x + 1 2) 2 = 2.94 (x + 1 2) 2 = 0.49 x + 1 2 = –0.7 atau/or x + 1 2 = 0.7 x = –1.2 x = 0.2 Jarak mengufuk titik G dari paksi simetri Horizontal distance of point G from the axis of symmetry = |0.2 – (– 1 2)| = |0.7| = 0.7 unit 2 (a) f(0) = –(0)2 + p(0) – 15 = –15 Koordinat F/Coordinates of F = (0, –15) (b) f(x) = –x2 + px – 15 = –(x2 – px + 15) = –[x2 – px + (– p 2) 2 – (– p 2) 2 + 15] = – [(x – p 2 ) 2 – p2 4 + 15] = – (x – p 2 ) 2 + p2 4 – 15 ∴ G = ( p 2 , p2 4 – 15) Diberi/Given G = (3, h) Bandingkan koordinat-x, Comparing x-coordinate, p 2 = 3 p = 6 Bandingkan koordinat-y, Comparing y-coordinate, p2 4 – 15 = h (6)2 4 – 15 = h 9 – 15 = h h = –6 (c) Apabila/When p = 6, f(x) = –x2 + 6x – 15 f(x) ⩾ –7 –x2 + 6x – 15 ⩾ –7 –x2 + 6x – 8 ⩾ 0 (–x + 2)(x – 4) ⩾ 0 2 4 x ∴ 2 ⩽ x ⩽ 4 3 (a) a = 2, b = –5, c = –3 a > 0, maka f(x) mempunyai titik minimum. a > 0, so f(x) has a minimum point. b2 – 4ac = (–5)2 – 4(2)(–3) = 25 + 24 = 49 (>0) Lengkung itu menyilang paksi-x pada dua titik yang berbeza. The curve intersects the x-axis at two different points. f(x) = 2x2 – 5x – 3 = 2(x2 – 5 2 x – 3 2) = 2[x2 – 5 2 x + (– 5 4) 2 – (– 5 4) 2 – 3 2] = 2[(x – 5 4) 2 – 49 16 ] = 2(x – 5 4) 2 – 49 8 ∴ Titik minimum = ( 5 4 , – 49 8 ) dan persamaan paksi simetri, x = 5 4 . ∴ Minimum point = ( 5 4 , – 49 8 ) and the equation of the axis of symmetry, x = 5 4 . f(x) = 0 2x2 – 5x – 3 = 0 (2x + 1)(x – 3) = 0 x = – 1 2 atau/or x = 3 f(0) = 2(0)2 – 5(0) – 3 = –3 3 –3 f(x) x 0 – 1 2 , – 49 8 5 4 (b) f(x) ⩽ 0 Daripada graf/From the graph, 3 –3 f(x) x 0 – 1 2 – 1 2 ⩽ x ⩽ 3 4 (a) g(x) = – 1 2 (x – 4)2 – 1 (b) Untuk/For f(x) = 1 2 (x – 4)2 + 1; a > 0, f(x) mempunyai titik minimum/has a minimum point (4, 1). f(0) = 1 2 (0 – 4)2 + 1 = 9 Untuk/For g(x) = – 1 2 (x – 4)2 – 1; a < 0, g(x) mempunyai titik maksimum/has a maximum point (4, –1). g(0) = – 1 2 (0 – 4)2 – 1 = –9 y x 0 1 9 –9 –1 4 g(x) = – (x – 4)2 – 1 1 2 f(x) = (x – 4)2 + 1 1 2 BAHAGIAN B 5 (a) (i) f(x) = 0 (3x2 – 12x – 36 = 0) ÷ 3 x2 – 4x – 12 = 0 (x + 2)(x – 6) = 0 x + 2 = 0 atau/or x – 6 = 0 x = –2 x = 6 P = (–2, 0), Q = (6, 0) (ii) f(x) = 3x2 – 12x – 36 = 3(x2 – 4x – 12) = 3[x2 – 4x + (–2)2 – (–2)2 – 12] = 3[(x – 2)2 – 4 – 12] = 3[(x – 2)2 – 16] = 3(x – 2)2 – 48 ∴ a = 3, h = –2, k = –48 (iii) f(x) = 3(x – 2)2 – 48 Oleh sebab a = 3 (> 0), f(x) mempunyai titik pusingan minimum./Since a = 3 (> 0), f(x) has a minimum turning point. Titik pusingan minimum/ Minimum turning point = (2, – 48) (b) (i) f(0) = 3(0 – 2)2 – 48 = –36 x y P Q –48 –36 –2 O 62 (ii) f(x) = 3(x – 2)2 – 48 r(x) = –3(x – 2)2 + 48 ∴ a1 = – 3, h1 = –2, k1 = 48 6 (a) (i) f(x) = m + nx – x2 = q – (x + p) 2 = q – (x2 + 2px + p2 ) = q – x2 – 2px – p2 = q – p2 – 2px – x2 Bandingkan pekali x, Compare the coeffi cient of x, –2p = n p = – n 2 (ii) Bandingkan pemalar, Compare the constant, q – p2 = m q – (– n 2 ) 2 = m q – n2 4 = m q = m + n2 4 PT SPM Add Math Tkt 4-Jawapan.indd 4 01/03/2023 9:17 AM PENERBIT ILMU BAKTI SDN. BHD.
J5 Praktis Topikal SPM: Matematik Tambahan Tingkatan 4 – Jawapan (b) n = 6, p = – (6) 2 p = –3 f(x) = q – (x – 3)2 Persamaan paksi simetri ialah x = 3. Equation of the axis of symmetry is x = 3. y = 13 menyentuh titik maksimum. y = 13 touches the maximum point. f(x) = 13 – (x – 3)2 = 13 – (x2 – 6x + 9) = 13 – x2 + 6x – 9 = –x2 + 6x + 4 a = –1, (a < 0, maka f(x) mempunyai titik maksimum.) a = –1, (a < 0, so f(x) has a maximum point). f(0) = 13 – (0 – 3)2 = 4 Graf itu menyilang paksi-y di (0, 4). The graph intersects the y-axis at (0, 4). (3, 13) y = 13 4 0 f(x) x ZON KBAT 1 (a) 5x2 = 1 280 (b) 5x2 = 1 280 x2 = 256 x = 16 cm Panjang sisi asal kadbod Original length of the sides of the cardboard = 16 + 5 + 5 = 26 cm BAB 3 Sistem Persamaan KERTAS 1 BAHAGIAN B 1 (a) 4x – 3y + z = 25 1 x + y – z = 4 2 3x – 2z = 17 3 1 : 4x – 3y + z = 25 2 × 3 : 3x + 3y – 3z = 12 (+) 7x – 2z = 37 4 4 : 7x – 2z = 37 3 : 3x – 2z = 17 (–) 4x = 20 x = 5 Daripada/From 3 , 3(5) – 2z = 17 –2z = 2 z = –1 Daripada/From 2 , (5) + y – (–1) = 4 5 + y + 1 = 4 y = –2 Maka/Hence, x = 5, y = –2, z = –1 (b) 3x + y + 2z = 2.8 1 5x – 2y + z = 2.5 2 4x + y – 3z = 1.9 3 3 – 1 : x – 5z = –0.9 4 2 × 1 : 6x + 2y + 4z = 5.6 2 : 5x – 2y + z = 2.5 (+) 11x + 5z = 8.1 5 4 + 5 : 12x = 7.2 x = 0.6 Daripada/From 4 , (0.6) – 5z = –0.9 5z = 1.5 z = 0.3 Daripada/From 1 , 3(0.6) + y + 2(0.3) = 2.8 y = 0.4 Maka/Hence, x = 0.6, y = 0.4, z = 0.3 2 (a) Pada titik/At point (p, 2); Garis lurus/The straight line, (2) = (p) + q q = 2 – p 4 Lengkung/The curve, (p) 2 + (2)2 – 3q(2) = 8 p2 + 4 – 6q = 8 2 Gantikan 1 ke dalam 2 , Substitute 1 into 2 , p2 + 4 – 6(2 – p) = 8 p2 + 4 – 12 + 6p = 8 p2 + 6p – 16 = 0 (p – 2)(p + 8) = 0 p – 2 = 0 atau/or p + 8 = 0 p = 2 p = –8 Apabila/When p = 2, q = 2 – (2) = 0 Apabila/When p = –8, q = 2 – (–8) = 10 (b) Diberi perimeter/Given perimeter = 56 m x + y + (y + 1) = 56 x + 2y = 55 1 Menggunakan teorem Pythagoras, Using Pythagoras’ theorem, x2 + (y) 2 = (y + 1)2 x2 + y2 = y2 + 2y + 1 x2 – 2y = 1 2 Daripada/From 1 , y = 55 – x 2 Gantikan y = 55 – x 2 ke dalam 2 , Substitute y = 55 – x 2 into 2 , x2 – 2( 55 – x 2 ) = 1 x2 – 55 + x = 1 x2 + x – 56 = 0 (x + 8)(x – 7) = 0 x + 8 = 0 atau/or x – 7 = 0 x = –8 x = 7 Oleh sebab/Since x > 0, x = 7 Daripada/From 1 , y = 55 – (7) 2 = 24 ∴ x = 7, y = 24 3 (a) Diberi/Given x2 – 2y + y2 = x + 3y = 12 Maka/Thus, x2 – 2y + y2 = 12 1 x + 3y = 12 2 Daripada/From 2 , x = 12 – 3y Gantikan x = 12 – 3y ke dalam 1 , Substitute x = 12 – 3y into 1 , (12 – 3y) 2 – 2y + y2 = 12 144 – 72y + 9y2 – 2y + y2 = 12 10y2 – 74y + 132 = 0 5y2 – 37y + 66 = 0 (5y – 22)(y – 3) = 0 5y – 22 = 0 atau/or y – 3 = 0 y = 22 5 y = 3 Apabila/When y = 22 5 , x = 12 – 3( 22 5 ) = – 6 5 Apabila/When y = 3, x = 12 – 3(3) = 3 (b) 2x + 3y + 6 = 0 1 2 x + 3 y = 4 2 Daripada/From 2 , 2y + 3x = 4xy 3 Daripada/From 1 , y = –6 – 2x 3 Gantikan y = –6 – 2x 3 ke dalam 3 , Substitute y = –6 – 2x 3 into 3 , 2( –6 – 2x 3 ) + 3x = 4x( –6 – 2x 3 ) 2(–6 – 2x) + 9x = 4x(–6 – 2x) –12 – 4x + 9x = –24x – 8x2 8x2 + 29x – 12 = 0 (x + 4)(8x – 3) = 0 x + 4 = 0 atau/or 8x – 3 = 0 x = –4 x = 3 8 Apabila/When x = –4, y = –6 – 2(–4) 3 = 2 3 Apabila/When x = 3 8 , y = –6 – 2( 3 8) 3 = – 9 4 Titik persilangan ialah The points of intersections are (–4, 2 3) dan/and ( 3 8 , – 9 4 ) KERTAS 2 BAHAGIAN B 1 (a) Diberi/Given 2x + y = x2 – y2 – 2xy – 1 = 1 2x + y = 1 1 x2 – y2 – 2xy – 1 = 1 2 Daripada/From 1 , y = 1 – 2x Gantikan y = 1 – 2x ke dalam 2 , Substitute y = 1 – 2x into 2 , x2 – (1 – 2x) 2 – 2x(1 – 2x) – 1 = 1 x2 – (1 – 4x + 4x2 ) – 2x + 4x2 – 1 = 1 x2 – 1 + 4x – 4x2 – 2x + 4x2 – 2 = 0 x2 + 2x – 3 = 0 (x + 3)(x – 1) = 0 x + 3 = 0 atau/or x – 1 = 0 x = –3 x = 1 Apabila/When x = –3, y = 1 – 2(–3) = 7 Apabila/When x = 1, y = 1 – 2(1) = –1 ∴ x = –3, y = 7 atau/or x = 1, y = –1 (b) Katakan/Let harga sebiji burger = RMx the price of a burger = RMx harga sebungkus nasi lemak = RMy the price of a packet of nasi lemak = RMy PT SPM Add Math Tkt 4-Jawapan.indd 5 01/03/2023 9:17 AM PENERBIT ILMU BAKTI SDN. BHD.
© Penerbit Ilmu Bakti Sdn. Bhd. (732516-M) 2023 J6 harga sepotong kek = RMz the price of a slice of cake = RMz Maka/Then x + y + 2z = 25 1 2x + y + z = 20 2 x + 2y + z = 19 3 2 – 3 : x – y = 1 4 2 × 2 : 4x + 2y + 2z = 40 5 5 – 1 : 3x + y = 15 4 : x – y = 1 (+) 4x = 16 x = 4 Daripada/From 4 , (4) – y = 1 y = 3 Daripada/From 1 , (4) + (3) + 2z = 25 2z = 18 z = 9 Maka, harga sebiji burger = RM4, harga sebungkus nasi lemak = RM3, harga sepotong kek = RM9. Therefore, the price of a burger = RM4, the price of a packet of nasi lemak = RM3, the price of a slice of cake = RM9. 2 (a) Katakan panjang segi empat tepat ialah x dan lebar ialah y. Let the length of the rectangle be x and the width be y. Perimeter/Perimeter = 44 cm 2x + 2y = 44 x + y = 22 1 Diberi pepenjuru/Given diagonal = 2 65 x2 + y2 = 2 65 x2 + y2 = (2 65 )2 x2 + y2 = 4 × 65 x2 + y2 = 260 2 Daripada/From 1 , y = 22 – x Gantikan y ke dalam 2 , Substitute y into 2 , x2 + (22 – x) 2 = 260 x2 + 484 – 44x + x2 = 260 2x2 – 44x + 224 = 0 x2 – 22x + 112 = 0 (x – 14)(x – 8) = 0 x – 14 = 0 atau or x – 8 = 0 x = 14 x = 8 ∴ Panjang = 14 cm dan lebar = 8 cm ∴ Length = 14 cm and width = 8 cm (b) Jumlah luas kolam = 9π Total area of the pool = 9π pq + 1 4 πp2 = 9π 1 DE – Lengkok EF/Arc EF = π q – p(π 4) = π q = π + πp 4 2 Gantikan q ke dalam 1 , Substitute q into 1 , p(π + πp 4 ) + 1 4 πp2 = 9π pπ + πp2 4 + 1 4 πp2 = 9π pπ + πp2 2 = 9π 2pπ + πp2 = 18π 2p + p2 = 18 p2 + 2p – 18 = 0 p = –(2) ± 2 2 – 4(1)(–18) 2(1) p = 3.359, –5.359 Oleh sebab/Since p > 0, p = 3.359 ZON KBAT 1 Diberi/Given y = (x + 1)(x2 – 6x + 8) 1 y = 4x – 16 2 Gantikan 1 ke dalam 2 , Substitute 1 into 2 , (x + 1)(x2 – 6x + 8) = 4x – 16 (x + 1)(x – 2)(x – 4) = 4(x – 4) (x + 1)(x – 2)(x – 4) – 4(x – 4) = 0 (x – 4)[(x + 1)(x – 2) – 4] = 0 (x – 4)(x2 – 2x + x – 2 – 4) = 0 (x – 4)(x2 – x – 6) = 0 (x – 4)(x + 2)(x – 3) = 0 x – 4 = 0, x + 2 = 0, x – 3 = 0 x = 4, x = –2, x = 3 Apabila/When x = 4, y = 4(4) – 16 = 0 Apabila/When x = –2, y = 4(–2) – 16 = –24 Apabila/When x = 3, y = 4(3) – 16 = –4 Maka, titik persilangan ialah Thus, the points of intersection are (–2, –24), (3, –4), (4, 0). BAB 4 Indeks, Surd dan Logaritma KERTAS 1 BAHAGIAN A 1 (a) 8(2x – 2) = 16x 23 (2x – 2) = (24 )x 2x – 2 + 3 = 24x 2x + 1 = 24x x + 1 = 4x 1 = 3x x = 1 3 (b) 2a = 7b 2 = 7 b a 1 14m = 7b (2 × 7)m = 7b 2 Gantikan 1 ke dalam 2 , Substitute 1 into 2 , (7 b a × 7)m = 7b (7 b a + 1 )m = 7b (7) m( b + a a ) = 7b Bandingkan/Compare, m( b + a a ) = b m = ab a + b 2 (a) 16(23x – 5) = 1 2x 24 × (23x – 5) = 2– x 23x – 5 + 4 = 2– x 3x – 1 = –x 4x = 1 x = 1 4 (b) (3x )2 – 4(3x )(31 ) + 27 = 0 (3x )2 – 12(3x ) + 27 = 0 Katakan/Let u = 3x , maka/then u2 – 12u + 27 = 0 (u – 3)(u – 9) = 0 u = 3 atau/or u = 9 3x = 3 atau/or 3x = 9 3x = 31 atau/or 3x = 32 x = 1 atau/or x = 2 3 (a) 3x = 162 – 3x 3x + 3x = 162 2(3x ) = 162 3x = 81 3x = 34 x = 4 (b) 52x – 1 = 32 log10 52x – 1 = log10 32 (2x – 1) log10 5 = log10 32 (2x – 1) 0.6990 = 1.5051 2x – 1 = 1.5051 0.6990 2x – 1 = 2.1532 2x = 3.1532 x = 1.5766 4 (a) 9u × 27u – 2 = 1 81 (32 )u × (33 )u – 2 = 3–4 32u + 3u – 6 =3–4 2u + 3u – 6 = –4 5u = 2 u = 2 5 (b) 33x 3y = 8 + 27x gh = 8 + (33 )x gh = 8 + 33x gh = 8 + g gh – g = 8 g(h – 1) = 8 g = 8 h – 1 5 (a) 2n – 1 + 2n = 48 2n ( 1 2) + 1(2n ) = 48 3 2 (2n ) = 48 2n = 48 × 2 3 2n = 32 2n = 25 n = 5 (b) 3n + 2 – 3n + 1 + 4(3n ) = 81 3n (32 ) – 3n (3) + 4(3n ) = 81 9(3n ) – 3(3n ) + 4(3n ) = 81 10(3n ) = 81 3n = 8.1 n = log10 8.1 log10 3 = 1.9041 6 (a) (3 – 5 )2 = (3 – 5 )(3 – 5 ) = 9 – 3 5 – 3 5 + ( 5 )2 = 9 – 3 5 – 3 5 + 5 = 14 – 6 5 (b) (3 2 + 3 )(3 2 – 3 ) = (3 2 )2 – ( 3 )2 = 9(2) – 3 = 18 – 3 = 15 7 (a) 3 3 + 6 = 3 3 + 6 × 3 – 6 3 – 6 = 3(3 – 6 ) (3)2 – ( 6 ) 2 = 3(3 – 6 ) 9 – 6 = 3(3 – 6 ) 3 = 3 – 6 PT SPM Add Math Tkt 4-Jawapan.indd 6 01/03/2023 9:17 AM PENERBIT ILMU BAKTI SDN. BHD.