[Lihat halaman sebelah3472/1 @2011 Trial SPM Hak Cipta BPSBPSK SULIT Kertas soalan ini mengandungi 24 halaman bercetak JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU 1. Tulis nama dan tingkatan anda pada ruangan yang disediakan. 2. Kertas soalan ini adalah dalam dwibahasa. 3. Soalan dalam bahasa Inggeris mendahului soalan yang sepadan dalam bahasa Melayu. 4. Calon dibenarkan menjawab keseluruhan atau sebahagian soalan sama ada dalam bahasa Inggeris atau bahasa Melayu. 5. Calon dikehendaki membaca maklumat di halaman belakang kertas soalan ini. BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN PERCUBAAN SPM ADDITIONAL MATHEMATICS Kertas 1 Ogos 2011 2 jam Dua jam 3472 / 1 Untuk Kegunaan Pemeriksa Soalan Markah Penuh Markah Diperolehi 1 3 2 3 3 3 4 3 5 3 6 3 7 3 8 4 9 4 10 4 11 3 12 3 13 4 14 3 15 3 16 3 17 4 18 2 19 3 20 3 21 3 22 3 23 3 24 3 25 4 TOTAL 80 Name : ………………..…………… Form : ………………………..…… http://chngtuition.blogspot.com SULIT 3472/1 [Lihat halaman sebelah 3472/1 @2011 Trial SPM Hak Cipta BPSBPSK SULIT 2 BLANK PAGE HALAMAN KOSONG http://chngtuition.blogspot.com SULIT 3472/1 [Lihat halaman sebelah 3472/1 @2011 Trial SPM Hak Cipta BPSBPSK SULIT 3 The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used. ALGEBRA 1 2 4 2 b b ac x a ÷ ± ÷ = 2 a m × a n = a m + n 3 a m ÷ a n = a m - n 4 (a m ) n = a nm 5 log a mn = log a m + log a n 6 log a n m = log a m - log a n 7 log a m n = n log a m 8 log a b = a b c c log log 9 T n = a + (n-1)d 10 S n = ] ) 1 ( 2 [ 2 d n a n ÷ + 11 T n = ar n-1 12 S n = r r a r r a n n ÷ ÷ = ÷ ÷ 1 ) 1 ( 1 ) 1 ( , (r = 1) 13 r a S ÷ = · 1 , r <1 CALCULUS 1 y = uv , dx du v dx dv u dx dy + = 2 v u y = , 2 v dx dv u dx du v dx dy ÷ = , 3 dx du du dy dx dy × = 4 Area under a curve = } b a y dx or = } b a x dy 5 Volume generated = } b a y 2 t dx or = } b a x 2 t dy 5 A point dividing a segment of a line ( x,y) = , 2 1 \ | + + n m mx nx | . | + + n m my ny 2 1 6 Area of triangle = ) ( ) ( 2 1 3 1 2 3 1 2 1 3 3 2 2 1 1 y x y x y x y x y x y x + + ÷ + + 1 Distance = 2 1 2 2 1 2 ) ( ) ( y y x x ÷ + ÷ 2 Midpoint (x , y) = \ | + 2 2 1 x x , | . | + 2 2 1 y y 3 2 2 y x r + = 4 2 2 ˆ xi yj r x y + = + GEOMETRY http://chngtuition.blogspot.com SULIT 3472/1 [ Lihat halaman sebelah SULIT 3472/1 @2011 Trial SPM Hak Cipta BPSBPSK 4 STATISTIC 1 Arc length, s = ru 2 Area of sector , L = 2 1 2 r u 3 sin 2 A + cos 2 A = 1 4 sec 2 A = 1 + tan 2 A 5 cosec 2 A = 1 + cot 2 A 6 sin 2A = 2 sinA cosA 7 cos 2A = cos 2 A – sin 2 A = 2 cos 2 A - 1 = 1 - 2 sin 2 A 8 tan 2A = A A 2 tan 1 tan 2 ÷ TRIGONOMETRY 9 sin (A± B) = sinA cosB ± cosA sinB 10 cos (A± B) = cosA cosB sinA sinB 11 tan (A± B) = B A B A tan tan 1 tan tan ± 12 C c B b A a sin sin sin = = 13 a 2 = b 2 + c 2 - 2bc cosA 14 Area of triangle = C absin 2 1 1 x = N x ¿ 2 x = ¿ ¿ f fx 3 o = N x x ¿ ÷ 2 ) ( = 2 _ 2 x N x ÷ ¿ 4 o = ¿ ¿ ÷ f x x f 2 ) ( = 2 2 x f fx ÷ ¿ ¿ 5 m = C f F N L m ( ( ( ( ¸ ( ¸ ÷ + 2 1 6 1 0 100 Q I Q = × 7 1 1 1 w I w I ¿ ¿ = 8 )! ( ! r n n P r n ÷ = 9 ! )! ( ! r r n n C r n ÷ = 10 P(AB) = P(A)+P(B)- P(A·B) 11 P (X = r) = r n r r n q p C ÷ , p + q = 1 12 Mean µ = np 13 npq = o 14 z = o µ ÷ x http://chngtuition.blogspot.com 5 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0,1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1) z 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 Minus / Tolak 0.0 0.1 0.2 0.3 0.4 0.5000 0.4602 0.4207 0.3821 0.3446 0.4960 0.4562 0.4168 0.3783 0.3409 0.4920 0.4522 0.4129 0.3745 0.3372 0.4880 0.4483 0.4090 0.3707 0.3336 0.4840 0.4443 0.4052 0.3669 0.3300 0.4801 0.4404 0.4013 0.3632 0.3264 0.4761 0.4364 0.3974 0.3594 0.3228 0.4721 0.4325 0.3936 0.3557 0.3192 0.4681 0.4286 0.3897 0.3520 0.3156 0.4641 0.4247 0.3859 0.3483 0.3121 4 4 4 4 4 8 8 8 7 7 12 12 12 11 11 16 16 15 15 15 20 20 19 19 18 24 24 23 22 22 28 28 27 26 25 32 32 31 30 29 36 36 35 34 32 0.5 0.6 0.7 0.8 0.9 0.3085 0.2743 0.2420 0.2119 0.1841 0.3050 0.2709 0.2389 0.2090 0.1814 0.3015 0.2676 0.2358 0.2061 0.1788 0.2981 0.2643 0.2327 0.2033 0.1762 0.2946 0.2611 0.2296 0.2005 0.1736 0.2912 0.2578 0.2266 0.1977 0.1711 0.2877 0.2546 0.2236 0.1949 0.1685 0.2843 0.2514 0.2206 0.1922 0.1660 0.2810 0.2483 0.2177 0.1894 0.1635 0.2776 0.2451 0.2148 0.1867 0.1611 3 3 3 3 3 7 7 6 5 5 10 10 9 8 8 14 13 12 11 10 17 16 15 14 13 20 19 18 16 15 24 23 21 19 18 27 26 24 22 20 31 29 27 25 23 1.0 1.1 1.2 1.3 1.4 0.1587 0.1357 0.1151 0.0968 0.0808 0.1562 0.1335 0.1131 0.0951 0.0793 0.1539 0.1314 0.1112 0.0934 0.0778 0.1515 0.1292 0.1093 0.0918 0.0764 0.1492 0.1271 0.1075 0.0901 0.0749 0.1469 0.1251 0.1056 0.0885 0.0735 0.1446 0.1230 0.1038 0.0869 0.0721 0.1423 0.1210 0.1020 0.0853 0.0708 0.1401 0.1190 0.1003 0.0838 0.0694 0.1379 0.1170 0.0985 0.0823 0.0681 2 2 2 2 1 5 4 4 3 3 7 6 6 5 4 9 8 7 6 6 12 10 9 8 7 14 12 11 10 8 16 14 13 11 10 19 16 15 13 11 21 18 17 14 13 1.5 1.6 1.7 1.8 1.9 0.0668 0.0548 0.0446 0.0359 0.0287 0.0655 0.0537 0.0436 0.0351 0.0281 0.0643 0.0526 0.0427 0.0344 0.0274 0.0630 0.0516 0.0418 0.0336 0.0268 0.0618 0.0505 0.0409 0.0329 0.0262 0.0606 0.0495 0.0401 0.0322 0.0256 0.0594 0.0485 0.0392 0.0314 0.0250 0.0582 0..0475 0.0384 0.0307 0.0244 0.0571 0.0465 0.0375 0.0301 0.0239 0.0559 0.0455 0.0367 0.0294 0.0233 1 1 1 1 1 2 2 2 1 1 4 3 3 2 2 5 4 4 3 2 6 5 4 4 3 7 6 5 4 4 8 7 6 5 4 10 8 7 6 5 11 9 8 6 5 2.0 2.1 2.2 2.3 0.0228 0.0179 0.0139 0.0107 0.0222 0.0174 0.0136 0.0104 0.0217 0.0170 0.0132 0.0102 0.0212 0.0166 0.0129 0.00990 0.0207 0.0162 0.0125 0.00964 0.0202 0.0158 0.0122 0.00939 0.0197 0.0154 0.0119 0.00914 0.0192 0.0150 0.0116 0.00889 0.0188 0.0146 0.0113 0.00866 0.0183 0.0143 0.0110 0.00842 0 0 0 0 3 2 1 1 1 1 5 5 1 1 1 1 8 7 2 2 1 1 10 9 2 2 2 1 13 12 3 2 2 2 15 14 3 3 2 2 18 16 4 3 3 2 20 16 4 4 3 2 23 21 2.4 0.00820 0.00798 0.00776 0.00755 0.00734 0.00714 0.00695 0.00676 0.00657 0.00639 2 2 4 4 6 6 8 7 11 9 13 11 15 13 17 15 19 17 2.5 2.6 2.7 2.8 2.9 0.00621 0.00466 0.00347 0.00256 0.00187 0.00604 0.00453 0.00336 0.00248 0.00181 0.00587 0.00440 0.00326 0.00240 0.00175 0.00570 0.00427 0.00317 0.00233 0.00169 0.00554 0.00415 0.00307 0.00226 0.00164 0.00539 0.00402 0.00298 0.00219 0.00159 0.00523 0.00391 0.00289 0.00212 0.00154 0.00508 0.00379 0.00280 0.00205 0.00149 0.00494 0.00368 0.00272 0.00199 0.00144 0.00480 0.00357 0.00264 0.00193 0.00139 2 1 1 1 0 3 2 2 1 1 5 3 3 2 1 6 5 4 3 2 8 6 5 4 2 9 7 6 4 3 11 9 7 5 3 12 9 8 6 4 14 10 9 6 4 3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00104 0.00100 0 1 1 2 2 2 3 3 4 Q(z) z f (z) O Example / Contoh: If X ~ N(0, 1), then P(X > k) = Q(k) Jika X ~ N(0, 1), maka P(X > k) = Q(k) | . | \ | ÷ = 2 2 1 exp 2 1 ) ( z z f t } · = k dz z f z Q ) ( ) ( http://chngtuition.blogspot.com 6 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT Answer all questions. Jawab semua soalan. 1. Diagram 1 shows the function 3 : , 0. x f x x x + ÷ = Rajah 1 menunjukkan fungsi 3 : , 0. x f x x x + ÷ = Find Cari (a) the value of m nilai m (b) image of -2 imej bagi -2 [ 3 marks ] [3 markah] Answer/Jawapan : (a) (b) For examiner’s use only 3 1 m -1 x f 3 x x + Diagram 1 Rajah 1 http://chngtuition.blogspot.com 7 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 2. Given that the function : 2 3 dan h x x ÷ ÷ 2 : ( 2) 1. k x x ÷ ÷ + Diberi fungsi dan 3 2 : ÷ ÷ x x h 2 : ( 2) 1. k x x ÷ ÷ + Find Cari (a) h - 1 (x) (b) h -1 k(3) [ 3 marks ] [3 markah] Answer/Jawapan : (a) (b) 3. Given that the functions 1 1 5 : , : and : 4 3 . 2 x f x ax b g x fg x x ÷ ÷ + ÷ + ÷ ÷ + Find the value of a and of b. [3 marks] Diberi fungsi-fungsi 1 1 5 : , : and : 4 3 . 2 x f x ax b g x fg x x ÷ ÷ + ÷ + ÷ ÷ + Cari nilai a dan nilai b. [ 3 markah ] Answer/Jawapan : 3 2 3 3 For examiner’s use only http://chngtuition.blogspot.com 8 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 4. Find the range of values of x for 2 ( 5) 5. x x x + > + [3 marks] Cari julat nilai x bagi 2 ( 5) 5 x x x + > + [ 3 markah ] Answer/Jawapan : 5. The quadratic equation 2 1 3 px x x p ÷ = ÷ + , where p is a constant, has no roots. Find the range of values of p. [3 marks] Persamaan kuadratik 2 1 3 px x x p ÷ = ÷ + , dengan keadaan p ialah pemalar ,tidak mempunyai punca. Cari julat nilai p. [ 3 markah ] Answer/Jawapan : 3 5 3 4 For examiner’s use only http://chngtuition.blogspot.com 9 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 6. Diagram 6 shows the graph of a quadratic function for 2 ( ) 2( ) 3 f x x m = ÷ ÷ . Rajah 6 menunjukkan graf fungsi kuadratik bagi 2 ( ) 2( ) 3 f x x m = ÷ ÷ . Diagram 6 Rajah 6 Find Cari (a) the value of m, nilai m, (b) the equation of the axis of symmetry, persamaan paksi simetri, (c) the coordinates of the minimum point. koordinat-koordinat titik minimum. [ 3 marks ] [3 markah] Answer/Jawapan : (a) (b) (c) For examiner’s use only 3 6 - f(x) 0 (3, 5) x - (÷1, 5) http://chngtuition.blogspot.com 10 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 7. Solve the equation 2 1 1 25 625 x x + ÷ = . [3 marks] Selesaikan persamaan 2 1 1 25 625 x x + ÷ = . [ 3 markah ] Answer/Jawapan : 8. Solve the equation 9 81 log log 3 1 x x + =÷ . [ 4 marks ] Selesaikan persamaan 9 81 log log 3 1 x x + =÷ . [4 markah] Answer/Jawapan : For examiner’s use only 3 7 4 8 http://chngtuition.blogspot.com 11 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 9. The first three terms of an arithmetic progression are 2h ÷ 4, 5h ÷ 1 and 6h + 4. Tiga sebutan pertama suatu janjang aritmetik adalah 2h ÷ 4, 5h ÷ 1 dan 6h + 4. Find, Cari, (a) the value of h, nilai h, (b) the sum of next 20 terms after the third term. hasil tambah 20 sebutan berikutnya selepas sebutan ketiga. [ 4 marks ] [4 markah] Answer/Jawapan : (a) (b) For examiner’s use only 4 9 http://chngtuition.blogspot.com 12 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 10. The sum to infinity of a geometric progression is 3 and the sum of its first two terms is 2 2 . 3 Hasil tambah hingga sebutan ketakterhinggaan bagi janjang geometri ialah 3 dan hasil tambah dua sebutan pertama ialah 2 2 . 3 Find the common ratio (r<0) and the first term. Cari nilai nisbah sepunya (r<0) dan sebutan pertama. [4 marks ] [4 markah] Answer/Jawapan 11. Aiman saves RM 300 from his salary in a certain month. In each succeeding month, he saves RM 50 more than the previous month. Aiman telah membuat simpanan RM 300 daripada wang gajinya pada bulan tertentu. Pada bulan yang berikutnya, beliau telah menambah simpanannya sebanyak RM 50 melebihi bulan sebelumnya. Calculate the amount of his saving in 3 years. Hitungkan jumlah simpanan beliau dalam masa 3 tahun. [3 marks] [3 markah] Answer/Jawapan : 3 11 For examiner’s use only 4 10 http://chngtuition.blogspot.com 13 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 12. The variables x and y are related by the equation . p x qy x + = . A straight line graph is obtained by plotting xy against x 2 , as shown in Diagram 12, where p and q are constants. Pemboleh ubah x dan y dihubungkan oleh persamaan . p x qy x + = . Graf garis lurus diperoleh dengan memplot xy melawan x 2 , seperti ditunjukkan pada Rajah 12 , dengan keadaan p dan q ialah pemalar. Find the value of p and of q. Cari nilai p dan nilai q. [ 3 marks ] [3 markah] Answer/Jawapan : Diagram 12 Rajah 12 x 2 xy (6,3) (4, 0) 0 For examiner’s use only 3 12 http://chngtuition.blogspot.com 14 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 13. In Diagram 13, the straight line PQ has an equation 1 8 2 x y t + = with gradient of 1 . 4 ÷ PQ intersects the x-axis at point P and intersects the y-axis at point Q. Dalam Rajah 13, garis lurus PQ mempunyai persamaan 1 8 2 x y t + = dengan kecerunan 1 . 4 ÷ PQ menyilang paksi-x di titik P dan menyilang paksi-y di titik Q. Find, Cari, (a) the value of t, nilai bagi t (b) the equation of the straight line that passes through Q and is perpendicular to PQ. persamaan garis lurus yang melalui Q dan berserenjang dengan PQ. [ 4 marks ] [4 markah] Answer/Jawapan : (a) (b) P (8, 0) Q (0, 2t) y x O Diagram 13 Rajah 13 For examiner’s use only 4 13 http://chngtuition.blogspot.com 15 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 14. Given that the point ( 4, 2) A ÷ and (4, 6) B . The point ( 2,3) P ÷ lies on the straight line of AB such that n m PB AP : : = . Diberi titik ( 4, 2) A ÷ dan (4, 6) B . Titik ( 2,3) P ÷ terletak pada garislurus AB dengan keadaan n m PB AP : : = . Find the value of m and of n. Cari nilai m dan nilai n . [ 3 marks ] [3 markah] Answer/Jawapan : For examiner’s use only 3 14 http://chngtuition.blogspot.com 16 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 15. Diagram 15 shows triangle ABC. K is a point on BC such that 2 : 1 : = KC BK . Given that 3 AB x = and 6 AC y = . Rajah 15 menunjukkan sebuah segitiga ABC. K ialah titik yang terletak digaris BC berkeadaan 2 : 1 : = KC BK . Diberi bahawa 3 AB x = dan 6 AC y = . Express in terms of x and y . Ungkapkan dalam sebutan x dan , y (a) BC (b) AK [ 3 marks ] [3 markah] Answer/Jawapan : (a) (b) A K C B Diagram 15 Rajah 15 For examiner’s use only 3 15 http://chngtuition.blogspot.com 17 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 16. Diagram 16 shows the vector and . OP OQ Given that ) 1 , 12 ( and ) 3 , 4 ( , ) 0 , 0 ( Q P O ÷ , find in terms of i and , j Diberi bahawa ) 1 , 12 ( and ) 3 , 4 ( , ) 0 , 0 ( Q P O ÷ , cari dalam sebutan i dan , j (a) PQ (b) the unit vector in the direction of . PQ vektor unit dalam arah . PQ [3 marks ] [3 markah] Answer/Jawapan : (a) (b) For examiner’s use only 3 16 x y Q (12,1) P (4,-3) O Diagram 16 Rajah 16 http://chngtuition.blogspot.com 18 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 17. Diagram 17 shows sector OAB and sector OCD, with centre O . Rajah 17 menunjukkan sector OAB dan sekitar OCD,yang berpusatkan O. Given that 0.8 u = rad and the area of sector OAB is 2 40cm . Diberi bahawa 0.8 u = rad dan luas bagi sektor OAB is 2 40cm . Find the value of Cari nilai bagi (a) k (b) the perimeter of the shaded region. perimeter kawasan yang berlorek [4 marks] [4 markah] Answer/Jawapan : (a) (b) 4 17 For examiner’s use only u O A B C D k cm Diagram 17 Rajah 17 3k cm http://chngtuition.blogspot.com 19 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 18. Given that 13 12 sin ÷ = A and , 5 4 cos = B where A and B are in the same quadrant. Find the value of ( ). sin B A ÷ [ 2 marks ] Diberi 13 12 sin ÷ = A dan , 5 4 cos = B dimana A dan B berada dalam sukuan yang sama. Cari nilai ( ). sin B A ÷ [ 2 markah ] Answer/Jawapan : 19. Given that } = 2 1 . 5 ) ( dx x f Find the value of ( ) . ) ( 1 2 dx x f x } + [ 3 marks ] Diberi } = 2 1 . 5 ) ( dx x f Cari nilai ( ) . ) ( 1 2 dx x f x } + [ 3 markah ] Answer/Jawapan : For examiner’s use only 2 18 3 19 http://chngtuition.blogspot.com 20 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 20. Given that 2 3 3 ( 4) , y x x = + find the value of dx dy when 1 x = ÷ . [ 3 marks ] Diberi 2 3 3 ( 4) , y x x = + cari nilai dx dy apabila 1 x = ÷ . [ 3 markah ] Answer/Jawapan : 21. The volume, V cm 3 of water in a vessel is given by 2 45 450 h h V ÷ = , where h cm is the height of the water in the vessel. Isipadu, V cm 3 air dalam takungan diberi oleh 2 45 450 h h V ÷ = , dimana h cm ialah tinggi air dalam takungan itu. . Calculate Kira (a) the value of h when V is maximum. nilai h apabila V maksimum. (b) the maximum volume of water in the vessel. isipadu maksimum air dalam takungan itu [ 3 marks ] [ 3 markah ] Answer/Jawapan : (a) (b) For examiner’s use only 3 20 3 21 http://chngtuition.blogspot.com 21 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 22. A set of positive integers consists of 3, 7, p, 5, 1 . Given that the mean for the set of data is 5. Satu set integer positif mengandungi 3, 7, p, 5, 1 . Diberi min untuk set data ini ialah 5. Find Cari (a) the value of p. nilai p. (b) the variance for the set of data. varian bagi set data ini. [ 3 marks ] [3 markah] Answer/Jawapan : (a) (b) 23. (a) Three letters are chosen from the word “BERHAD” . Calculate the number of different ways in which the three letters can be chosen if there is no restriction. Tiga huruf dipilih daripada perkataan “BERHAD” . Hitung bilangan cara berlainan huruf ini boleh dipilih jika tiada sebarang syarat dikenakan. (b) In how many ways can the word “BERHAD” be arranged if the vowel are arranged side by side? . Dalam berapa carakah huruf-huruf dalam perkataan “BERHAD” dapat disusun jika huruf vokal hendaklah disusun sebelah menyebelah. [ 3 marks ] [3 markah] Answer/Jawapan : (a) (b) 3 23 For examiner’s use only 3 22 http://chngtuition.blogspot.com 22 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 24. The probability that Rajesh qualifies for the final 400m event is 1 4 while the probability that Razali qualify is 1 3 . Kebarangkalian bahawa Rajesh layak kepertandingan akhir acara 400m ialah 1 4 manakala kebarangkalian untuk Razali layak ialah 1 . 3 Find the probability that Hitung kebarangkalain bahawa (a) both of them did not qualify for the final. kedua-dua mereka tidak layak kepertandingan akhir. (b) only one of them qualifies for the final. hanya salah seorang daripada mereka layak kepertandingan akhir. [ 3 marks ] [3 markah] Answer/Jawapan : (a) (b) For examiner’s use only 3 24 http://chngtuition.blogspot.com 23 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT 25. Diagram 25 shows a normal distribution graph. Rajah 25 menunjukkan graf taburan normal . X is a continuous random variable which is normally distributed with a mean of 55 kg and a standard deviation of 3 kg. Given that the area of the shaded region is . X ialah pembolehubah rawak selanjar yang tertabur secara normal dengan min 55 kg dan sisihan piawai 3 kg. Diberi bahawa luas bagi kawasan berlorek ialah . Find Cari (a) P(X (b) the value of k . nilai bagi k. [ 4 marks ] [4 markah] Answer/Jawapan : (a) (b) For examiner’s use only 4 25 END OF QUESTION PAPER KERTAS SOALAN TAMAT k f (x) x Diagram 25 Rajah 25 http://chngtuition.blogspot.com 24 SULIT 3472/1 [Lihat halaman sebelah 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK SULIT INFORMATION FOR CANDIDATES MAKLUMAT UNTUK CALON 1. This question paper consists of 25 questions Kertas soalan ini mengandungi 25 soalan 2. Answer all questions. Jawab semua soalan 3. Write your answers in the spaces provided in the question paper. Tulis jawapan anda dalam ruang yang disediakan dalam kertas soalan. 4. Show your working. It may help you to get marks. Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini boleh membantu anda untuk mendapatkan markah. 5. If you wish to change your answer, cross out the answer that you have done. Then write down the new answer. Sekiranya anda hendak menukar jawapan, batalkan jawapan yang telah dibuat. Kemudian tulis jawapan yang baru. 6. The diagrams in the questions provided are not drawn to scale unless stated. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. 7. The marks allocated for each question are shown in brackets. Markah yang diperuntukkan bagi setiap soalan ditunjukkan dalam kurungan. 8. A list of formulae is provided on pages 3 to 5. Satu senarai rumus disediakan di halaman 3 hingga 5. 9. A booklet of four-figure mathematical tables is provided. Sebuah buku sifir matematik empat angka disediakan. 10. You may use a non-programmable scientific calculator. Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram. 11. Hand in this question paper to the invigilator at the end of the examination. Serahkan kertas soalan ini kepada pengawas peperiksaan di akhir peperiksaan. http://chngtuition.blogspot.com SULIT 1 3472/2 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 3472/2 Matematik Tambahan Kertas 2 2 ½ jam Ogos 2011 BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2011 ADDITIONAL MATHEMATICS Kertas 2 Dua jam tiga puluh minit JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU 1. This question paper consists of three sections : Section A, Section B and Section C. 2. Answer all questions in Section A , four questions from Section B and two questions from Section C. 3. Give only one answer / solution to each question. 4. Show your working. It may help you to get marks. 5. The diagram in the questions provided are not drawn to scale unless stated. 6. The marks allocated for each question and sub-part of a question are shown in brackets. 7. A list of formulae and normal distribution table is provided on pages 2 to 4. 8. A booklet of four-figure mathematical tables is provided. 9. You may use a non-programmable scientific calculator. Kertas soalan ini mengandungi 20 halaman bercetak http://chngtuition.blogspot.com SULIT 3472/2 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 2 The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used. ALGEBRA 1 x = a ac b b 2 4 2 ÷ ± ÷ 2 a m × a n = a m + n 3 a m ÷ a n = a m - n 4 (a m ) n = a nm 5 log a mn = log a m + log a n 6 log a n m = log a m - log a n 7 log a m n = n log a m 8 log a b = a b c c log log 9 T n = a + (n-1)d 10 S n = ] ) 1 ( 2 [ 2 d n a n ÷ + 11 T n = ar n-1 12 S n = r r a r r a n n ÷ ÷ = ÷ ÷ 1 ) 1 ( 1 ) 1 ( , (r = 1) 13 r a S ÷ = · 1 , r <1 CALCULUS 1 y = uv , dx du v dx dv u dx dy + = 2 v u y = , 2 du dv v u dy dx dx dx v ÷ = , 3 dx du du dy dx dy × = 4 Area under a curve = } b a y dx or = } b a x dy 5 Volume generated = } b a y 2 t dx or = } b a x 2 t dy 5 A point dividing a segment of a line ( x,y) = , 2 1 \ | + + n m mx nx | . | + + n m my ny 2 1 6. Area of triangle = ) ( ) ( 2 1 3 1 2 3 1 2 1 3 3 2 2 1 1 y x y x y x y x y x y x + + ÷ + + 1 Distance = 2 2 1 2 2 1 ) ( ) ( y y x x ÷ + ÷ 2 Midpoint (x , y) = \ | + 2 2 1 x x , | . | + 2 2 1 y y 3 2 2 y x r + = 4 2 2 xi yj r x y . + = + GEOM ETRY http://chngtuition.blogspot.com SULIT 3472/2 [ Lihat halaman sebelah 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 3 STATISTIC TRIGONOMETRY 1 Arc length, s = ru 2 Area of sector , A = 2 1 2 r u 3 sin 2 A + cos 2 A = 1 4 sec 2 A = 1 + tan 2 A 5 cosec 2 A = 1 + cot 2 A 6 sin2A = 2 sinAcosA 7 cos 2A = cos 2 A – sin 2 A = 2 cos 2 A-1 = 1- 2 sin 2 A 8 tan2A = A A 2 tan 1 tan 2 ÷ 9 sin (A± B) = sinAcosB ± cosAsinB 10 cos (A± B) = cos AcosB sinAsinB 11 tan (A± B) = B A B A tan tan 1 tan tan ± 12 C c B b A a sin sin sin = = 13 a 2 = b 2 +c 2 - 2bc cosA 14 Area of triangle = C absin 2 1 1 x = N x ¿ 2 x = ¿ ¿ f fx 3 o = N x x ¿ ÷ 2 ) ( = 2 _ 2 x N x ÷ ¿ 4 o = ¿ ¿ ÷ f x x f 2 ) ( = 2 2 x f fx ÷ ¿ ¿ 5 M = C f F N L m ( ( ( ( ¸ ( ¸ ÷ + 2 1 6 100 0 1 × = P P I 7 1 1 1 w I w I ¿ ¿ = 8 )! ( ! r n n P r n ÷ = 9 ! )! ( ! r r n n C r n ÷ = 10 P(AB)=P(A)+P(B)-P(A·B) 11 p (X=r) = r n r r n q p C ÷ , p + q = 1 12 Mean , µ = np 13 npq = o 14 z = o µ ÷ x http://chngtuition.blogspot.com SULIT 4 3472/2 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0,1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1) z 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 Minus / Tolak 0.0 0.1 0.2 0.3 0.4 0.5000 0.4602 0.4207 0.3821 0.3446 0.4960 0.4562 0.4168 0.3783 0.3409 0.4920 0.4522 0.4129 0.3745 0.3372 0.4880 0.4483 0.4090 0.3707 0.3336 0.4840 0.4443 0.4052 0.3669 0.3300 0.4801 0.4404 0.4013 0.3632 0.3264 0.4761 0.4364 0.3974 0.3594 0.3228 0.4721 0.4325 0.3936 0.3557 0.3192 0.4681 0.4286 0.3897 0.3520 0.3156 0.4641 0.4247 0.3859 0.3483 0.3121 4 4 4 4 4 8 8 8 7 7 12 12 12 11 11 16 16 15 15 15 20 20 19 19 18 24 24 23 22 22 28 28 27 26 25 32 32 31 30 29 36 36 35 34 32 0.5 0.6 0.7 0.8 0.9 0.3085 0.2743 0.2420 0.2119 0.1841 0.3050 0.2709 0.2389 0.2090 0.1814 0.3015 0.2676 0.2358 0.2061 0.1788 0.2981 0.2643 0.2327 0.2033 0.1762 0.2946 0.2611 0.2296 0.2005 0.1736 0.2912 0.2578 0.2266 0.1977 0.1711 0.2877 0.2546 0.2236 0.1949 0.1685 0.2843 0.2514 0.2206 0.1922 0.1660 0.2810 0.2483 0.2177 0.1894 0.1635 0.2776 0.2451 0.2148 0.1867 0.1611 3 3 3 3 3 7 7 6 5 5 10 10 9 8 8 14 13 12 11 10 17 16 15 14 13 20 19 18 16 15 24 23 21 19 18 27 26 24 22 20 31 29 27 25 23 1.0 1.1 1.2 1.3 1.4 0.1587 0.1357 0.1151 0.0968 0.0808 0.1562 0.1335 0.1131 0.0951 0.0793 0.1539 0.1314 0.1112 0.0934 0.0778 0.1515 0.1292 0.1093 0.0918 0.0764 0.1492 0.1271 0.1075 0.0901 0.0749 0.1469 0.1251 0.1056 0.0885 0.0735 0.1446 0.1230 0.1038 0.0869 0.0721 0.1423 0.1210 0.1020 0.0853 0.0708 0.1401 0.1190 0.1003 0.0838 0.0694 0.1379 0.1170 0.0985 0.0823 0.0681 2 2 2 2 1 5 4 4 3 3 7 6 6 5 4 9 8 7 6 6 12 10 9 8 7 14 12 11 10 8 16 14 13 11 10 19 16 15 13 11 21 18 17 14 13 1.5 1.6 1.7 1.8 1.9 0.0668 0.0548 0.0446 0.0359 0.0287 0.0655 0.0537 0.0436 0.0351 0.0281 0.0643 0.0526 0.0427 0.0344 0.0274 0.0630 0.0516 0.0418 0.0336 0.0268 0.0618 0.0505 0.0409 0.0329 0.0262 0.0606 0.0495 0.0401 0.0322 0.0256 0.0594 0.0485 0.0392 0.0314 0.0250 0.0582 0..0475 0.0384 0.0307 0.0244 0.0571 0.0465 0.0375 0.0301 0.0239 0.0559 0.0455 0.0367 0.0294 0.0233 1 1 1 1 1 2 2 2 1 1 4 3 3 2 2 5 4 4 3 2 6 5 4 4 3 7 6 5 4 4 8 7 6 5 4 10 8 7 6 5 11 9 8 6 5 2.0 2.1 2.2 2.3 0.0228 0.0179 0.0139 0.0107 0.0222 0.0174 0.0136 0.0104 0.0217 0.0170 0.0132 0.0102 0.0212 0.0166 0.0129 0.00990 0.0207 0.0162 0.0125 0.00964 0.0202 0.0158 0.0122 0.00939 0.0197 0.0154 0.0119 0.00914 0.0192 0.0150 0.0116 0.00889 0.0188 0.0146 0.0113 0.00866 0.0183 0.0143 0.0110 0.00842 0 0 0 0 3 2 1 1 1 1 5 5 1 1 1 1 8 7 2 2 1 1 10 9 2 2 2 1 13 12 3 2 2 2 15 14 3 3 2 2 18 16 4 3 3 2 20 16 4 4 3 2 23 21 2.4 0.00820 0.00798 0.00776 0.00755 0.00734 0.00714 0.00695 0.00676 0.00657 0.00639 2 2 4 4 6 6 8 7 11 9 13 11 15 13 17 15 19 17 2.5 2.6 2.7 2.8 2.9 0.00621 0.00466 0.00347 0.00256 0.00187 0.00604 0.00453 0.00336 0.00248 0.00181 0.00587 0.00440 0.00326 0.00240 0.00175 0.00570 0.00427 0.00317 0.00233 0.00169 0.00554 0.00415 0.00307 0.00226 0.00164 0.00539 0.00402 0.00298 0.00219 0.00159 0.00523 0.00391 0.00289 0.00212 0.00154 0.00508 0.00379 0.00280 0.00205 0.00149 0.00494 0.00368 0.00272 0.00199 0.00144 0.00480 0.00357 0.00264 0.00193 0.00139 2 1 1 1 0 3 2 2 1 1 5 3 3 2 1 6 5 4 3 2 8 6 5 4 2 9 7 6 4 3 11 9 7 5 3 12 9 8 6 4 14 10 9 6 4 3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00104 0.00100 0 1 1 2 2 2 3 3 4 4 Q(z) z f (z) O Example / Contoh: If X ~ N(0, 1), then P(X > k) = Q(k) Jika X ~ N(0, 1), maka P(X > k) = Q(k) | . | \ | ÷ = 2 2 1 exp 2 1 ) ( z z f t } · = k dz z f z Q ) ( ) ( http://chngtuition.blogspot.com SULIT 3472/2 [ Lihat halaman sebelah 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 5 Section A Bahagian A [40 marks] [40 markah] Answer all questions. Jawab semua soalan. 1 Solve the following simultaneous equations: Selesaikan persamaan serentak berikut: 4x + 3y = x 2 – xy = 8 Give your answer correct to 3 significant figures. Beri jawapan betul kepada 3 angka bererti. [5 marks] [5 markah] 2 (a) Prove that 0 sin(2 90 ) cos 2 u u + = [2 marks] Buktikan 0 sin(2 90 ) cos 2 u u + = [2 markah] (b) i. Sketch the graph of u 2 cos 2 = y for . 2 0 t u s s [3 marks] Lakar graf bagi of u 2 cos 2 = y untuk . 2 0 t u s s [3 markah] ii. Hence, using the same axes, sketch a suitable straight line to find the number of solution for the equation 0 sin(2 90 ) 1 . u u t + = ÷ State the number of solutions. [3 marks] Seterusnya, dengan menggunakan paksi yang sama, lakar satu garis lurus yang sesuai untuk mencari bilangan penyelesaian bagi persamaan 0 sin(2 90 ) 1 . u u t + = ÷ Nyatakan bilangan penyelesaian itu. [3 markah] http://chngtuition.blogspot.com SULIT 3472/2 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 6 3 Ravi and Hamid were given a piece of wire each, it is to be bend into several parts successively as shown in Diagram 3. Ravi dan Hamid diberi seutas dawai seorang untuk dibengkokkan kepada beberapa bahagian berturut-turut seperti yang ditunjukkan dalam Rajah 3. The first part must be measured A cm and every part should be shortened by D cm respectively. Ravi’s wire which is 720 cm was bent exactly into 10 parts and Hamid’s which is 1040 cm was bent exactly 20 parts. Bahagian pertama mesti berukuran A cm and setiap bahagian seterusnya dipendekkan sebanyak D cm masing-masing. Dawai Ravi, yang panjangnya 720 cm dibengkokkan tepat kepada 10 bahagian dan Dawai Hamid, yang panjangnya 1040 cm dibengkokkan tepat kepada 20 bahagian. Calculate, Hitung, (a) the value of A and of D, [4 marks] nilai bagi A dan nilai bagi D. [4 markah] (b) the difference of the length of the last part of the both wire. [2 marks] beza di antara panjang bahagian terakhir bagi kedua-dua dawai tesebut. [2 markah] A cm Diagram 3 Rajah 3 http://chngtuition.blogspot.com SULIT 3472/2 [ Lihat halaman sebelah 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 7 4. Diagram 4 shows the straight line y = x + 9 intersecting the curve y = (x – 3) 2 at points M and N (7,16). Rajah 4 menunjukkan garis lurus y = x + 9 bersilang dengan lengkung y = (x – 3) 2 pada titik-titik M dan N(7,16). Diagram 4 Rajah 4 Find Cari (a) the area of the shaded region K, [4 marks] luas rantau berlorek K, [4 markah] (b) the volume generated when the shaded region A is rotated through 360° about x-axis. [3 marks] isipadu janaan apabila rantau berlorek A diputarkan 360 0 pada paksi-x. [3 markah] ( 7,16 ) http://chngtuition.blogspot.com SULIT 3472/2 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 8 5. Solution by scale drawing will not be accepted. Penyelesaian secara lukisan berskala tidak akan diterima. Diagram 5 shows quadrilateral PQRS. Rajah 5 menunjukkan sisi empat PQRS. Diagram 5 Rajah 5 Given that the coordinates of point P(÷3, 3) and point S(2, 6). Diberi bahawa koordnat bagi titik P(÷3, 3) dan titik S(2, 6). (a) Find the equation of PQ, [3 marks] Cari persamaan garis lurus PQ, [3 markah] (b) A point W moves such that its distance from point Q is always twice its distance from point S. Find the equation of the locus of point W. [4 marks] Titik W bergerak dengan keadaan jaraknya dari titik Q adalah sentiasa dua kali jaraknya dari titik S. Cari persamaan lokus bagi W. [4 markah] O S(2, 6) Q y x P(÷3, 3) R http://chngtuition.blogspot.com SULIT 3472/2 [ Lihat halaman sebelah 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 9 6. Table 6 shows the marks obtained by a group of students in a class. Marks Number of students 25 – 39 10 40 – 54 15 55 – 69 k 70 – 84 16 85 – 99 6 Table 6 Jadual 6 7 (a) Given that the median mark of a student is 60 , find the value of 26 k [3 marks] 7 Diberi markah median pelajar ialah 60 , cari nilai bagi 26 k [3 marks] (b) Calculate the standard deviation of the distribution. [3 marks] Kira sisihan piawai taburan tersebut. [ 3 marks ] (c) If each marks of the students in the class is multiplied by 2 and then subtracted by 3. For the new set of marks, find the standard deviation. Jika setiap markah pelajar dalam kelas didarab dengan 2 dan kemudian di tolak dengan 3. Bagi set baru markah, cari sisihan piawai. [1 mark] [1 markah] http://chngtuition.blogspot.com SULIT 3472/2 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 10 Section B Bahagian B [40 marks] [40 markah] Answer any four questions from this section. Jawab mana-mana empat soalan daripada bahagian ini. 7 Use graph paper to answer this questions. Gunakan kertas graf untuk menjawab soalan ini. Table 7 shows the values of two variables x and y obtained from an experiment. Variable x and y are related by equation , where k and p are constants. Jadual 7 menunjukkan nilai-nilai bagi dua pembolehubah, x dan y, yang diperoleh daripada satu eksperimen. Pembolehubah x and y dihubungkan oleh persamaan , dengan keadaan k dan p adalah pemalar. x 0.5 0.7 0.9 1.0 1.2 y 4 7 13 20 50 Table 7 Jadual 7 (a) Plot log 10 y against x 2 , using a scale of 2 cm to 0.2 unit on both axes. Hence, draw the line of best fit. [6 marks] Plot log 10 y melawan x, dengan menggunakan skala 2 cm mewakili 0.2 unit pada kedua-dua paksi. Seterusnya, lukis garis lurus penyuaian terbaik. [6 markah] (b) Use your graph in 7(a) to find the value of Gunakan graf anda di 7(a) untuk mencari nilai (i) k (ii) p (iii) x when y = 10 x apabila y = 10 [4 marks] [4 markah] 2 x y pk = 2 x y pk = http://chngtuition.blogspot.com SULIT 3472/2 [ Lihat halaman sebelah 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 11 8. (a) Given that the equation of the curve 2 4 3 2 ÷ + = x x y is passing through point ( ) 3 , 1 ÷ . Diberi bahawa persamaan lengkung 2 4 3 2 ÷ + = x x y melalui titik ( ) 3 , 1 ÷ . Find Cari (i) the equation of normal to the curve at point (-1,3). [3 marks] persamaan normal kepada lengkung pada titik (-1,3). [3 markah] (ii) the approximate value in y, when x decrease from 2 to 1.99. [3 marks] nilai hampir bagi y, apabila x menyusut dari 2 kepada 1.99. [3 markah] (b) Given that the gradient function of a curve 5 3 2 8 x x ÷ which has a turning point at ) 3 , ( p . Diberi bahawa fungsi kecerunan kepada lengkung 5 3 2 8 x x ÷ yang mempunyai titik pusingan pada titik ) 3 , ( p . Find Cari (i) the value of p nilai p. (ii) the equation of the curve. persamaan lengkung itu. [4 marks] [ 4 markah ] http://chngtuition.blogspot.com SULIT 3472/2 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 12 9 Diagram 10 shows triangles OAB and APR. Rajah 10 menunjukkan segi tiga OAB dan APR. OPA, OQB, PQR and ABR are straight lines. Given , ~ a OA = , 2 ~ b OB = OA OP= 3 and OB OQ= 2 . OPA, OQB, PQR dan ABR ialah garislurus. Diberi , ~ a OA = , 2 ~ b OB = OA OP= 3 dan OB OQ= 2 (a) Express each of the following vectors in terms of ~ a and/ or ~ b . Ungkapkan dalam sebutan ~ a dan/ atau ~ b . (i) AB (ii) PQ [3 marks] [3 markah] (b) Given AB m AR= and PQ n QR= , where m and n are constants. Diberi AB m AR= dan PQ n QR= , dengan keadaan m dan n ialah pemalar. Express OR in terms of Ungkapkan OR dalam sebutan (i) m, ~ a and ~ b , m, ~ a dan ~ b , (ii) n, ~ a and ~ b . n, ~ a dan ~ b . [4 marks] [4 markah] (c) Hence, find the value of m and of n. Seterusnya, cari nilai m dan nilai n. [3 marks] [ 3 markah ] P R B Q O Diagram 10 Rajah 10 A http://chngtuition.blogspot.com SULIT 3472/2 [ Lihat halaman sebelah 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 13 10 (a) At SMK Bandar Baru, 30 out of 120 form 4 students cycle to school. If 10 students are chosen at random, calculate the probability that Di SMK Bandar Baru, 30 daripada 120 orang pelajar tingkatan 4 mengayuh basikal ke sekolah. Jika 10 orang pelajar dipilih secara rawak, hitung kebarangkalian bahawa (i) exactly 2 of them cycle to school tepat 2 orang pelajar mengayuh basikal ke sekolah. (ii) more than 2 students cycle to school lebih daripada 2 orang pelajar mengayuh basikal ke sekolah. [4 marks] [4 markah] (b) In a fresh water fish pond, the weight of one type of fish is normally distributed with a mean of 1.0 kg and a standard deviation of 0.3 kg. Dalam sebuah kolam ikan air tawar, berat sejenis ikan adalah tertabur secara normal dengan min 1.0 kg dan sisihan piawai 0.3 kg. (i) Find the probability of a fish chosen randomly which is less than 0.8 kg. Cari kebarangkalian bahawa seekor ikan yang dipilih secara rawak mempunyai berat kurang daripada 0.8 kg. (ii) If there are 100 fish which has weight less than 0.8 kg, find the total number of fish in the pond. Jika terdapat 100 ekor ikan yang mempunyai berat kurang daripada 0.8 kg, cari jumlah ikan di dalam kolam itu. (iii) Given that 20% of the fish has the weight more than m kg. Find the value of m . Diberi bahawa 20% daripada ikan tersebut mempunyai berat lebih daripada m kg. Cari nilai bagi m . [6 marks] [6 markah] http://chngtuition.blogspot.com SULIT 3472/2 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 14 11 Diagram 11 shows a semicircle ABE and a sector OCD, with centre O. Rajah 11 menunjukkan semibulatan ABE dan sektor OCD, yang berpusatkan O. Given that the length of AE = 16 cm and OB : BC = 2 : 1 . Diberi bahawa panjang AE = 16 cm dan OB : BC = 2 : 1 . [Use ] [Guna ] Calculate Hitung (a) , in radian. , dalam radian. [3 marks] [3 markah] (b) the area, in cm 2 , of the shaded region, luas, dalam cm 2 , bagi kawasan berlorek,. [3 marks] [3 markah] (c) perimeter, in cm, for the whole diagram. perimeter, dalam cm, untuk seluruh rajah. [4 marks] [4 markah] O A B D E C Diagram 11 Rajah 11 http://chngtuition.blogspot.com SULIT 3472/2 [ Lihat halaman sebelah 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 15 Section C Bahagian C [20 marks] [20 markah] Answer any two questions from this section. Jawab mana-mana dua soalan daripada bahagian ini. 12. A particle moves along a straight line from a fixed point O, which situated 3 meter on the right of point O. Its velocity,V ms -1 is given by , 12 8 t v 2 + ÷ = t where t is the time in seconds, after leaving O. Suatu zarah bergerak disepanjang suatu garis lurus daripada satu titik tetap O yang berada 3 meter di kanan O. Halajunya.V ms -1, diberi oleh , 12 8 t v 2 + ÷ = t dengan keadaan t ialah masa dalam saat.selepas melalui O. [Assume motion to the right is positive] [Anggapkan gerakan ke arah kanan sebagai positif] Find Cari (a) (i) the minimum velocity, [2 marks] halaju minimum, [2 markah] (ii) the range of values of t when the particle moving to the left, [ 3 marks ] julat nilai t apabila zarah bergerak ke kiri, [ 3 markah ] (b) Sketch the velocity-time graph for 5. t 0 s s Hence or otherwise calculate the total distance travelled during the first 5 seconds after leaving O. [5 marks] Lakarkan graf halaju-masa untuk 5. t 0 s s Seterusnya atau dengan cara lain hitung jumlah jarak yang dilalui dalam 5 saat yang pertama selepas melalui O. [5 markah] http://chngtuition.blogspot.com SULIT 3472/2 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 16 13 Diagram 13 shows two triangles ACD and DBC, where AEC and BED are straight lines. Rajah 13 menunjukkan segitiga ACD dan DBC di mana AEC dan BED adalah garis lurus. Given that AE = 5 cm, AD = 4 cm, EC = 8 cm, , 38 0 = ZADE and DEC Z is an obtuse angle . Diberi bahawa AE = 5 cm, AD = 4 cm, EC = 8 cm, , 38 0 = ZADE dan DEC Z ialah sudut cakah. . Calculate Kira (a) (i) DEC Z [3 marks] [3 markah] (ii) the length of DC [3 marks] panjang DC [3 markah] (iii) the area of the triangle ADC. [2 marks] luas segitiga ADC. [2 markah] (b) Sketch triangle C’ B’ D’ which has a different shape from triangle CBD such that D’C’ = DC , B’C’ = BC and . ' ' ' CDB B D C Z = Z [ 2 marks ] Lakar segitiga C’B’D’ yang mempunyai bentuk yang berlainan dengan segitiga CBD dengan keadaan D’C’ = DC , B’C’ = BC dan . ' ' ' CDB B D C Z = Z [2 markah] Diagram 13 Rajah 13 D 38 0 A B C E http://chngtuition.blogspot.com SULIT 3472/2 [ Lihat halaman sebelah 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 17 14 Use the graph paper provided to answer this question. Gunakan kertas graf yang disediakan untuk menjawab soalan ini. Mr. Ridhuan intends to plant banana trees and papaya trees on a piece of 80 hectares land. He employed 360 labours and allocated a capital of at least RM 24 000. Mr. Ridhuan used x hectares of land to plant banana trees and y hectares to plant papaya trees. Each hectare of banana trees farm is supervised by 3 labours while each hectare of papaya trees farm is supervised by 6 labours. The cost of consumption for a hectare of banana trees farm are RM 800 and a hectare of papaya trees farm are RM 300. En. Ridhuan ingin menanam pokok pisang dan pokok betik di atas sebidang tanah seluas 80 hektar. Dia mempunyai 360 orang tenaga pekerja dan modal sekurang- kurangnya RM 24 000. En. Ridhuan menggunakan x hektar tanah untuk menanam pokok pisang dan y hektar tanah untuk menanam pokok betik. Setiap hektar ladang pokok pisang diselia oleh tiga orang pekerja sementara enam orang pekerja untuk setiap hektar ladang pokok betik. Kos perbelanjaan untuk sehektar ladang pokok pisang ialah RM 800 dan sehektar ladang pokok betik ialah RM 300. (a) State three inequalities, other than x > 0 and y > 0 which satisfy the above conditions. Nyatakan tiga ketaksamaan selain x > 0 dan y > 0 yang memuaskan syarat-syarat di atas. [3 marks] [3 markah] (b) Using the scale of 2 cm to 10 hectares on both axes, draw and shade a region R which satisfies all of the above conditions. Dengan menggunakan skala 2 cm kepada 10 hektar pada kedua-dua paksi, lukis dan lorekkan rantau R yang memuaskan semua syarat-syarat di atas. [3 marks] [3 markah] (c) Based on your graph, answer the following questions: Berdasarkan graf anda, jawab soalan-soalan berikut : (i) If the area of land allocated for planting banana trees is twice the land for papaya trees, find the maximum area of land for each type of fruit. Jika luas kawasan tanah untuk menanam pokok pisang adalah 2 kali luas kawasan tanah untuk menanam pokok betik, cari keluasan maksimum tanah yang digunakan untuk menanam setiap tanaman. (ii) The profit gained by selling bananas are RM 700 and by selling papayas are RM 250 for each hectare. Find the minimum profit gained. Keuntungan hasil jualan pisang ialah RM 700 dan hasil jualan betik ialah RM 250 bagi setiap hektar. Cari keuntungan minimum yang diperolehi. [4 marks] [4 markah] http://chngtuition.blogspot.com SULIT 3472/2 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 18 END OF QUESTION PAPER KERTAS SOALAN TAMAT 15. Table 15 shows the price indices of the monthly expenditures for Muhammad’s family in the year 2000 based on the year 1996, the change in price index from the year 2000 to the year 2004 and the percentage of expenditure respectively. Jadual 15 menunjukkan indeks harga bagi perbelanjaan bulanan keluarga Muhammad pada tahun 2000 berasaskan tahun 1996, perubahan indeks harga dari tahun 2000 ke tahun 2004 dan peratus perbelanjaan masing-masing. Expenditure Perbelanjaan Price index in the year 2000 Indeks harga pada tahun 2000 Change of price index from the year 2000 to the year 2004 Perubahan indeks harga dari tahun 2000 ke tahun 2004 Percentage of expenditure (%) Peratus perbelanjaan (%) P 110 Increased 15 % x Q 120 Decreased 10 % 30 R 125 Unchanged y S 150 Unchanged 15 Table 15 Jadual 15 (a) Calculate the expenditure of item P in the year 1996 if its expenditure in the year 2000 is RM 150.65. Hitungkan perelanjaan bagi item P pada tahun 1996 jika perbelanjaannya pada tahun 2000 ialah RM 150.65. [2 marks] [2 markah] (b) The composite index number in the year 2000 based on the year 1996 is 121.25. Nombor indeks gubahan pada tahun 2000 berasaskan tahun 1996 ialah 121.25. Calculate Hitungkan (i) the value of x and of y, nilai x dan nilai y, (ii) the total monthly expenditure in the year 2000 if the total monthly expenditure for Muhammad’s family in the year 1996 is RM 260. jumlah perbelanjaan pada tahun 2000 jika perbelanjaan keluarga Muhammad pada tahun 1996 ialah RM 260. [5 marks] [5 markah] (c) Find the composite index number in the year 2004 based on the year 1996. Cari nombor indeks gubahan pada tahun 2004 berasaskan tahun 1996. [3 marks] [3 markah] http://chngtuition.blogspot.com SULIT 3472/2 [ Lihat halaman sebelah 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 19 INFORMATION FOR CANDIDATES MAKLUMAT UNTUK CALON 1 This question paper consists of three sections : Section A, Section B and Section C. Kertas soalan ini mengandungi tiga bahagian Bahagian A, Bahagian B dan Bahagian C 2 Answer all questions in Section A, four questions from Section B and two questions from Section C. Jawab semua soalan dalam Bahagian A, mana-mana empat soalan daripada Bahagian B dan mana- mana dua soalan daripada Bahagian C 3 Write you answer on the ‘buku jawapan’ provided. If the buku jawapan is insufficient, you may ask for ‘helaian tambahan’ from the invigilator. Jawapan anda hendaklah ditulis di dalam buku jawapan yang disediakan. Sekiranya buku jawapan tidak mencukupi, sila dapatkan helaian tambahan daripada pengawas peperiksaan. 4 Show your working. It may help you to get marks. Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini boleh membantu anda untuk mendapatkan markah. 5 The diagrams in the questions provided are not drawn to scale unless stated. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. 6 The marks allocated for each question and sub-part of a question are shown in brackets. Markah yang diperuntukan bagi setiap soalan dan cerian soalan are shown in brackets. 7 A list of formulae is provided on pages 2 and 3. Satu senarai rumus disediakan di halaman 3 hingga 5 8. Graph paper and booklet of four – figure mathematical tables is provided. Kertas graf dan sebuah buku sifir matematik empat angka disediakan. 9. You may use a non-programmable scientific calculator. Anda dibenarkan menggunakan kalkulator scientific calculator yang tidak boleh diprogramkan. 10. Tie the ‘ helaian tambahan’ and the graph papers together with the ‘buku jawapan’ and hand in to the invigilator at the end of the examination. Ikat helaian tambahan dan kertas graf bersama-sama dengan buku jawapan dan serahkan kepada pengawas peperiksaan pada akhir peperiksaan. http://chngtuition.blogspot.com SULIT 3472/2 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT 20 NO.KAD PENGENALAN ANGKA GILIRAN Arahan Kepada Calon 1 Tulis nombor kad penganalan dan angka giliran anda pada petak yang disediakan. 2 Tandakan ( / ) untuk soalan yang dijawab. 3 Ceraikan helaian ini dan ikat sebagai muka hadapan bersama-sama dengan buku jawapan. Kod Pemeriksa Bahagian Soalan Soalan Dijawab Markah Penuh Markah Diperoleh ( Untuk Kegunaan Pemeriksa) A 1 5 2 8 3 6 4 7 5 7 6 7 B 7 10 8 10 9 10 10 10 11 10 C 12 10 13 10 14 10 15 10 JUMLAH http://chngtuition.blogspot.com 1 3472/1 Matematik Tambahan Kertas 1 2 jam Ogos 2011 BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2011 ADDITIONAL MATHEMATICS Paper 1 Skema Pemarkahan ini mengandungi 6 halaman bercetak MARKING SCHEME http://chngtuition.blogspot.com 2 PERATURAN PEMARKAHAN- KERTAS 1 No. Solution and Mark Scheme Sub Marks Total Marks 1(a) (b) 2 3 ÷ B1: 1 3 ÷ = + m m 2 1 ÷ 2 1 3 2(a) 2 3 + x 1 3 (b) 2 5 B1: k(3) = 2 2 3 a = 6 and b = -11 B2: a = 6 or b = -11 B1: b x a x fg + + = ÷ ) 2 5 ( ) ( 1 3 3 4 2 1 , 5 > ÷ s x x B2: (2 1)( 5) 0 or x x ÷ + > B1: 0 5 9 2 2 > ÷ + x x 3 3 5 1 11 p ÷ < < B2 : ( 1)( 11) 0 p p + ÷ < B1: 2 ( 1) 4(3)( 1) 0 p p ÷ ÷ ÷ + < 3 3 6(a) (b) (c) m = 1 x = 1 (1, ÷3) 1 1 1 3 5 ÷ 1 2 1 ÷ 11 http://chngtuition.blogspot.com 3 7 2 5 x = B2: 1 2( 2) 4( 1) or 2 4 4 2 x x x x + = ÷ ÷ + = ÷ + B1: 1 2( 2) 2 4( 1) 1 5 5 x x + ÷ = 3 3 8 1 3 x = B3: 3 2 3 9 x ÷ = B2: 2 9 log (3 ) 2 x x = ÷ B1: 9 9 log 3 log 81 x (for change base) 4 4 9(a) (b) h = 1 B1: 5h ÷ 1 ÷ (2h – 4 ) = 6h + 4 – (5h – 1) 1460 B1: | | 20 20 2(16) 19(6) 2 s = + OR 23 3 s s ÷ = | | | | 23 3 2( 2) 22(6) 2( 2) 2(6) 2 2 ÷ + ÷ ÷ + OR | | 20 20 16 130 2 s = + 2 2 4 10 1 3 r = ÷ and a = 4 B3 : 1 3 r = ÷ or a = 4 B2 : 8 3(1 )(1 ) 3 r r + ÷ = B1 : 3 1 = ÷ r a or 3 8 = + ar a 4 4 11 42300 B2 : | | 36 36 2(300) 35(50) 2 S = + B1 : 36 or 300 or 300, 350, 400 S a = 3 3 http://chngtuition.blogspot.com 4 12 2 and 4 3 q p = = ÷ B2: 2 or 4 3 q p = = ÷ B1: 2 1 3 0 3 or 6 4 2 p xy x q q ÷ = + = ÷ 3 3 13(a) (b) t = 1 B1: 4 1 8 0 0 2 ÷ = ÷ ÷ t y = 4x + 2 B1: Using m 1 • m 2 = -1 and m 2 = 4 2 2 4 14 m = 1 and n = 3 1 2: 3 ( 4) (4) (2) (6) 1: 2 or 3 m B n n m n m B n m n m = ÷ + + = ÷ = + + 3 3 15(a) (b) x 3 ÷ + y 6 2x + y 2 2 1 B1 : (3 6 ) or (6 3 ) or equivalent 3 3 x y y x ÷ ÷ 1 2 3 16(a) (b) 8 4 i j + 1 1 (8 4 ) or (2 ) 80 5 i j i j + + B1: 2 2 4 8 + = AB = 80 1 2 3 http://chngtuition.blogspot.com 5 17(a) (b) r = k =10 B1: 2 1 (0.8) 40 2 r = 72 cm ) 8 . 0 ( 30 S or 10(0.8) : 1 CD = = AB S B 2 2 4 18 65 33 ÷ B1 : 5 3 sin ÷ = B OR 13 5 cos = A 2 2 19 2 13 ÷ B2 : ) 5 ( 2 1 2 2 ÷ + ( ¸ ( ¸ x B1 : } } + 1 2 1 2 ) ( dx x f dx x OR 2 2 x OR } ÷ = 1 2 5 ) ( dx x f 3 3 20 27 B2 : 2 2 3 2 3 3 ( 1) (9( 1) )(( 1) 4) (( 1) 4) (2( 1)) ÷ ÷ ÷ + + ÷ + ÷ B1 : 2 2 3 2 3 3 (9 )( 4) ( 4) (2 ) x x x x x + + + or equivalent 3 3 21(a) (b) 5 B1 : 450 90 0 h ÷ = 1125 2 1 3 http://chngtuition.blogspot.com 6 22(a) (b) p = 9 8 B1 : 2 2 2 2 2 2 2 3 7 9 5 1 (5) 5 o + + + + = ÷ 1 2 3 23(a) (b) 20 240 B1 : 5! X 2! 1 2 3 24(a) (b) 1 2 5 12 B1 : 1 2 3 1 4 3 4 3 × + × 1 2 3 25(a) (b) 0.1515 58.09 B2 : 55 1.03 3 k ÷ = B1 : z = 1.03 1 3 4 http://chngtuition.blogspot.com 1 3472/2 Matematik Tambahan Kertas 2 Ogos 2011 2 ½ jam BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2011 TINGKATAN 5 ADDITIONAL MATHEMATICS Paper 2 Skema Pemarkahan ini mengandungi 9 halaman bercetak MARKING SCHEME http://chngtuition.blogspot.com 2 http://chngtuition.blogspot.com 3 No Solution and Mark Scheme Sub Marks Total Marks 1 3 4 8 x y ÷ = or 4 3 8 y x ÷ = 8 3 4 8 2 = | . | \ | ÷ ÷ x x x or 8 4 3 8 4 3 8 2 = | . | \ | ÷ ÷ | . | \ | ÷ y y y 7x 2 – 8x – 24 = 0 or 21y 2 - 80y - 64 = 0 ) 7 ( 2 ) 24 )( 7 ( 4 ) 8 ( ) 8 ( 2 ÷ ÷ ÷ ± ÷ = x or ) 21 ( 2 ) 64 )( 21 ( 4 ) 80 ( ) 80 ( 2 ÷ ÷ ÷ ± ÷ = y x = 2.51 , -1.37 OR y = 4.49 , -0.679 y = -0.680 , 4.49 x = -1.37 , 2.51 5 5 2 (a) 0 0 (sin 2 cos90 cos 2 sin90 ) u u + u 2 cos (b) i) ii) Number of solutions = 4 2 6 8 t u 2 2 ÷ = y t u 2 2 ÷ = y 1 y 0 t 2 t u 2 cos 2 = y -1 2 2 u P1 K1 K1 N1 K1 N1 N1-cosine curve N1-amplitude N1- 2 cycle K1-straight line N1 K1-for equation N1 http://chngtuition.blogspot.com 4 3 (a) a = A , d = -D S 10 = 10 [2 99( )] 720 2 A D + ÷ = or S 20 = 20 [2 19( )] 1040 2 A D + ÷ = K1 Solving the simultaneous in linear equation K1 A = 90 or D = 4 N1 A = 90 and D = 4 N1 (b) T 10 = 90 + 9(-4) or T 20 = 90 + 19(-4) K1 Difference = 40 N1 4 2 6 4(a) Area of trapezium – area under curve } + 7 0 ) 9 ( dx x or ) 7 )( 9 16 ( 2 1 + OR dx x } ÷ 7 0 2 ) 3 ( 7 0 2 9 2 ( ¸ ( ¸ + x x OR 7 0 3 3 ) 3 ( ( ¸ ( ¸ ÷ x = 2 175 - 3 91 = 2 6 1 57 / 6 343 unit 4 7 (b) y = 0 , x = 3 OR ( ) | | dx x } ÷ 3 0 2 2 ) 3 ( t 3 0 5 5 ) 3 ( ( ¸ ( ¸ ÷ x t 3 5 243 unit t 3 K1 K1 K1 N1 K1 K1 N1 http://chngtuition.blogspot.com 5 5(a) 5 3 PQ m = ÷ K1 5 3 ( 3) 3 y x ÷ = ÷ + K1 3y = ÷ 5x ÷ 6 N1 3 7 (b) Q(0, ÷2) P1 2 2 2 2 ( 0) ( 2) ( 2) ( 6) x y or x y ÷ + + ÷ + ÷ K1 2 2 2 2 ( 0) ( 2) 2 ( 2) ( 6) x y x y ÷ + + = ÷ + ÷ K1 0 156 52 16 3 3 2 2 = + ÷ ÷ + y x y x N1 4 6(a) (b) (c) L = 54.5 OR 25 and k P1 47 25 7 2 60 54.5 15 26 k k + | | ÷ | = + | | \ . K1 k = 13 N1 Mean, 3563 60 x ÷ = P1 2 232 755 3563 (* ) 60 60 18.79 o = ÷ = N1 37.58 N1 3 3 1 7 K1 http://chngtuition.blogspot.com 6 8(a) equivalent or x y m x dx dy i normal 7 2 2 1 4 6 ) ( + = = + = ( ) ( ) 84 . 17 2 ) 2 ( 4 ) 2 ( 3 ) 01 . 0 )( 16 ( ) ( 2 y y y ii new o o + ÷ + = ÷ = 6 10 (b) 3 5 8 ( ) 0 2 2 x i x p ÷ = = 16 45 2 1 1 2 1 1 ) ( 4 4 + + ÷ = + + ÷ = x x y c x x y ii 4 9(a) (b) (i) 2 a b ÷ + (ii) ÷ ÷ + ÷ OB OA 2 1 3 1 1 3 a b ÷ + ( ) ( 2 ) (1 ) 2 i a m a b m a mb + ÷ + ÷ + 1 ( ) ( ) 3 1 (1 ) 3 ii b n a b na n b + ÷ + ÷ + + 3 4 3 10 K1 K1 N1 K1 K1 N1 K1 N1 K1 N1 N1 K1 N1 K1 N1 K1 N1 http://chngtuition.blogspot.com 7 (c) n m or m m 3 1 1 1 2 ÷ = ÷ + = K1 Solve the simultaneous linear equation K1 m = 2, n = 3 (Both correct) N1 10(a) (b) (i) 10 2 8 2 ( 2) (0.25) (0.75) P x C = = = 0.2186 (ii) ( 2) 1 ( 0) ( 1) ( 2) P x P x P x P x > = ÷ = ÷ = ÷ = 10 0 10 10 1 9 0 1 1 (0.25) (0.75) (0.25) (0.75) 0.2186 C C = ÷ ÷ ÷ = 0.4744 (i) 0.8 1 ( 0.8) ( ) 0.3 P x P z ÷ < = < ( 0.667) P z = < ÷ = 0.2524 (ii) Total 100 0.2524 = = 396 (iii) ( ) 0.2 P x m > = Z = 0.842 1.0 0.842 0.3 m÷ = m = 1.253 kg 4 6 10 11(a) (b) BC = 4 , OC = 12 P1 8 cos 12 u = K1 0.8411 u = rad N1 Area shaded = Area triangle OCE – Area sector OBE = 2 1 1 (8)(12)sin0.8411 (8) (0.8411) 2 2 ÷ K1K1 = 8.863 cm 2 N1 3 3 10 K1 N1 K1 N1 K1 N1 N1 P1 K1 N1 http://chngtuition.blogspot.com 8 (c) OR Area shaded = 2 1 1 (8)( 80) (8) (0.8411) 2 2 ÷ K1K1 = 8.863 N1 2.3009 AOB Z = rad P1 Perimeter = arc AB + BC + arc CD + AE + ED = 8(2.3009) + 4 + 12(0.8411) + 16 + 4 K1K1 = 52.50 cm N1 4 12(a) (i) 2t - 8 = 0 t = 4 V = -4 2 10 (ii) (b) 2 8 12 0 t t ÷ + < ( 2)( 6) 0 t t ÷ ÷ < 2 6 t < < 3 Total distance = meter dt v dt v 3 59 2 32 3 5 3 32 2 0 5 2 = | . | \ | ÷ + = + = } } 5 K1 N1 K1 K1 N1 t = 0 .s= 3 t = 2 ,s = 3 2 13 t = 5, s = 3 2 4 meter 3 59 ) 3 2 4 3 2 13 ( 3 3 2 13 = ÷ + ÷ OR K1 K1 N1 K1 K1 N1 P1-shape P1-passing through point (0,12), (2,0) and (5,-3) v t 12 0 2 5 http://chngtuition.blogspot.com 9 13(a) (i) ° = Z ° = Z = Z 493 . 150 507 . 29 5 38 sin 4 DEC DEA DEA Sin (ii) o DAE 493 . 112 507 . 29 38 180 = ÷ ÷ = Z P1 2 2 2 4 13 2(4)(13) cos112.49 DC = + ÷ ° K1 14.99 DC = N1 (iii ) Area of ADC = ° × × × 49 . 112 sin 13 4 2 1 K1 = 24.02 N1 (b) 10 14 (a) x + y s 80 N1 3x + 6y s 360 or equivalent N1 800x + 300y > 24 000 or equivalent N1 3 10 (b) Refer graph 3 (c) (i) Draw x = 2y K1 x = 52 , y = 26 N1 (both) (ii) K min = 700(10) + 250(54) K1 (substitud the point in R) = RM 20 500 N1 4 K1 K1 N1 D' B' C' N1-shape N1-label http://chngtuition.blogspot.com 10 15 (a) 96 150.65 100 110 p = × K1 = RM136.95 N1 2 10 (b) (i) x + y = 55 110x + 120(30) + 125y + 150(15) 121.25 100 = K1 15y = 225 (Solve simultaneous equation) K1 y = 15 , x = 40 (both) N1 (ii) 2000 260 121.25 100 p = × K1 = RM315.25 N1 3 2 (c) 2004 1996 126.50(40) + 108(30) + 125(15) + 150(15) 100 I = K1 126.50, 108, 125, 150 ( 2004 1996 I for every item) P1 = 124.25 N1 3 http://chngtuition.blogspot.com SULIT 2 BLANK PAGE HALAMAN KOSONG 3472/1 3472/1 @2011 Trial SPM Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT http://chngtuition.blogspot.com SULIT 3 3472/1 The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used. ALGEBRA 1 2 3 4 5 6 7 x b b 4ac 2a 2 8 logab = logc b logc a am an = a m + n am an = a m (am) n = a nm loga mn = log am + loga n loga n 9 Tn = a + (n-1)d 10 Sn = 11 Tn = ar n-1 n [2a (n 1)d ] 2 log a mn = n log a m m = log am - loga n n a(r n 1) a(1 r n ) 12 Sn = , (r 1) r 1 1 r a 13 S , r <1 1 r CALCULUS 1 y = uv , dy dv du u v dx dx dx v du dv u dx dx , 2 v 4 Area under a curve = y a b dx or 2 u dy y , v dx dy dy du dx du dx = x dy a b 5 Volume generated 3 = y 2 dx or a b b = x 2 dy a GEOMETRY 1 Distance = 2 Midpoint (x , y) = ( x2 x1 ) 2 ( y 2 y1 ) 2 5 y y2 x1 x2 , 1 2 2 2 A point dividing a segment of a line nx mx 2 ny1 my 2 ( x,y) = 1 , mn mn 3 4 r x y 2 6 Area of triangle 1 = ( x1 y 2 x2 y3 x3 y11 ) ( x2 y1 x3 y 2 x1 y3 ) 2 ˆ r xi yj x2 y 2 3472/1 @2011 Trial SPM Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT http://chngtuition.blogspot.com com .blogspot. p + q = 1 4 = f ( x x) f 2 = fx f x 2 11 5 1 2 N F m = L C fm 12 13 14 Mean µ = np 6 I Q1 100 Q0 npq x z= TRIGONOMETRY 1 Arc length.P(A B) P (X = r) = nCr p r q n r .SULIT 4 STATISTIC 3472/1 1 x = x N 7 8 9 2 I 2 x = fx f (x x ) N 2 3 = = x N x _2 w1 I1 w1 n! n Pr (n r )! n! n Cr (n r )!r! 10 2 P(A B) = P(A)+P(B).2bc cosA Area of triangle = 1 absin C 2 2 tan A 1 tan 2 A [ Lihat halaman sebelah SULIT 3472/1 @2011 Trial SPM Hak Cipta BPSBPSK http://chngtuition.1 = 1 .2 sin2A 8 tan 2A = 14 2 2 9 sin (A B) = sinA cosB cosA sinB 1 2 r 2 10 cos (A B) = cosA cosB sinA sinB 11 tan (A B) = tan A tan B 1 tan A tan B 12 a b c sin A sin B sin C a2 = b2 + c2 . L = 3 sin 2A + cos 2A = 1 4 sec2A = 1 + tan2A 5 cosec A = 1 + cot A 6 sin 2A = 2 sinA cosA 13 7 cos 2A = cos2A – sin2 A = 2 cos2A . s = r 2 Area of sector . 3 0.00676 0.0244 0.0869 0.0 0.3372 0.1379 0.2483 0.2546 0.0778 0.3156 0.1401 0.1230 0.1949 0.0764 0.0516 0.2420 0.1 0.00131 0.00100 0.0418 0.00415 0.00272 0.2877 0.1867 0.00264 0.00523 0.0139 0.00219 0.3015 0.1292 0.2912 0.0268 0.00427 0. 1).00466 0.2090 0.00639 0.00149 0.0 1.0 0.0344 0.2578 0.0222 0.3974 0.0668 0.5 0.5 2.00336 0.3085 0.3936 0.1056 0.blogspot.00226 0.0594 0.3783 0.2236 0.00126 0.3632 0.00144 0.1685 0.com .3859 0.0655 0.4286 0.3557 0.0606 0.1922 0.00181 0.2981 0.1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0.0359 0.4 0.4168 0.00604 0.4090 0.3821 0.00508 0.7 1.1112 0.00776 0.4641 0.0708 0.1357 0.. 1).3446 0.1170 0.0170 0.2389 0.00440 0.0455 0.0166 0.00379 0.3 0 0.00621 0.0188 0.00233 0.00118 0.0465 0.3050 0.0901 0.2206 0.00587 0.0643 0.0154 0.2119 0.4522 0.3336 0.0618 0.3745 0.3897 0.00280 0.0918 0.2 1.1736 0.0485 0.00307 0.00357 0.1894 0.4 0.00154 0.00193 0.2451 0. then P(X > k) = Q(k) Jika X ~ N(0.0392 0.9 1.1977 0.00169 0.3520 0.0207 0.4801 0.0202 0.1038 0.5 1.0681 0.00159 0.00368 0.4247 0.00298 0.00889 2.1446 0.00317 0.00248 0.1251 0.4761 0.0436 0.1611 0.00111 0.2177 0.0125 5 0.00391 0.2810 0.1635 0.0823 0.0281 0.4013 0.00842 3 0.0329 0.00122 0.4207 0.2296 0.6 2.0526 0.6 1.3192 0.4052 0.0233 0.0314 0.1314 0.3300 0.0630 0.8 0.0146 0.0110 1 2 3 4 5 6 7 8 9 Minus / Tolak 4 4 4 4 4 3 3 3 3 3 2 2 2 2 1 1 1 1 1 1 0 0 0 0 3 2 2 2 2 1 1 1 0 0 8 8 8 7 7 7 7 6 5 5 5 4 4 3 3 2 2 2 1 1 1 1 1 1 5 5 4 4 3 2 2 1 1 1 12 12 12 11 11 10 10 9 8 8 7 6 6 5 4 4 3 3 2 2 1 1 1 1 8 7 6 6 5 3 3 2 1 1 16 16 15 15 15 14 13 12 11 10 9 8 7 6 6 5 4 4 3 2 2 2 1 1 10 9 8 7 6 5 4 3 2 2 20 20 19 19 18 17 16 15 14 13 12 10 9 8 7 6 5 4 4 3 2 2 2 1 13 12 11 9 8 6 5 4 2 2 24 24 23 22 22 20 19 18 16 15 14 12 11 10 8 7 6 5 4 4 3 2 2 2 15 14 13 11 9 7 6 4 3 2 28 28 27 26 25 24 23 21 19 18 16 14 13 11 10 8 7 6 5 4 3 3 2 2 18 16 15 13 11 9 7 5 3 3 32 32 31 30 29 27 26 24 22 20 19 16 15 13 11 10 8 7 6 5 4 3 3 2 20 16 17 15 12 9 8 6 4 3 36 36 35 34 32 31 29 27 25 23 21 18 17 14 13 11 9 8 6 5 4 4 3 2 23 21 19 17 14 10 9 6 4 4 f ( z) 1 exp z 2 2 2 1 f (z) Q(z) Example / Contoh: If X ~ N(0.00402 0.00114 0.4325 0.1762 0.0968 0.4880 0.1 1.1190 0.0162 0.0409 0.00866 0.0537 0.00175 0.2514 0.0582 0.00714 2.1335 0.0559 0.4681 0.00326 0.1003 0.0143 0.0721 0.1814 0.0228 0.2 0.0694 0.00104 0.0136 0.0301 0.0212 0.2 2.1271 0.2148 0.00212 0.2327 0.1075 0.0158 0.0885 0.00494 0.0446 0.5000 0.2946 0.2676 0.00347 0.0132 0.7 0.0129 4 0.00289 0.3409 0.0116 8 0.0495 0.0505 0.3264 0.0934 0.3707 0.00657 0.4443 0.1788 0.0375 0.4404 0.0113 9 0.0401 0.1492 0.1020 0.00570 0.00256 0.0427 0.0571 0.8 1.0192 0.1562 0.4920 0.00798 0.8 2.00755 0.00734 0.2843 0.1210 0.4364 0.2005 0.00205 0.1093 0.1539 0.00199 0.1841 0.0749 0.0183 0.2033 0.00939 0.0104 2 0.0351 0.00139 0.00695 0.1151 0.0122 6 0.0307 0.0735 0.0197 0.4129 0.00820 0.3228 0.3 1.0174 0.00453 0.3594 0.00164 0.4 1.00554 0.0256 0.0294 0.1 2.2776 0.00240 0.1587 0.0179 0.00964 0.0853 0.1515 0.4721 0.1131 0. 1) z 0.0475 0.0239 0.0367 0.3669 0.2266 0.4602 0.0548 0.0102 0.00480 0.0107 1 0.4840 0.0287 0.0 2.1469 0.0262 0.0274 0.4960 0.0793 0.0150 0.3483 0.0384 0.1423 0.9 3.4562 0.00539 0.0838 0.2743 0.4483 0.00187 0.2061 0.0250 0.6 0.2611 0.0217 0.9 2.0119 7 0.0336 0.3121 0.00914 0. maka P(X > k) = Q(k) z [Lihat halaman sebelah SULIT Q( z ) f ( z ) dz k O 3472/1 @ 2011Trial SPM Hak Cipta BPSBPSK http://chngtuition.00990 0.1660 0.0951 0.1711 0.0985 0.7 2.00135 0.2643 0.2358 0.2709 0.00107 0.5 SULIT 3472/1 THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0.0808 0.0322 0. x 0. x3 . x 0. x x3 Rajah 1 menunjukkan fungsi f : x .6 SULIT For examiner’s use only 3472/1 Answer all questions.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT .blogspot. x Diagram 1 shows the function f : x x f x3 x -1 m Diagram 1 Rajah 1 Find Cari (a) the value of m nilai m (b) image of -2 imej bagi -2 [ 3 marks ] [3 markah] Answer/Jawapan : (a) (b) 1 3 3472/1 @ 2011Trial SPM http://chngtuition. 1. Jawab semua soalan. Given that the functions f : x ax b. g 1 : x Cari nilai a dan nilai b.1 (x) (b) h -1 k(3) [ 3 marks ] [3 markah] Answer/Jawapan : (a) (b) 2 3 3. 2 Diberi fungsi h : x 2 x 3 dan k : x ( x 2) 1. 2 [3 marks] x5 and fg 1 : x 4 3x. 3472/1 For examiner’s use only Find Cari (a) h . [ 3 markah ] Answer/Jawapan : x5 and fg 1 : x 4 3x. 2 3 3 3472/1 @ 2011Trial SPM http://chngtuition. 2 Given that the function h : x 2 x 3 dan k : x ( x 2) 1. Diberi fungsi-fungsi f : x ax b.blogspot.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT .7 SULIT 2. g 1 : x Find the value of a and of b. Find the range of values of x for 2x(x 5) x 5. 2 The quadratic equation px 1 3 x x p . [3 marks] 2 Persamaan kuadratik px 1 3 x x p . [ 3 markah ] Answer/Jawapan : 5 3 3472/1 @ 2011Trial SPM http://chngtuition. Find the range of values of p.tidak mempunyai punca. dengan keadaan p ialah pemalar .8 For examiner’s use only SULIT 4.blogspot. Cari julat nilai x bagi 2x(x 5) x 5 Answer/Jawapan : 3472/1 [3 marks] [ 3 markah ] 4 3 5. Cari julat nilai p.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT . has no roots. where p is a constant. the equation of the axis of symmetry.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT . persamaan paksi simetri. nilai m. For examiner’s use only f(x) (1. 3472/1 2 Diagram 6 shows the graph of a quadratic function for f ( x) 2( x m) 3 .9 SULIT 6. koordinat-koordinat titik minimum. 2 Rajah 6 menunjukkan graf fungsi kuadratik bagi f ( x) 2( x m) 3 .blogspot. 5) x 0 Diagram 6 Rajah 6 Find Cari (a) the value of m. the coordinates of the minimum point. [ 3 marks ] [3 markah] Answer/Jawapan : (a) (b) (c) (b) 6 (c) 3 3472/1 @ 2011Trial SPM http://chngtuition. 5) (3. Solve the equation log9 x log81 3x 1 .10 For examiner’s use only SULIT 7. 3472/1 Solve the equation 25x 2 1 . [4 markah] Answer/Jawapan : 8 4 3472/1 @ 2011Trial SPM http://chngtuition. 625x 1 [3 marks] Selesaikan persamaan 25x 2 1 .com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT . [ 4 marks ] Selesaikan persamaan log9 x log81 3x 1 . 625x 1 [ 3 markah ] Answer/Jawapan : 7 3 8.blogspot. hasil tambah 20 sebutan berikutnya selepas sebutan ketiga.blogspot. Cari. 5h 1 and 6h + 4. [ 4 marks ] [4 markah] Answer/Jawapan : (a) For examiner’s use only (b) (b) 9 4 3472/1 @ 2011Trial SPM http://chngtuition.11 SULIT 9. nilai h. 5h 1 dan 6h + 4.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT . (a) the value of h. 3472/1 The first three terms of an arithmetic progression are 2h 4. the sum of next 20 terms after the third term. Find. Tiga sebutan pertama suatu janjang aritmetik adalah 2h 4. Aiman saves RM 300 from his salary in a certain month.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT . Find the common ratio (r<0) and the first term. Hitungkan jumlah simpanan beliau dalam masa 3 tahun. Aiman telah membuat simpanan RM 300 daripada wang gajinya pada bulan tertentu. In each succeeding month. Cari nilai nisbah sepunya (r<0) dan sebutan pertama.12 For examiner’s use only SULIT 10. he saves RM 50 more than the previous month. beliau telah menambah simpanannya sebanyak RM 50 melebihi bulan sebelumnya. 3472/1 The sum to infinity of a geometric progression is 3 and the sum of its first two terms is 2 . [3 marks] [3 markah] Answer/Jawapan : 11 3 3472/1 @ 2011Trial SPM http://chngtuition. [4 marks ] [4 markah] Answer/Jawapan 2 3 2 3 10 4 11. Calculate the amount of his saving in 3 years.blogspot. Pada bulan yang berikutnya. Hasil tambah hingga sebutan ketakterhinggaan bagi janjang geometri ialah 3 dan hasil tambah dua sebutan pertama ialah 2 . xy (6. [ 3 marks ] [3 markah] Answer/Jawapan : 12 3 3472/1 @ 2011Trial SPM http://chngtuition. A straight line graph x p qy. The variables x and y are related by the equation x 3472/1 p qy. . Graf garis lurus x For examiner’s use only is obtained by plotting xy against x2 . as shown in Diagram 12. 0) Diagram 12 Rajah 12 x2 Find the value of p and of q. Pemboleh ubah x dan y dihubungkan oleh persamaan x diperoleh dengan memplot xy melawan x2. where p and q are constants. Cari nilai p dan nilai q. dengan keadaan p dan q ialah pemalar.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT .13 SULIT 12.3) 0 (4.blogspot. seperti ditunjukkan pada Rajah 12 . . (a) the value of t. garis lurus PQ mempunyai persamaan 1 dengan 8 2t 1 kecerunan .blogspot. Cari. the straight line PQ has an equation x y 1 with gradient of 8 2t 1 . 2t) O x P (8. nilai bagi t (b) the equation of the straight line that passes through Q and is perpendicular to PQ. persamaan garis lurus yang melalui Q dan berserenjang dengan PQ. 4 y Q (0.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT . 3472/1 In Diagram 13. PQ intersects the x-axis at point P and intersects the y-axis at point Q.14 For examiner’s use only SULIT 13. [ 4 marks ] [4 markah] Answer/Jawapan : (a) (b) 13 4 3472/1 @ 2011Trial SPM http://chngtuition. 0) Diagram 13 Rajah 13 Find. 4 x y Dalam Rajah 13. PQ menyilang paksi-x di titik P dan menyilang paksi-y di titik Q. 15 SULIT 3472/1 For examiner’s use only 14. 2) and B(4. Cari nilai m dan nilai n .blogspot.3) lies on the straight line of AB such that AP : PB m : n . [ 3 marks ] [3 markah] Answer/Jawapan : 14 3 3472/1 @ 2011Trial SPM http://chngtuition.6) .6) .3) terletak pada garislurus AB dengan keadaan AP : PB m : n . Given that the point A(4. Titik P(2.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT . Find the value of m and of n. 2) dan B(4. Diberi titik A(4. The point P(2. com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT . C BC (b) AK (a) [ 3 marks ] [3 markah] Answer/Jawapan : (a) (b) 15 3 3472/1 @ 2011Trial SPM http://chngtuition. B K A Diagram 15 Rajah 15 Express in terms of x and y . K is a point on BC such that BK : KC 1 : 2 . K ialah titik yang terletak digaris BC berkeadaan BK : KC 1 : 2 . 3472/1 Diagram 15 shows triangle ABC.blogspot. Diberi bahawa AB 3x dan AC 6 y . Given that AB 3x and AC 6 y . Rajah 15 menunjukkan sebuah segitiga ABC. Ungkapkan dalam sebutan x dan y .16 For examiner’s use only SULIT 15. 17 SULIT 16. 3472/1 y Diagram 16 shows the vector OP and OQ. For examiner’s use only Q (12,1) O x P (4,-3) Diagram 16 Rajah 16 Given that O(0,0) , P(4,3) and Q(12,1) , find in terms of i and j , Diberi bahawa O(0,0) , P(4,3) and Q(12,1) , cari dalam sebutan i dan j , (a) PQ (b) the unit vector in the direction of PQ. vektor unit dalam arah PQ. [3 marks ] [3 markah] Answer/Jawapan : (a) (b) 16 3 3472/1 @ 2011Trial SPM http://chngtuition.blogspot.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT 18 For examiner’s use only SULIT 3472/1 17. Diagram 17 shows sector OAB and sector OCD, with centre O . Rajah 17 menunjukkan sector OAB dan sekitar OCD,yang berpusatkan O. 3k cm k cm O A C B D Diagram 17 Rajah 17 Given that 0.8 rad and the area of sector OAB is 40 cm 2 . Diberi bahawa 0.8 rad dan luas bagi sektor OAB is 40 cm 2 . Find the value of Cari nilai bagi (a) k (b) the perimeter of the shaded region. perimeter kawasan yang berlorek [4 marks] [4 markah] Answer/Jawapan : (a) 17 4 (b) 3472/1 @ 2011Trial SPM http://chngtuition.blogspot.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT 19 SULIT 18. Given that sin A 3472/1 4 12 and cos B , where A and B are in the same 13 5 quadrant. Find the value of sin A B . [ 2 marks ] For examiner’s use only 4 12 dan cos B , dimana A dan B berada dalam sukuan yang 13 5 sama. Cari nilai sin A B . Diberi sin A [ 2 markah ] Answer/Jawapan : 18 2 19. Given that 1 2 f ( x) dx 5. Find the value of 1 x f ( x) dx. 2 1 [ 3 marks ] Diberi f ( x) dx 5. Cari nilai x f ( x) dx. 1 2 2 [ 3 markah ] Answer/Jawapan : 19 3 3472/1 @ 2011Trial SPM http://chngtuition.blogspot.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT cari nilai apabila x 1 . dx [ 3 markah ] Answer/Jawapan : 20 3 21. V cm3 air dalam takungan diberi oleh V 450h 45h 2 .com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT . dx [ 3 marks ] dy 2 3 3 Diberi y x ( x 4) . V cm3 of water in a vessel is given by V 450h 45h 2 . isipadu maksimum air dalam takungan itu [ 3 marks ] [ 3 markah ] Answer/Jawapan : (a) 21 (b) 3 3472/1 @ 2011Trial SPM http://chngtuition. Isipadu.20 For examiner’s use only SULIT 20. nilai h apabila V maksimum.blogspot. dimana h cm ialah tinggi air dalam takungan itu. find the value of 3472/1 dy when x 1 . Calculate Kira (a) the value of h when V is maximum. where h cm is the height of the water in the vessel. 2 3 3 Given that y x ( x 4) . (b) the maximum volume of water in the vessel. . The volume. (a) Three letters are chosen from the word “BERHAD” . [ 3 marks ] [3 markah] Answer/Jawapan : (b) (a) 23 (b) 3 3472/1 @ 2011Trial SPM http://chngtuition. Hitung bilangan cara berlainan huruf ini boleh dipilih jika tiada sebarang syarat dikenakan. Tiga huruf dipilih daripada perkataan “BERHAD” . varian bagi set data ini.blogspot. nilai p. 7. 1 . the variance for the set of data. 3472/1 A set of positive integers consists of 3. 1 . [ 3 marks ] [3 markah] Answer/Jawapan : (a) For examiner’s use only (b) (b) 22 3 23. 7. p. Given that the mean for the set of data is 5.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT . p. Calculate the number of different ways in which the three letters can be chosen if there is no restriction. Diberi min untuk set data ini ialah 5. In how many ways can the word “BERHAD” be arranged if the vowel are arranged side by side? . Dalam berapa carakah huruf-huruf dalam perkataan “BERHAD” dapat disusun jika huruf vokal hendaklah disusun sebelah menyebelah. 5. Satu set integer positif mengandungi 3. Find Cari (a) the value of p. 5.21 SULIT 22. 3 1 4 Kebarangkalian bahawa Rajesh layak kepertandingan akhir acara 400m ialah manakala kebarangkalian untuk Razali layak ialah Find the probability that Hitung kebarangkalain bahawa (a) both of them did not qualify for the final.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT . hanya salah seorang daripada mereka layak kepertandingan akhir. kedua-dua mereka tidak layak kepertandingan akhir. 3 [ 3 marks ] [3 markah] Answer/Jawapan (a) : (b) 24 3 3472/1 @ 2011Trial SPM http://chngtuition. (b) only one of them qualifies for the final. 3472/1 The probability that Rajesh qualifies for the final 400m event is probability that Razali qualify is 1 while the 4 1 . 1 .blogspot.22 For examiner’s use only SULIT 24. Find Cari (a) P(X (b) the value of k . Rajah 25 menunjukkan graf taburan normal .23 SULIT 25. Given that the area of the shaded region is . X ialah pembolehubah rawak selanjar yang tertabur secara normal dengan min 55 kg dan sisihan piawai 3 kg. nilai bagi k. Diberi bahawa luas bagi kawasan berlorek ialah . 3472/1 For examiner’s use only f (x) k Diagram 25 Rajah 25 x X is a continuous random variable which is normally distributed with a mean of 55 kg and a standard deviation of 3 kg. [ 4 marks ] [4 markah] Answer/Jawapan (a) : (b) 25 4 END OF QUESTION PAPER KERTAS SOALAN TAMAT 3472/1 @ 2011Trial SPM http://chngtuition. Diagram 25 shows a normal distribution graph.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT .blogspot. Then write down the new answer. The marks allocated for each question are shown in brackets. 7. A booklet of four-figure mathematical tables is provided. Sekiranya anda hendak menukar jawapan. Satu senarai rumus disediakan di halaman 3 hingga 5. Serahkan kertas soalan ini kepada pengawas peperiksaan di akhir peperiksaan. Tulis jawapan anda dalam ruang yang disediakan dalam kertas soalan. cross out the answer that you have done. Kemudian tulis jawapan yang baru. 5. 8. batalkan jawapan yang telah dibuat.24 SULIT INFORMATION FOR CANDIDATES MAKLUMAT UNTUK CALON 3472/1 1. Ini boleh membantu anda untuk mendapatkan markah. 3472/1 @ 2011Trial SPM http://chngtuition. Tunjukkan langkah-langkah penting dalam kerja mengira anda. Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram. Jawab semua soalan 3.blogspot. A list of formulae is provided on pages 3 to 5. The diagrams in the questions provided are not drawn to scale unless stated. Markah yang diperuntukkan bagi setiap soalan ditunjukkan dalam kurungan. 4. Show your working. 10.com Hak Cipta BPSBPSK [Lihat halaman sebelah SULIT . Answer all questions. 6. Hand in this question paper to the invigilator at the end of the examination. 11. 9. You may use a non-programmable scientific calculator. It may help you to get marks. Sebuah buku sifir matematik empat angka disediakan. This question paper consists of 25 questions Kertas soalan ini mengandungi 25 soalan 2. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. If you wish to change your answer. Write your answers in the spaces provided in the question paper. 6. 8.blogspot. This question paper consists of three sections : Section A. Kertas soalan ini mengandungi 20 halaman bercetak 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT http://chngtuition. Give only one answer / solution to each question. 3. The marks allocated for each question and sub-part of a question are shown in brackets. 2. Section B and Section C. It may help you to get marks. Answer all questions in Section A . 5. 9.com . A booklet of four-figure mathematical tables is provided. The diagram in the questions provided are not drawn to scale unless stated. 4. You may use a non-programmable scientific calculator. 7. A list of formulae and normal distribution table is provided on pages 2 to 4.SULIT 3472/2 Matematik Tambahan Kertas 2 2 ½ jam Ogos 2011 1 3472/2 BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2011 ADDITIONAL MATHEMATICS Kertas 2 Dua jam tiga puluh minit JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU 1. Show your working. four questions from Section B and two questions from Section C. 2 y . r <1 1 r CALCULUS dy dv du u v dx dx dx du dv v u u dy dx 2 dx .y) = nx1 mx 2 ny1 my 2 . Area of triangle = r x2 y 2 4 r xi yj x2 y 2 1 ( x1 y 2 x2 y3 x3 y11 ) ( x2 y1 x3 y 2 x1 y3 ) 2 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK SULIT http://chngtuition. (r 1) r 1 1 r a 13 S . y) = 3 ( x1 x2 ) 2 ( y1 y 2 ) 2 5 A point dividing a segment of a line ( x. mn mn y y2 x1 x2 .blogspot.SULIT 2 3472/2 The following formulae may be helpful in answering the questions.com . ALGEBRA 1 x= b b 4ac 2a 2 8 logab = logc b logc a 2 am an = a m + n 3 am an = a m . 1 2 2 6. 3 4 Area under a curve = y a b dx or = x dy a b dy dy du dx du dx 5 Volume generated = y 2 dx or a b a b = x 2 dy GEOM ETRY 1 Distance = 2 Midpoint (x . v v dx 1 y = uv . The symbols given are the ones commonly used.n 9 Tn = a + (n-1)d 10 Sn = 11 4 (am) n = a nm 5 6 7 loga mn = log am + loga n n [2a (n 1)d ] 2 Tn = ar n-1 log a mn = n log a m m loga = log am .loga n n a(r n 1) a(1 r n ) 12 Sn = . p + q = 1 Mean . s = r 2 Area of sector . A = 3 sin 2A + cos 2A = 1 4 sec2A = 1 + tan2A 5 cosec2 A = 1 + cot2 A 6 sin2A = 2 sinAcosA 7 cos 2A = cos2A – sin2 A = 2 cos2A-1 = 1.SULIT 3 STATISTIC 3472/2 1 x = x N 7 8 9 2 I 2 x = fx f (x x ) N 2 3 = = x N x _2 w1 I1 w1 n! n Pr (n r )! n! n Cr (n r )!r! 10 2 P(A B)=P(A)+P(B)-P(A B) p (X=r) = nCr p r q n r .blogspot.2 sin2A 9 sin (A B) = sinAcosB cosAsinB 1 2 r 2 10 cos (A B) = cos AcosB sinAsinB 11 tan (A B) = tan A tan B 1 tan A tan B 12 a b c sin A sin B sin C 13 14 a2 = b2 +c2 . = np 4 = f ( x x) f 2 = fx f x 2 11 5 1 2 N F M = L C fm I P1 100 P0 12 13 14 npq x z= 6 TRIGONOMETRY 1 Arc length.com .2bc cosA Area of triangle = 1 absin C 2 8 tan2A = 2 tan A 1 tan 2 A 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK [ Lihat halaman sebelah SULIT http://chngtuition. 4129 0.0838 0.00219 0.3156 0.4721 0.7 0.0526 0.0359 0.2296 0.00539 0.0721 0.4840 0.00820 0.00289 0.2420 0.6 2.1711 0.0465 0.00379 0.4207 0.00866 0.9 3.0951 0.0436 0.00657 0.0968 0.1 1.0918 0.00508 0.0655 0.0268 0.0618 0.3974 0.0301 0.2266 0.1190 0.00100 2 2 1 1 1 0 0 2.0853 0.4 1.00494 0.2389 0.00402 0.0516 0.00240 0.00523 0.3821 0.0102 3 0.0139 0.0158 0.0116 8 0.0606 0.0694 0.0455 0.00248 0.00604 0.0188 0.0427 0.4681 0.0505 0.4641 0.00154 0.1685 0.6 1.00368 0.1841 0.0174 0.4325 0.6 0.2206 0.0668 0.0630 0.2946 0.0823 0.2119 0.0764 0.1814 0.0351 0.0281 0.1292 0.4013 0.00122 0.00570 0.00126 0.0749 0.00149 0.0708 0.4960 0.0 1.1314 0.1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0. then P(X > k) = Q(k) Jika X ~ N(0.0162 0.1131 0.0 2.3745 0.2358 0.0885 0.0217 0.0735 0.2611 0.00842 3 2 2 2.SULIT 4 4 3472/2 THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0.0495 0.9 1.1635 0.4168 0.1401 0.4443 0.00391 0.00695 0.4483 0.0869 0.2148 0.00480 0.8 0.0643 0.00776 0.4522 0.00131 0.00466 0.00144 0.00264 0.com .00415 0.00440 0.0793 0.0179 0.0228 0.1335 0.00104 0.8 1.00714 0.00280 0.0314 0.2843 0.0166 0.00187 0.00676 0.3446 0.3897 0.0119 7 0.1020 0.0778 0.0 0.4404 0.2090 0.3936 0.0122 6 0.2578 0.2177 0.2743 0.1423 0.0375 0.1357 0.3 0 0.3557 0.00181 0.0537 0.00554 0.1922 0.00939 0.3707 0.0222 0.2676 0.0485 0.1112 0.1736 0.3300 0.0475 0.00357 0.00193 0.00621 0.0207 0.3 1.1093 0.7 2.1 0.5 0.0901 0. 1).4562 0.7 1. 1) 1 2 3 4 5 6 7 8 9 z 0.00964 0.0401 0.00175 0.0202 0.1949 0.00990 0.4364 0.2451 0.3336 0.3192 0.00336 0.1379 0.00139 0.1170 0.3632 0.1611 0.0392 0.0681 0.2912 0.00114 f ( z) 1 exp z 2 2 2 1 f (z) Q(z) Example / Contoh: If X ~ N(0.8 2.3520 0.0256 0.1788 0. 1).2005 0.0307 0.3859 0.0125 5 0.4761 0.1003 0.0146 0.2 2.0985 0.3409 0.0582 0.4286 0.0571 0.00107 0.1977 0.00453 0.0170 0.blogspot.00307 0.0192 0.0262 0.0132 0.2 0.3050 0.0239 0.00317 0.2877 0.00427 0.9 2.0336 0. maka P(X > k) = Q(k) z SULIT Q( z ) f ( z ) dz k O 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK http://chngtuition.4801 0.0110 Minus / Tolak 4 4 4 4 4 3 3 3 3 3 2 2 2 2 1 1 1 1 1 1 0 0 0 0 8 8 8 7 7 7 7 6 5 5 5 4 4 3 3 2 2 2 1 1 1 1 1 1 5 5 4 4 3 2 2 1 1 1 12 12 12 11 11 10 10 9 8 8 7 6 6 5 4 4 3 3 2 2 1 1 1 1 8 7 6 6 5 3 3 2 1 1 16 16 15 15 15 14 13 12 11 10 9 8 7 6 6 5 4 4 3 2 2 2 1 1 10 9 8 7 6 5 4 3 2 2 20 20 19 19 18 17 16 15 14 13 12 10 9 8 7 6 5 4 4 3 2 2 2 1 13 12 11 9 8 6 5 4 2 2 24 24 23 22 22 20 19 18 16 15 14 12 11 10 8 7 6 5 4 4 3 2 2 2 15 14 13 11 9 7 6 4 3 2 28 28 27 26 25 24 23 21 19 18 16 14 13 11 10 8 7 6 5 4 3 3 2 2 18 16 15 13 11 9 7 5 3 3 32 32 31 30 29 27 26 24 22 20 19 16 15 13 11 10 8 7 6 5 4 3 3 2 20 16 17 15 12 9 8 6 4 3 36 36 35 34 32 31 29 27 25 23 21 18 17 14 13 11 9 8 6 5 4 4 3 2 23 21 19 17 14 10 9 6 4 4 0.1 2.0244 0.0344 0.0250 0.00326 0.3 0.00272 0.00164 0.1762 0.2 1.0183 0.1446 0.0287 0.1539 0.1271 0.4602 0.3085 0.3015 0.1660 0.0329 0.3264 0.3783 0.1151 0.0233 0.0594 0.4090 0.1515 0.1230 0.0367 0.5 1.0274 0..0129 4 0.3594 0.00118 0.00755 0.2483 0.0418 0.2061 0.00914 0.2546 0.00111 0.4 0.0934 0.4920 0.0 0.0808 0.1469 0.00798 0.3372 0.0154 0.00639 0.0197 0.4880 0.5 2.00233 0.2643 0.1492 0.4 0.1894 0.1562 0.00205 0.2514 0.00256 0.00169 0.00347 0.00159 0.2709 0.00298 0.0113 9 0.4247 0.0446 0.0384 0.0409 0.3669 0.00889 0.0212 0.00587 0.0150 0.1075 0.5000 0.0143 0.1056 0.1251 0.0548 0.3483 0.3228 0.00226 0.00212 0.2327 0.1867 0.0559 0.1038 0.0294 0.0107 1 0.00135 0.2776 0.3121 0.0322 0.2810 0.4052 0.2236 0.0104 2 0.2033 0.00199 0.00734 0.1587 0.2981 0.0136 0.1210 0. [3 markah] 3472/2 @ 2011 Trial SPM Hak Cipta http://chngtuition. sketch a suitable straight line to find the number of solution for the equation sin(2 900 ) 1 State the number of solutions. 3472/2 1 Solve the following simultaneous equations: Selesaikan persamaan serentak berikut: 4x + 3y = x2 – xy = 8 Give your answer correct to 3 significant figures. using the same axes. Jawab semua soalan. Lakar graf bagi of y 2 cos 2 untuk 0 2 . Sketch the graph of y 2 cos 2 for 0 2 .SULIT 5 Section A Bahagian A [40 marks] [40 markah] Answer all questions. Nyatakan bilangan penyelesaian itu.com BPSBPSK [ Lihat halaman sebelah SULIT . Beri jawapan betul kepada 3 angka bererti. Hence. dengan menggunakan paksi yang sama. .blogspot. [5 marks] [5 markah] 2 (a) Prove that sin(2 900 ) cos 2 [2 marks] Buktikan sin(2 90 ) cos 2 0 [2 markah] (b) i. [3 marks] Seterusnya. [3 marks] [3 markah] ii. lakar satu garis lurus yang sesuai untuk mencari bilangan penyelesaian bagi persamaan sin(2 900 ) 1 . it is to be bend into several parts successively as shown in Diagram 3.blogspot. Ravi dan Hamid diberi seutas dawai seorang untuk dibengkokkan kepada beberapa bahagian berturut-turut seperti yang ditunjukkan dalam Rajah 3.SULIT 6 3472/2 3 Ravi and Hamid were given a piece of wire each. Bahagian pertama mesti berukuran A cm and setiap bahagian seterusnya dipendekkan sebanyak D cm masing-masing. Dawai Ravi. [4 marks] nilai bagi A dan nilai bagi D. (a) the value of A and of D. Calculate. Ravi’s wire which is 720 cm was bent exactly into 10 parts and Hamid’s which is 1040 cm was bent exactly 20 parts. [2 markah] 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK http://chngtuition. Hitung. yang panjangnya 720 cm dibengkokkan tepat kepada 10 bahagian dan Dawai Hamid. yang panjangnya 1040 cm dibengkokkan tepat kepada 20 bahagian. [2 marks] beza di antara panjang bahagian terakhir bagi kedua-dua dawai tesebut. [4 markah] (b) the difference of the length of the last part of the both wire.com SULIT . A cm Diagram 3 Rajah 3 The first part must be measured A cm and every part should be shortened by D cm respectively. 16). Rajah 4 menunjukkan garis lurus y = x + 9 bersilang dengan lengkung y = (x – 3)2 pada titik-titik M dan N(7. 2 ( 7.SULIT 7 3472/2 4.16).blogspot. [4 markah] (b) the volume generated when the shaded region A is rotated through 360° about x-axis. [3 marks] isipadu janaan apabila rantau berlorek A diputarkan 3600 pada paksi-x.16 ) Diagram 4 Rajah 4 Find Cari (a) the area of the shaded region K.com BPSBPSK [ Lihat halaman sebelah SULIT . [4 marks] luas rantau berlorek K. [3 markah] 3472/2 @ 2011 Trial SPM Hak Cipta http://chngtuition. Diagram 4 shows the straight line y = x + 9 intersecting the curve y = (x – 3) at points M and N (7. (a) Find the equation of PQ. Solution by scale drawing will not be accepted. 6) P(3. 6). 6).blogspot.com SULIT . 3) and point S(2. Diagram 5 shows quadrilateral PQRS. [4 markah] 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK http://chngtuition. Cari persamaan lokus bagi W. Rajah 5 menunjukkan sisi empat PQRS. [3 marks] Cari persamaan garis lurus PQ. y S(2. Find the equation of the locus of point W. Diberi bahawa koordnat bagi titik P(3. Penyelesaian secara lukisan berskala tidak akan diterima. 3) dan titik S(2. 3) R O Q x Diagram 5 Rajah 5 Given that the coordinates of point P(3. [3 markah] (b) A point W moves such that its distance from point Q is always twice its distance from point S.SULIT 8 3472/2 5. [4 marks] Titik W bergerak dengan keadaan jaraknya dari titik Q adalah sentiasa dua kali jaraknya dari titik S. 9 3472/2 Table 6 shows the marks obtained by a group of students in a class. find the standard deviation. cari nilai bagi k 26 [3 marks] (b) Calculate the standard deviation of the distribution. cari sisihan piawai.blogspot. [ 3 marks ] (c) If each marks of the students in the class is multiplied by 2 and then subtracted by 3. [1 mark] [1 markah] 3472/2 @ 2011 Trial SPM Hak Cipta http://chngtuition.com BPSBPSK [ Lihat halaman sebelah SULIT . [3 marks] Kira sisihan piawai taburan tersebut. find the value of k 26 [3 marks] 7 Diberi markah median pelajar ialah 60 . Jika setiap markah pelajar dalam kelas didarab dengan 2 dan kemudian di tolak dengan 3. For the new set of marks. Marks 25 – 39 40 – 54 55 – 69 70 – 84 85 – 99 Number of students 10 15 k 16 6 Table 6 Jadual 6 (a) Given that the median mark of a student is 60 7 .SULIT 6. Bagi set baru markah. x dan y. 3472/2 7 Use graph paper to answer this questions. Jadual 7 menunjukkan nilai-nilai bagi dua pembolehubah. Hence.5 4 0.9 13 Table 7 Jadual 7 1.2 unit pada kedua-dua paksi.SULIT 10 Section B Bahagian B [40 marks] [40 markah] Answer any four questions from this section. lukis garis lurus penyuaian terbaik. Variable x and y are related by equation y pk x . Gunakan kertas graf untuk menjawab soalan ini. Jawab mana-mana empat soalan daripada bahagian ini. Table 7 shows the values of two variables x and 2 y obtained from an experiment.com SULIT .2 50 (a) Plot log 10 y against x2 . where k and p are constants. x y 0. using a scale of 2 cm to 0. dengan menggunakan skala 2 cm mewakili 0. yang diperoleh daripada2 satu eksperimen.7 7 0. [6 markah] (b) Use your graph in 7(a) to find the value of Gunakan graf anda di 7(a) untuk mencari nilai (i) k (ii) p (iii) x when y = 10 x apabila y = 10 [4 marks] [4 markah] 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK http://chngtuition.0 20 1.2 unit on both axes.blogspot. dengan keadaan k dan p adalah pemalar. Pembolehubah x and y dihubungkan oleh persamaan y pk x . Seterusnya. draw the line of best fit. [6 marks] Plot log 10 y melawan x. [3 marks] persamaan normal kepada lengkung pada titik (-1. [3 marks] nilai hampir bagi y. (b) Given that the gradient function of a curve at ( p. 3 . [4 marks] [ 4 markah ] 3472/2 @ 2011 Trial SPM Hak Cipta http://chngtuition. (ii) the equation of the curve. 8 x3 which has a turning point 2x5 8 x3 yang mempunyai 2x5 Diberi bahawa fungsi kecerunan kepada lengkung titik pusingan pada titik ( p.3) .99. persamaan lengkung itu.3).99.3). [3 markah] 1.SULIT 8.blogspot. Find Cari (i) the value of p nilai p. Find Cari (i) the equation of normal to the curve at point (-1.3) . when x decrease from 2 to 1. apabila x menyusut dari 2 kepada 1.com BPSBPSK [ Lihat halaman sebelah SULIT . 3 . 11 3472/2 (a) Given that the equation of the curve y 3 x 2 4 x 2 is passing through point Diberi bahawa persamaan lengkung y 3 x 2 4 x 2 melalui titik 1. [3 markah] (ii) the approximate value in y. Seterusnya. where m and n are constants. OB 2 b. OB 2 b. Diberi OA a. ~ ~ ~ (ii) n. OQB. [3 marks] [ 3 markah ] 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK http://chngtuition. dengan keadaan m dan n ialah pemalar. ~ ~ m. 3 OP OA dan ~ ~ 2 OQ OB (a) Express each of the following vectors in terms of a and/ or b . ~ ~ [4 marks] [4 markah] (c) Hence. PQR dan ABR ialah garislurus. Diberi AR m AB dan QR n PQ . ~ n. 3 OP OA and 2 OQ OB .blogspot. Given OA a. Express OR in terms of Ungkapkan OR dalam sebutan (i) m. ~ ~ OPA. a and b .SULIT 9 12 3472/2 Diagram 10 shows triangles OAB and APR. a dan b . a and b . Rajah 10 menunjukkan segi tiga OAB dan APR. ~ ~ Ungkapkan dalam sebutan a dan/ atau b . OQB. A B R P Q Diagram 10 Rajah 10 O OPA. ~ ~ (i) AB (ii) PQ [3 marks] [3 markah] (b) Given AR m AB and QR n PQ .com SULIT . a dan b . PQR and ABR are straight lines. find the value of m and of n. cari nilai m dan nilai n. Dalam sebuah kolam ikan air tawar. the weight of one type of fish is normally distributed with a mean of 1. cari jumlah ikan di dalam kolam itu. (iii) Given that 20% of the fish has the weight more than m kg.8 kg. [6 marks] [6 markah] 3472/2 @ 2011 Trial SPM Hak Cipta http://chngtuition.blogspot. Cari nilai bagi m . calculate the probability that Di SMK Bandar Baru.0 kg and a standard deviation of 0. Find the value of m . 30 out of 120 form 4 students cycle to school. 30 daripada 120 orang pelajar tingkatan 4 mengayuh basikal ke sekolah. Jika 10 orang pelajar dipilih secara rawak.8 kg.0 kg dan sisihan piawai 0. more than 2 students cycle to school lebih daripada 2 orang pelajar mengayuh basikal ke sekolah. Diberi bahawa 20% daripada ikan tersebut mempunyai berat lebih daripada m kg.3 kg.SULIT 10 13 3472/2 (a) At SMK Bandar Baru. berat sejenis ikan adalah tertabur secara normal dengan min 1.3 kg. (i) Find the probability of a fish chosen randomly which is less than 0.com BPSBPSK [ Lihat halaman sebelah SULIT . Jika terdapat 100 ekor ikan yang mempunyai berat kurang daripada 0. Cari kebarangkalian bahawa seekor ikan yang dipilih secara rawak mempunyai berat kurang daripada 0. If 10 students are chosen at random. find the total number of fish in the pond.8 kg. [4 marks] [4 markah] (ii) (b) In a fresh water fish pond. hitung kebarangkalian bahawa (i) exactly 2 of them cycle to school tepat 2 orang pelajar mengayuh basikal ke sekolah. (ii) If there are 100 fish which has weight less than 0.8 kg. bagi kawasan berlorek. perimeter. Rajah 11 menunjukkan semibulatan ABE dan sektor OCD. .blogspot. with centre O.. in cm.SULIT 11 14 3472/2 Diagram 11 shows a semicircle ABE and a sector OCD. yang berpusatkan O. in cm2 . Diberi bahawa panjang AE = 16 cm dan OB : BC = 2 : 1 .com SULIT . untuk seluruh rajah. dalam radian. C B A O E Diagram 11 Rajah 11 D Given that the length of AE = 16 cm and OB : BC = 2 : 1 . in radian. luas. [4 marks] [4 markah] 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK http://chngtuition. [Use ] [Guna ] Calculate Hitung (a) . of the shaded region. [3 marks] [3 markah] (c) perimeter. dalam cm2 . dalam cm. [3 marks] [3 markah] (b) the area. for the whole diagram. Seterusnya atau dengan cara lain hitung jumlah jarak yang dilalui dalam 5 saat yang pertama selepas melalui O. which situated 3 meter on the right of point O. A particle moves along a straight line from a fixed point O.blogspot. 3472/2 12. [5 marks] Lakarkan graf halaju-masa untuk 0 t 5.V ms-1 is given by v t 2 8t 12. Its velocity. [ 3 markah ] (b) Sketch the velocity-time graph for 0 t 5. Hence or otherwise calculate the total distance travelled during the first 5 seconds after leaving O. dengan keadaan t ialah masa dalam saat.selepas melalui O.V ms-1. where t is the time in seconds. diberi oleh v t 2 8t 12. Halajunya. [5 markah] 3472/2 @ 2011 Trial SPM Hak Cipta http://chngtuition. [2 marks] halaju minimum.SULIT 15 Section C Bahagian C [20 marks] [20 markah] Answer any two questions from this section. after leaving O. Suatu zarah bergerak disepanjang suatu garis lurus daripada satu titik tetap O yang berada 3 meter di kanan O. [ 3 marks ] julat nilai t apabila zarah bergerak ke kiri. Jawab mana-mana dua soalan daripada bahagian ini. [2 markah] (ii) the range of values of t when the particle moving to the left.com BPSBPSK [ Lihat halaman sebelah SULIT . [Assume motion to the right is positive] [Anggapkan gerakan ke arah kanan sebagai positif] Find Cari (a) (i) the minimum velocity. [ 2 marks ] Lakar segitiga C’B’D’ yang mempunyai bentuk yang berlainan dengan segitiga CBD dengan keadaan D’C’ = DC .blogspot.com SULIT .. Diberi bahawa AE = 5 cm. B’C’ = BC and C ' D ' B ' CDB. AD = 4 cm. D 380 A C E B Diagram 13 Rajah 13 Given that AE = 5 cm. [2 markah] 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK http://chngtuition. ADE 380 . luas segitiga ADC. where AEC and BED are straight lines. and DEC is an obtuse angle. dan DEC ialah sudut cakah. AD = 4 cm. [3 marks] [3 markah] [3 marks] [3 markah] [2 marks] [2 markah] (b) Sketch triangle C’ B’ D’ which has a different shape from triangle CBD such that D’C’ = DC . B’C’ = BC dan C ' D' B ' CDB. Calculate Kira (a) (i) DEC (ii) the length of DC panjang DC (iii) the area of the triangle ADC. ADE 380 . EC = 8 cm. Rajah 13 menunjukkan segitiga ACD dan DBC di mana AEC dan BED adalah garis lurus.SULIT 13 16 3472/2 Diagram 13 shows two triangles ACD and DBC. EC = 8 cm. Gunakan kertas graf yang disediakan untuk menjawab soalan ini. Dia mempunyai 360 orang tenaga pekerja dan modal sekurangkurangnya RM 24 000. Dengan menggunakan skala 2 cm kepada 10 hektar pada kedua-dua paksi.SULIT 14 17 3472/2 Use the graph paper provided to answer this question. Ridhuan ingin menanam pokok pisang dan pokok betik di atas sebidang tanah seluas 80 hektar. Nyatakan tiga ketaksamaan selain x 0 dan y 0 yang memuaskan syarat-syarat di atas. Find the minimum profit gained. [3 marks] [3 markah] (c) Based on your graph. [3 marks] [3 markah] (b) Using the scale of 2 cm to 10 hectares on both axes. draw and shade a region R which satisfies all of the above conditions. other than x 0 and y 0 which satisfy the above conditions. Mr. find the maximum area of land for each type of fruit. En. Mr. Ridhuan menggunakan x hektar tanah untuk menanam pokok pisang dan y hektar tanah untuk menanam pokok betik. Ridhuan intends to plant banana trees and papaya trees on a piece of 80 hectares land. The cost of consumption for a hectare of banana trees farm are RM 800 and a hectare of papaya trees farm are RM 300. cari keluasan maksimum tanah yang digunakan untuk menanam setiap tanaman. Keuntungan hasil jualan pisang ialah RM 700 dan hasil jualan betik ialah RM 250 bagi setiap hektar. Jika luas kawasan tanah untuk menanam pokok pisang adalah 2 kali luas kawasan tanah untuk menanam pokok betik. En. Each hectare of banana trees farm is supervised by 3 labours while each hectare of papaya trees farm is supervised by 6 labours. lukis dan lorekkan rantau R yang memuaskan semua syarat-syarat di atas. Setiap hektar ladang pokok pisang diselia oleh tiga orang pekerja sementara enam orang pekerja untuk setiap hektar ladang pokok betik. Cari keuntungan minimum yang diperolehi. Kos perbelanjaan untuk sehektar ladang pokok pisang ialah RM 800 dan sehektar ladang pokok betik ialah RM 300. jawab soalan-soalan berikut : (i) If the area of land allocated for planting banana trees is twice the land for papaya trees. The profit gained by selling bananas are RM 700 and by selling papayas are RM 250 for each hectare.com BPSBPSK [ Lihat halaman sebelah SULIT . [4 marks] [4 markah] (ii) 3472/2 @ 2011 Trial SPM Hak Cipta http://chngtuition. He employed 360 labours and allocated a capital of at least RM 24 000. Ridhuan used x hectares of land to plant banana trees and y hectares to plant papaya trees.blogspot. (a) State three inequalities. answer the following questions: Berdasarkan graf anda. Cari nombor indeks gubahan pada tahun 2004 berasaskan tahun 1996.25. perubahan indeks harga dari tahun 2000 ke tahun 2004 dan peratus perbelanjaan masing-masing. Price index in the year 2000 Indeks harga pada tahun 2000 Change of price index from the year 2000 to the year 2004 Perubahan indeks harga dari tahun 2000 ke tahun 2004 Percentage of expenditure (%) Peratus perbelanjaan (%) Expenditure Perbelanjaan P Q R S 110 120 125 150 Increased 15 % Decreased 10 % Unchanged Unchanged Table 15 Jadual 15 x 30 y 15 (a) Calculate the expenditure of item P in the year 1996 if its expenditure in the year 2000 is RM 150. Hitungkan perelanjaan bagi item P pada tahun 1996 jika perbelanjaannya pada tahun 2000 ialah RM 150.blogspot. jumlah perbelanjaan pada tahun 2000 jika perbelanjaan keluarga Muhammad pada tahun 1996 ialah RM 260. Calculate Hitungkan (i) the value of x and of y.SULIT 18 3472/2 15. (ii) the total monthly expenditure in the year 2000 if the total monthly expenditure for Muhammad’s family in the year 1996 is RM 260.25. Nombor indeks gubahan pada tahun 2000 berasaskan tahun 1996 ialah 121. the change in price index from the year 2000 to the year 2004 and the percentage of expenditure respectively. [2 marks] [2 markah] The composite index number in the year 2000 based on the year 1996 is 121. Table 15 shows the price indices of the monthly expenditures for Muhammad’s family in the year 2000 based on the year 1996. nilai x dan nilai y. Jadual 15 menunjukkan indeks harga bagi perbelanjaan bulanan keluarga Muhammad pada tahun 2000 berasaskan tahun 1996.65.com SULIT . [3 marks] [3 markah] END OF QUESTION PAPER KERTAS SOALAN TAMAT 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK http://chngtuition.65. [5 marks] [5 markah] (b) (c) Find the composite index number in the year 2004 based on the year 1996. Graph paper and booklet of four – figure mathematical tables is provided. Tie the ‘ helaian tambahan’ and the graph papers together with the ‘buku jawapan’ and hand in to the invigilator at the end of the examination. mana-mana empat soalan daripada Bahagian B dan manamana dua soalan daripada Bahagian C Write you answer on the ‘buku jawapan’ provided. you may ask for ‘helaian tambahan’ from the invigilator.blogspot. Markah yang diperuntukan bagi setiap soalan dan cerian soalan are shown in brackets. It may help you to get marks. 3472/2 @ 2011 Trial SPM Hak Cipta http://chngtuition. Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan. Sekiranya buku jawapan tidak mencukupi. Jawab semua soalan dalam Bahagian A. 6 The marks allocated for each question and sub-part of a question are shown in brackets. 7 A list of formulae is provided on pages 2 and 3.com BPSBPSK [ Lihat halaman sebelah SULIT . four questions from Section B and two questions from Section C. Ikat helaian tambahan dan kertas graf bersama-sama dengan buku jawapan dan serahkan kepada pengawas peperiksaan pada akhir peperiksaan. Kertas soalan ini mengandungi tiga bahagian Bahagian A. Section B and Section C. You may use a non-programmable scientific calculator. 9. sila dapatkan helaian tambahan daripada pengawas peperiksaan. Bahagian B dan Bahagian C 2 Answer all questions in Section A. Satu senarai rumus disediakan di halaman 3 hingga 5 8.SULIT 19 INFORMATION FOR CANDIDATES MAKLUMAT UNTUK CALON 3472/2 1 This question paper consists of three sections : Section A. 3 4 Show your working. 5 The diagrams in the questions provided are not drawn to scale unless stated. If the buku jawapan is insufficient. Tunjukkan langkah-langkah penting dalam kerja mengira anda. 10. Ini boleh membantu anda untuk mendapatkan markah. Jawapan anda hendaklah ditulis di dalam buku jawapan yang disediakan. Kertas graf dan sebuah buku sifir matematik empat angka disediakan. Anda dibenarkan menggunakan kalkulator scientific calculator yang tidak boleh diprogramkan. KAD PENGENALAN 20 3472/2 ANGKA GILIRAN Arahan Kepada Calon 1 2 3 Tulis nombor kad penganalan dan angka giliran anda pada petak yang disediakan. Ceraikan helaian ini dan ikat sebagai muka hadapan bersama-sama dengan buku jawapan.SULIT NO. Kod Pemeriksa Bahagian Soalan Soalan Dijawab Markah Penuh Markah Diperoleh ( Untuk Kegunaan Pemeriksa) 1 A 5 8 6 7 7 7 10 10 10 10 10 10 10 10 10 JUMLAH 2 3 4 5 6 7 B 8 9 10 11 12 C 13 14 15 3472/2 @ 2011 Trial SPM Hak Cipta BPSBPSK http://chngtuition.blogspot.com SULIT . Tandakan ( / ) untuk soalan yang dijawab. 3472/1 Matematik Tambahan Kertas 1 2 jam Ogos 2011 BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2011 ADDITIONAL MATHEMATICS Paper 1 MARKING SCHEME Skema Pemarkahan ini mengandungi 6 halaman bercetak 1 http://chngtuition.com .blogspot. com .KERTAS 1 No.blogspot. x B1: 2x 9x 5 0 2 3 3 or 5 1 2 5 1 p 11 B2 : ( p 1)( p 11) 0 B1: ( p 1) 4(3)( p 1) 0 2 3 1 11 3 6(a) (b) (c) m=1 x=1 (1. 3) 1 1 1 3 2 http://chngtuition. 1(a) Solution and Mark Scheme Sub Marks 2 Total Marks 3 3 2 B1: (b) m3 1 m 1 1 2 2(a) x3 2 5 2 B1: k(3) = 2 a = 6 and b = -11 B2: a = 6 or b = -11 B1: fg 1 ( x) a( 1 3 (b) 2 3 3 3 x5 )b 2 4 1 2 B2: (2x 1)( x 5) 0 x 5.PERATURAN PEMARKAHAN. com . 400 3 http://chngtuition.7 x B2: 2 5 2( x 2) 1 4( x 1) 2 or x 2 4 x 4 3 3 B1: 8 5 1 3 2( x 2) 1 2 1 5 4( x 1) x B3: B2: B1: 4 4 3x3 92 log9 x2 (3x) 2 log 9 3x (for change base) log 9 81 2 5h 1 (2h – 4 ) = 6h + 4 – (5h – 1) 2 4 9(a) h=1 B1: (b) 1460 B1: 20 2(16) 19(6) 2 23 3 s23 s3 OR 2(2) 22(6) 2(2) 2(6) 2 2 20 OR s20 16 130 2 s20 4 4 10 1 and a = 4 3 1 B3 : r or a = 4 3 8 B2 : 3(1 r )(1 r ) 3 a 3 1 r B1 : or r 42300 a ar 8 3 3 3 11 B2 : S36 36 2(300) 35(50) 2 B1 : S36 or a 300 or 300.blogspot.350. blogspot.12 2 and p 4 3 2 B2: q or p 4 3 1 p 30 3 B1: xy x 2 or q q 64 2 q t = 1 B1: 3 3 13(a) 2 4 2t 0 1 08 4 2 (b) y = 4x + 2 B1: Using m1• m2 = -1 and m2 = 4 14 m = 1 and n = 3 3 3 m 1 n 3 n( 4) m(4) n(2) m(6) B1: 2 or 3 nm nm B2 : 15(a) (b) 3x + 6 y 2x + 2 y B1 : 2 1 (3x 6 y) or (6 y 3x) or equivalent 3 3 1 2 3 16(a) 8i 4 j 1 3 (b) 1 (8i 4 j ) or 80 1 (2i j ) 5 80 2 2 2 B1: AB 8 4 = 4 http://chngtuition.com . 8) 40 2 2 (b) 72 cm B1 : S AB 10(0.8) or SCD 30(0.8) 18 33 65 : 2 2 B1 19 sin B 3 5 OR cos A 5 13 3 3 13 2 : B2 x2 (5) 2 2 1 B1 : x dx f ( x) dx 2 2 1 1 OR x2 2 OR f ( x) dx 5 2 1 20 27 B2 B1 : (1) 2 (9(1) 2 )((1)3 4) 2 ((1)3 4)3 (2(1)) : 3 3 x 2 (9 x 2 )( x 3 4) 2 ( x3 4)3 (2 x) or equivalent 21(a) 5 B1 : 2 3 450 90h 0 1 (b) 1125 5 http://chngtuition.blogspot.com .17(a) r = k =10 B1: 2 4 1 2 r (0. 03 6 http://chngtuition.03 1 3 4 k 55 3 B1 : z = 1.1515 (b) 58.09 B2 : 1.com .22(a) (b) p=9 8 B1 : 2 1 2 3 3 7 9 5 1 (5)2 5 2 2 2 2 2 23(a) (b) 20 240 B1 : 1 2 5! X 2! 1 3 24(a) 1 2 5 12 B1 : 3 (b) 2 1 2 3 1 4 3 4 3 25(a) 0.blogspot. blogspot.com .3472/2 Matematik Tambahan Kertas 2 Ogos 2011 2 ½ jam BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2011 TINGKATAN 5 ADDITIONAL MATHEMATICS Paper 2 MARKING SCHEME Skema Pemarkahan ini mengandungi 9 halaman bercetak 1 http://chngtuition. 2 http://chngtuition.blogspot.com . 679 x = -1.37 .49 OR y = 4. y = -0.80y . -1.680 . -0.No Solution and Mark Scheme Sub Marks Total Marks 1 y 2 8 4x 8 3y x 3 4 or 2 P1 5 5 8 4x x x 8 3 7x2 – 8x – 24 = 0 or or 21y2 .blogspot.51 .49 .com .2 cycle -1 2 ii) y 2 y 2 2 2 K1-straight line K1-for equation Number of solutions = 4 N1 3 http://chngtuition.51 N1 N1 K1 2 (a) (sin 2 cos 900 cos 2 sin 900 ) cos 2 N1 K1 2 8 y (b) i) 2 1 0 y 2 cos 2 6 N1-cosine curve N1-amplitude 2 N1.64 = 0 8 3y 8 3y y 8 4 4 K1 x (80) (80) 2 4(21)(64) (8) (8) 2 4(7)(24) y 2(21) 2(7) or x = 2.37 4. 2. blogspot.3 (a) a = A .com . d = -D S10 = 4 S20 = 6 10 [2 A 99( D)] 720 2 or 20 [2 A 19( D)] 1040 2 K1 K1 N1 N1 2 Solving the simultaneous in linear equation A = 90 or D =4 A = 90 and D = 4 (b) T10 = 90 + 9(-4) Difference = 40 or T20 = 90 + 19(-4) N1 K1 4(a) Area of trapezium – area under curve 4 7 ( x 9) dx 0 7 or 1 (16 9)(7) 2 OR ( x 3) 0 7 2 dx 7 K1 K1 x2 9 x 2 0 = 7 OR ( x 3) 3 3 0 175 2 343 1 / 57 unit 2 6 6 - 91 3 N1 K1 = (b) y=0.x= 3 OR ( x 3) 2 dx 3 2 0 K1 3 ( x 3) 5 5 0 3 K1 243 unit 3 5 4 N1 http://chngtuition. 58 N1 1 5 http://chngtuition.5 OR 25 and k N1 P1 3 7 47 k 2 25 7 60 54.5(a) mPQ 5 3 K1 3 7 5 y 3 ( x 3) 3 3y = 5x 6 (b) Q(0.79 N1 (c) 37.blogspot.5 15 26 k k = 13 (b) Mean. x K1 N1 3563 60 3 P1 K1 232 755 3563 2 (* ) 60 60 18. 2) K1 N1 P1 4 ( x 0)2 ( y 2)2 or ( x 2) 2 ( y 6) 2 K1 K1 ( x 0)2 ( y 2)2 2 ( x 2) 2 ( y 6) 2 3x 2 3 y 2 16x 52 y 156 0 6(a) L = 54.com . com .blogspot.84 (b) K1 N1 8 x3 (i ) 0 2 x5 p2 (ii) y 1 1 c 4 2x x 1 1 45 y 4 2 x 16 x (i) a 2b (ii) K1 N1 4 K1 N1 9(a) N1 3 10 1 1 OA OB 3 2 1 ab 3 K1 N1 4 K1 N1 K1 N1 3 (b) (i) a m(a 2b) (1 m)a 2mb 1 (ii ) b n( a b) 3 1 na (1 n)b 3 6 http://chngtuition.01) y new 3(2) 2 4(2) 2 y 17.8(a) (i ) dy 6x 4 dx 1 2 K1 6 10 m normal K1 2 y x 7 or equivalent N1 K1 (ii) y (16)(0. 75)10 10C1 (0.8411) 2 2 2 = 8.com . n = 3 (Both correct) K1 K1 N1 10(a) (i) P( x 2) 10C2 (0.8411 (8)2 (0.0 0.842 m 1.(c) 1 2m 1 m or 1 m n 3 Solve the simultaneous linear equation m = 2.3 m = 1.2186 (ii) P( x 2) 1 P( x 0) P( x 1) P( x 2) K1 N1 4 10 1 10C0 (0.8411 rad (b) Area shaded = Area triangle OCE – Area sector OBE = 1 1 (8)(12)sin 0.2186 K1 = 0.8) P( z N1 0.75)8 = 0.4744 (b) (i) P( x 0. OC = 12 P1 K1 N1 3 cos 8 12 0.253 kg 11(a) BC = 4 .667) K1 N1 6 = 0.25)2 (0.blogspot.2 Z = 0.25)1 (0.2524 N1 P1 K1 N1 10 = 396 (iii) P( x m) 0.3 P( z 0.25)0 (0.842 0.863 cm K1K1 N1 3 7 http://chngtuition.75)9 0.8 1 ) 0.2524 (ii) Total 100 0. 8 = 0 t =4 V = -4 K1K1 N1 2 K1 N1 3 K1 K1 N1 10 (ii) t 2 8t 12 0 (t 2)(t 6) 0 2t 6 v (b) 12 P1-shape P1-passing through point (0.s = 13 3 2 t = 5. (2. s = 4 3 2 2 2 13 3 (13 4 ) 3 3 3 59 meter 3 8 K1 K1 N1 http://chngtuition.s= 3 K1 OR K1 N1 32 5 32 3 3 2 59 meter 3 2 t = 2 .3009 rad P1 Perimeter = arc AB + BC + arc CD + AE + ED = 8(2.blogspot.8411) + 16 + 4 = 52.OR (c) Area shaded = 1 1 (8)( 80) (8)2 (0.8411) 2 2 = 8.3009) + 4 + 12(0.com .0) and (5.-3) 5 0 2 5 t Total distance = v dt v dt 0 2 2 5 t = 0 .50 cm 12(a) (i) 2t .863 K1K1 N1 4 AOB 2.12). 507 112.13(a) (i) 4 sin 38 5 DEA 29.507 SinDEA DEC 150.49 2 (iii ) Area of ADC = = 24.493 (ii) 10 K1 K1 N1 DAE 180 38 29.99 1 4 13 sin112.02 (b) D C N1-shape N1-label B K1 N1 14 (a) x + y 80 3x + 6y 360 or equivalent 800x + 300y 24 000 or equivalent N1 N1 N1 3 10 (b) Refer graph 3 (c) (i) Draw x = 2y x = 52 .com .493o DC 4 13 2(4)(13) cos112. y = 26 (ii) Kmin = 700(10) + 250(54) (substitud the point in R) = RM 20 500 K1 N1 (both) K1 4 N1 9 http://chngtuition.blogspot.49 2 2 2 P1 K1 N1 DC 14. 15 (a) p96 150. 150 ( I 2004 for every item) 1996 K1 P1 N1 3 = 124.blogspot. x = 40 (both) (ii) p2000 260 121.25 100 15y = 225 (Solve simultaneous equation) y = 15 .50(40) + 108(30) + 125(15) + 150(15) 100 126.25 (c) I 2004 1996 126. 108.65 100 110 10 K1 N1 2 = RM136. 125.com .50.25 100 K1 N1 2 = RM315.95 (b) (i) x + y = 55 3 K1 K1 N1 110x + 120(30) + 125y + 150(15) 121.25 10 http://chngtuition.