47/ NO. KAD PENGENALAN Additional Mathematics Paper ANGKA GILIRAN Hours JABATAN PELAJARAN NEGERI PULAU PINANG ADDITIONAL MATHEMATICS Paper Two Hours J
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1 PROGRAM DIDIK CEMERLANG AKADEMIK SPM ADDITIONAL MATHEMATICS MODULE 9 MODEL SPM QUESTIONS ( PAPER ) ORGANISED BY: JABATAN PELAJARAN NEGERI PULAU PINANG
2 47/ NO. KAD PENGENALAN Additional Mathematics Paper ANGKA GILIRAN Hours JABATAN PELAJARAN NEGERI PULAU PINANG ADDITIONAL MATHEMATICS Paper Two Hours JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU. Tuliskan angka giliran dan nombor kad pengenalan anda pada ruang yang disediakan.. Calon dikehendaki membaca arahan di halaman. Kod Pemeriksa Questions Marks Actual Marks Total Jabatan Pelajaran Pulau Pinang
3 INFORMATION FOR CANDIDATES. This question paper consists of 5 questions.. Answer all questions.. Give only one answer for each question. 4. Write your answer clearly in the spaces provided in the question paper. 5. Show your working. It may help you to get marks. 6. If you wish to change your answer, cross out the work that you have done. Then write down the new answer. 7. The diagrams in the questions provided are not drawn to scale unless stated. 8. The marks allocated for each question and sub-figure mathematical tables is provided. 9. You may use a non-programmable scientific calculator. 0.A booklet of four-figure mathematical tables is provided..you may use a non programmable scientific calculator..this question paper must be handed in at the end of examination. Jabatan Pelajaran Pulau Pinang
4 Answer all questions. Diagram, the function h maps x to y and the function k maps y to z. x y z h k 4 DIAGRAM Determine (a) hk ( 4), (b) kh ( ). [ marks] Answer : (a) (b). The function p is defined as p (x) = Find x, x 4. 4 x (a) p(x), (b) p(5). [ marks] Answer : (a) (b) Jabatan Pelajaran Pulau Pinang 4
5 . The following information refers to the functions h and g. g (x) = 4 x fg (x) = x + 5 Find f (x). [ marks] Answer : 4. The straight line y = t intercept with the curve y = + 5x 4x at the two points A and B. Find the range of values of t. [ marks] Jawapan : 5. Solve the quadratic equation places. x x 5 Give your answer correct to three decimal [ marks] Answer : x = Jabatan Pelajaran Pulau Pinang 5
6 6. Diagram shows the graph of a quadratic functions g(x) = 4 (x + h) where h is constant. y (,k) y = g(x) O DIAGRAM x The curve y = g(x) has maximum point (, k), where k is a constant.state (a) the value of h, (b) the value of k,. (c) the equation of the axis of symmetry. [ marks] Answer : (a) h = (b) k = (c) 7. Solve the equation 64 x. x x (8 )(4 ) [ marks] Answer : x = Jabatan Pelajaran Pulau Pinang 6
7 8. Solve the equation log (x ) + = log x [ marks] Answer : x = x 9. Given that log x = p and log y = r, express log in term p and r. y [ marks] Answer : 0. The sum of the first n term of an arithmetric progression is given by Sn= 5n. Find the fifth term. [ marks] Answer : Jabatan Pelajaran Pulau Pinang 7
8 . The first three terms of an arithmetic progression are,,, Find (a) the common ratio, (b) the sum of the first 0 terms after the 5th term. [ 4 marks] Answer : (a) (b). The sum of the first n terms of an arithmetric progression 47, 44, 4, is 5. Find (a) the common difference of the progression, (b) the value of n. [4 marks] Answer : (a) (b) n = Jabatan Pelajaran Pulau Pinang 8
9 . Diagram shows a straight line graph of log 0 y againts log 0 x. The variables x and y are related by equation y = p x q, where p and q are contants. log 0 y 4 O log 0 x (a) Calculate the value of p and q, DIAGRAM (b) Find the value of y if x = 0. [4 marks] Answer : (a) p = q = (b) y = 4. The following information refers to the equations of two straight lines, AB and CD, which parallel to each other. AB : y = p x + q CD : y = (q + ) x + Where p and q are constants Express p in terms q. [ marks] Answer : p = Jabatan Pelajaran Pulau Pinang 9
10 5. Given that B point is ( 8, 5) and O is origin. (a) Express OB in terms of i and ~ j. ~ (b) Find the unit vector in the direction of OB. [ marks] 6. Given that OA = i + j, ~ ~ AB = :. Find (a) (b) AB, OR. OB = 0 Answer : (a) (b) i + 6 j and R is a point on AB such that AR : ~ ~ [ marks] Answer : (a) (b) 7. Solved the equation sek x cot x for 0 x 60. [4 marks] Answer : x = Jabatan Pelajaran Pulau Pinang 0
11 8. Diagram 4 shows the sector OPQ, centre O with a radius of 5 cm. Line PR is perpendicular to the line OQ and QR = cm. Q R P RAJAH 4 O [4 marks] Find ( a ) POR in radians, ( b ) the perimeter of the shaded region. Answer : ( a ). ( b ). 9. Given that f ( x) 4x(5 x), find f "( x) [4 marks] Answer : Jabatan Pelajaran Pulau Pinang
12 0. Given that x = t + t and y = t (a) find in term of t. dy (b) If y decreases from 4.0 to.98, Find the corresponding small change in t. [4 marks] Answer : (a) (b) x. Given that y and dy 4 f ( x) with f (x ) is a function in x. x Calculate the value of f ( x). [ marks] Answer : Jabatan Pelajaran Pulau Pinang
13 . A chess team consists that of 5 students. The team will be chosen from a group of 6 boys and 4 girls. Calculate the number of teams that can be formed such that each team consists of (a) (b) 4 boys, Not more than girls. [4 marks] Answer : (a) (b). The mean of five numbers is m. The sum of the squares of the numbers is 70 and the standard deviation is 9h. Express m in term of h. [ marks] Answer: m = Jabatan Pelajaran Pulau Pinang
14 4. Ali has 6 accessories of the item which consists of P, Q, R, S, T and U. He wants to arrange 4 of the items in a row. Find the probability that (a) the arrangment is TUPS, (b) the arrangment does not incule P item [ marks] Answer : (a) (b) 5. X is a random variable of normal distribution with a mean of 0 and a variance 9, find the value of r such that P ( X < r ) = [4 marks] Answer : END OF QUESTION PAPER Jabatan Pelajaran Pulau Pinang 4
15 Answers. ( a ) h k ( 4) = ( b ) k h ( ) = 4. ( a ) p x ( x ) = 4 x y p ( y ) = 4 y x = y 4 x ( 4 y ) = y 4 x x y = y y ( x + ) = 4 x y = p ( x ) = y 4x x 4x x, x, 4(5) p(5) ( b ) 5 = 5. g ( x ) = 4 x y = 4 x x = 4 y x = g (x) = 4 y 4 x f ( x ) = f g g ( x ) = f g 4 x 4 x = 5 = x 4. y = t, y = + 5 x 4 x + 5 x 4 x = t Jabatan Pelajaran Pulau Pinang 5
16 4 x 5 x + t = 0 Intercept at two points, use b 4 a c > 0 ( 5 ) 4 ( 4 ) ( t ) > 0 6 t < 7 t < 5. 5 ( x ) = ( x ) x 0 x = Use x = b b 4ac a x = ( 0) ( 0) () 4()( ) x = 0 6 x =.4, x = g (x) = 4 ( x + h ) x + h = 0, x = h ; y = 4 ( h, 4 ) = (, k ) (a) h = (b) k = 4 (c) symmetry Axis : x = 7. 6 ( x+ ) x ½ = x. ( x + ) ( x + ) = x + [ x ] 8 x = x = 8 8. log ( x ) + = log x = log = log 4 log 4 ( x ) = log x 4 x 4 = x x 4 x + 4 = 0 ( x ) = 0 x = Jabatan Pelajaran Pulau Pinang 6
17 9. log x y log log x log y 0. S n = 5 n 5 = log log x log y = 5 + p r T n = S n S n T 5 = S 5 S 4 = [ 5 ( 5 ) ] [ 5 ( 4 ) ] = 7 = 5. (a) r = = (b) T, T, T 5, T 6,.. T 5 the sum of = S 5 S 5 S 5 = S 5 = 5 ( ) ( ) 5 ( ) ( ) S 5 S 5 = 5456 = ; = , 44, 4, ( a ) d = = n ( b ) S n = [a ( n ) d] n [(47) ( n )( )] n 9 7 n = 0 ( n + 5 ) ( n 5 0 ) = 0 = 5 n = 50. y = p x q ( a ) log 0 y = log 0 p + q log 0 x log 0 p = p = 0 = 00 Jabatan Pelajaran Pulau Pinang 7
18 q = ( b ) log 0 y = + ½ log 0 x Alternatif method 4. y = y = x = 0, log 0 y = + ½ log 0 0 y = p x q p q = 00 x ( 0 ) ½ = 6. p q x q x p = ( q ) log 0 y =.5 y = 0.5 y = ( a ) OB = 8 i 5 j ~ ~ 6. ( a ) ( b ) the unit vector in the direction of OB = AB = AO + OB 8i 5 j 89 0 = 6 ( b ) OR = ( OB) ( OA) 0 OR = 6 6 OR = 5 Jabatan Pelajaran Pulau Pinang 8
19 7. cot. + sek x = x tan x + sek x = tan x + + tan x = x tan x = tan x = tan x = ± = 45 o, 5 o, 5 o, 5 o 8. ( a ) Cos POR = 5 4 = 0. 8 POR ( b ) s PQ = j 9. f ( x ) = = r r = 5 x =. 75 RQ = cm, PR = cm ( Pythagoras Teorem ) The perimeter of the shaded region = x(5 x) u = 4x v = ( 5 x ) du 4 dy dv u v dv du ( )(5 x)( ) f '( x) 4x[ 4(5 x)] (5 x) (4) = 4 8 x 60 x + 00 f ''( x) 96x 60 = (a) x = t + t y = t dt dy dy 6t dt dt dt dy ( 6t) ( 6t) (b) y.98 4 = -0.0 Jabatan Pelajaran Pulau Pinang 9
20 dy dt dy dt t. 4 y t y dy dt dy 4 f ( x) dy f ( x) f ( x) 4 = = x x ( a ) C C 0 6 = 0 4 ( b ) Total of player boys and girls = = 0 0 The number of ways of selecting from 5 player = C 5 The number of ways of selecting player consisting of boy and 4 girls 6 4 = C C The number of team is less than boys = 5 6 = x m ; x 70 ; 9h x n 9 h = 70 5 x m m = 44 9h m = 44 9h 4. ( a ) The probability of arrangement is TUPS Jabatan Pelajaran Pulau Pinang 0
21 = ( b ) The probability does not included the P accessory = = 5 6 C C variance = = 9 Standard deviation = = mean = = 0 P ( X < r ) = Z = P X X 0 r 0 = P r 0 Z = r 0 P Z = P r 0 Z = r 0 = =. 9 6 r = Jabatan Pelajaran Pulau Pinang
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