Nama Pelajar : 347/ Additional Mathematics Paper September 00 Tingkatan 5 :. PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA SEKOLAH MENENGAH NEGERI KEDAH DARUL AMAN PEPERIKSAAN PERCUBAAN SPM 00 ADDITIONAL MATHEMATICS MARKING SCHEME Paper.

SULIT 347/ SPM Trial Examination 00 Kedah Darul Aman Marking Scheme Additional Mathematics Paper Question Solution/ Marking Scheme Answer Marks (a) 8 3 + n (a) 3 (a) B: = 5 3 (a) B: 4g ( x) + = 5x (b) 0 f ( x) 8 (b) 3x 3 (a) 5x 4 4 5(4x + ) 4 (b) B: = 4 4 (b) 4 B: 9 = k 9 3 9 3 B : α + α = or α α = 9 k 9 5 (a) 3 (b) (c) -7 6 B: - 6 5 6 x 3 5 B: ( 5x 6)( x + ) 7 B: x + 3 = 4(x + 3) B : x+ 3 4(x+ 3) or x = 5 3 3 347/ Additional Mathematics Paper SULIT

SULIT 3 347/ Question Solution/ Marking Scheme Answer Marks 8 7 7 B: 4 3 9 9 9 3 4 B: 9 or 3 5 B3: + 3 6 B: log log p p p log + log or 33 4 or 7 q q q or log p + log q 7 6 4 0 B: log pq = log p + log q (a) - 0 (b) B: S 0 = [( 6) + 9(3)] (b) 75 3 B: a = 6 and d = 3 (a) B: r 3 = 7 B: ar = 6 or ar 4 = 6 (a) 3 (b) 3 3 B: B: S 3 8 = ( r = 3x 4 ) + 8 + 3y B : = and = 7 4 4 3 (, 0) 3 3x + 8 + 3y B : = or = 7 4 4 4 B : p =. or q = 6. 3 B : 5..5 p = 4 or 5. = q. or.5 = q.(4) p = -. q = 6.3 3 5 B: h=7 or k=8 B: BA = hi ( k 3) j h=7 k=8 3 347/ Additional Mathematics Paper SULIT

SULIT 4 347/ Question Solution/ Marking Scheme Answer Marks 6 B: 3 5h = 6 h 8 3 5h B: = k or AB // AC 6h 8 3 3 7 B : B : 0 75 or 0 95 0 0 x 45 = 30 or 50 0 75 0, 95 0 3 8 B : (3) (4 ) () = θ.5 rad 3 B: (3) () or (4) θ 9 (a) B: m = (b) B: 0 = x + 8 (a) y = x + 8 (b) (6, 0) 0 B3: 6(0 + )(4 0) + (4 0) B: (x + )(4 3x)( 3) + (4 3x) 8 4 B: (x + )(4 3x)( 3) or () B : = B: x 8() or + c 8x (4 3x) x 8 y = + + 5 3 x 347/ Additional Mathematics Paper SULIT

SULIT 5 347/ Question Solution/ Marking Scheme Answer Mark 3 B : δ y = 4() p -3p 3 B : dy dx = 4 x 3 or δx = p 3 B: 0 3 3 7 C (0.5) (0.85) 0.98 4 (a) 35 7 5 (b) B: C 3 C or C + C3 5 5 (b) 30 5 (a) B : 00 9 5 (a).6 (b) B : 88 9 5 (b)0.788 END OF MARKING SCHEME 347/ Additional Mathematics Paper SULIT

MARKING SCHEME ADDITIONAL MATHEMATICS PAPER SPM TRIAL EXAMINATION 00 N0. SOLUTION MARKS x = 0 y P y + 0 y y = 4 K Eliminate x y ( ) 0y + 4 = 0 ( y ) ( y ) 4 6 = 0 y = 4 or y = 6 x = or x = K Solve quadratic equation (a) k = 6 P 5 (b) Mid point 3, 8, 33, 38, 43 P (i) Mean fx 3 + 4 8 + 7 33 + 5 38 + 3 43 = = f + 4 + 7 + 5 + 3 685 = = 34.5 0 K Use formula and calculate (ii) Varian fx = x f 3 + 4 8 + 7 33 + 5 38 + 3 43 = 34.5 0 4055 34.5 = 0 = 9.69 K Use formula and calculate (iii) Median, m N F (0) 5 L C 30.5 = + = + 5 fm 7 = 34.07 K Use formula and calculate 8

N0. SOLUTION MARKS 3 3 y = x x + 3 SMKDarulaman j*k (a) dy = x x = 3 dx x x 3 = 0 ( x ) ( x ) + 3 = 0 x =, 3 x = y = 3 x = 3 y = K Equate and solve quadratic equation, 3 and ( 3, ) (b) Equation of normals : m normal = 3 y = ( x + ) 3 3 y = x + or equivalent 3 3 y = ( x 3) 3 y = x + 3 or equivalent 3 K Use m normal to form equations 4 (a) y 6 y = 3sin x P Modulus sine shape correct. P Amplitude = 3 [ Maximum = and Minimum = -] O - π π 3π π x P Two full cycle in 0 x π - 3x y = π P Shift down the graph

N0. SOLUTION MARKS 4 3x (b) 3sin x = or 3x y = π π For equation SMKDarulaman j*k Draw the straight line 3x y = π K Sketch the straight line Number of solutions = 5. 5 7 (a) Common ratio, r = 4 (b) A 6 = π 4 = 56π ( 3) OR 5 T6 = ar = π 4 = 56π ( 4) 5 K (c) S S 6 6 ( 4 ) π ( 4 ) π = 4 4 4 4 = 34.5π.5π = 340π K Use S 6 or S K Use S 6 - S 6

N0. SOLUTION MARKS 6 (a) K for using vector (i) uuur uuur uuur triangle for a(i) or OD = OC + CD a(ii) = 6a + b % % (ii) uuur uuur uuur AB = OB OA uuur uuur = OD OA = 3a + 6b 3a % % % = 6b % OR uuur uuur AB = CD = 6b % [ K ] (b) uuur AE uuur uuur = AB + BE uuur = 6b + h OD % = 6b + h( 3a + 6b) % % % ( ) a + kb = 3ha + 6 + 6h b % % % % K for using vector uuur triangle and BE K 3h = h = 3 k = 6 + 6h = 6 + 6 3 = 8 K for equating coefficients correctly 8

N0. SOLUTION MARKS 7 (a) x 3 4 5 6 SMKDarulaman j*k log 0 y 0.65 0.87.08.30.5.74 6 correct values of log y (b) log 0 y K Plot log 0 y vs x Correct axes & uniform scale 6 points plotted correctly 0.43 Line of best-fit 0 x (c) (i) (ii) log 0 y = ( k log 0 A ) x + log 0 A x =.6 y-intercept = log 0 y A =.69 gradient = gradient k = * log A = 0.5 0 k log 0 A P K K 0

N0. SOLUTION MARKS SMKDarulaman j*k 8 (a) P(5, ) P Q(, 0 ) P (b) A = ( 4y + ) dy 0 K use x dy = = 7 3 4y 3 3 + y 0 OR equivalent K correct limit K integrate correctly (c) V = π 5 x 4 dx K integrate π y dx = π x 4 x 5 K correct limit K integrate correctly = π 0

N0. SOLUTION MARKS 9 (a) cos 4 POQ = 0 K Use ratio of trigonometry or equivalent POQ =.6 rad. (b) ( π.6 ) rad PQ = 0 ( π.6 ) P K Use s = rθ = 5.4 cm (c) 0 4 = 9.7 cm P Area of trapezium POQR = (6 + 0) 9.7 = 73.36 cm * K Area of sector POQ = (0) (.6) = 58 cm K Use formula A = r θ Area of shaded region = 73.36 58 = 5.36 cm K 0

N0. SOLUTION MARKS 0. (a) Equation of str. line PQR : SMKDarulaman j*k m = K y = x + (b) x + 6 = P(, ) x + K solving simultaneous equation (c) ( x) + ( ) + = 0 ( y) + () or = + K Use the ratio rule R( 4, ) (d) (i) ( x 4) + ( y + ) = ( x + ) + ( y ) K Use distance formula 4 [ x 8x + 6 + y + y + ] = x + 4x + 4 + y 4y +4 x + y x + 4y + 5 = 0 (ii) Substitute x = 0, y + 4y + 5 = 0 b 4ac = (4) 4()(5) = 44 < 0 No real root for y, The locus does not intercept the y-axis. K Substitute x = 0 and use b 4ac to make a conclusion if b 4ac = -44

0 N0. SOLUTION MARKS (a) µ = 80, σ = SMKDarulaman j*k 65 80 P ( X 65) = P ( Z ) = P ( Z.5) K Use Z = X µ σ = 0.056 = 0.8944 K Use Q(Z) (b) 0.056 4000 = 4 or 43 K (c) 00 = 0.05 4000 Q( Z ) = 0.05 Z =.645 P K Find value of Z m 80 =.645 m = 60.6 g K K m µ Use σ Use negative value 0

N0. SOLUTION MARKS (a) - 5 ms - (b) v < 0 t - 4t - 5 < 0 (t 5) (t +) < 0 K K 0 < t < 5 (c) v 7 8 7 6 4 P (for shape ) 0-5 0 5 6 0 5-6 t -5-9 -4-5 -6-8 -9-0 P min(,-9), (6,7) &(0,-5) must be seen - (d) Total dis tan ce 5 6 = vdt + 0 5 vdt 5 6 3 t 3 t = t 5t + t 5t 3 3 0 5 K for 5 6 and 0 5 K (for Integration; either one) 3 3 5 6 5 ( 5) 5( 5) 0 + ( 36) 30 ( 5) 5( 5) 3 3 3 = ( ) = 33 + 30 ( 33 ) 3 3 K (for use and summation) = 36 m 3

0 N0. SOLUTION MARKS 3 P09 (a) 00 = 5 60 K SMKDarulaman j*k P 09 = RM75 (b) (i) 0 = ( 5 4) + ( 0m) + ( 80 5) + 50m + 450 + m K 440 + 40m = 350 + 70m K (use formula) m = 3 (ii) P 07 00 = RM 30 0 K = RM 5 (c) 0 + (0 0.5) = 38 K I 0 / 07 = ( 5 4) + ( 38 3) + ( 80 5) + ( 50 6) 8 K = 3

0 N0. SOLUTION MARKS 4 (a) y 00 x + y 800 4x + y 400 (b) y 000 900 4x + y = 400 800 700 600 (00,600) 500 400 R 300 y = 00 00 00 x + y = 800 00 00 300 400 500 600 700 800 900 x At least one straight line is drawn correctly from inequalities involving x and y. All the three straight lines are drawn correctly K (c) (i) 650 Region is correctly shaded (ii) Maximum point (00, 600) Maximum profit = 0(00) + 6(600) K = RM 7600

0 N0. SOLUTION MARKS 5 (a) TQ = 9 + 6 (9)(6)cos56 o K SMKDarulaman j*k TQ = 7.54 cm (b) 0 sin QTR sin 56 = 6 7. 54 K QTR = 4 o 3 (c) 4.8 = ( RS)( 6)sin 56 o K RS = 7 ST = 7 9 (or ST + 9 in formula of area) K (d) = 8 cm 0 Area QTR = (9)(6)sin 56 =.38 cm K Area of quadrilateral PQTS = (4.8).38 K = 6.8 cm 0 END OF MARKING SCHEME