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39 SPM Trial Examination 0 Mark Scheme Addition Mathematics Paper No Marking Scheme Total Mark. (a) 7 (b) 8 (c) many to many banyak dengan banyak. (a) 5 (b) h and k B: h or 9h k 5 or g(x) = x - B: h( x) k. (a) 4 ( ) x g x 7 x 4 Accept: 7 (b) p 0 p 4 B: 7 or 7 () 4 p 4. p 6 5. B: ( ) 5( ) p 0 OR ( x )( x ) 0 OR h 5 and h p x B: ( indicated the correct region ) - B: ( x )(x ) 6. (a) x (b) f ( x) ( x ) 6 B: b or c 6
40 7. 8. n B: n n 4 or equivalent 4 B: n n or equivalent x = 64 B: log x or log 4 x log B: x log x 4 log 4 or log 4 9. x = 40 x x 8 B: x 0 x 0.. (a) (b) 8 4 B: n = 0 B: 56 + ( n ) ( ) < 0. a = 7 and b = 4 () B: a = 7 or b = OR = a () or 5 = a 5 B: 5 a b() or a b() OR b B: log y bx a. (a) h B: h 0 6 or equivalent (b) : B: n() m(6) n m
41 4. 6, 6 B: p p B: m or m p 5. ( a) k 6 B: k 8 0 * 6i 8 j 0 i + 4 j 5 5 (b) 6. (a) x y (b) x y 4 B: x ( x y) or y ( x y) 4 4 B : ( x y) or ( x y) 4 4 p 7. (a) sec k (b) B: k k k 8. ( a) 5 (b) B: () (.9) (0 ) 5 B : () (.9) or (0 ) B: 4 ()( )(x ) () B: (x ) () 0. B : 5 dx dt 5 (4 x ) or equavalent
42 B: dy dx 4 x dy or 5 dt. y = x 4x 5 B: = () 4() + c B: y = x 4x + c. (a) x = 4 (b) x = 4 x B: 5= 7 6. (a) 76 (b) 008 B : 8 C 4 x 5 C + 8 C 5 x 5 C + 8 C 6 x 5 C 0 B: 8 C 4 x 5 C 4. (a) x = 5 B : x 6 9 (b) B : or (a) B: P ( Z ) 8 (b) B: P( < Z < ) Total 80 4
43 SPM TRIAL EXAM 0 Marking Scheme Additional Mathematics Paper Section A Question Part Solution Marks x = 5 + y () x + y = () (5 + y) + y = y + 0y 6 = ()( 6) y () y = 0.87, -.94 x = 5 + (0.87), x = 5 + (-.94) = 5.56 = -.74 or -.74 (a) A(0, -) (b) k k f ( x) ( x kx ) 4 4 k k ( x ) 4 k 0 k 6 p 4 (c) x ( x 6) 0 0 x 6 (a) L r, L r, L r 6 L L ( r ) r L L ( r 6) ( r ) Common difference, d (b)(i) 4 ( n )( ) n.4 n 8 (b)(ii) 0 S S cm or cm (using in calculator) 0
44 4 (a) (b) sin xcos x LHS = y cos x = sin x cos x = sin x 4 y = sin x O y x x 4 (c) 5 (a) (b) Graph y sin x cycle or amlitude 4 All correct y x Straight line y x Number of solution = (9.5) 7(9.5) 6(9.5) (9.5) 6(49.5) 0(59.5) 8(69.5) x 0 = 6.5 (0) 9 Q (0) 86 Q Q Q = =.
45 6 (a)(i) u 0v (ii) EF EC 5 = ( ED DC) 5 = (4u4u5 v) 5 = 5v BF BA AE EF = 0v8u 5v = 8u 5v (b) BF 8u 5v Section B BD u 0v = 4(8u-5v) BF BD 4 BF // BD and B is the common point Thus B, F and D are collinear 7 (a) f ( x) x c c f ( x) x 4 (b) ( + ½ ()() x 4x 4() 4() (c) 9 unit 6 4 ( 4 y) dy = 4 y y 4 4 = 4(4) 4() =
46 8 (a) y x x Refer to graph Using the correct, uniform scale and axes All points plotted correctly Line of best fit (b)(i) y x.0.98 h = 7 =.0 (ii) k = - (iii) when x = 4, y x y = 7x = 7(4) p =
47 Answer for No. 8(a) y x 4 x x 0 x 8 6 x 4 x x -
48 9 (a) or 4.8 o o or PR = (0) sin or (0)sin(0.r) = 5.90 (b) S 0(0.6) or S 5.90(0.6) (c) PQR RST 4.8 OT = 0 (5.90) sin or Perimeter = 0 + 0(0.6) + 5.9(0.6) = 6.05 (5.90) 0.6 (0) 0.6 sin 4.8 o (5.90) (0) 0.6 sin 4.8 o =.4 0 (a)(i) m BD = y 5 = ( x ) y = x + (ii) (x + ) + x = 7 or equivalent M(, ) x 4 y 0 or, =x()+(), + =y()+(5) + D(, ) ( x) ( y) or () () (b) ( x ) + ( y ) = ( + ) + ( + ) x + y x 6y 0 = 0
49 (a)(i) C (ii) C 4 0 or C or C or equivalent C 0 C C or equivalent 0.89 (b) PX 70 PZ p 76 PX p % or PZ 0. 5 p p 8.60 Section C (a) v 8 ms a 0 6t (b) 0 6(0) = 0 cm s 0 6t 0 5 t s (c) 5 5 v 8 0( ) 6 cm s v 0 8 0t t 0 (d) ( 4 t )( t) 0 t 4 0t s 8t t c
50 s 8t 5t t 48 cm 8(4) 5(4) 4 (a) x 0 y z 60 (b) h = (c) RM78 (d) (a)(i) ½(6)(AK)( ) = 4 Using formula of area of triangle 5 AK = 5 cm (ii) 4 cos AKB 5 AB = (5)(6)( ) 5 Using cosine rule AB = 0. cm (b) = x + 6 (x)(6)( 4 ) 5 Using cosine rule (c)(i) 5x 8x = 0 Simplify to general form Draw obtuse triangle or shows point C on KC and side BC B 6 cm K C B or A K C C
51 (ii) sin K' CB ' ' 5 6 Using sine rule K C B = (a) x.5y 0 or 4x 5y 40 40x 80y 640 or x y 6 x y (b) Refer to graph x and y axes with correct scales At least two lines drawn correctly Correct region shaded (c)(i) {4,5,6} or 4 x 6 (ii) Maximum point ( 8, 4 ) 5 (8 ) + 45 ( 4 ) RM 80
52 y Graph for Question x = y 6 5 R 4 ( 8, 4 ) 0 4x + 5y = x + y = 6 x
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