ACJC Solutions Paper 2

1 Anglo-Chinese Junior College H2 Mathematics 9740 Qn Paper 2 Solution 1 (i) (ii) 2.5 3 4 3 0.5 7 2 0.5 5 , , , 2.5 4 3 0.5 3 7 0 1.5 5 0 ,: i j k ---(1) ---(2) ---(3) (1) (2), (3) 2 k kk kk Equating components k k k Solving Equations k                                     . 1 2. ,            2, 2, 2.5 (2) 2 7 2 3 2 8 2 2 2 6 7,8,6 r Intersection point is When k           2 Let  y f x         1sin 2 1 2 2 2 2 22 2 22 2 2 2 2 2 2 2 ln sin 2 diff. w.r.t. 12 14 1 4 2 diff. w.r.t. 4 1 4 2 14 4 1 4 2 1 4 1 4 4 2 2 1 4 4 4 (shown) xye yx x dy y dx x dyxy dx x x dy d y dy xdx dx dxx dy d y dyx x xdx dx dx d y dyx x y dxdx d y dyx x y dxdx                                3 2 2 2 3 2 2 32 2 32 diff. w.r.t. 1 4 8 4 4 4 1 4 12 8 x d y d y d y dy dyx x x dx dxdx dx dx d y d y dyxx dxdx dx         

2         01 ' 0 2 '' 0 4 ''' 0 16 f f f f       23 81 2 2 ... 3f x x x x     1sin 2 6 1 23 6 sin 2 6 12 2 1 4 1 1 8 11 2 2 4 4 3 4 21 3 xee x x x e                             3i 1 sin diff. w.r.t. cos xu x dx udu           2 2 2 2 2 21 2 2 1 sin 1 sin cos 2 2sin 1 2sin sin cos 1 sin cos cos cos 2 1 2 sin 2 1 42 2sin cos 1 42 1 12 sin 122 x x dx u u u du u u u u du u u du u du u du u uc uu uc x x x x c                               

3 ii a) b)   11 92 00 11 29 00 12 10 2 21 0 Area =2 2 2 =2 2 2 112 2 2 sin 12 2 2 10 2 122 10 4 19 52 R x x x dx x x dx x x x dx x xdx x x x xx x x x                                 211 229 00 Vol. of revolution formed when R is rotated completely about -axis 2 2 2 14.995 x x x x dx x x dx         4 (i) Method 1: 11 : 1 2 03 rline l                    Since (1,1,0) lies on 2p , 11 11 01 11 2 a a a                       Method 2: Since direction vector of line l is perpendicular to normal of 2p ,

4 4(ii) 11 20 31 1 2 3 0 2 a a a                        Method 3: 1 1 1 12 1 1 1 11 2 a a a a a                               Method 1: Normal method 1 1 : 1 2 1 01 : 0 1 11 01 0 1 2 11 11 1 3 r r p line AN                                                   0 1 1 110 1 1 331 1 4 ON                                        Method 2: Projection Method 11 11. 1 1 33 11 1 1 1 111 1 1 331 1 1 11 1 1 1 1 1133 11 AN AB                                                                                       

5 1 0 1 111 0 1331 1 4 AN ON OA ON AN OA                                4 (iii) Let the acute angle between planes 1p and 2p be  . 11 12 11sin 11 12 11 1 2 1 2 3 6 18 28.1                                           4 (iv) Distance from Origin to p1 2 2 1 3 1 1     Distance from Origin to p3 ( 0)2 12 2 2 b b b      2 3 12 3 2 3 12 3 4 6 10 b b b    4 (v) 1p , 2p and 4p meet in a line l. Hence (1,1,0) lies on 4p . 1 12 03 2 c d cd                    Normal of 4p is perpendicular to line l.

6 (vi) 1 2 2 0 33 4 9 0 5 7 c c c d                        5 , 7 cd 5 Randomly choose 10 programmers, 1 secretary and 1 section head for the sample. Stratified sampling guarantees a representative sample of each group (i.e. programmers, secretaries and section heads) in the population. CANNOT accept any of the following answers:  “… allows the opinions of different strata to be considered separately.”  “… accurate …”  “… unbiased …” 6 (a) (i) (ii) (b) 4 boys and 3 girls _ B _ B _ B _ B _ 4! × 5P3 = 24 × 60 = 1440 Or 4! × (5C3 × 3!) = 24 × 60 = 1440 G BBBB GG type so 4! × 4! = 576 Case (i): 3 boys and 3 girls 4C3 × (3  1)! × 3! = 4 × 2 × 6 = 48 Case (ii): 4 boys and 2 girls (4  1)! × 3C2 × 4P2 = 6 × 3 × 12 = 216 Or (4  1)! × 3C2 × 4C2 × 2! = 216 Total number of arrangements = 48 + 216 = 264 7 Plot of y against x. Highest y = 6.3; Lowest y = 2.7 Highest x = 12.3; Lowest x = 4.4 Correlation coefficient = r = 0.263 A linear model is not appropriate as the scatter diagram shows that the points are not close to a straight line and the value of r is quite close to 0. B B B G G G B B B G B G x y 4.4 12. 3 2.7 6.3

7 7 For y = ax2 + b, value of r = 0.913 For y = a ln x + b, value of r = 0.971 Therefore, (b) y = a ln x + b is a better model as the value of r in (b) is closer to 1 than for (a). Line of regression is y = 11.042 948 39  2.687 231 256 ln x When y = 6.1, ln x = 1.839 420 548  x = 6.29 (3 sf) Accept x = 6.3 (1 dec place) This estimate is valid since the value of r is close to 1 and the value of y used in within the range of experimental data (4.5 ≤ y ≤ 6.3). Or, may say that x comes from interpolation instead of “within the range of experimental data”. 8(a) Let X be the number of cars arriving at the jetty in 30 mins     (6) ( ) 2 0.0446 ( ) 2 0.0620 X Po i P X ii P X   8(b) Let Y be the number of cars arriving at the jetty in 20 mins     (4) 0.1 0.9 7 Y Po P Y k P Y k Least k    k  P Y k 6 0.88933 7 0.94887 8 0.97864 9 (a) Let X be the time of journey from Town A to Town B.       2Let , 1160 70 4 20 60 1 70 1 4 20 60 70 0.67449 1. 64485 Solve: 53.0491 53 10.3054 10 (nearest minute) XN P X P X P Z P Z                             (b) Let Y be the time of journey from Town B to Town C.     80,9 0.02 86.1612 Min 87 mins Last time of departure 10 33h YN P Y k k k   

8     980, 20 80 2 78 82 0.997 YN P Y P Y         10 Let L be the length and B be the breadth of a tile             22 22 1 2 3 10 1 2 3 10 18.9,0.3 8.9,0.1 22 2(18.9) 2(8.9) 55.6 4(0.3 ) 4( 0.1 ) 0.4 ............ 10(55.6) 556 ............ 10(0.4) 4 L N B N Let P L B EP Var P E P P P P Var P P P P                    Let W be the number of tiles with a red tint.           500,0.6 0.95 300 5 200 5, is large Use Normal approximation: 300, 120 0.95 0.05 1 0.052 1 281.982 282.48 Greatest = 282 WB P W k np and nq n W N approximately P W k P W k P W k k k k              11   0 1 0 0 To test: H : 27 H : 27 at 2%level 27Under H : 26 27 Test statistic = 1.22474 p-value = 0.1158 >0.02 Do not reject H XTt s      There is insufficient evidence at the 2% level of significance to conclude the mean foot length of an 18 year old man of high intelligence is more than 27cm Assumption: Assume the population of foot length of 18 year old man of high intelligence is normally distributed

9 0 1 0 2 2 0 To test: H : H : at 4%level Under H by Central limit theorem, since is large 1, , (123.20) 2.08813660 59 To reject H : 2 0.04 60 0.02 60 2. 60 k k n sX N k approx where s xkPZ s xkPZ s xk s          