DHS H2 MATH P2 Mark Scheme

1 2015 Year 6 H2 Math Prelim Exam Paper 2 Mark Scheme Qn Suggested Solution 1(i) Given 1 1 4u  and 1 22 4 ,1 .(1 ) (2 ) nnuu n nn n     Let P(n) be the proposition that 22 1 (1 ) nu nn  , 1.n Consider P(1): LHS = 1 1 (given)4u  RHS = 22 11 LHS1( 1 1 ) 4 Hence P(1) is true. Assume P(k) is true for some ,k  i.e. 22 1 (1 ) ku kk  Consider P(k+1). 22 22 2 2 2 22 2 2 22 2 22 1LHS 4 (1 ) (2 ) 14 (1 ) (1 ) (2 ) (2 ) 4 (1 ) (1 ) (2 ) (1 ) (2 ) 1 = RHS(1 ) (2 ) k k u u kk k kk kk k kk kk k k kk k kk             Hence P(k) is true  P(k+1) is true. Since P(1) is true & P(k) is true  P(k+1) is true, by mathematical induction, 22 1 (1 ) nu nn  for all 1n .

2 (ii) 122 11 11 ()(1 ) (2 ) 4 NN n nn nuunn n            1 12 23 1 1 11 22 22 1 1 ()4 1 [4 () () ] 1 4 11 1 44 (1 ) (2 ) 11 16 4( 1) ( 2) n N NN N n NN n uu uu uu uu uu uu NN NN                      (iii) As 22 1,04( 1) ( 2)NNN   So 2 1 2 11 1 0( 1)( 2) 16 16 N n nn n   Hence, the series is convergent and 1 22 11 ( 1)( 2) 16n nn n   (iv) 22 2 2 22 22 2 22 1 1 2 11 (1 ) ( 1 ) ( 1 ) 1 (Let 1)(1 ) (2 ) 11 16 4( 1 1) ( 1 2) 11 16 4 ( 1) NN nn N k nn n nn knkk k NN NN                        Qn Suggested Solution

3 2 Surface area of cone 2π(12 )(13 ) 156 πx xx Surface area of side and bottom of model 22π(12 ) 2 π(12 ) 144 π 24πx xy x x y   Cost of making the cone in dollars   22156π 0.05 7.8 πx x Cost of making the side and bottom of model in dollars  22(144π 24π ) 0.02 2.88 π 0.48πx xy x xy  Total cost of making model in dollars = 22 27.8 π 2.88 π 0.48 π 10.68 π 0.48 πx xx yxx y  Given that the model costs $100 to make, 2 2 10.68π 0.48π 100 100 10.68 π ...... (1)0.48π xx y xy x   V, Volume of the model     22 21π 12 π 12 5 48 π (5 3 )3x yx x x x y  Substituting equation (1) into V, 2 2 33 3 100 10.68 π48π 53 0.48π 240π 30000 3204 π 30000 2964 π (shown) xVx x x xx x xx        2d 30000 8892 π 0d 30000 1.0363 (5 s.f.) = 1.04 (3 s.f.)8892π V xx x     30000100 10.68 π 63.9688892π 40.9 (3 s.f.)0.48π 1.0363300000.48π 8892π y    2 2 d 17784π 0 0 d maximum volume occurs when 1.04 and 40.9. V xxx xy      

4 Qn Suggested Solution 3(i) 2 4 * *4 4 2 0 (shown) z z zz z zz      (ii) πi 313 i 2 e  (iii) Let A, B, P, Q be points that represent the complex numbers 2 , 13 i , 1z and 2z respectively. (iv) Method 1 12 12 Since ( ) and ( ) lie on circle (centred at origin) of radius 2, 2 Pz Qz zz   2πi 3 2 πi 3 2 1 1 2π 3 π angle bisector3 2πarg ( ) 3 2e 1 3i πarg ( ) (angle on straight line)3 2e 1 3i AOB AOQ QOB z z z z                Method 2 2zz Im (Qz

5 Cartesian equation of circle: 222 2 ---(1)xy  30 3 1Gradient of line : 12 3 3 1Gradient of perpendicular bisector 3 1 3 AB       Cartesian equation of perpendicular bisector:  12 03 0 3 ---(2) Solving (1) and (2), 1 or 1 3 3 13 i a n d 13 i yx yx xx yy zz            (v) Let R be the point that represent the complex numbers w. Note that PQ forms the diameter of the circle centred at origin with radius 2 units.      12arg arg π right angle in semicircle2 zw zw PRQ        ()Rw

6 Qn Suggested Solution 4(i)(a) Area covered by weed at the end of the first week 2 0.9(500 80) 0.9(500) 80(0.9) m   Area covered by weed at the end of the second week 2 22 0.9(0.9(500) 80(0.9) 80) 0.9 (500) 80(0.9 0.9 ) 541.8 m      (b) Area covered by weed at the end of the nth week     2 2 0.9 500 80(0.9 0.9 ... 0.9 ) 10 . 90.9 500 0.9(80) 0.1 0.9 500 720 1 0.9 m n n n n n n          Therefore, 720.k (c) As , 0.9 0 nn  So 0.9 500 720(1 0.9 ) 0(500) 720(1 0) 720nn    Hence the area covered with weed at the end of the week in the long run is 720 m2. (ii)(a) Change in area covered with weed in the nth week 2 80 (50 10( 1)) 40 10 m n n     (b) Area covered with weed at the end of the nth week 1 1 2 1 2 500 (40 10 ) 500 40 10 500 40 10 (1 ) 2 500 35 5 m n r nn rr r r nnn nn            

7 Qn Suggested Solution 5(i) Each level/strata is not proportionally represented and hence not stratified sampling. (ii) Divide the students of the school into strata in terms of age or gender. Select the number of students in each stratum e.g. 20 male and 20 female students for the survey in order to meet the quota. Then stand at the entrance of the school at the start of a school day to survey the first 20 male and first 20 female students that enter the school. One disadvantage of quota sampling is that the sample obtained is likely to be biased as students who come to school later will not have a chance to be considered (or selected).

8 Qn Suggested Solution 6(i) Given ( 32) ( 67),PM PM  By symmetry, 32 67 49.52  (ii) Given that M and F are times spent in minutes by a randomly chosen male and female customer, respectively. i.e. 2~N ( 4 9 . 5 ,1 8 )M , 2~N ( 7 1 ,3 5)F Then 2 12345 35~N ( 7 1 , )55 FFFFFF  2 22353 ~ N(71 3(49.5), 3 (18 ))5FM  i.e. 3 ~ N( 77.5, 3161)FM ( 3 ) ( 3 0) 0.084033 0.0840 (3 s.f.)PF M PF M    (iii) For 2~ N(71, 35 ), ( 0) 0.0213FP F  The probability that the time spent by a female customer being less than zero is not negligible. OR The probability 0.0213 suggests that for every 100 customers, more than 2 spend less than zero minutes at the salon. Since time is a non -negative quantity, the normal distribution with the given mean and standard deviation is not an appropriate model.        x | 32 µ 67

9 Qn Suggested Solution 7(i) Let X denote the number of bits that are corrupted during the transmission process, out of 8 bits. B(8,0.03).X P( 2) 1P ( ) 0.022341 (to 5 s.f.) = 0.0223 (3 s.f.) 1 (shown) X X    (ii) Required probability P( 0 | 2 P({ 0} { 2} P( 2 P( 0 ) ) ) ) 1) 0.78374 1 0.0233 0.802 (3 s.f P( .) X X XX X X X             (iii) Let Y denote the number of bytes that are corrupted during the transmission process, out of 100 bytes. B(100,0.0223).Y Since 50100n is large and p is small enough such that 0.0223(100) 2.23 5,np   oP( 2 . 2 3 )Y approximately. P(5 10) P( ) P( ) 0.0263 (3 s.f.) 95 Y YY    

10 Qn Suggested Solution 8(i) Required number of ways 2!! 2 5 40   (ii) Required number of ways 3 1C2 ! 720 5!   Explanation: Number of ways to choose a brother = 3 1C Number of ways to arrange the 2 parents in the group of chosen brother and parents = 2! Number of ways to arrange the group of 3 with the remaining 4 people =5! (iii) Required number of ways 4 3(4 1)! P 144    Explanation: Number of ways to arrange the remaining 4 people at a round table = (4 1)! Number of ways to slot in the brothers = 4 3P