RI 2018 P2 A-Level H2 Math Solution

1 | P a g e Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student (i)    1 3 1 3 2 3 2 3 2 3 2 3 3 2 3 2 d 1 15d 3 1 13 15 d d3 3 9 1 152 3 9Curve passes through 0,69 8 182 9 1 15 182 3 1 215 43 9 1 215 43 9 2f 3 4 45 9 y yx y y x y x C C y x y x y x x y x                                                   (ii)   1 3 2 3 d 1 15 4d 3 237 9 1 237 15 18 542 3 y yx y x               Coordinates are (54, 237) Raffles Institution H2 Mathematics (9758) Solution for 2018 A-Level Paper 2

2 | P a g e Question 2 No. Suggested Solution Remarks for Student (a)        4 3 2 2 4 3 2 2 2 4 20 56 0 Since the coefficients are real, 2 3i is also a root. 2 3i 2 3i 4 13 4 20 56 4 13 4 comparing coefficients: 20 16 4 56 13 4 1 13 13 4 52 69 x x sx x t x x x x x x sx x t x x x ax b a a a b b t b s b a                                         224 4 1 0 2 1 0 1Thus, the other roots are 2 3i and .2 x x x       (b) 3 27w  (i) 3 2 2 27 3 Comparing constant terms, 27 3 9 Comparing coefficients of , 0 3 3 3 9 0 3 9 36 3 3 3 3 3 3i or i2 2 2 2 2 w w w cw d d d w c d c w w w                        (ii) i0 1 2i 32 2i 33 3 3e 3 3 3i 3e2 2 3 3 3i 3e2 2 w w w          

3 | P a g e (iii) Sum of roots = 0 Product of roots = 27

4 | P a g e Question 3 No. Suggested Solution Remarks for Student (i) 5 5 5 4 4 4 1 2 0 10 5 5 0 4 4 2 1 3 AD BC OD OD                                                                      D is (–5, –4, 3) (ii) 10 0 2 0 5 5 0 4 4 10 0 10 10 5 8 0 4 90 2 10 40 8 0 8 : . 90 0 . 90 400 40 10 40 8 90 BC BE BC BE BCE r x y                                                                                                         40 400 4 45 20 200z x y z    

5 | P a g e (iii) 2 2 2 2 2 10 0 2 5 5 0 4 4 8 1 0 1 10 0 16 8 0 8 10 2 5 2 1 80 40 8 4 5 . 45 40 20cos 8 5 40 4 45 BC BA BC BA                                                                                                             220 58.630 58.6    (iv)   Let be the mid-point of 5 5 01 1 4 4 42 2 1 3 2 Required distance ˆ. 0 0 4 0 4 . 45 10 2 20 2441 34 M AD OM OA OD ME n                                                                                  0 6.882441

6 | P a g e Question 4 No. Suggested Solution Remarks for Student (i)      2 4 6 4 6 2 24 6 4 6 2 2 34 6 2 4 6 6 6 2 4 2 2 2ln cos 2 ln 1 ...2! 4! 6! 2 4ln 1 2 ... 3 45 2 4 1 2 42 ... 2 ...3 45 2 3 45 1 2 42 ...3 3 45 2 4 1 8 82 4 3 45 2 3 3 x x xx x xx x x x xx x x xx x x x xx x                                                            4 6 2 4 642 3 45 x xx   Not valid for 4x  since ln cos 2 ln 0 which is undefined4           (ii)    4 62 2 2 2 4 3 5 0.53 50.5 20 0 4 642ln cos 2 3 45d d 4 642 d3 45 4 642 9 225 ln cos 2 4 64d 2 1.0644 (4 d.p.)9 225 x xxx x xx x x x x x xx C x x xx xx                            (iii) 0.5 20 ln cos 2d 1.0670 (4 d.p.)x xx  To remind us that answers from GC is an approximation.

7 | P a g e Section B: Statistics Question 5 No. Suggested Solution Remarks for Student (i) As the manager does not know how the MTTF is distributed, he needs to have a random sample of size large enough so that he could apply central limit theorem on the sample mean on MTTF. In general, 30 is considered large. The fans have to chosen randomly, example he could label the N number (assume very large population) of fans manufactured in the day from 1 to N and generate n (sample size of at least 30) distinct numbers from {1, 2, … , N) using random function of a calculator. The fans labeled according to the numbers generated will be the sample. (ii) Null hypothesis, 0H : 65000 Alternative hypothesis, 1H : 65000 where  is the population MTTF. (iii) Perform an one-tailed test at 5% significance level. Under H0, 2 ~ N 65000, approximately by Central Limit 43 Theorem since 43 is large sX n      No reason to reject H0 (that is do not reject H0),   2 2 -value 0.05 P 64230 0.05 64230 65000P 0.05 43 770 43P 0.05 3069.7127 9423136.061 that is, 9420000 (3 s.f.) p X Z s Z s s s s                     

8 | P a g e Question 6 No. Suggested Solution Remarks for Student (i) Note that for any path allowing the bug to move from S to D, the bug has to take 5 left forks and 3 right forks. The required probability is 8 5 3 5 3 5 56 .C p q p q (ii) This is a binomial distribution in disguise. Let X = no. of left forks out of 8. (the rest will be right forks) X ~ B(8, p)                      8 5 3 5 8 4 4 8 5 3 8 5 3 8 6 2 4 5 5 6 P 0 P 1 ... P 4 P 5 P 6 P 7 P 8 We want P 5 to be the largest, so P 4 P 5 P 5 P 6 70 56 56 28 70(1 ) 56 56(1 ) 28 5 2 9 3 5 2Thus, 9 3 X X X X X X X C p q X X X and X X C p q C p q C p q C p q q p q p p p p p p p p                                  (iii) 80.9 0.430 (3 s.f.)

9 | P a g e Question 7 No. Suggested Solution Remarks for Student P ,P ,PA a B b C c   (i)                  P , P , P P ' ' 1 P 1 P P P 1 since and are independent, so P P P 1 (1 ) (1 )(1 ) P ' P ' Thus, ' and ' are independent. A a B b C c A B A B A B A B a b ab A B A B A B a b a a b A B A B                           (ii)        P ' ' 1 P 1 P P P 1 since and are mutually exclusive, so P0 A C A C A C A C a c A C A C               Below is a possible venn diagram where with ' and ' ' ' A C A C C A A C C A          A C Note that if ' and ' are mutually exclusiveA C , P ' ' 0A C      1 0 P P 1 a c A C      Note also   ' (1) ' ' ' ' ' (2) (1),(2) ' C A C A A C A C C A C                  2 1 1P , P , P ' ' '5 5 10 3P ' ' 5 A B C A B C A C c         

10 | P a g e (iii) P P 1( ) 10 11 ( ) 10 9( ) (1) 1P 0 A B C A B C A B C              ( ) ( , are independent) 2= (2) P 5 A B ab A B b   Method 1 Maximum value of P( )A B occurs when b is maximum and c is minimum, ie when C  B. 2 2 9( ) ( ( ( ) 5 5 10 1 5 5 2 3 6 P P ) P ) P AA B C A B b B b b                 So maximum value of 2 5 1( ) 5 6 3P A B      Minimum value of P( )A B occurs when b is minimum and c is maximum, ie when  B A C    . So 2 1( ) ( ) ( ) 5 5P P PB A B B C b b      1 3b  Minimum value of 2 1 2( ) 5 3P 15A B      C B A  C B A  0.4 x