RI 9758/02 2018

1 | Page 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 d1 15d3 113 15 d d33 91 1523 9Curve passes through 0,69 8 18 2 91 15 1823 12 15 439 12 15 439 2f 3 4 45 9 y yx y yx y xC C yx yx yx xy x − = −  −=   −= +  ⇒= =    −= +  −= +  −= +   = = ++  ∫∫ (ii) ( ) 1 3 2 3 d1 15 4d3 237 91 237 15 18 5423 y yx y x =−= = = − −=  Coordinates are (54, 237) Raffles Institution H2 Mathematics (9758) Solution for 2018 A-Level Paper 2

2 | Page Question 2 No. Suggested Solution Remarks for Student (a) ( )( ) ( )( ) ( )( ) 4 32 2 4 32 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 xx x x s x xt x x x a xb aa abb tb sb a − + − += + −− −+ =−+ − + − += − + + + −=− ⇒= − − = − ⇒= = = = −+= ( ) 224 4 10 2 1 0 1Thus, the other roots are 2 3i and . 2 xx x− +=⇒ − = + (b) 3 27w = (i) ( ) ( ) 32 2 27 3 Comparing constant terms, 27 3 9 Comparing coefficients of , 0 3 3 3 90 3 9 3 6 3 33 3 33 i or i2 22 22 w w w cw d dd w cd c ww w −=− ++ − = −⇒= = −+⇒= + += −± −= = −− −+ (ii) i0 1 2i 3 2 2i 3 3 3 3e 3 33i 3e22 3 33i 3e22 w w w π π− = = = −+ = = −− =

3 | Page (iii) Sum of roots = 0 Product of roots = 27

4 | Page Question 3 No. Suggested Solution Remarks for Student (i) 5 55 4 44 1 20 10 5 5 0 44 213 AD BC OD OD = −     −− = −       −−       = +− =−                  D is (–5, –4, 3) (ii) 10 0 2 05 5 04 4 10 0 10 10 5 8 0 4 90 2 10 40 8 08 : . 90 0 . 90 400 40 10 40 8 90 BC BE BC BE BCE r xy − = −     = −= −       −−       × = ×− =             = =    ++          40 400 4 45 20 200z xyz= ⇒+ + =

5 | Page (iii) 2 2 22 2 10 0 2 550 44 8 101 10 0 16 8 0 8 10 2 5 2 1 80 40 84 5 . 45 40 20cos 8 5 40 4 45 BC BA BC BA θ − =      = −− = −       −           × = ×− = =                       = ++ + +          220 58.630 58.6θ = =  (iv) ( ) Let be the mid-point of 550 11 44 422 13 2 Required distance ˆ. 004 0 4 . 45 10 2 20 2441 34 M AD OM OA OD ME n  −   = + = − +− =−     =        −−          = =        0 6.88 2441 ≈

6 | Page Question 4 No. Suggested Solution Remarks for Student (i) ( ) ( ) ( ) ( ) 246 46 2 246 46 22 346 2 46 6 6 24 222ln cos 2 ln 1 ...2! 4! 6! 24ln 1 2 ...3 45 24 1 242 ... 2 ...3 45 2 3 45 1 242 ...3 3 45 241 8 824 3 45 2 3 3 xxxx xxx xx xxxx xxx xx x xxx  = −+−+  = +− + − +    = −+ − +−−+ − +      +−+ − +  ≈− + − − − −   = 46 2 4 642 3 45 xxx−− − Not valid for 4x π= since ln cos2 ln0 which is undefined4 π  =  (ii) ( ) ( ) 46 2 22 24 35 0.5350.5 20 0 4 642ln cos 2 3 45dd 4 642d 3 45 4 642 9 225 ln cos 2 4 64d 2 1.0644 ( 4 d.p.)9 225 xxxx xxxx xx x xxxC x xxxxx −− − ≈ = −− − = −− − + ≈− − − = − ∫∫ ∫ ∫ (iii) ( )0.5 20 ln cos 2 d 1.0670 (4 d.p.)x xx ≈−∫ To remind us that answers from GC is an approximation.

7 | Page 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 : 65000µ = Alternative hypothesis, 1H : 65000µ < where µ 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 t hat is, 9420000 (3 s.f.) p X Z s Z s s s s > <>  − <>  −<> > > ≥

8 | Page 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 53 53 5 56 .Cpq pq= (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 53 5 8 44 8 53 8 53 8 62 45 56 P 0P 1 . . . P 4P 5P 6P 7P 8 We want P 5 to be the largest, so P 4P 5 P 5P 6 70 56 56 28 70(1 ) 56 56(1 ) 28 52 93 52Thus, 93 XX XXXXX Cpq X X X and X X Cpq Cpq Cpq Cpq qp qp pp pp pp p = ≤= ≤= <= >= ≥= ≥= = = = <= = >= <> <> −< −> >< << (iii) 80.9 0.430 (3 s.f.)=

9 | Page Question 7 No. Suggested Solution Remarks for Student ( ) ( ) ( )P ,P ,PAa BbCc= = = (i) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) P ,P ,P P ' ' 1P 1P P P 1 since and are independent, so P P P 1 (1 ) (1 )(1 ) P' P' Thus, ' and ' are independent. Aa BbCc A B AB A B AB a b ab A B A B A B ab a ab AB AB = = = ∩= − ∪ = −−+ ∩ =−−+ ∩= =−− − = −− = (ii) ( ) ( ) ( ) ( ) ( ) ( ) P ' ' 1P 1P P P 1 si nce and are mutually exclusive, so P 0 A C AC A C AC ac A C AC ∩= − ∪ = − − +∩ =−− ∩= Below is a possible venn diagram where with ' and ' '' AC A C C A AC CA φ ∪= Ω = = ∩ =∩= A C Note that if ' and ' are mutually exclusiveAC , ( )P' '0AC∩= ( ) ( ) 10 PP1 ac AC −−= += Note also ( ) ' (1) '' ' ' ' (2) (1), (2) ' CA CA AC AC C AC φ φ ∩= ⇒⊆ ∩= ⇒⊆ = = ( ) ( ) ( ) ( ) 21 1P ,P ,P ' ' '5 5 10 3P' ' 5 A BC A B C AC c = ∩ = ∩∩ = ∩= −

10 | Page (iii) P P 1() 10 11( ) 10 9( ) (1) 1P 0 ABC ABC A BC ′′ ′∩∩ ∪∪ = ⇒− = ∪⇒= ∪  ( ) ( , are independent) 2= (2) P 5 AB ab A B b = ∩   Method 1 Maximum value of P()AB∩ occurs when b is maximum and c is minimum, ie when C ⊆ B. 2 29( )( ( ( ) 5 5 10 15 5 23 6 P P) P) P AABC A B b B b b = =+− = ⇒= − = ∪∪ + ∩ So maximum value of 25 1() 56 3P AB = = ∩ Minimum value of P()AB∩ occurs when b is minimum and c is maximum, ie when ( )B AC ′′∩∩= ∅ . So 21() ( ) ( ) 55PP PB AB BC bb= ∩+ ⇒= +∩ 1 3b⇒= Minimum value of 21 2() 53P 15AB = = ∩ C B A ∅ C B A ∅ 0.4 x−