2020 RI H2 Math Prelims P2 Solutions

2020 RI H2 Mathematics Prelim Paper 2 Solutions Section A: Pure Mathematics [40 marks] 1 [5] Given that 1 3 ,2u  let the common difference of the arithmetic series be d and the common ratio of the geometric series be r. 2 2 3 3 3 21and (1)2 2 2 2 60 ( 3)( 2) 0 2, since r is non negative. rr rr rr r            333 322 dr -----------------(2) Substitute r = 2 into (2), 7 .2d  Sum of first 10 odd numbered terms of AP 10 32 9(7) 33022S    2 [5] 4 4 8 8 8 4 ( 1) (1)1 ( 1) (2)1 (2) 1 17 (1) 1 16 arS r arS r r r          44 4 4 12 11 17 1 16 1 1 1 ,16 2 2 rr r r r r        When 11 1 ,2 12 1 2 ar S a    When 22 12, 123 1 2 br S b    Ratio is a : 1 3 b or 3a : b

3(i) [4]  gR , 3   fD 0, Since gfRD , fg does not exist.  fR 1, 4  gD , 4  Since fgRD , gf exists.       gfR g 2 , g 4 1, 3   (ii) [4] Largest possible value of c is 1. For 1x , 11xx       g 1 3x x x          2 1 13 12 2 1 1 g 2 1 , , 0. x x y xy x y x x x x x                4 [2]     1 1 1 1 2 2 2 2 tan 2 tan 2 tan 2 tan 2 e d 1 1ed2 1 2 2e 4 d4 2e d d4 2 (Shown)d x x x x y y x x x yx x yxy x                   

4 (i) [5]         2 2 2 2 3 2 2 2 2 3 2 2 2 3 2 2 2 3 2 2 d4 2 ---------- (1)d d d d4 2 2 ---------- (2)d d d d d d d d4 2 2 2 2d d d d d d d d d4 4 2 2 ---------- (3)d d d d yxy x y y yxx x x x y y y y yx x x x x x x x y y y yxx x x x x              When 0, 1.xy From equations (1), (2), (3) we get 23 23 d 1 d 1 d 1, , .d 2 d 4 d 8 y y y x x x   By Maclaurin’s Theorem, 1 23 23 tan 2 1 1 1e1 2 4 2! 8 3! 1 1 11 ...2 8 48 x xxx x x x                       (ii) [3] 1 23tan 2 1 1 1e 1 ... 2 8 48 x x x x                 1 1 22 2 22 2 2 2 tan 2 tan 2 e 1 1 1 ... 1281 111 ... 1 2 3 ...28 111 2 1 3 ...28 e 3 17 1 ...281 x x x x x x x x x x xx xx x                                        

5(i) [2] 2 3OC a , 3OD b (ii) [4] E on BC :  1OE OB OC   E on AD :  1OE OA OD      21 1 3 3OE           b a a b 3 3 (1) 2 2 3 (2)     (1) + (2): 2 3 1    4 3OE ab (iii) [3] Area of CDE is 1 1 2 2 32 2 3 3 1 2 2 3 since 2 3 3 1 4 2 2 3 3 DC CE                            a b a b a b a b a a b b 0 a b a b (iv) [3] 4 3OE ab is the diagonal of the rhombus with sides 4 3 a and b . So 43 34     aab b , i.e. :4: 3OA OB  Alternatively,     22 4 cos cos 44 33 44 cos cos33 44 cos cos33 4 1 cos 1 cos , 1 cos 03 3 i.e. :4 3: AOE FOE OE OE OA OB                                     a a b b a b ab a a b a b b ab a b a b ab a b

Section B: Probability and Statistics [60 marks] 6(i) [2]     P( ')P( ') P( ') P( ')P( ') P( ') 12 5063 63 100 XYXY Y XYY XY     (ii) [2]                 P ' P( ) P ' P ' P ' P ' P 'P 3 1 63 4 2 100 31P 50 X Y X Y X Y X Y X Y YX X               (iii) [3] P( ) P( ')P( ) P( ) P( ) P( ') 31 1 50 2 3 25 X Y X YX X Y X X Y           Since 3 31 37P( ) =P( ) P( ),25 50 100X Y X Y     X and Y are not independent events OR Since 31 50P( ) P( | ')50 63X X Y , X and 'Y are not independent events So X and Y are not independent events

7(i) [5] Let X be the mass of a bag of sugar in kg. The necessary assumption is X follows a normal distribution. 0 1 H : 1 H : 1     Under H0, 20.08~ N 1, 8X   8.4 1.058x Test Statistic: 1 0.08 8 XZ  Level of significance: 5% Reject H0 if p-value < 0.05 Using GC, p-value 0.0385 0.05 Since p-value = 0.0385 < 0.05, we reject 0H and conclude there is sufficient evidence, at the 5% significance level, to support the manufacturer’s concern. (ii) [3] Under H0, 2 ~ N 1, 8X   H0 not rejected -value 0.05p   2 2 2 2 2 P 1.05 0.05 1.05 1P 0.05 8 80.05 1.64485 0.058 1.64485 0.007392 X Z               Assumption that X follows a normal distribution is required as question did not state the distribution of X and n = 8 is too small to use Central Limit Theorem.

8 (i) [3] Let X and Y be the masses in grams of a randomly chosen apple and a randomly chosen potato in grams respectively. i.e. 2N(90,13 )X , 2N(170,30 )Y .  E 2 170 2( 0 19 0)YX      2 2 2Var 2 30 2 ) 576(13 1YX     2 N 10,1576YX Required probability P( 2 )YX P( 2 0)YX   0.401 (3 s.f.) (ii) [3] Let 1 2 5 1 2 6 ... ...T X X X Y Y Y        .  E 5(90) 6(170) 1470T      22Var 5(13 ) 6(30 ) 6245T     N 1470, 6245T Required probability P(1200 1500)T   0.648 (3 s.f.) (iii) [3] Let    1 2 5 1 2 60.85 ... 0.75 ...W X X X Y Y Y         E (0.85)(5)(90) (0.75)(6)(170) 1147.5W      2 2 2 2Var (0.85 )(5)(13 ) (0.75 )(6)(30 ) 3648.0125W     N 1147.5, 3648.0125W Required probability P( 1200)W 0.808 (3 s.f.)

9 (i) [2] P( 7) P( 7) P( ) 0.8 0.2 0.6k X X X k         (shown) P( 7) 0.3X (ii) [1] Since P( ) P( 7) 0.3,k X X       by symmetry 7 .2 k  (iii) [2] 7P( 7) 0.8 P( ) 0.8 12 XZ      Therefore 7 0.8416(4 d.p) 12   7 0.8416 4.0845 12      (iv) [3] 2 7 1.169k    P( ) 0.135(3 s.f)Xk (v) [4] 2P( ) 3P( ) P( ) 0.4 X r X r Xr       Let Y be the number of observations out of 10 with values greater than r. B(10,0.4)Y P( 6) 1 P( 5) 0.166(3s.f)YY     10(i) [1] number of committees if there is no restrictions in the selection = 13 7 1716C  (i) [2] There are 3 cases: Case 1: 4 women 3 men Number of committees = 67 43 525CC Case 2: 5 women 2 men Number of committees = 67 52 126CC Case 3: 6 women 1 man Number of committees = 67 61 7CC Total number of committees = 658 (Shown) (iii) [1] P(the committee will consist of 4 single women and 3 single men) 56 43 100 50 658 658 329 CC   (iv) [2] Number of committees with no married member 5 6 5 6 4 3 5 2 100 15 115 C C C C     

Number of committees with at least one married member 658 115 543   P(the committee will contain at least one married member) 543 658 Alternative Solution We consider 3 cases: Case 1: Husband in, wife out Number of committees 5 6 5 6 4 2 5 1 75 6 81C C C C       Case 2: Wife in, husband out Number of committees 5 6 5 6 5 6 3 3 4 2 5 1 200 75 6 281C C C C C C          Case 3: Both husband and wife are in Number of committees 5 6 5 6 5 3 2 4 1 5 150 30 1 181C C C C C         Total number of such committees = 543 P(the committee will contain at least one married member) 543 658 (v) [2] Number of ways to sit 4 women around the table = 3! Number of ways to slot the 3 men = 4 3 2 24   Number of sitting arrangements where the all the men are separated 3! 24 144   Required probability 144 6! 1 5   (vi) [3] Number of ways to sit 4 women around the table = 3! Number of ways to sit the man next to his wife = 2 Number of ways to sit the other 2 men = 3 2P Number of ways to have all the men separated from each other and the married couple sit next to each other 3! 2 3 2 72     Required probability 72 144 1 2  

11 (i) [2] x 2 5 6 8 9 10 P( )Xx 1 4 =0.25 3 10 =0.3 1 5 =0.2