RVHS 2023 JC1 Promos 9649 (Solutions)

1 9649 / 01 / 2023 2023 RVHS JC1 H2 FM Promo Exam (Solutions) 1 Solution [7] (a) Let Pn be the statement “52nn is divisible by 3” for n   . When n = 1: 525 2 3nn   , which is divisible by 3. Hence, 1P is true. Assume Pk is true for some k   , i.e. 52kk is divisible by 3, or equivalently 523kk q , where q  . Want to show 1Pk is true , i.e. 1152kk is divisible by 3.    11525 5 2 2 55 2 32 5 3 3 2 (by inductive hypothesis) 35 2 , kk k k kk k k k q q           which is divisible by 3 since 52 kq  .  1Pk is true. Since 1P is true, and Pk is true  1Pk is true, by mathematical induction, Pn is true for all n   . (b)           1155 2 5 5 5 2 2 10 5 2 10 3 , whe 2 r 3 2 e 0 nn n n nn pp p          Since any number in the set is of the form 30 p, the greatest common divisor is 30.

2 9649 / 01 / 2023 2 Solution [5 marks] (i) Let  2 2f. 87 x xx   By Simpson’s rule,       0.25 f 4 4f 4.25 2f 4.5 +4f 4.75 f 53 0.6796763482 0.6797 4 d.p. I        (ii)  5 24 5 24 5 1 4 1 2 d 87 12 d 94 42s i n 3 12sin 3 Ix xx x x x               1 1 12sin 0.6796763 1sin 0.33983      

3 9649 / 01 / 2023 3 Solution [7 marks] Characteristic equation: 2 40m  2m C.F. is 22xx czA e B e  Let P.I. be 2 cos 2 sin 2x pzk x e x x   22 22 '2 2 s i n 22 c o s 2 '' 4 4 4 cos 2 4 sin 2 xx p xx p z kxe ke x x zk x e k e x x         Sub into DE: 2244 4 c o s 2 4 s i n 2xxkxe ke x x    24c o s 2 s i n 2xkxe x x  28e cos 2 4sin 2x x x  Comparing coefficients: 48 2 181 8 184 2 kk            222 112c o s 2 s i n 282 xxxzA e B e x e x x    when 0x , 0z : 1 8AB  22 2 2d1 22 4 2s i n 2 c o s 2d4 xx x xz AeB e x ee x xx    when 0x , d 1d z x  : 1222 2 1 2 0 AB AB     1 16AB   22 211 1 1 2c o s 2 s i n 216 16 8 2 xx xzee x e x x      

4 9649 / 01 / 2023 4 Solution [7 marks] (i) 15 3(' ) 25 5 10 2() 25 5 6(' ) ( ' ) ( ) 25 PB PG P GB P B P G     Thus G and B’ are independent. (ii) P(exactly 1 A and at least 1 R) = P(1AR, 1A'R') + P(1AR', 1A'R) + P(1AR, 1A'R) = 21 4 27 27 1 1 11 11 25 2                = 56 300 P(exactly 1 A | at least 1 Red) P(exactly 1 A and at least 1 R) P(at least 1 R) 56 300 16 21 25 2          56 300 1201 300   140.311 or 45 5 Solution [7] 12 944 ( * ) 2 nn nnxx x     Substitute 2 n nnxy  into (*):     12 12 12 12 222 24 8 2 944 2 4 1 62 92 44 0 ( s h o w n ) nnn nn n nn n n nn n n n nn yy y yy y yy y                   The characteristic equation of the recurrence relation is

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