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

5 9649 / 01 / 2023    2 2 44 0 20 2 (repeated roots) () 2 2( ) 22 n n nn n nn mm m m yA B n xy A B n           0 1: 1 1 0xA A    1 3: 2 0.5 3 1.25xB B    Hence, 1.25 2 2nn nxn  6 Solution [8] (i) Let the equation of 1H be 22 22 1xy ab . Difference in distances of ship from the two stations 0.3 140 42 km   Hence, 42 212a 50 252c 22 2 2 25 21 184 b b    Equation of 1H is 22 1441 184 xy  (ii) The equation of 2H with reference to the coordinate axes of A and B is, 22 2(4 5 ) (4 5 ) 17 6 1324 76 324 xy x y        22 2 1 : 1 184 1441 184 441 xy xHy       Solving 2(4 5 )76 1 324 xy    and 2 184 1441 xy   , 23.027586 and 6.1029768xy Hence, the coordinates of the location of the ship with respect to coordinate axes of A and B is  23.0,6.10 (to 3sf) .

6 9649 / 01 / 2023 7 Solution [8] 32 s i n 2 0 3sin 2 2 20 2 or 2 233 0 or 63             (i) (ii) Required area    3 2 4 3 3 24 3 3 4 3 3 4 3 2 1 32 s i n 2 d2 1 34 3 s i n 2 4 s i n 2 d2 1 34 3 s i n 2 22 c o s 4 d2 11 52 3 c o s 2 s i n 422 11 5 5 3 324 3 4 25 9 3 units24                                

7 9649 / 01 / 2023 8 Solution [8] (i) 22π 2 0 π 222 0 π 22 22 0 π 22 0 dd2π ddd 2π 4sin 16cos 4sin 2 d 8π sin 16cos 16sin cos d 32π sin cos 1 sin d (shown) xyAx t tt tt t t ttt t t tt t t          Using GC to solve, 261.3 unitsA (ii) Required volume    4 120 π 2 120 π 2 0 π 32 0 π 4 2 0 3 2π d d2π dd 2π 4sin cos 2 ( 1) 4cos d 64π sin cos d cos64π 4 16π units xy y x xxy y t t tt t t tt t t             (4, 1) (0,1) 0

8 9649 / 01 / 2023 9. Solution [9 marks] (i) N is the carrying capacity, which is the largest population of white-tailed deer the ecosystem of the state of Kentucky can support. (ii) (iii) d1 ( ) 100d5 P PN P Ht   2 2 2 d1 (900 ) 100 0d5 900 500 0 900 500 0 900 900 4(500 ) 810000 200045022 P PPHt PP H PP H H HP           For population to not become extinct, there should be at least 1 equilibrium point, i.e. 810000 2000 0 H 405H and 0 810000 2000150 450 2 810000 2000 600 810000 2000 360000 225 HP H H H     Thus, max H is 225. 10 Solution [9 marks] (i)  4e 5d g( , )d3 x xyy xyxx    0.25h nx 11 1 g( , )nn n nyy h xy  0 2 3 0.25 0.722222 0.5 0.670187 y  0.670 P t O

9 9649 / 01 / 2023 (ii) 1 24521 21 2 7 3 32 3 32 9 9u         0.5 745 . 5 92 21 93 . 5 0.59032 2 e y       (iii) The Euler method is com putationally simpler but it lacks accuracy due to less accurate gradients used. The improved Euler’s method, while computationally more tedious, is more accurate as it takes the mean of the initial and next gradient, giving a better approximation to the gradient. (iv) d5 4 e d3 3 xyx yx xx   I.F. 5d3 21+ d 3 2ln( 3) 2 e e e e( 3 ) x xx xx xx x x         22 2 4ee( 3 ) e( 3 ) d 3 4( 3 ) d 43 2 x xxyx x x x xx x xC          Since y(0) = 2 3 , 2 963 CC 22e( 3 ) 2 1 2 6xyx x x 2 2 21 2 6 e( 3 )x xxy x   Thus, p(x) = 221 2 6x x

10 9649 / 01 / 2023 11 Solution [13 marks] (a) By ratio theorem, 1OR   qp  0OR PQ         22 22 1 11 10                  qp q p qq p p q p qp  22 2 2 2 22 1        qp p p p pq (b)(i) 22 1 0 22 7 xy z xyz     By GC, 1 2 11 4 01              r Thus 1l : 12 4' 1 , ' 02               r  (ii) 21 10 21 22 0 4 a aa               11 4 01 14 1 7 ab ab b               (iii) Let the foot of perpendicular of B on 2π be N. Let 11 :0 2 52 BNl               r 11 1 02 2 7 52 2 11 9 7 2                          