TJC 2023 JC1 Promos 9649 (Solutions)

2023 JC 1 H2 Further Mathematics (9649) Promotion Examination Solution 1 1(a) Consider  2,2,0 and  3, 6,0  W.      2,2,0 3, 6,0 2 3,2 6,0       22 22 3 2 6 2 2 26 3 4 46 6 86 0       W is not closed under addition. W is not a subspace of 3 . (b) Consider  11 1,,x yz and  22 2,,x yz V     11 1 2 22 1 2 1 21 2,, ,, , ,x yz x yz x x y y z z    Since    12 1 2 12 12 1 2 1 1 22 2 2 2 and 000 xx y y yy yy zz yz yz           V is closed under addition. Consider    11 1 1 1 1, , , , where k x y z kx ky kz k  Since      11 1 1 1 1 12 2 and 0kx k y ky ky kz k y z     V is closed under scalar multiplication Also (0, 0, 0) belongs to V V is a subspace of 3 . 2 d = 2r2 – 2 = 2(r2 – 1) Since S exists, |r| < 1 Hence r2 – 1 < 0 and d < 0 (shown) 2r6 – 2r2 = 2r2 – 2 r6 – 2r2 + 1 = 0 Let u = r2: u3 – 2u + 1 = 0 (u – 1)(u2 + u – 1) = 0 Hence u = 1 or 15 2u  Since |r| < 1, 0 < u < 1; hence 51 2u  d = 2(r2 – 1) = 2(u – 1) = 53

2023 JC 1 H2 Further Mathematics (9649) Promotion Examination Solution 2 3(a) (i) (a) (i) Since P is a point on E, F1P + F2P = 2a for all points P. F1 is (c, 0) and F2 is (−c, 0), where 22ca b Hence the perimeter is 2222aa b (a)(ii) If F1PF2 forms an equilateral triangle, F1F2 = F1P = F2P = a. Hence 2ae = a 1 2e . Therefore equations of directrices is x = 1 2 2a a  (b) (i) 33 2 22 s i n 1s i nr    . Hence e = 1 Therefore the conic is a parabola. (ii) The equation of the directrix is 3 2y . 4(a) k is the net growth rate and N is the carrying capacity of the environment. (b) d1 1d dd4 ( 4 ) 4 P PkkP P ttP P      11 1 dd44 4 kP tPP    ln ln 4P Pk t C   where C is a real number e w h e r e e4 kt CP AAP   When t = 0, P = 2  A = 1 (4 ) ktPP e 4e 1e kt ktP  (c) 4e 4 1e e 1 kt kt ktP  If k = 0.6, as t  , e −kt  0, so P  4 In the long run, the population of birds will increase and stabilize at 4,000.

2023 JC 1 H2 Further Mathematics (9649) Promotion Examination Solution 3 5(a) [Solution] 12 12 3 , 34 40, 50 nn nuu u n uu    (b) 12 3 4 nn nuu u  Characteristic equation is 244 3 0mm  13 or 22m Thus, the general solution is 13 22 nn nuA B         .   1 22 2 13Since 40, 40 1 22 13Since 50, 50 2 22 uA B uA B                    Solving by GC, 7010, 3AB  17 0 310 , 123 2 nn nun          

2023 JC 1 H2 Further Mathematics (9649) Promotion Examination Solution 4 (c) Total amount save in n months 123 12 1 2 ...... 11 1 7 0 3 3 310 ..... .....22 2 3 2 2 2 11 7 0 3 310 1 122 3 2 2 31 11 22 n nn nn uuu u                     