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                                                                 10 1 317 0 132 2 nn                 Using GC, n Total amount in n months 21 $349085 22 $523661 Hence she needs 22 months to save at least $520 000.

2023 JC 1 H2 Further Mathematics (9649) Promotion Examination Solution 5 6(a) Consider where is eigenvector of and λ is corresponding eigenvalue. Av = v v A  02 2 1 1 2 1 11 2 1 1 2 1 11 0 0 0 0 12 3 S h o w n kk k k kk                                          (b)   det     AI v 0 AI0 22 det 3 1 3 0 11 3        202 2 3 det 3 1 3 0 11 3                       2 2 29 3 3 2 3 3 1 0 23 6 2 3 2 0                             2 2 23 6 2 3 0 26 8 0 224 0                 2, 4 and 2  When 2,   2AI v 0 22 2 31 3 11 1    v0 11 1 1 1 0 eigenvector = 11 1                     1 0 1    When 4,   4AI v 0 42 2 33 3 11 1    v0

2023 JC 1 H2 Further Mathematics (9649) Promotion Examination Solution 6 110 1 1 2 eigenvector = 112            0 1 1    200 110 02 0 a n d 1 0 1 00 4 0 1 1        D Q (c)         4 411 41 1111 4 1 3 3 3 3333 3             BAI QDQ Q IQ QD IQ QD IQ QD IQ QD IQ QD IQ QD I Q Hence 110 101 011    PQ  4 4 500 30 1 0 00 1      CDI 625 0 0 =0 1 0 00 1   

2023 JC 1 H2 Further Mathematics (9649) Promotion Examination Solution 7 7(a) As , cos 2 and sin2 rr    . Hence there is a vertical asymptote x = 2. (b) (c) When y = 0, r sin  = 0   = 0. Area = 2 0 21 (tan +sec ) d2       2 0 221 tan 2 tan sec sec d2        2 0 221 (sec 1) 2 tan sec sec d2         2 01 2(tan sec )2     1 2122 4    units2 (d) 2d sec sec tand r    Arc length = 24 2 4 d dd rr      4 22 2 4 (tan sec ) (sec sec tan ) d          = 2.67 (3sf) x = 2 1

2023 JC 1 H2 Further Mathematics (9649) Promotion Examination Solution 8 8(a)  21 2 31 3 42 4 32 3 :2 : : :5 12 1 1 23 2 6 15 21 1 010 4 12 1 1 010 4 15 21 1 010 4 12 1 1 010 4 05 0 0 000 0 12 1 1 010 4 000 5 4 000 0 RR R RR R RR R RR R                                      A     Since dim of null space of T = 1, Rank(A) = 4 – 1 = 3  54 0  0, 4,     (b) When 0  , we have 12 11 010 4 0000 0000       A basis for range space of T is 12 23,12 01                .

2023 JC 1 H2 Further Mathematics (9649) Promotion Examination Solution 9 (c) Since 7 3 p q      lies in the range space of T, 12 23 7 12 01 3 p ab q                       has unique solution for a and b. Since b = 3 and a = 1, hence solving the equations we get 5pq  (d) 12 11 0 010 4 0 0000 0 0000 0 x y z w                               20 40 xy z w yw     19 04 10 01 x y zwz w                         Since 19 04 for any 10 01 kk                   , 19 04 and 10 01                   are linearly independent. Hence a basis for the null space of T is 19 04 , 10 01                (e) 2 3 2 1     Ax has a particular solution 0 1 0 0       . Thus the general solution is 01 9 104 where ,010 001                           x 

2023 JC 1 H2 Further Mathematics (9649) Promotion Examination Solution 10 9(a) 44 22 00 tan d tan .tan dnn nIx x x x x   4 22 0 tan .(sec 1) dn x xx   4412 0 0 1 tan tan d1 nn x xxn     2 1 1 nIn   (shown) I1 =  4 4 0 0 1tan d ln(sec ) ln 2 ln 2 2xx x     31 1 31II   = 11 ln 222 (b) Using shell method, Volume =  5 22 2 0 5 52 ( s i n 1 ) d2 x xx      55 224 22 00 125 2s i n d 2 d4 x xx x x      55 22 34 2 0 0 1252 d 2 31 2 xxx     5 2 2 0 2s i n dx xx    5 25 22 0 0 2 cos 2 cos dx xx x x     