NJC 2023 JC1 Promos 9649 (Solutions)

National Junior College Mathematics Department Page 1 of 13 2023 SH1 H2 Further Mathematics Promotional Examination Solutions Qn Solution 1 Method 1: Standard check for linear independence. Consider the equation  uA u 0 . (1)  2 2 () 0 ( ).            A u Au A0 Au A u 0 Au 0 0 A O Au 0 Au 0   Sub into (1): 0 ( ).     Au 0 Au 0  Equation (1) has only the trivial solution. Therefore, u and Au are linearly independent. Method 2: Proof by contradiction. Suppose u and Au are linearly dependent. Version 1 Then, kuA u for some k  .   2 , k k k k      Au A Au Au Ou 0 0 contradicting Au 0 . Version 2 Then, kAu u for some k  . (1)  2 . k k k k     Au u Au A u Ou Au Au 0 Since Au 0 , it must be the case that 0k  . Then, from (1), 0Au u 0 , contradicting Au 0 . Therefore, u and Au are linearly independent. 2(i) Consider the interval 02 π . When cos 2 0  , π 3π or 44  .  π 24 0 π 42 0 2 2 2 1Total area 4 cos 2 d (by symmetry)2 12s i n 22 πsin sin 02 10 . a a a a a          

National Junior College Mathematics Department Page 2 of 13 (ii) Differentiating both sides of 22 cos 2ra  w.r.t.  : 2 2 2 42 42 22 22 d22 s i n 2d ds i n 2 d d sin 2 sin 2 sin 2 .d cos2 cos2 rra ra r ra a a ra            π 224 2 0 π 224 0 π 4 0 π 4 0 sin 2Total length 4 cos 2 d cos 2 cos 2 sin 24 d cos 2 14 d cos 2 4 sec 2 d (shown) aa a a a                  3(i) (a) When 1 2c , the recurrence relation becomes 21nnuu  which results in a constant sequence. (i) (b) When 0c , the recurrence relation becomes 1 21 4 0nn nuu u  . Auxiliary equation: 2 1 4 0mm   21 2 1 2 0 . m m   General solution:   1 2 n nuA B n . As n , 1 2 0 n  ,   1 2 0 n nuA B n  . The sequence converges to 0 as n becomes very large. (ii)   21 21 4 0nn nuu c u    Auxiliary equation: 22 1 4 0mm c   21 4 2 11 4 2 14 2 1 .2 c m c c      Since we have shown the sequence converges when 0c in part (i)(b), we may consider the case where 0c here. General solution:    11 22 nn nuA c B c . Method 1: Magnitude of both roots are at most 1.

National Junior College Mathematics Department Page 3 of 13 11 22 11 22 311 22 2 311 1 22 2 2 33 22 1a n d 1 11 11 11 cc cc cc cc c               11 22 c . Method 2: Observe that both roots are equidistant from 1 2 . Observe that the larger root, 1 2 c , is at least 1 2 . Thus, for the sequence to converge, we must have 1 12 1 2 c c   11 .22 c Also, check that when 1 2c  , the smaller root 1 2 c satisfies 110 22 c and will thus have a magnitude of at most 1. Hence, the required range of values of c is 11 22 c . 4(i)   null space of nn nS  XM v X  Take  nnOM  . Then, null space of Ov 0 v O . nS O . Take , nSAB and k  . Then, null space of vA and null space of vB  Av 0 and Bv 0 .      AB v A vB v000 . null space of  vA B nSAB . nS is closed under addition.  kk k  Av A v 0 0 null space of kvA nkSA . nS is closed under scalar multiplication. Therefore, nS is a subspace of  nnM  .

National Junior College Mathematics Department Page 4 of 13 (ii)     22 2 22 22 22 1 null space of2 10 20 20 , 2 0 2, 2 2 ab abS cd cd ab ab cd cd ab ab cdcd ab ab cdcd bb                                           M M M M     ,2 21 0 0 ,00 2 1 21 0 0span , . 00 2 1 bddd bd b d                   Since 21 00   and 00 21   are not scalar multiples of each other, they are linearly independent. Therefore, a basis for 2S is 21 0 0,00 2 1        . (iii) Take 11 1 1 n n nn n aa S aa   A    and denote 1 n n v v   v  . Then, 11 1 1 1 0 0 n nn n n aa v aa v                             11 1 12 2 1 11 2 2 ... 0 ... 0 nn nn n n n av av av av av av      Each equality reduces the degree of freedom of the entries in a matrix by 1. Since A has 2n entries and there are n linearly independent equalities, the entries in A have a degree of freedom of   2dim nn nn n  M  . Thus, the dimension of nS , which is the degree of freedom of the entries of a matrix from nS , is 2nn . (iv) Since nn Ox0v for all nx  , column space of vO .    column space of nn OX M v X  . Therefore,   column space of nnXM v X  is not a subspace of  nnM  .

National Junior College Mathematics Department Page 5 of 13 5(i)   1 1 16 151 116 4 60nn nn aa aa     General solution: 1 16 n naA B , 0n . 0 00aA B . Since   151 1 16 406 0a   ,  115 1 41 6 A B . By GC, 4, 4.AB  1 1644 , 0 . n nan    (ii) lim 4nn a   . (iii) Let mb denote the amount of drug, in mg, at the end of the mth day of taking d mg of the drug a day. Then, 0 4b  .   1 1 16 11 116 16 mm mm bbd bb d     General solution: 1 16 m mbC D , 0m . 0 44bC D . Since    41 1 16 164 dbd   , 4 1 16 16 d CD  15 154, .ddCD    1 15 16 154 mdd mb   , 0.n Since 1 16 1 and 15 1560 4 4 0ddd      ,  nb is an increasing sequence converging to 15 d . So, for an overdose to never occur, 15 80d d . Thus, we have 16 15 80 75. d d   Therefore, the maximum value of d is 75. 6(i) When 0y , cos 2 0tt  , 0t or cos 2 0t  , π 3π or 44t  . d1 d 2 x t t  d cos 2 2 sin 2d y tt tt  .

National Junior College Mathematics Department Page 6 of 13  3π 224 π 4 3π 24 2 2 π 4 2 2 Surface area dd 3 ππ2π d πdd 4 4 1 π2π cos 2 2 sin 2 d 22 31.958 units (to 5 s.f.) 32.0 units (to 3 s.f.) xyxt tt tt t t t t                      (ii)    3π 4 π 4 3π 4 π 4 3π 4 π 4 3π 3π 4 4 ππ 44 3π 4 π 4 2 2 3 Volume d2π dd 12π cos 2 d 2 π cos 2 d sin 2 sin 2π d22 3ππ cos 2π 88 4 π π 002 π units .2 xxy t t tt t t t tt t tt t t t                         (iii) 3π 4 π 4 2 π cos 2 d 1 ππ 3ππ 5π 3ππ f4 f + 2 f4 f f3 8 48 28 4 π 3π 5π0 π 024 22 22 4.94605 (to 5 d.p.) tt t                                    2 2 π 2 π 2 4.94605Percentage error 0.228%  which is very small. Thus, the approximation is very accurate.

National Junior College Mathematics Department Page 7 of 13 7(i) 12 11  A . 2 12 0 ... n nn n   vA v A v A v . nBA . (ii) det 0 IA  2 12 011 12 0 12 12             For 12  , we solve the system   12  IA x0 . 2 12 1 222 0022 12 12 0 012 R RR R R                .  A corresponding eigenvector is 2 1   . For 12  , we solve the system   12  IA x0 . 2 12 1 222 0022 12 12 0012 R RR R R               .  A corresponding eigenvector is 2 1    . 1 12 022 22 . 11 11 01 2            A   1 1 12 022 22 11 11 01 2 12 022 22 . 11 11 01 2 n n n n                     A

National Junior College Mathematics Department Page 8 of 13 (iii)               0 1 0 1 11 1 12 0 122 22 111 11 01 2 21 2 21 2 12 11 122 1212 12 21 2 21 2 121 22 1212 12 21 2 21 21 22 12 12 n n n n nn nn nn nn nn n                                   