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