NYJC 2023 JC1 Promos 9649 (Solutions)

1 Suggested Solution for 2023 NYJC JC1 FM EOY 1 Suggested Solution (a)  2 2 2 0 10 0 or 1      AA AA AA AA A (b) 2 12 1 since the matrix is invertible.    AA AA AA AI (c) A is non-invertible: 0wz xyA -- (1) 2 2 2 wxwx y w x x z yzwy zy xy z       AA 2 2 (using (1)) 1 wx y w ww z w wz    Note that solving wy zy y or wx xz x is also possible. As qn did not mention y or x is non-zero, need to at least mention that if x or y is zero, then A will be the identity matrix whose determinant is not zero. Note: 0x and 0y  . This is because if 0x or 0y , det (A) = 0, means 0wz , which is a contradiction.

2 2 Suggested Solution          2 2 2 2 . When 1, LH dLet P be the propo S sition that f 4 sin 2 , for d d sin 2d d 2cos 2d 22 s i n 2 4 s i n 2 R H S P1 i s e tru . = n n nnx x n x xx xx n xx                    2 2 21 1 21 dAssume P is true for some . i.e. f 4 sin 2 To show P 1 is true. i.e. .d d f4 s i n 2 . d k k k k k k k x x k kx x x x                         22 2k+2 2 2 1 2 2 dLHS = sin 2d d 4 sin 2 By inductive hypothesis d d 42 c o s 2d 422 s i n 2 4s P true P 1 true and since P 1 is true, by math in 2 RHS d f4 s i n 2 , f o r d ematical induction . Sin k k k k k n n n xx xx xx x x xx nx kk                            2 2 2 ce Maclaurin's expansion of f is given by f f 0 f ' 0 f '' 0 ... f 0 ... 2! ! and from the result of induction and sin 0 0 f 0 =0 and all even derivatives are d f4 s i n 2 , f o coeff zero r d ici n n n n n x xxxx n xx nx           ent of even powers of is equal t o 0x    21 21 d f2 4 c o s 2d n n n x xx    When 0x ,      21 21 d f2 4 c o s 0 2 4 0d n n n n xx      Therefore, all the odd powers of x (3 )n has non-zero coefficient. Clearly the first derivative, . f' 0 2 c o s ( 0 ) 0 Thus the Maclaurin expansion only contains odd powers of x.

3 3 Suggested Solution (a) 2secd 2sind 2t a n2 t x at a tt   sin sin sin2sin cos22 aaat at tt t    d cosd y att  Arc length 3 4 22 dd ddd xy ttt        2 22 2 2 2 2 3 4 sin 2 cos d sin aaa a t t t       4 2 3 11d sinat t      3 4 cos dsin tat t      3 4 31 3ln sin ln ln ln 22 2 2 aat a       (b) 2 1 0 d 4 n nxI x x     0 1 2 1 d 4 n xx x x       12 1 2 1 00 241 4 dnnx xn x x x        221 0 31 4 d nnx x x       2 2 2 1 0 431 d 4 n xnx x x         

4   11 0 2 22 0 3 4 1d 1d 44 nnxxnx nx xx            234 11nn n nII I n     241 3nnnI n I   (i) Area 3 2 1 0 3 111d 3 33 1 x x I x       Using  241 3nnnI n I   3134 2 3II  31 81 333II  Area 11 18 1 2 833 333 3 3 3 I I   1 2 1 2 1 00 d4 3 2 2 3 4 xIx x x          Area =  2 8 10 1632 3 333 3 3  (ii) Volume  1 3 2 2 0 1 2d 3 1 4 xxx x          42 3 I   241 3nnnI n I   4244 3 3II  42 133 4II  2 024 3II  20 123 2II  1 0 0 2 11 0 1 ds i n 264 xIx x       

5 2 1 362 1123 23I      4 17 3324 1133 34I     Volume 4233 I  2723 3 2 4 23336         

6 4(a) Suggested Solution (i) As 1, and nnnu l u l    2 (since the sequence is positive) kl al lk lk l k l a la     (i)     1 2 2 0 Since and 0 n nn n n nn n n n n nn n n ku auu u uk ku a u ku uk au uk au au uk uak              We have 1 0nnuu  . Hence 1nnuu  . 5(b) Suggested Solution (i) 110.8 0.1 --- (1)nn nxx y  110.2 0.9 --- (2)nn nyx y  From (1), 11 11 0.1 0.8 10 8 nn n nn n yx x yx x     Substitute into (2),  11 1 0.2 0.9 10 8 97 nn n n nn n yx x x yxx      1 1 1 19 7 10 7 1 0.7 0.1 nn n nn nn x xx xx xx        Alternatively,  11 11 1 From (1): 0.8 0.1 0.8 0.1 1 using given info 0.7 0.1 nn n nn n xx y xx x        (ii) General solution: 0.7 n nx AB

7 1 0.7 0.9 xA B  0.7 0.1BB Solving, 1 3B , 17 21A Thus 17 10.721 3 n nx  21 710 . 7 32 1 n nnyx   5 Suggested Solution (a) Since 4dim( ) 4 and there are 4 vectors in the set. It is sufficient to prove that if the vectors are linearly independent, the set is a basis for 4 . Using GC, 111 1 01 1 0det 6 0110 2 01 3 4       , the vectors are linearly independent. Hence the set forms a basis for 4 . (b) 1 111 1 4161 1 01 1 0 1 3 5 1 3 2 7 110 2 9471 6 01 3 4 1 0 6 7 1 7 4161 1111 1 1 2 3 1 1 351 32 7 0 110 371 0 1 9471 6110 2 2 5 7 0 10 6 7 17 0 1 3 4 3 4 7 0                                               M M (c) 10 10 011 0rref ( ) 0001 0000      M Hence basis for range space of T = 121 371,,250 340              

8 Alternatively, 10 10 01 1 0rref ( ) 00 01 00 00 T      M Hence basis for range space of T = 10 0 01 0,,110 00 1                     Dim(Range space of T) + Dim(Kernel of T) = 4 Dim(Kernel of T) = 4 -3 = 1 (d) 123 2 7 121 0 371 =250 340 121 0 1 2 10 371 0 1 2 250 0 1 2 340 0 2 3 12 1 0 12 1 0 01 2 01 2 00 0 00 1 00 7 2 00 0 db b c d bb cc dd bb cb db c                                                    b       Alternatively, 12 3 10 0 0 01 0 =110 00 1 1 00 0 100 0 100 0 01 0 0 1 0 0 1 0 110 010 001 00 1 0 0 1 0 0 0 b c d bb b cc d ddc b                                                                 Hence cb , d for 0 b c d       to be in the range space of T.

9 6 Suggested Solution  22 2d1 1 1 sec 1 tan 1d2 22 2 2 tx x tx          2 2d 1 dxt t  tan 2 xt   2 2tan 1 tx t  ; 2 2 1cos 1 tx t   ; 2 2sin 1 tx t  1 d32 s i n c o s xx x  2 2 22 12 d21 132 11 ttt t tt        22 2 d33 4 1 ttt t     2 1 d22 ttt    2 1 d 11 t t      1tan 1 tc  1tan tan 1 2 x c      (i)  1cos3 sin 3 d sin 3 3 sin 3 3 d2mx nx x m n x m n x x   1 cos(3 3 ) 1 cos(3 3 ) 2( 3 3 )2 ( 3 3 ) mn x mn x Cmn mn            (ii) 22 0 cos (3 ) sin (3 ) 2cos3 sin 3 dmx nx mx nx x   0 1c o s 6 1c o s 6 sin(3 3 ) sin(3 3 ) d22 mx nx mn x mn x x       0 0 sin 6 sin 6 cos(3 3 ) cos(3 3 ) 12 12 3( ) 3( ) mx nx m n x m n xx mn m n m n           cos(3 3 ) cos(3 3 ) 1 1 3( ) 3( ) 3( ) 3( ) mn mn mn mn mn mn                  Since sin 0k  for all integer k. If m is even and n is odd, then 3 mn and 3 mn will be

10 odd, hence cos(3 3 ) co 3 1 s( 3 )mn mn    Answer: 22 22 2( ) 4 3 22 3( ) 3( ) ( 3 )( ) mnmn n mmmn mn n n        