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