NYJC 2023 FM CT2 P1 - modified (Solutions)

2023 NYJC JC2 CT2 9649/1 Solutions Page 1 of 23 SOLUTIONS: Question 1 Qn Key Points 1 2025 NJC Promo Q7(a) Consider the equation ( ) ( ) ( ) ( ) ( ) ( ) 0 ( ) 0.             + − + + − + + − =  − + + + − + − + + = u v w v w u w u v u v w Since u, v and w are linearly independent, we have 0, 0, 0.          − + = + − = − + + = By GC, 0  = = = . Thus, () has only the trivial solution. Therefore, +−u v w , +−v w u , and +−w u v are linearly independent.

2023 NYJC JC2 CT2 9649/1 Solutions Page 2 of 23 Question 20 Key Points 2 22 22 d d d d d d d d d d d d d d y z y y y z yzy x x x x x x x= −  = −  = + . Substitute into the given DE gives ( )dd 2 3 2edd xzy z y yxx+ + + − = dd 2 2edd d 3 2ed x z x zy zyxx z zx + + − = +=  Multiply DE by IF = 3d 3ee x x = gives ( ) 34 34 d e 2ed 1e e on integrating2 xx xx zx zC = =+ When 0x= , d 2 1 1d yzy x= − = − = . Thus 111 22 CC= +  = . Therefore ( ) ( ) 34 3 1e e 1 2 1 ee2 xx xx z z − =+ =+ ( ) 3d1 eed2 xxy yx −− = + Multiply DE by IF = d ee x x− − = gives ( ) ( ) 4 4 d1 e 1 ed2 11e e on integrating24 xx xx yx y x C −− −− =+ = − + When 0x= , 1y= . Thus 191 88 CC=− +  = . Therefore Differentiate Simplify to linear form Obtain PS Use d d yzy x=−

2023 NYJC JC2 CT2 9649/1 Solutions Page 3 of 23 4 3 1 1 9ee 2 4 8 1 1 9e e e2 4 8 xx x x x yx yx −− − = − + = − + Obtain final PS (cao)

2023 NYJC JC2 CT2 9649/1 Solutions Page 4 of 23 Question 3 2025 NJC Promo Q7(b) Key Points (a) Take ( )22,a b e f c d g h             M  and k  . ( ) ( ) ( ) ( ) 1 2 1 11 22 11 F F d d d FF a b e f c d g h a e b f c g d h a e x b f c g x d h x ax b c x d x ex f g x h x a b e f c d g h − −−    +        + + =  ++ = + + + + + + + = + + + + + + +       =+                 F preserves addition. ( ) ( ) 1 2 1 1 2 1 FF d d F. a b ka kbk c d kc kd kax kb kc x kd x k ax b c x d x abk cd − −        =             = + + + = + + + =     F preserves scalar multiplication. Therefore, F is a linear transformation.

2023 NYJC JC2 CT2 9649/1 Solutions Page 5 of 23 (b) ( ) ( ) 1 2 1 123 1 F0 d 0 032 03 2 3 2 2 203 3 pq rs px q r x s x q r xpx sx p q r p q r ss p s ps − −  = + + + =  ++ + =  ++   + + − − + − =       += =−   The null space of F is 3 , , 3 ,, 3 0 0 1 0 0 ,,0 1 0 0 1 0 3 0 0 1 0 0span , , .0 1 0 0 1 0 pq p s q rrs sq q r srs s q r q r s  =−  −=  −     = + +             −      =               Since the three matrices above are linearly independent, a basis for the null space of F is 3 0 0 1 0 0,,0 1 0 0 1 0  −                  . (c) The range space of F is  . (d) For pq rs   to be antisymmetric, T p q p q p r r s r s q s −−     =− =      −−     

2023 NYJC JC2 CT2 9649/1 Solutions Page 6 of 23  0ps== and qr=− . Thus, antisymmetric matrices are of the form 0 , 0 r rr −   . To show null space of FW  : Method 1: ( ) 1 2 1 1 1 0F 0 0 d 0 0 d 0. r x r r x xr x − −  −  = + − + + = =   null space of FW . Method 2: Since 0 3 0 0 1 0 000 0 1 0 0 1 0 r rrr −−       = − +               , null space of FW  . Note that question is not asking to show that W is a subspace only. It needs you to show that it is a subspace of the nullspace of F. Hence, do not just show that the set of matrices contains the zero matrix and closed under vector addition and scalar multiplication (not required). All we need to show is that W is a subset of the nullspace of F (which is already a subspace) and the “zero vector” is in W, we are done. Also, since 01span 10W  − =  , W is a subspace of ( )22M  . Hence, W is a vector space. Therefore, W is a subspace of the null space of F. The dimension of W is 1. 01span 10W  − =  implies that zero matrix is in W already.

2023 NYJC JC2 CT2 9649/1 Solutions Page 7 of 23 Question 4 Key Points (a) 3d e2d xy xyx =− 0 1 2 0, 0.5, 1x x x= = = 0 0y = ( ) 0 1 0 0.5 0e 0 0y  + − = ( ) ( ) 3 0.5 2 0 0.5 0.5e 0 1.12042 1.120 (3dp) y  + − = = (b) 01 0, 1xx== ( ) 0 1 0 1 0e 0 0u = + − = ( ) 3 1 0 1e 0 0 1 10.04272 10.043 (3dp) y +−  + = = Initial approximation (c) 3 3 2d 2 3 2 e e ed 25 5 25 x x xy xx −= + − 2 3 3 2 2 d 21 9 4 e e ed 25 5 25 x x xy xx −= + + Since 2 2 d 0d y x  , the curve of y against x is concave upwards. The approximations in (i) and (ii) are underestimates. (d) When 3 3 21 1 11, e e e 3.219095 25 25xy −= = − +   (i) gives a better estimate.

2023 NYJC JC2 CT2 9649/1 Solutions Page 8 of 23 Question 5 (Source : 2025 VJC Promo Qn 11) (a) (a) 2 2/3 f ( , ) 7 0.05 3x y y x= − − 1/3f 2f ( , ) f 0.1 x y xxy y − − = =  −  When 0, f xx= is undefined. Hence points on the canopy where 27 0.05zy=− (b)(i) Instantaneous rate of descent 2/3 2f ( , ) f ( , ) 4 0.01D x y x y x y −= =  = +u (ii)the least instantaneous rate of descent occurs where 1 and 0.xy= = Least rate of descent f ( , ) 2 3D x y= = u  the canopy is considered inefficient in shedding rainwater. (c) Length of 2 vertical beams ( )f ( , ) f ( , ) 2f ( , ) f is even wrt. a b a b a b x= + − =  The vertical plane is .yb=  the vertical trace is 2 2/37 0.05 3z b x= − − . Length of arc PQ 2d1d d a a z xx− =+  ( ) 21/3 0 2/3 0 2/3 1/3 0 2 1 2 d (by symmetry of z about = 0) 42 1 d 42 d a a a x x x x x x x x −= + − =+ +=    2/3 1/3 0 4( , ) 2f ( , ) 2 d (shown) a xL a b a b x x + = + 

2023 NYJC JC2 CT2 9649/1 Solutions Page 9 of 23 (i) 1/3 2/3 0 2( , ) 2f ( , ) 3 4 d 3 a L a b a b x x x −= + +  ( ) ( ) 3/22/3 2/3 2/3 3/2 2/3 3/2 2/3 0 2 2 2f ( , ) 2 4 2 7 0.05 3 2( 4) 16 2( 4) 0.1 6 2 a a b x b a a a b a = + + = − − + + − = + − − − For a fixed a, the largest possible ( , )L a b occurs when 0.b=  ( ,0)La is largest. (ii) From GC, largest 2/3 3/2 2/3( ,0) 2( 4) 6 2L a a a= + − − occurs when 1.a= 3/2(1,0) 2(5) 6 2 10 5 8 14.4 (3sf)L = − − = − =

2023 NYJC JC2 CT2 9649/1 Solutions Page 10 of 23 Question 6 Key Points (i) Let P and Q be the points representing the complex numbers z and iez  respectively. Then OQ is an anticlockwise rotation of OP about O through an angle . (ii) Since π 3BAC= , ( ) ( ) ππarg arg arg 33       −− − − =  =  − .  ABC (in anticlockwise order) is equilateral if and only if ( ) πi 3e   − = − from (i). (iii) ( ) ππii33ee      −− = −  = − πi 3e  −−= − (take reciprocal) Thus ππii33 πe e 2cos 1 3         −−− + = + = =−− . (iv) Let z   −= − . Then 1 z   −= − . By (iii), 111 z z         −− + =  + =−− 2 10zz − + = (*) So (*) has a root   − − . By symmetry,   − − is also a root of (*) 2 1 1 4 1 31 0 i 2 2 2z z z −− + =  = =  . Since πarg 3    − = − , 13 i22   − =+− and 13 i22   − =−− . Minus 1m if 13 or 2 3 2 2 2 1 i i     − =−− − =+−