NYJC 2022 FM CT2 P2 - modified (Solutions)

2022 NYJC JC2 Common Test 2 Exam 9649/2 Solution Page 1 of 13 SOLUTIONS: Qn Source: VJC Promo 2025 Qn 1 1 Qn Suggested Solution 1 Perimeter of C = ( ) ( ) 24 2 4 4 cos 2 8 sin 2 daa      − +− 24 2 4 4sin4 cos 2 2 da      − = + 2 4 44 1 2 d 3sina    − +=  24 0 3sin8 1 2 da  = + (as 21 3sin 2 + is a even fn) Let 2= 22 0 3si1d n 18 2a w w  = + 22 0 3sin4 1 da w w  +=  Perimeter of E 22( 2 sin ) c ) d( osaa      − =− + 2)3(sin d 1a    − = + The shell method uses vertical strips, each of which gives rise to a distinct shell. The disc method involves horizontal strips which overlap due to the turning points. As such, turning points need to be determined and the resulting volume is a combinatio n of adding and subtracting volumes formed by different regions under the graph. 2 0 2 0 2 d 2 ( sin 2 ) d 499.77... which is just under 500 (shown) V xy x x x x x     = =+ =  

2022 NYJC JC2 Common Test 2 Exam 9649/2 Solution Page 2 of 13 2 0 )12 3(sin da   +=  (as 21 3sin + is a even fn) 22 0 )4 3(si 1nda   +=  (as sin( )sin 2 =+ for 0 2  ) Thus perimeter of C = perimeter of E. Qn Suggested Solution 2 Let 2π 2πcos isin99 =+ . (a) (i) 99 1 0 1zz− =  = ( ) 2 πi 9 2πi 9 0 2 3 4 5 6 7 8 e , 0,1, 2, ,8 , 0,1, 2, ,8 since e 1 , , , , , , , , k k zk k z                = = = = =  = =   (ii) 9 2 3 4 5 6 7 8 110 1          −+ + + + + + + + = = − 9since 1, 1= . (b) (i) 18 5 27 3 4 6 ww ww ww ww − − − − = = = = (Since 9 1w = ) 2 3 4 4 3 2 1       − − − −+ + + + + + + 2 3 4 5 6 7 8 1        = + + + + + + + =− (ii) ( ) ( ) ( ) 1 2 3 4 2 3 4 21zz           − − − −  = + + + = + + + = . From (b)(i), ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 1 1 234 234 11 2 Re 1 1 Re 2 1 Re Re Re Re 2 2π 4π 6π 8π 1 cos cos cos cos9 9 9 9 2 z z z z z         + =−  + =−  =−  + + + =−  + + + =−  + + + =− (iii) ( )( ) 2 3 4 1 2 3 4 12zz        − − − −= + + + + + + ( )( ) ( ) ( ) 2 3 4 1 2 3 4 1 2 2 3 34 3 2               − − − − − − − + + + + + + = + + + + + +

2022 NYJC JC2 Common Test 2 Exam 9649/2 Solution Page 3 of 13 ( ) ( )( ) ( ) ( )( ) ( )( ) ( ) 2 2 3 3 23 4 3 2 4 3 2 Re 2 2 Re 2 Re 2π 4π 6π4 6cos 4cos 2cos9 9 9          = + + + + + + = + + + = + + + Qn Suggested Solution 3 2 0 1 2 1 0 0 1 0 0 1 0 0 0 1 2 0 1 2 2 2 0 2 2 0 0 1 2 2 1 0 1 2 0 1 2 0 0 0 0 kkk k k k k k k k k k k k k k k −            − − − −      →→      − + − −      − − − − −      If dim(N) = 1 and dim(R) = 3, 0 and 1kk . Case of dim(N) = 0 and dim(R) = 4 is not possible as there are at most 3 pivots. (i) 12 2 3 4 1 1 0 0 0 1 2 0 1 1 2 0 1 2 , ,20 0 0 0 0 1 0 0 0 0 0 0 0 1 xx x x x      −−        =−−−      = +     =+         x  . A basis for N is 12 12 ,10 01 −−                    . A basis for R is 01 11 ,21 10  −                −    . Since 2 2 0 1 R −      . A basis for NR is 1 1 1 0 −  . (ii) Note that ( ) Range space of TAx Hence 01 11() 21 10 where ,  −           =+          −    Ax 

2022 NYJC JC2 Common Test 2 Exam 9649/2 Solution Page 4 of 13 ( ) 01 11 21 10 01 11 21 10 1 1 1 0       −             =+           −    −           =+        −    −  =+    A Ax A AA 0 3 =Ax 1 1 1 0   −     =     A0 . Hence m = 3. Source 2025 ACJC Promo Qn 8 4(i) From diagram, 13,22B  . ( ) ( ) 2 2 2 2 222 2 2 2 2 2 1cos 1 sin cos 2 3sin 2 132 cos 1 cos 3 sin 44 3 3cos 3 sin 2 S OP AP BP r r r r r r r r r r r rr         = + + = + − + + −  +− = + − + + − − + + = − + + 6 3cos 3 sinS rr  = − + and 3sin 3 cosS r  =− − + For stationary points, 0S r  = and 0S   = . 3sin 3 cos 0 r− − + =

2022 NYJC JC2 Common Test 2 Exam 9649/2 Solution Page 5 of 13 As 0r  : 3sin 3 cos= 1 πtan 63 =  = 6 3cos 3 sin 0 ππ6 3cos 3 sin66 3 3 36 22 3 3 r r r r − + = =+ =+ = Thus, the centroid ( ) 3 π 3 π 1 3, cos , sin ,3 6 3 6 2 6P x y P P    ==           . To prove that this gives a minimum point at the centroid: 2 2 6S r  = 2 3sin 3 cos ππ3sin 3 cos66 0 S r   =− =− = 2 2 3cos 3 sin π π 33cos 3 sin6 6 3 323 3 2 S r  =+ =+    =   = As 22 2 2 22 12 0 12 0S S S rr     − = − =      and 2 2 60S r  = (or equivalently, 2 2 20S   = ), then this is a minimum point by the second derivative test. Alternatively, Provided that S is first expressed in terms of r and  as asked, this problem can be solved by recasting S in Cartesian form and finding , , , ,x y xx xy yyS S S S S respectively. This is left as an exercise to the reader. (ii) 0.5 1 3 3 3,,2 4 4 4GG   + =         

2022 NYJC JC2 Common Test 2 Exam 9649/2 Solution Page 6 of 13 3 3 OGm= 1 3 3 3,,2 6 6 6MM    =          3 3 OMm= As both lines have the same gradient with O as a common point, O, M and G are collinear. Alternatively, A “vectors” argument can be used to show that 2 3OM OG=  with O as a common point, or use a “polar coordinates” argument to say that the polar angle of M and G is 6  . Qn Suggested Solution 5 Let be a Markov matrix, then the sum of entries of the rows in equal to 1. 1 1 1 1 1 1 has eigenvalue 1 and corresponding eigenvector . 1 1 1 and have sam T TT T                   =                   AA AA AA  e eigenvalues since . Thus, has eigenvalue 1. T− = −A I A I A (i) The matrix representing the migration, 0.95 0.02 0.05 0.03 0.90 0.05 . 0.02 0.08 0.90  = M (ii) The characteristic polynomial: ( ) ( ) ( )( ) ( ) ( ) 2 0.90 0.05 0.02 0.05 0.02 0.050.03 0.020.08 0.90 0.08 0.95 0.0 0.90 0.90 0.05 0.95 0.806 1.8 0.03 0.0 2 0.05 det det 0.03 0.90 0.05 . 0.02 0.08 0 14 0.02 0.02 0.044 0. .90 0.9 05 0.7644 2. 5             − − = −  − = − −+− − − = − − + − − + − + − =− M I 235144 2.75  +− (iii) After 1 day, the distribution

2022 NYJC JC2 Common Test 2 Exam 9649/2 Solution Page 7 of 13 0.35 0.95 0.02 0.05 0.35 0.35 0.03 0.90 0.05 0.35 0.3 0.02 0.08 0.90 0.3 0.355 0.341 0.305          ==               = M The distribution of caravans after a day is 0.355, 0.341 and 0.305 for the airport, Dandeton and Morningong respectively. Using GC, eigenvalues are 0.84, 0.91 and 1. 0.11 0.02 0.05 When 0.03 0.06 0.05 0.02 0.08 0.06 1 Using GC, an associated eigenvector is 2 . 3 0.84,  −=    − =  M I 0.04 0.02 0.05 When 0.03 0.01 0.05 0.02 0.08 0.01 3 Using GC, a 0.91, n associated eigenvector is 1 . 2  − = − − −   =  M I 0.05 0.02 0.05 When 0.03 0.10 0.05 0.02 0.08 0. 1, 10  − − = − − = M I 15 Using GC, an associated eigenvector is 1 0. 11    1 1 3 15 0.84 0 0 Diagonalising, where 2 1 10 and 0 0.91 0 . 3 2 11 0 0 1 − −       = = =       −    M QDQ Q D

2022 NYJC JC2 Common Test 2 Exam 9649/2 Solution Page 8 of 13 1 1 0.35 0.84 0 0