HCI 2022 FM Prelim P2 - modified (Solutions)

Page 1 of 12 2022 Preliminary Examination H2 Further Mathematics Paper 2 (9649/02) Solutions s/n Solution 1(i) Method 1: (R-formula) ( ) ( ) ( ) ( ) ( ) 22 22 2 2 22 2 44 44 44 2 4 1 3 44 3 22 c 0 n , si s os 1 si sin cos nce , inc 1 2 sin 1 sin ( e ) 0 x x x x x xx x x x        − = − − − + − =  − + = =   =  = Method 2: (sum and difference identities) ( ) ( ) 22 2 2 2 2 2 2 2 2 2 2 22 2 44 4 4 4 4 2 2 2 2 2 3 4 4 4 3 2 1 1 1 1 1 2 2 , since , since 0 cos 1 sin cos( ) cos sin( )sin 1 sin( ) cos cos( )sin cos( ) sin( ) 1 sin( ) cos( ) sin( ) 1 sin( ) 0 x x x xx x x x x x x x x x x x          − = − − + = − + + = − + = =   = = (ii) Method 1: (standard form) ( ) ( )( ) ( ) ( ) ( ) ( ) 12 22 22 2 2 2 3 2 2 3 2 2 3 2 2 3 2 2 44 44 44 2 ( )d 2 cos 1 sin d 2 cos 2 sin 2 d sin cos x y y x x x x x x x x x x x x x x                − = − − − −  = − + − − = − − − −    3 2 2 4 4sin cos sin co (0) (0) s    −  −−= − − 32 3 4 2 40 ( 1 1) 0 2 unit     =  −− −− −−  =− Method 2: (R-formula)

Page 2 of 12 s/n Solution ( ) ( )( ) ( ) ( )( ) ( ) 12 22 22 2 2 22 3 2 2 3 2 2 3 2 2 3 2 2 33 22 22 3 32 2 22 44 44 33 4 4 4 4 2 ( ) d 2 cos 1 sin d 2 cos sin 1 d 2 sin( ) 1 d 2 sin( ) d s 2d () 2 2 2 co 2 cos cos x y y x x x x x x x x x x x x x x x x x xx                           − = − − −  = − +  −− =− =−   − − − − − − =−  =      32 22 unit  =− 2(a) True. Let ( ) 0p  = be the characteristic equation of A. When 0 = , ( )(0) det A I(0) det(A) 0p = − = = Since det(A) 0= , A is not invertible. (b) True. Let I be the identity matrix. Then 1A I AI−= A is diagonalisable (c) False. Method 1: Let 1 1 1 1 0 1 1 1 0 0 1 0 1A 1 0 0 0 1 0 1 0 1 1 0 1 −         = = =         −          A is diagonalisable, but A is not diagonal. Method 2: Let 11A 02 −= Then 1 1 1 1 1 1 1 1 1 1 2 1 0 0 1 0 2 0 1 0 1 0 2 0 2 − − − − −          ==                    A is diagonalisable, but A is not diagonal. Method 3: Let 11A 00 =  characteristics equation is 11 00   −− = ( 1) 0 −=

Page 3 of 12 s/n Solution 0= or 1 = When 0 = , 1 1 1 0 1 0 0 0 1 xx kyy         =  =         −         When 1 = , 2 0 1 0 1 0 1 0 0 xx kyy −        =  =                 Then 1 1 1 0 0 1 1A 1 0 0 1 1 0 −    =   −−    is diagonalisable, but A is not diagonal. (d) False. Consider the identity matrix I. Its eigenvalue 1 is repeated, but I is diagonalisable since 1I I (I)I−= . 3(i) Let iz x y=+ 2Re( ) 0zz+ 2Re( i ( i) ) 0x y x y + + +  22Re( i 2 i ) 0x y x xy y+ + + −  22 0x x y+ −  ( ) ( ) 2 22 2 1 211 24 11 44 1 x yxy +  + −   −  (ii) 1 2 1 2 1 2 1 2 2 2 2 2z z z z+ − − − − ++ − + = − − − Using relationship 2 2 2c a b=+ for hyperbola, 2 21 1 1 4 4 2 2cc = + =  = Hence foci of hyperbola are at ( )21 22 ,0−− and ( )21 22 ,0−+ 1 2 1 2 1 2 2 2 2 2( ) 1z z a− − − + − − − = = = 4(i)

Page 4 of 12 s/n Solution Let 11 12 21 22 A aa aa = , 11 12 21 22 B bb bb = 11 12 11 12 11 21 12 22 21 22 21 22 AB | a a a ab b b ba a a a          = + +                Hence 11 12 11 12 11 21 12 22 21 22 21 22 11 12 21 22 colsp(A) |, |, T a a a ap b b q b b p qa a a a aa aa              + + +                         +      =   =   Since p and q can both be 0, the zero column belongs to T. Hence T  . Let 11 12 11 12 1 11 21 1 12 22 21 22 21 22 a a a ap b b q b ba a a a           + + +                     , 11 12 11 12 2 11 21 2 12 22 21 22 21 22 a a a ap b b q b ba a a a           + + +                     be two vectors from T. Since ( ) 11 12 11 12 1 11 21 1 12 22 21 22 21 22 11 12 11 12 2 11 21 2 12 22 21 22 21 22 11 12 1 2 11 21 21 22 a a a ap b b q b ba a a a a a a ap b b q b ba a a a aap p b b aa           + + +                               + + + +                         = + +         ( ) 11 12 1 2 12 22 21 22 aaq q b b aa     + + +            T is closed under vector addition Let k be a constant. Since 11 12 11 12 11 21 12 22 21 22 21 22 11 12 11 12 11 21 12 22 21 22 21 22 , a a a ak p b b q b ba a a a a a a akp b b kq b ba a a a            + + +                                 = + + +                     T is closed under scalar multiplication.  T is a subspace of colsp(A) . (shown) Alternative Solution ( To show that T is a subset of colsp(A)) If Tv then there exist 2u  such that ( ) ( ) ( )AB A B colsp A= = v u u Therefore ( )colsp AT  (ii) From (ii), colsp(AB) is a subspace of colsp(A) .

Page 5 of 12 s/n Solution  by part (i), ( ) ( )dim colsp(AB) dim colsp(A) . Hence ( ) ( )rank(AB) dim colsp(AB) dim colsp(A) rank(A)=  = (shown) (iii) If B is invertible, then 1rank(A) rank(ABB ) rank(AB) rank(A)−=   Hence rank(AB) rank(A)= (shown) 5(i) ( )1 1001 900 900 9kuk= + = + ( ) ( ) 1 100 100 1 ( 900) 1 900 9 nn n k k uu uk + = + + = + + + (ii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 32 1 ) 1 3 (1 3 100 100 100 100 100 100 100 100 100 100 1 (900 9 ) 1 1 (900 9 ) (900 9 ) 1 1 1 (900 9 ) (900 9 ) (900 9 ) 1 1 (900 9 ) 1 (900 9 ) (900 9 ) 1 (900 9 ) nn n n n n nn k kk kkk k k k k u u k u k k u k k k u k k k uk − − − − − − − = + + + = + + + + + +  = + + + + + + + + + = + + + + + + + + + = = + + +  ( ) ( ) ( ) ( ) ( ) ( ) 1 11 1 11 100 100 100(900 9 ) 100 100 100(900 9 ) 100 100 11 11 1 1 1 1 (900 9 ) 1 1 n nn nn k k kkk k kkk k u k − −− −− + +  +−  +− = + + + −  = + + + + −  (iii) 10 years 10(12) 120n = = ( ) ( ) 119 119 100(900 9 ) 100 1001 (900 9 ) 1 1 1000000 kkk kk +  + + + + −   Using GC graph, 2.9533231k  minimum value of k is 2.96 . (3 s.f.) (iv) Adnan should not take up the offer. An interest of 2.96% per month is already very attractive, which makes the monthly 2.96% compounded interest too good to be true. To invest an amount of $1000 12 10 $120000  = and turn it into over a million dollars in 10 years is rather incredulous. This is likely a scam. 6(a) 2 2 d d d dd d y y ppxx x=  = 2 2 dd 1 1dd d 1 1 ...(1)d yy xxx p pxx  − = −= Integrating factor: d1 1e xx x − = k 1000 000 ($) 2.9533231

Page 6 of 12 s/n Solution Multiply (1) throughout by 1 x : 2 d1 1 1 d d1 ...(2)d p px x x x p x x x −=  = Integrating (2) w.r.t. x : ln (since 0) ln p x C xx p x x Cx = +  =+ d ln ...(3)d y x x Cxx =+ Integrating (3) w.r.t. x : 22 2 2 2 2 2 ln d ln d2 2 2 ln2 4 2 ln2 y x x Cx x x x Cxxx x x CxxE x x Dx E =+ = − + = − + + = + +   (b) 2 2 d d d d d d d d d d d d y y p p y pppx x y x y x  =  = = =   32 3 2 d d d ee ddd yyy y p ppxyx  =  = 2 shown)   (*)d1 e(d yp yp = Integrating (*): 2 1 d e d 1 e 1 e d 1 d e y y y y pyp Cp p C y x C = − = + =− +  =− +  When 0x= , 0y= and d 1d y x =− . 0 110 e CC − =−  =+ d 1 d e y y x =−

Page 7 of 12 s/n Solution e d 1d e ln y y yx xD y x D =− =− + = − +  When 0x= , 0y= . 0 ln 0 1 DD = +  = ln 1yx = − Hence ln (1 )yx=− (since 01 x ) 7(a) Let dm be the difference between the rating before the HBL and after the HBL 0H : 0dm = 1H : 0dm  Student A B C D E F G Before HBL 4 8 7