2020 A Levels H3 Math Paper 1 Soln

Solutions to 2020 GCE ‘A’ Level H3 Mathematics (9820/01) 1 (i) For any positive integer n and positive numbers x and y, prove that ( )( ) 11 n nnn x y n x y −− +  . [3] (ii) Hence, for any positive numbers a, b and c such that 1abc= , prove that ( ) ( ) ( ) 2 3 4 1 1 1 256a b c+ + +  . [7] 1 (i) By AM-GM inequality applied to the n-sequence 1 , , , , n x x x y    , ( )( ) 1 11 1 1 Since both sides are positive, raising both sides to the power of gives 1 n nn n nn x x x y x x x yn n n x y n x y −− −    + + + +           − +    Equality is achieved when xy= . (ii) We have for and , 0n x y +  , ( )( ) 11 n nnn x y n x y −− +  ---- (*) Let , , 0abc  such that 1abc= . Put 2, 1,n x y a= = = into (*) gives ( ) 2 14 aa+ with equality if 1ax== . Put 13, , 2n x y b= = = into (*) gives ( ) 3 271 4bb+ with equality if 1 2bx== . Put 14, , 3n x y c= = = into (*) gives ( ) 4 2561 27cc+ with equality if 1 3cx== . Since 1 1 1112 3 6 abc  =  = , at least one of the above inequalities is strict. Hence multiplying gives ( ) ( ) ( ) 2 3 4 27 2561 1 1 4 256 4 27a b c a b c   + + +  =      as desired.

2 2 For constants 0a and b, the function f is defined by ( ) 1f , bxx ax b a= −+ . (i) Find ( ) 1f x− . [3] (ii) It is given that 3f has a fixed point, i.e. a number p such that ( ) 3f pp= . Prove that either p is a fixed point of f or 3f fixes all values of x for which ( ) 3f x exists. [7] (iii) Hence write down two examples of recurrence relations of the form 11 1n n nAx x Bx+++= which, for almost all initial values 1x , generate periodic sequences with periods 2 and 3, respectively. In each case, state any initial values which have to be excluded. [4] 2 ( ) 1f , bxx ax b a= −+ . (i) 1 1 1 1 ,0y ax b x b yax b y a y =  + =  = −  +  ( ) 1 1 1 1f , 0 byy x b y a y ay−  −= = − =  . Replacing dummy y by x, ( ) 1 1f , 0 bxxx ax− −= . (ii) Given that for some p  , ( ) 3f pp= . Clearly ( ) ( ) 2,f ,f bp p p a− so that ( ) 3f p exists. Moreover, ( ) ( )( ) ( ) 32 2 1f f f 0 fp p p a p b= = =  + . Thus ( ) ( ) ( ) 3 2 1f f fp p p p −=  = ( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 3 2 2 2 11 1 1 11 bp apab ap b ap ap b bp a abp b a p abp a abp b abp ab p b p ap a b b bp a bp −=  + +  + = − + +  + = + + − − −  + = − + −

3 2020 GCE ‘A’ Level H3 Mathematics 9820/01 ( ) ( )( ) ( )( ) 2 2 2 22 22 1 10 1 or 1 0 ap a b bp a b a b bp ap a b bp ap p ap b  + = − +  + − − =  =− − − =  = + Case 1: Since ( ) 1f p ap b= + , ( )1 fp p pap b =  =+ . That is, p is a fixed point of f. Case 2: ( ) ( ) 2 2 11f 1a b x b x b b bx=−  = = − + − , 1x b . ( ) ( )( ) ( ) ( ) 2 2 1 1 1f f f f 1 11 1 bxxx b bx b x bb b bx −= = = =  −   −  −  . ( ) ( )( ) 32 2 2 11f f f f 11bxx x x bx bxbb bx−= = = =   −  −   . Since x was arbitrary, 3f fixes all 1x b for which ( ) 3f x exists. (iii) ( )1 1 1 1 11 1 , n n n n n nn n Ax x Bx x Ax B Bxx Ax B A + + + + + =  + =  = − + Letting ( )1 fnnxx+ = , we have ( ) 1f n n x Ax B= + , n Bx A− . Put A = 1, B = 0 into above gives ( )1 1f , 0n n n n x x x x+ = =  for n = 1, 2, …. 2 1 11 1nn n n xx x x + + = = = for 1,2,n=  . That is, 1 3 5x x x= = =  which implies that the sequence is periodic with period 2 where 1 0x  . Also, 1 2 31 1 1x x x=  =  =  which is a sequence with period 1. So 1 1x  . So the recurrence relation 11 1, 0,1nnx x x+ = generates a periodic sequence with period 2.

4 From (ii), ( ) 23 1f, n n nA B x x x B=−  =  . Choose 1, 1AB=− = so that we obtain the recurrence relation 1 1 1 1, 1n n nx x x x++− + =  . 1 1 1 1 1 2 11 1 1 11 1 1 1 1 1 1 1 n n n n n n n n x x x x x x x x + + + − − − − + =  = − = − − =− =− − 2 nx −= for n = 3, 4, …. and 1 1x  . Also, if 1 0x = , then 2 2 3 3 3 31 1 1 0 1x x x x x x= − + = − + =  = , a contradiction. So 1 0x  . So the recurrence relation 1 1 1 1, 0,1n n nx x x x++− + =  generates a periodic sequence with period 3. 3 For any non-negative integer n, the function Pn is defined by ( ) 0 P ! n i n i tt i== . (i) Use mathematical induction to prove that ( )( ) 0 e d ! 1 e P t n x t nx x n t−− =− . [7] (ii) State the value of 0 ednxxx  −  and briefly justify your answer. [2] (iii) For 0nt , prove that ( )1 P 1 nn n tt tnn −    +   −       . [6]

5 2020 GCE ‘A’ Level H3 Mathematics 9820/01 3 ( ) 0 P ! n i n i tt i== where 0n . (i) To prove ( )( ) 0 e d ! 1 e P t n x t nx x n t−− =− for 0n . For n = 0, LHS = 0 000 e d e d e 1 e tt tx x x tx x x− − − − = =− = −  . RHS = ( )( ) 0 0 0 0! 1 e P 1 e 1 e ! i t t t i tt i− − − =− = − = −  = LHS. So the statement holds for n = 0. Assume the statement holds for n = k for some 0k  . That is, ( )( ) 0 e d ! 1 e P t k x t kx x k t−− =− for some 0k  . For n = k + 1, using integrating by parts with 1 d,edkx vux x+−== gives ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 11 000 1 1 1 0 e d e 1 e d e 1 ! 1 e P by (IH) e 1 ! 1 e P 1! 1 ! 1 e 1 ! ! tt tk x k x k x k t t k kt t k kki t i x x x k x x t k k t tkt k ttk ki + − + − − + − − +− − + − = =− + + =− + + − = + − + − +   = + − +  +   ( ) ( ) 0 1 0 since P ! 1 ! 1 e ! k i k i k i t i tt i tk i = + − =  =  = + −     which implies the statement also holds for n = k + 1 and so the statement holds for all 0n by induction. (ii) ( )  00 0 e d lim e d !lim 1 e P e !lim 1 ! ! t n x n x t t nt n it t i x x x x nt tn i n  −− → − → − → = = =− =−   =   since for each 0 in and a given finite n, e 0 as ittt − → → and therefore 0 elim 0 ! n it t i t i − → = = . (iii) For 0nt ,

6 ( ) ( ) 0 0 0 0 1 11 ! 11 ! 11 since 1! ni n i in i n i i n i i ntt inn n n n i t in n n n i t n n n i t n n n i i n n n = = = =    +=         − − + =  − − +    =          − − +                     with equality being achieved when n = 1. Since 0,0 1 tnt n    . Therefore we have ( ) ( ) ( )( ) ( ) 2 2 2 2 111 2! 1 1 2! 1 2! 1 for a given finite and since 02! ! P n n n nnt t t nn n n n n tt nn tt ttt n t n t − − − +     − = + − − + − +           +  = + + +      + + +  + + + +  =     Putting both inequalities together, we have for 0nt , ( )1 P 1 nn n tt tnn −    +   −       .

7 2020 GCE ‘A’ Level H3 Mathematics 9820/01 4 Consider integer solutions of the equation 39 23 65 1872x y z+ + = . (i) Given that y is prime, prove that 13y= . [2] (ii) Given that 13y= , (a) show that ( )2 modulo 5x and ( )2 modulo 3z , (b) hence find the solution for which zx− is minimal. [3] [2] (iii) Find all solutions such that each of x, y and z is prime. [4] 4 39 23 65 1872x y z+ + = (i) ( ) 39 23 65 1872 23 13 3 48 5 13 where x y z y x z k k + + + = = − − =   That is, 23y is a multiple of 13. Since 23 and 13 are coprime, y must be a multiple of 13. But since y is prime, 13y= . (ii) ( )39 23 13 65 1872 39 65 1573x z x z+ + =  + = ---- (*) (a) Modulo 5, 39 4xx , 65 0z , 1573 3 . So (*) implies ( )4 3 8 4 8 8 8 4 2 0x x x   −  −  −  (modulo 5).