NYJC TJC VJC 2024 H3 Math Prelim (Solutions)

Qn Suggested Solution Mark Scheme 1a Use Cauchy-Schwartz Inequality 2 22 1 1 1 n n n i i i i i i i a b a b = = =                Let iiax= and 1ib = for i = 1,2,3,….,n 2 2 2 2 1 1 1 1 1 n n n n i i i i i i i x x n x = = = =                       Divide both sides by n2, 2 2 11 nn ii ii xx nn ==                          OR Let iiax= and 1 ib n = for 1, 2,3,...,in= . M1 A1: [AG] 1b Let i i iax = and iib = for 1, 2,3,...,in= . 2 2 1 1 1 n n n i i i i i i i i xx   = = =                The necessary and sufficient condition for the equality to hold is 1 1 2 2 12 12 i.e. nn n n xxx .... x x .... x     = = = = = = M1 A1 B1 1c In (ii), let 1 i it = and iixy= for i = 1,2,3,….,n 2 2 1 1 1 11n n n i ii i i i i i yyt t t= = =                ( )( ) ( )( ) ( ) 11 1 1 11 1 11 1 11 (sum of geometric series)1 1 1 as 2 1 and 0 < 1 1 n n tt i i t n tt n tt t t, = − − − = − =−    −   Thus 2 2 11 1nn i iii ii yytt==   . OR ( )( ) ( ) ( ) 11 11 22 1 21 2 1 2 11 22 1 11 1 since 0 1 nn ii ii n n n tt==  − = = −−     Thus 2 2 11 1nn i iii ii yytt==   . M1: Replacement M1: Sum of GP A1: Show sum is less than 1, leading to answer [AG] [A1]: Show sum is less than 1, leading to answer [AG]

2a i Let , for some 0 1x x b b= +   . Then 1 1 1 1 1x x b x x b x+ = + +  + = − + = +           . ( )f ( 1)= 1 1x x x + + − + = +1 1xx −− = f ( )x x x=− Hence, f is periodic with a period of 1. B1 M1 A1: [AG] 2a ii B1: Shape of f B1: Open circles indicated B1: Intersections and endpoints indicated 2a iii ( )( ) 1.8 2 1 15 41 11 1Area d 0.8 0.84 4 2 2 1.04 0.32 0.72 x x x=− − + − =− =  M1 M1 A1 2b i 3log 2k =  32 log 3 k Hence 2333 k  k = 9, 10, …, 26. M1 A1 2b ii x1 = 2 + 1 = 3; x2 = 3; x3 = x4 = … = x7 = x8 = 4; x9 = 5. 0 1 2 3 3 1 for 1, 2 3 for 1, 2 1 for 3, 4,5 4 for 3, 4,5 1 1 for 6,7,8 4 for 6,7,8 5 for 9,10,111 for 9,10,11 n n xn n xn n x x x n n nxn   += =   += = = + = + = = =  =+=   Observe that 33 0 for 1, 2 3 for 1, 2 1 for 3, 4, ,8 4 for 3, 4, ,8log 3 log 2 for 9,10, , 26 5 for 9,10, , 26 nn nnnn nn ==  ===  + =        ==  Hence, 33 lognxn=+  . B1 B1 B1

3a ( ) ( ) ( ) ( ) 11 2 2 2 ! ! !! ! ! 2 ! 2 ! n nn nn n n n n n −−  = = =  . B1: Apply definition 3b π 1 2 21 0 24sin d 21 n n n nI x x nn − + == +  for all 0n + . For n = 0, LHS = π 2 0 sin d 1xx = . RHS = 10 04 102(0) 1 −  =+  whereby definition ( ) 0 0! 10 0! 0 0 !  == − . So statement holds for n = 0. Assume π 1 2 21 0 24sin d 21 k k k kI x x kk − + == +  for some 0k + . ππ 22 2 3 2 2 1 00 sin d sin sin dkk kI x x x x x ++ + = =  . Let ( ) 2 2 2 1 dsin 2 2 sin cosd kk uu x k x x x ++=  = + d sin cosd v x v xx =  =− ( ) ( ) ( ) ( ) ( ) ( ) ( ) π 2 23 1 0 ππ 22 2 2 1 2 2 0 0 π 2 2 1 2 0 1 1 1 1 sin d sin cos 2 2 sin cos d 2 2 sin 1 sin d 2 2 2 2 2 3 2 2 22 23 22 2 4 b 2 3 2 1 k k kk k kk kk kk k I x x x x k x x x k x x x k I k I k I k I kII k kk kkk + + ++ + + + + − = =− + + = + − = + − +  + = + += + +=  ++     ( ) 11 y (IH) 241 2 3 2 2 1 k kk kkk −+ += ++  ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 !!11 2 2 1 2 2 1 2 ! 2 2 1 ! 1 ! 2 1 2 2 ! k kkkk kk k k k k k kk − ++ =++  + + += ++ 1 22 1 k k − += + Hence 11 1 224 123 k k kI kk −+ + + =  ++  proving the statement by induction. B1: Base case M1: Split integrand M1: Integrate by parts M1: Ik+1 in terms of Ik M1: Use IH M1: Apply nCr formula and simplify A1: Final form

3c For 1x  , ( ) ( ) 1 π 22 1 2 1 2 1 000 π 212 00 π(*) 212 0 0 24 sin d 21 sin d sin d n n n n nn n n n n n x t t xnn x t t x t t − + + + ==  + =  + =  = +    =   =        (*) we have swapped the summation and integral symbols. ( ) ( ) π 2 2 0 π 2 220 sin d since sin 1 1 sin sin d 1 1 cos xt t x t xt xt t xt  =  −  = −−   M1: Use part (b) M1: Sum of GP A1: [AG] 3d ( ) ( ) π 0cos2 2 2 2 201 1 2 0 2 2 1 1 22 0 1 22 sin 1 dd 1 1 cos 1 1 11 d1 1 tan 11 1 tan 11 utxt t x u x t x u uxx ux x xu x xx x xx = − −         =− − − − −         =  − +  =   −−   =  −−    Let 1 22 tan tan 11 xxtt xx − =  = −− sin tx= 1sintx −= 1 1 21 2 0 24 sin 21 1 n n n n xxnn x − − + =  =+ − . M1: Substitution M1: Integrate A1: Justify 1 2 1 tan 1 sin x x x − −   − = to get answer [AG] 3e 1 1 21 2 0 24 sin 21 1 n n n n xxnn x − − + =  =+ − . Differentiate w.r.t. x the equation in (c) and LHS term by term for 1x  1 1 21 2 0 2d 4 d sin d 2 1 d 1 n n n n xxnx n x x − − + =   = + −   M2: Differentiate both sides (M1 for each side)

( ) ( ) ( ) ( ) 1 3 2 1 2 1 2 22 0 1 1 2 32 2 20 24 d 1 1 1 1 2 sin 2 1 d 2 11 2 1 sin4 for 1 1 1 n n n nn n n x x x xnnx xx n xxxxn x x − −+− = − − =   =  − − − + −−  = +  − −   Putting 1 2x= into above, 11 32 20 1 0 11sin2 11 224 1 2 1 114 4 2 42 π 3 93 n n n n n n n n −− = − =  =+ − −  = +   Hence by (a), ( ) ( ) 12 00 2! 42 π 2 ! 3 93nn nn nn − == = = + . M1: Substitute 1 2x= to get 1 0 2 n n n − =   A1

4a 3 3 3x y z (mod p) ( ) ( )( ) 33 22 | | p x y p x y x y xy − − + + Since 0p x y−  −  , p does not divide xy− and so ( ) 22|p x y xy++ . B1: p divides difference M1: Factorisation A1: Explain to show result [AG] 4b WLOG, by the result in (a), ( ) 22|p y z yz++ too. Hence, p divides ( ) ( ) ( )( ) 2 2 2 2 2 2x y xy y z yz x z xy yz x z x y z+ + − + + = − + − = − + + . Again, 0p x z−  −  , p does not divide xz− and so ( )|p x y z++ . B1: p divides 22y z yz++ B1: p divides difference to show result [AG] 4c By part (b), p divides x y z++ . Furthermore, 03 x y z p + +  and so, x y z p+ + = or 2x y z p+ + = . Case 1: x y z p+ + = Here, p divides ( ) ( ) 2 2 2 2 2x y z x y z xy xz yz+ + = + + + + + . By the result in (a) again, WLOG, we see that p divides 22x y xy++ , 22x z xz++ and 22y z yz++ . Hence, p divides the sum given by ( ) ( ) 2 2 222x y z xy xz yz+ + + + + . Therefore, x y z p+ + = divides ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 . x y z xy xz yz x y z xy xz yz x y z  + + + + + − + + + + +  = + + Case 2: 2x y z p+ + = Similarly, ( ) ( ) ( ) 2 2 2 2| | |p x y z p x y z p x y z+ +  + +  + + by the same argument given in Case 1. Since 2x y z p+ + = is even, ( ) ( ) 22 2 2 2x y z x y z xy xz yz+ + = + + − + + is even too. Now since ( ) 2 2 22| x y z++ , ( ) 2 2 2|p x y z++ , and ( )gcd 2, 1p = , 2x y z p+ + = must divide 2 2 2x y z++ . B1 M1: Attempt to show p divides a sum/diff involving 2 2 2x y z++ A1: Show p divides ( ) ( ) 2 2 22 2 x y z xy xz yz ++ + + + B1: Show p divides 2 2 2x y z++ B1: Explain that ( ) 2 2 22| x y z++ B1: Show 2p divides 2 2 2x y z++

5a Suppose that ( )0f1 x  fo