TMJC 2026 H3 Math Prelim (Ans)

TMJC/2026 JC2 Preliminary Examination Marking Scheme/H3 Math (9820/01)/Math Dept Page 1 of 11 2026 H3 MATH (9820) JC 2 PRELIMINARY EXAMINATION – MARKING SCHEME Qn Solution Mark Scheme 1 Number Theory [10] (a) Suppose n is a composite number. Let ,n xy= where x, y are integers greater than 1. ( ) ( ) ( ) 12 2 1 2 1 (2 1) 2 2 1 yxy x yyx x x −− − = − = − + + +  Since 1x and 1y , (2 1)x− and ( ) ( ) 12 2 2 1 yyxx −− + + +  are integers more than 1 Therefore, 2 1 2 1n xy− = − is a composite number which contradicts with 21n − being a prime number. Hence, n is a prime number. M1 Let n xy= B1 Factorise 2𝑥𝑦 − 1 A1 factors are integers more than 1, therefore leading to a contradiction (b) For any integer 1n , since 2n is an even integer that is not prime, using the contrapositive statement from part (a), 221n − is not a prime number. 2 12 12 2 1 4 1 (4 1) 4 4 1 3 4 4 1 nn nn nn −− −− − = − = − + + + = + + +   For any integer 1n , 124 4 1nn−−+ + +  is an integer. Therefore, 221n − is divisible by 3. B1 (using contrapositive of statement in (a)) M1 3[4𝑛−1 + 4𝑛−2 + ⋯ + 1] A1 (state 124 4 1nn−−+ + +  is an integer and prove result) (c) Let p and q be two consecutive prime numbers, When 1n , 221n − is an odd number. Suppose 221npq+ = − , Then pq+ must be odd. Therefore, one of the prime numbers must be odd and the other must be even. Therefore, 2p= and 3q= . When 22 3 2 1 n+ = − 262 n= However, since 6 is not a power of 2, there is no integer solution for this equation. Therefore, 221n − cannot be expressed as a sum of two consecutive prime numbers. (shown) B1 Sum of prime numbers is odd B1 One of the prime numbers must be 2 M1 22 3 2 1 n+ = − A1 Explain that there are no integer solution

TMJC/2026 JC2 Preliminary Examination Marking Scheme/H3 Math (9820/01)/Math Dept Page 2 of 11 Qn Solution Mark Scheme 2 Integration, Inequalities [13] (a) 2 2 22 0 cos 1 cos 10 for 1 x x xxx     2211 1 11d lim d 1lim 1lim 1 1 n n n n n xxxx x n  → → → = =−  = − + =  Since 21 1 dxx converges, by Comparison Test, 2 21 cos dx xx converges. M1 – appropriate comparison B1 - Integration M1 – convergence A1 – answer and reason (b) 2201 21 1 1 11 1 e1 d d , where ee 3 3 1lim d 3 1lim tan 33 11lim tan tan 3 3 3 1 πlim tan 633 1 ππ 263 π 33 x x x n n n n n n x u u u uu u n n  → − → −− → − → ==++ = +  =       =−          =−   =−  =   B1 – correct substitution M1 – integration M1 – taking limits A1 (c)(i) Using Cauchy Schwarz Inequality ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 since 02 xyz y z x z x yy z x z x y x y z y z x z x y y z x z x y x y z xyz x y z x y zy z x z x y x y z x y z x y zy z x z x y  + + + + + + + +++   + + + + + + + + = + +  + + + +  + +  +++  ++ + +  + +  +++ M1 – splitting using Cauchy Schwarz A1 – leading to show form

TMJC/2026 JC2 Preliminary Examination Marking Scheme/H3 Math (9820/01)/Math Dept Page 3 of 11 (c)(ii) 2 2 2 3 2 3 23 3 using AM-GM inequality2 x y z x y z y z x z x y x y z xyz +++ + +++ ++  For equality, x y z== 22 313 1 1 2 xy xyy x x y+ + =+++ Since 1, 1z x y z= = = = This also satisfies criteria for equality for Cauchy Schwarz inequality: 1 2 xz y y z x y xz y z x z x y             ++ +     = = = +++ M1 – AMGM M1 – condition for equality A1

TMJC/2026 JC2 Preliminary Examination Marking Scheme/H3 Math (9820/01)/Math Dept Page 4 of 11 Qn Solution Mark Scheme 3 Differential Equations [12] (a) dd dd yvy vx x v xx=  = + ( ) dd dd d Fd Fd d vyxv xx vyxv xx vvv xx =−  = −  −= B1 d𝑦 d𝑥 = 𝑥 d𝑣d𝑥 + 𝑣 B1 (must lead to result) (b) 22 222 2 d d d 1d y x yxy xx y y x y y y x x x x x +=+ + = + = + +  Therefore, 2F( ) 1v v v= + + ( ) 2 2 1d d d1 d v v vv xx vv xx + + − = += 2 2 1 1 1 d 1 1d 11 dd1 tan ln tan ln v v x x vxvx v x C y xCx − − =+ =+ =+  =+  When 1x= , 0y= ( ) 1tan 0 ln1 0 C C − =+ = ( ) ( ) 1tan ln tan ln tan ln y xx y xx y x x −  = = = M1 (apply result from (a) and replace 𝑦 𝑥 with v) A1 M1 Variable separable A1 (integration) B1 A1 (c) Since ( )tan lny xx = , As π 4ex→ , π 4 πtan ln e tan 1 4yx → = = Therefore, 2 2d 1 1 1 1 3dy y yx x x = + + → + + = M1 (find limit of 𝑦 𝑥) A1 M1, A1

TMJC/2026 JC2 Preliminary Examination Marking Scheme/H3 Math (9820/01)/Math Dept Page 5 of 11 Qn Solution Mark Scheme 4 Complex Numbers [13] (a) AQ  is parallel to AC  So ()q a c a− = − where  is real. qa ca − =− which is real as required. Hence, ( ) ( ) ( )( ) ( )( ) * ** ** * * * ** ** * * * * q a q a c a c a qaq a q a ca caca q a q a ca ca c a q a c a q a −− =−−  −−− == − −− −− =− −  − − = − − Since ** 1,aa cc== ( ) ( ) ( ) * * * 1 1 1 Multiplying by , as 0 and so c a q q a a c a ac acq c q a a c q acq a c    − − = − −       − =− − −  + = + B1 M1 M1 AG M1 AG (b) (i) Consider 2 1 , where . n kk k z tw t = −  Since it is a sum of squared moduli, ( )f 0.t  Summing gives ( ) 222f   k k k k k kz t w z t z w t wt − − +=    Let 22,,k k k kA z B z w C w= = =   Then ( ) 2 .f  A tB t tt BC= − − + (1) Let ( )0.BtC C= Substituting in (1), we get 2 .f   CB BAC = − Since ( )f 0,t  2 2 0BA C AC B −  M1 M1 M1

TMJC/2026 JC2 Preliminary Examination Marking Scheme/H3 Math (9820/01)/Math Dept Page 6 of 11 2 22 1 1 1 2 22 1 1 1                                                                                    n n n k k k k k k k n n n k k k k k k k z w z w z w z w = = = = = =                           AG Equality holds when 312 1 2 3 ... n n zzzz w w w w= = = = (or kkzw = ) A1 (ii) Using Cauchy-Schwarz, ( )( ) 22 22 222 1 2 3 1 2 3 1 2 3 2 2 1 2 2 108 Largest possible value of 2 2 6 3. z z z z z z z z z + +  + + + + =  + + = M1 A1

TMJC/2026 JC2 Preliminary Examination Marking Scheme/H3 Math (9820/01)/Math Dept Page 7 of 11 Qn Solution Mark Scheme 5 Counting, Inclusion-Exclusion Principle, Pigeonhole Principle [14] (a)(i) In the table, the 10 rows denote the 10 flavours, while the 3 columns denote the 3 friends. A ‘1’ would indicate that the flavour is chosen for a friend while a ‘0’ indicates otherwise. Since no row is completely filled with ‘1’s, no flavour is chosen for all 3 friends at the same time. No row is completely filled with ‘0’s, so every flavour is chosen at least once. No column is completely filled with ‘0’s, so every bowl has at least one scoop of ice- cream. B1 (no row ‘1’) B1 (no row ‘0’) B1 (column) (ii) There are 32 2 6−= possible ways to fill up a row with ‘0’s and ‘1’s so that it has neither 3 ‘0’s nor 3 ‘1’s. Hence, there are 106 ways to fill up the table. If one column contains all ‘0’s, there are 103 ways to fill up the table. If two columns contain all ‘0’s, there is exactly 1 way to fill up the table. Hence, by the Principle of Inclusion and Exclusion, the number of ways to fill up the table such that no column is completely filled with ‘0’s is 10 10 336 3 1 6028903212   −  +  =       There are 90289032 ways in which the girl may buy the bowls of ice - cream. M1 ( 106 ) M1 ( 103 ) M1 (PIE) A1 (b)(i) Distribute 6 identical balls into 4 distinct boxes ,,A B C and D . (a) Boxes B and C each contain at least 1