NASS 2025 AMATH PRELIM P2 MS

1 Ngee Ann Secondary School Secondary 4 Additional Mathematics-O 2025 Preliminary Examination Paper 2 Marking Scheme Qn No. Qn Part Solutions Marks (Remarks) Total 1 (a) 1925 108 243 2 64 35 36 3 81 3 2 3 03 93 43 25 3 −+ ×= ×− ×+ = −+ = M1: At least 2 correct A1 2 (b) 2 522 1 5 21 54 51 5 4 51 51 51 5 35 4 51 31 544 p p + ++=− +− += + +−= × +− +−= − = + M1: Attempt to rationalise M1: Expansion of surds A1 3

2 Qn No. Qn Part Solutions Marks (Remarks) Total 2 (a) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 22 2 22 2 2 2 d cos 2 d 1 sin 2 2sin 2 1 sin 2 cos 2 2cos 2 1 sin 2 2sin 2 2sin 2 2cos 2 1 sin 2 2sin 2 2(sin 2 cos 2 ) 1 sin 2 2sin 2 2 1 sin 2 2 sin 2 1 1 sin 2 2 1 sin 2 x xx xx xx x xx x x x xx x x x x x x   + − +−= + −− −= + −− += + −−=− + −+= + −= + B1: numerator B1 for denominator M1: Realising 22sin 2 cos 2 1xx+= A1 AG 4 (b) ( ) π π 4 4 0 0 π π 4 4 0 0 π 4 0 2 cos 2 d1 sin 2 1 sin 2 1 2 1 cos 2 d2 8 1 sin 2 16 1 sin 2 1 101 d8 1 sin 2 16 2 1 1 16 xxxx xxxx xx − =++  − =− −× + + = −− +  = ∫ ∫ ∫ M1: Reverse statement M1: Multiply by 1 16− M1: Correct use of limits A1 4

3 Qn No. Qn Part Solutions Marks (Remarks) Total 3 (a) ( ) 2 3212xx x x x+=+ + 2 32 2x xx++ 322 12 14 5xxx+ ++ ( ) 32242xxx− ++ 28 12 5xx++ ( ) 32 2 2 12 14 5 1 xxx xx + ++ + = ( ) 2 2 8 12 52 1 xx xx +++ + B1 1 (b) ( ) ( ) ( ) ( ) 2 22 22 8 12 5 111 8 12 5 1 1 x x AB C xxxx x x x A x Bx x Cx ++ = ++ +++ + += + + ++ When 0x= , 5A= . When 1x=− , 1C =− . When 1x= , 26B= B = 3 ( ) ( ) 32 22 2 12 14 5 5 3 1 2 111 xxx xxxx x + ++ =++ − +++ M1: Correct form of partial fractions M1: Remove denominator A1: V alue of A A1: Value of C A1: Value of B 5

4 Qn No. Qn Part Solutions Marks (Remarks) Total 4 (a) ( ) 2 2 10 2 6 10 6 0 x nx m m x x nx m −+ =− +− += Since line is tangent to curve, ( ) ( ) ( ) ( ) 2 2 2 6 4 10 0 6 40 Since 6 0, 40 0, 0 nm n m n mm −− = −= −≥ ≥ ≥ Hence m cannot be negative. M1: Eliminate x or y M1: Use of discriminant M1 A1 4 (bi) ( ) ( )( ) 2 2 2 61 0 6 42 1 0 28 8 0 3.5 xx k k k k − ++ > −− +< −< > M1: Use of discriminant A1 2 (bii) ( ) ( ) ( )( ) 2 2 2 2 2 10 Discriminant 0 14 0 3 2 10 3 2 10 31 10 kx k x k kk kk kk kk + + +< < +− < − + +< − −> + −> 1 3k <− or 1k > Since 0k < , 1 3k <− . M1: Use of discriminant M1: Factorise quadratic A1 3 5 (a) 2 bc at t c