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 ct at b = + = + 2Draw against .ct t 2t 0.64 1 1.44 1.96 3.24 ct 16.8 15.0 13.0 10.5 3.8 See graph next page B1: Correct statement M1: Table of values B1: Plot points B1: Draw best fit line 4

5 (b) 2ct at b= + gradient 20 4 5 a= =− =− vertical intercept 20 b= = M1 A1 (Accept 5.13 4.88a− < <− ) B1 (Accept 19.5 20.5b<< ) 3 (c) 2When 0.5, 0.25.tt= = 2When 0.25, 18.75 0.5 18.75 37.5 30 t ct c c = = = = > Since the concentration is greater than 30 mg/L, the chemical is not safe for plants when 0.5h.t = M1 (Accept 18.25 19.25ct<< ) (36.5 38.5c<< ) A1 2 Graph for 5a 5 ct t2 1 2 3 4 15 10 20 18.75 0.25

6 Qn No. Qn Part Solutions Marks (Remarks) Total 6 (a) ( )2 log log 3 2 2wwy wx x= + −− For y to be defined, 0 0 wx x > > and 3 20 2 3 x x −> > Therefore, 2 3x> . B1: Final answer only 1 (b) ( ) ( ) ( ) ( ) 2 2 22 2 32 2log log 3 2 2 log log 3 2 log 32log log 3 2 ww ww w w w y wx x wx x w wx x w xx = + −− = + −−  −=   = −  M1: Use of power law M1: Realise 22 log w w= M1: Use of product and quotient laws A1: Manipulation to correct format 4 (c) ( ) ( ) ( ) 32 3 20 32 32 2 log 3 2 0 32 321 3 2 10 13 1 0 w xx x xw xx xx x xx −= −= −= − −= − ++= ( )( ) ( ) 2 2 3 10 1 1 43 1 23 1 11 (no real solutions)6 xx x ++= −± −= −± −= or 1x= (only one real root) M1 M1: Factorisation M1: Using formula or discriminant A1 A1: For value of x 5

7 Qn No. Qn Part Solutions Marks (Remarks) Total 7 (a) ( )( )( ) 32 32 3 13 15 9 9 3 22 24 0 6 4 10 xx x x xx x xxx + − −=+ + − −= +−+ = 6, 4 or 1xx x= −= = − ( )6, 45A −− and ( )1, 0B − M1: Eliminate x or y M1: Factorise A1, A1 4 (b) Shaded region ( ) ( ) ( ) ( ) ( ) ( ) 1 32 6 3 32 0 1 32 6 3 32 0 14 32 6 342 3 0 3 13 15 9 9 3 13 15 3 22 24 3 13 15 11 244 13 1542 1 1 11 24 324 216 396 1444 81 11727 45 042 x x x x dx x x x dx x x x dx x x x dx x xx x xx xx − − − − − − = +−−−+ − +−− = +−− − +−− = +− − − +− −  = −− + − − − + − +− − − ∫ ∫ ∫ ∫ 2212.5 units  = M1 M1 M1: Correct integral M1: Correct integral M1: Substituting limits A1 6

8 Qn No. Qn Part Solutions Marks (Remarks) Total 8 (a) (alt. s)ECD AFE θ∠= ∠= ∠ sin 6sin6 cos 4cos4 sin 4sin4 AE AE CD CD ED ED θθ θθ θθ = ⇒= =⇒= = ⇒= 4 6sin 4cos 4sin 6sin 4 16sin 4cos P P θθθθ θθ = ++++ = ++ M1 A1 2 (b) 2216 4 4 17R= += 4tan 16 14.036 14.0 α α = = °≈ ° ( )4 4 17 sin 14.036P θ= + +° B1 M1 A1 3 (c) Max P = 4 4 17+ ( )sin 14.036 1 14.036 90 75.964 76.0 θ θ θ + °= + °= ° = °≈ ° B1 M1 A1 3 (d) ( ) ( ) 4 4 17 sin 14.036 15 sin 14.036 0.66697 14.036 41.834 27.8 θ θ θ θ + + °= + °= + °= ° = ° M1 A1 2 (e) Since the maximum P = 4 4 17 20.492+≈ , the perimeter of ABCE could be 20 cm. B1 1 θ

9 Qn No. Qn Part Solutions Marks (Remarks) Total 9 (a) ( ) 2 32 5 xexy += ( ) 22 22 d3 2 32d5 5 76 55 xx xx y e exx e xe = ++ = + At stationary points, ( ) 22 2 76 055 76 0 xx x e xe ex += += 2 0 (N.A) orxe = 7 6x=− M1, M1 M1 A1, A1 5 (b) 2 22 2 2 22 d 14 6 12 d 555 12 20 5 xx x xx y e e xex xe e = ++ += When 7 6x=− , 77 33 2 2 712 20d 6 d5 0.116 0 eey x −−−+= = > The stationary point is a minimum point. M2 M1 A1 4 (c) ( ) 2d1 76d5 xy exx = + When 1x>− , 21 05 xe > , 76 0 x+> and ( ) 21 76 05 xex +> . Since d 0d y x > , the gradient of the curve is positive when 1x>− . B1 B1 2

10 Qn No. Qn Part Solutions Marks (Remarks) Total 10 (a) Gradient of DE 37 1 44 2 −+= =−−− Gradient of normal to DE = 2 ( ) 44 37,22 0, 5 Midpoint of DE −+ −−  = = − Equation of normal: 25yx= − ( )4 2 5 3 15 55 1 25 3 xx x x y −=− = = =−= − Centre = ( )1, 3− Radius = ( ) ( ) 22 1 4 3 3 5 units+ +−+ = Equation of circle: ( ) ( ) ( ) ( ) 22 2 22 1 35 1 3 25 xy xy −++ = −++ = M1 M1 M1 M1: Eliminate x or y B1 B1 A1 7 (b) Distance between ( )1,1− and ( )1, 3− ( ) ( ) 22 11 31 4.47 5 = + +−− = < Since the distance between ( )1,1− and ( )1, 3− is lesser than the radius of the circle, the point ( )1,1− lies inside the circle. M1 A1 2 (c) Solving simultaneously 4 3 15yx= − and 3yx= − : ( )4 3 3 15 3 6 xx x y −