SCSS 2024 AM Prelim P1 MS

Preliminary Examinations 2024 Paper 1 Marking Scheme Level: Sec 4E/5N 1 Additional Mathematics (90 marks) Qn. # Solution Mark Allocation 1 2d 92d y x axx =+ 29 2 0x ax+ (9 2 ) 0x x a+ 2 9xa− or 0x M1 (Find d d y x ) M1 ( d 0d y x  ) A1 2 ( )( )5 2 4 2 7 2ab− + = + 5 20 2 2 8 7 2a a b+ − − = + (5 8) (20 ) 2 7 2a a b− + − = + 5 8 7a−= and 20 ab−= 3a= , 17b= M1 (expansion) M1 (compare coefficient) A2 3(a) 222 12 11 2( 6 ) 11x x x x+ + = + + 222 ( 3) 3 11x= + − + 22( 3) 7x= + − B1 (either 2( 3)x+ or 7− correct) B2 (all correct) 3(b) 22 12 11 11x x px+ + = + 22 (12 ) 0x p x+ − = 2(12 ) 4(2)(0) 0p− −  2(12 ) 0p− 12p M1 (sim eqn) M1 (Find discriminant) A1 4(a) Let ABC x= . AB AC= (tangents from external point) ACB ABC x = = (base angles of isosceles triangle) CEB ACB x = = (alternate segment theorem) 180CED ACB = − (adj. angles on straight line) 180 x=− (180 )ABC CED x x + = + − 180= M1 ( ACB ABC = ) M1 ( CEB ACB = ) Note: If first M1 not awarded, maximum 2 out of 3 marks A1 4(b) Suppose there exists a circle that passes through A, B, E and C. 180BAC ABC ACB = − − (sum of angles of triangle) 180 2 x=− 180BAC CEB = − (opp angles of cyclic quad) 180 x=− M1 (opp angles of cyclic quad)

Preliminary Examinations 2024 Paper 1 Marking Scheme Level: Sec 4E/5N 2 Qn. # Solution Mark Allocation For 0x , 180 2 180xx−  − Hence, there is no circle that passes through A, B, E and C. A1 (contradiction) 5(a) 5log 2 3log 5 xx+= 5 5 3log 2 logx x+= Let 5logux= 32u u+= 2 2 3 0uu+ − = ( 1)( 3) 0uu− + = 1u= or 3u=− 5log 1x= or 5log 3x=− 5x= or 35x −= 1 125x= M1 (change of base) M1 (form quad eqn) M1 (solve quad eqn) A2 5(b) B1 (shape) B1 (x-int and y-axis asymptote) 6(a) Least value 1= Greatest value 7= B1 B1 6(b) Period 4= or 720 B1 6(c) B1 (shape + correct number of cycles) B1 (coordinates of start/end point + max/min points)

Preliminary Examinations 2024 Paper 1 Marking Scheme Level: Sec 4E/5N 3 Qn. # Solution Mark Allocation 6(d) sin 24 xx   =− 33sin 24 x x  =− 33sin 4 424 x x  + =− + 3 44yx =− + After drawing line: Number of solutions 3= M1 (find eqn of line) A1 (draw line + number of solutions) 7 2d 3 cos 2 dd xy e x xx −=+ 231 sin 222 xe x c−=− + + Sub 0x= , d 5d y x = 35 2 c=− + 13 2c= 2d 3 1 13 sin 2d 2 2 2 xy exx −=− + + 23 1 13 sin 2 d2 2 2 xy e x x −= − + + 2 1 3 1 13 cos 24 4 2 xe x x c−= − + + Sub (0, 3) 1 313 44 c= − + 1 5 2c = 23 1 13 5 cos 24 4 2 2 xy e x x −= − + + M1 (integrate 23 xe− ) M1 (integrate cos 2x ) M1 (find c) M1 (integrate 23 2 xe−− ) M1 (integrate 1 sin 22 x ) M1 (integrate 13 2 ) A1 8(a) 1 2 2(3 ) r nr r nTx r x − +  =−  23 ( 2) ( )n r n r r rn xxr − − −=− 33 ( 2)n r r n rn xr −−=− M1 (general term) M1 (simplification)

Preliminary Examinations 2024 Paper 1 Marking Scheme Level: Sec 4E/5N 4 Qn. # Solution Mark Allocation 30nr−= 3nr= where r is a positive integer Thus n is a multiple of 3. A1 (explanation) 8(b) Term independent of x: 93 r= 3r = 63 4 9 3 ( 2)3T =− 489888=− For 6 1 x term: 9 3 6r− =− 5r = 4 5 6 6 9 3 ( 2)5Tx −=− 6 326592 x=− 6 coefficient of term independent of 1coefficient of x x 489888 326592 −= − 3 2= B1 (Obtain 489888− ) M1 (Find r for 6 1 x term) M1 (Find 6 1 x term) A1 9(a) 2 22 9 4 8 ( 2)( 1) 2 ( 1) 1 x x A B C x x x x x −+ = + +− + − + + 229 4 8 ( 1) ( 2) ( 2)( 1)x x A x B x C x x− + = + + − + − + Sub 1x=− 29( 1) 4( 1) 8 ( 1 2) B− − − + = − − 7B=− Sub 2x= 229(2) 4(2) 8 (2 1) A− + = + 4A= Sub 0x= 229(0) 4(0) 8 4(1) 7( 2) ( 2)(1) C− + = − − + − 5C = 2 22 9 4 8 4 7 5 ( 2)( 1) 2 ( 1) 1 xx x x x x x −+ = − +− + − + + M1 (form 3 fractions) M1 (form identity) M2 (A, B, C correct) M1 (1 of 3 constants correct) A1 9(b) 2 22 9 4 8 4 7 5 d d( 2)( 1) 2 ( 1) 1 xx xxx x x x x −+ = − +− + − + + 74ln( 2) 5ln( 1) 1x x c x= − + + + ++ B3 (B1 for each term) Note: Subtract 1 mark if there is no “+ c”

Preliminary Examinations 2024 Paper 1 Marking Scheme Level: Sec 4E/5N 5 Qn. # Solution Mark Allocation 10(a) d d sv t= 213 2 t ve=− + 2103 2 t e=− + 26 t e= ln 6 2 t= 2ln 6t = M1 (find v) M1 (v = 0) A1 10(b) At 0t= , 1s= At 1t = , 1.35s=− (3sf) Since displacement changes from positive to negative, the particle passes through 0s= some time between 0t= and 1t = . Hence particle passes through O in first second. M1 (both values of s) A1 (explanation) 10(c) At 2ln 6t = , 4.7506s=− At 4t= , 4.6109s=− Total distance (1 4.7506) (4.7506 4.6109)= + + − 5.89= cm (3sf) M1 (both values of s) M1 (sum of distances) A1 11(a) Refer to attached graph B1 (table of values) B1 (plot points) B1 (draw line) 11(b) Using points (0, 4.17) and (2, 3.78), Gradient 4.17 3.78 02 −= − 0.195=− (accept 0.225− to 0.165− ) 15ktC Ae −=+ ln( 15) lnC A kt− = − 0.195k = (3 s.f.) (accept 0.165 to 0.225 ) ln 4.17A= (accept 4.14 to 4.2) 64.7A= (3 s.f.) (accept 62.8 to 66.7) 0.19564.7 15 tCe −=+ OR ln( 15) 0.195 4.17Ct− =− + 0.195 4.1715 tCe −+−= 0.195 4.1715 tC e e −− =  0.19564.7 15 tCe −=+ B1 (Gradient) M1 (Form linear eqn) A1 (Find A) A1 M1 (remove ln) A2 (A1 to find A, A1 for eqn) 11(c) 0.19564.7155 15 35te− + M1 (accept 35= )

Preliminary Examinations 2024 Paper 1 Marking Scheme Level: Sec 4E/5N 6 Qn. # Solution Mark Allocation 0.195 20 64.7155 te−  200.195 ln 64.7155t −  6.02t  Year 2030 M1 (apply ln) A1 (Year) 12(a) Let 1,12B x x + ( ) 2 22 1( 2) 1 5 5 2xx + + + = 22 14 4 1 1254x x x x+ + + + + = 25 5 120 04 xx+ − = 2 4 96 0xx+ − = ( 8)( 12) 0xx− + = 8x= or 12x=− (rej) 5y= (8,5)B M1 (form eqn using length) M1 (simplification) M1 (solve quad eqn) A1 12(b) Gradient of 75 78BC −= − 2=− 1Gradient of Gradient of 2 2AB BC = − 1=− Therefore 90ABC= Since ABCD is a parallelogram with int angle 90= , ABCD is a rectangle. M1 (Gradient of BC) M1 (Show right angle) A1 (explanation) 12(c) Length 22(8 7) (5 7)BC= − + − 5= units Area of 5 5 5ABCD= 25= units2 M1 (Find BC) A1 13(a) 4d 6( 3)(2 5) (2)d y xx −= − − ( ) 4 36 25x =− − For 2.5x , since numerator of d 0d y x  , d 0d y x  Therefore there are no stationary points. M1 ( d d y x without 2 ) M2 (correct d d y x ) A1 (with explanation)

Preliminary Examinations 2024 Paper 1 Marking Scheme Level: Sec 4E/5N 7 Qn. # Solution Mark Allocation 13(b) At 1x= , 2 9y=− At 1x= , d4 d9 y x =− Gradient of normal 9 4= Eqn of normal: 29 ( 1)94yx+ = − 9 89 4 36yx=− 36 81 89yx=− Points of intersection: 2 90 78 81 89x x x+ − = − 2 9 11 0xx+ + = 29 9 4(1)(11) 2(1)x −  −= 9 37 22=−  Difference between x-coordinates 9 37 9 37 2 2 2 2 =− + − − − 37= B1 (y – coordinate) M1 (gradient of