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