SCSS 2023 4NA AM Prelim P1 MS

Preliminary Examination 2023 Solution Level: Sec 4G2 Additional Mathematics Paper 1 (70 marks) Qn. # Solution Mark Allocation 1 ( ) ( ) ( ) ( ) ( )( ) ( )( ) 3 3 3 3 2 2 2 54 128 2 27 64 2 3 4 2 3 4 3 3 4 4 2 3 4 9 12 16 x x x x x x x x x + =+ =+  = + − +  = + − + B1: ( ) 3327 3xx= B1: 364 4= B1: All correct 2i 2 4 1 3 0x kx k+ + − = For two real and distinct roots, discriminant 0 ( ) ( )( ) ( ) 2 2 2 4 4 1 1 3 0 16 4 12 0 4 4 3 1 0 kk kk kk − −  − +  + −  24 3 1 0kk+ −  (shown) M1 A1 2ii 24 3 1 0kk+ −  ( )( )4 1 1 0kk− +  11 or 4kk−  M1 A1 3i Period 2 1 2 = 4= Amplitude = 4 B1 B1

Preliminary Examination 2023 Solution Level: Sec 4G2 Qn. # Solution Mark Allocation 3ii B1: Correct cycle B1: Correct ( , 1) − B1: Correct maximum and minimum points 4i ( ) ( ) 2 2 3 d 23d 2 3 d 2 6 3 y xx y x x x x c =+ =+ = + +  At (1, 12), ( ) ( ) 3212 1 6 13 15 3 c c = + + = 321 6533y x x = + + M1: Attempt to integrate A1 M1: Substitution to obtain arbitrary constant A1 4ii 2d 26d y xx =+ Since 222 0, 2 0, 2 6 0,xxx  +  d 0d y x  , therefore this curve has no stationary point. B1: Show d 0d y x  4iii 2d 26d y xx =+ Since 222 0, 2 0, 2 6 0,xxx  +  d 0d y x  , therefore P( )yx= is an increasing function. B1: Show d 0d y x 

Preliminary Examination 2023 Solution Level: Sec 4G2 Qn. # Solution Mark Allocation 5a sin 30 cos 45 cos30 sin 45 1 2 3 2 2 2 2 2 26 4  +       =+                += M1 A1 5b 2cos 2 32x  += Reference angle 1 2cos 2 − =  4 = 2 , 2 , 2 , 2 23 4 4 4 4x         + = − + − +  17 23 412 (N.A.), , , (N.A.)12 12 12 12x    =− 17 23, 24 24x = M1 M1 A2 6i ( ) ( ) ( ) ( ) 32 32 42 42 d 1 d 1dd 1 3 1 ( 2 ) 6 1 xxx x xx x x − −   =− − =− − − = − M1 A1 6ii ( ) ( ) 2 2 430 22 0 61 d 11 x x xx  =−−   ( ) ( ) 2 2 430 22 0 16 d 11 x x xx  =−−   ( ) ( ) 2 2 430 22 0 11 d 611 x x xx  = −−   ( ) ( ) 3322 1 1 1 6 1 2 1 0  =−  −− M1 M1 M1: Substitute limits

Preliminary Examination 2023 Solution Level: Sec 4G2 Qn. # Solution Mark Allocation 11 1 6 27 = − − 14 81=− A1 7i 2.7 0.1 1.32a −== 2 12 6b == 0.1 2.7 1.42c +== B1 M1, A1 B1 7ii 1.3sin 1.46yt =+  1.3sin 16 1.46y =  +  2.53= m M1 A1 7iii 06 t B1 8a ( )( ) ( ) ( ) 2 2 2 2 3 2 3 2 5 3 1 5 3 1 5 3 1 5 3 1 3 2 5 3 1 5 3 1 15 3 10 3 2 5 3 1 17 11 3 74 + + += − − + ++ = − + + += − += M1 B1: Correct numerator B1: Correct denominator A1 8b ( )7 2 1xx+ = − ( ) 7 1 2 7 7 1 1 2 7 1 2 7 71 1 2 7 7 1 7 1 7 1 xx x x + = − + = − −= + −−= +− M1: Rearrange to make x the subject M1: Attempt to rationalise

Preliminary Examination 2023 Solution Level: Sec 4G2 Qn. # Solution Mark Allocation ( ) 2 2 7 1 14 2 7 71 15 3 7 6 − − += − −+= 51 7 22=− + M1: Numerator or denominator correct A1 9i 2 2 d 3d y x =− d 3 dd y xx =− 3xc=− + At ( )2, 1− , d 2d y x = ( )2 3 2 c=− + 8c= d 38d y xx =− + 2 3 8 d 3 8 2 y x x x xd = − + −= + +  At ( )2, 1− , ( ) ( ) 2 321 8 2 2 11 d d −− = + + =− 23 8 112y x x=− + − M1 M1: Substitution to obtain arbitrary constant A1 M1 M1: Substitution to obtain arbitrary constant A1 9ii Discriminant ( ) 2 38 4 11 2 = − − − 20=−  Since discriminant is less than zero, 23 8 112y x x=− + − has no real roots, therefore the curve does not intersect the x-axis. M1: Find discriminant A1

Preliminary Examination 2023 Solution Level: Sec 4G2 Qn. # Solution Mark Allocation 10i ( ) ( ) 2 1118 8 8 8 8222A k k k k= −   − −   − −   ( ) 2 2 164 2 4 8 2 164 64 8 2 kk kk = − − − = − + − 218 2kk=− (shown) M1: Find area of triangle PQX or PSY M1: Area of square subtract 3 triangles A1 10ii d 8d A kk =− For stationary value of A, d 0d A k = 80 k−= 8k = ( ) ( ) 218 8 8 2A=− 32A= M1 M1 A1 M1 A1 10iii 2 2 d 1d A k =− Since 2 2 d 0d A k  , A is a maximum value. B1 11i 22 6 8 39x y x y+ − + = 22 6 8 39 0x y x y+ − + − = 26 3 28 4 g g f f =− =− = = Centre of circle: (3, 4)− Radius ( ) ( ) 2 23 4 39= − + − − = 8 units M1 A1 M1 A1 11ii At ( )5, 4A −− , ( ) ( ) ( ) ( ) 22 5 4 6 5 8 4 39 LHS RHS = − + − − − + − = = B1

Preliminary Examination 2023 Solution Level: Sec 4G2 Qn. # Solution Mark Allocation Therefore, 5x=− and 4y=− satisfy the equation of circle and point ( )5, 4A −− lies on the circle. At ( )11, 4B − , ( ) ( ) ( ) ( ) 22 11 4 6 11 8 4 39 LHS RHS = + − − + − = = Therefore, 11x= and 4y=− satisfy the equation of circle and point ( )11, 4B − lies on the circle. Length of AB ( ) ( ) 22 11 5 4 4= − − + − − −       256 16 units = = Since the length of AB is 2 the radius of the circle, and the points ( )5, 4A −− and ( )11, 4B − lie on the circle, AB is a diameter of the circle B1 M1 A1