2022 ACSBR AMath Prelims P2 Ans

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 2 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) 1 (a) 22 3 2x x   2 32 1 2x x      2 2 3 32 1 4 4x                  M1 Use of completing the square/ comparing 2 3 252 4 16x           2 3 252 4 8x      A1 (b) 2 3 252 4 8y x       Max value of 25 3.1258y   B1 Max value 2 2( 3) ( 8) 20x y    Centre  3,8 Radius 20 Lowest point of the circle 8 20 B1 Lowest point = 3.528 Since the max value of 2 2 3 2 3.528y x x    , R1 the curves will not intersect. [5] 2     2 1 2 1 1 1 1 .. 1 22 2 2 n n nn n nx x x                                    211 .. 2 8 n nn x x     M1 Attempt to use binomial theorem seen.  1 1 2 n xax           211 1 .. 2 8 n nnax x x           2 2 11 ... 2 2 8 n nn anax x x x       M1 Expand   211 ... 2 8 2 n nn ana x x              

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 2 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) Comparing coefficient of x 02 na  M1 2 na  Comparing coefficient of x2  1 15 8 2 4 n n an    M1  1 4 30n n an    1 4 30 2 nn n n       2 30 0n n     6 5 0n n   6n (NA) or 5n A1 5 2a  A1 12 2a [6] 3 (a) Gradient AB = 5 12 4 3    = 1 Gradient of perpendicular bisector = 1 M1 Midpoint AB = 3 4 12 5,2 2       = 1 17,2 2     M1 17 1 2 2y x       8y x  ---- (1) 3 4y x  ----(2) 8 3 4x x    M1 4 12x  3x Subst. 3x into (1) 3 8 5y     M1 Center C  3, 5 shown AG (b) Radius   22(3 3) 5 12  

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 2 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) = 7 units B1 Use A(3,12) and C(3,5) o.e     2 2 23 5 7x y    M1 or use general form equation of circle. 2 2 6 9 10 25 49x x y y      2 2 6 10 15 0x y x y     A1 [7] 4 3 23 6y x x  2d 9 12d y x xx   M1 First derivative 2 2 d 18 12d y xx   18 12 0x  2 3x A1 Subst. 2 3x 3 2 2 2 163 63 3 9y             Coordinates of P 2 16,3 9      Subst. 0y  3 23 6 0x x    23 2 0x x   0 or 2x x  A1 2x Q (2, 0)