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) Shaded area = 2 3 2 2 3 1 2 163 6 2 2 3 9x x dx             M1 M1 Area under curve Area of triangle or finding the equation of line PQ = 2 4 3 2 3 3 32 24 27x x      =   4 3 3 3 2 2 3216 16 24 4 3 3 27                     M1 Subst. =   4 3 3 3 2 2 3216 16 24 4 3 3 27                    

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 2 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) 64 27 units2 A1 Must be exact [7] 5 (a) 1 1lg lg lg2 2 2 x y x y       1lg lg lg2 2 x y x y        1 2lg lg 2 x y xy     M1 Use of product law 2 x y xy  2x y xy    2 4x y xy  2 2 2 4x xy y xy   M1 2 2 2 0x xy y     2 0x y  A1 0x y  x y (shown) AG (b)     2 2 4log 3 log 15 1x x         2 2 2 2 2 log 15 log 3 1 log 2 x x     M1 Change of base     2 2 22 log 3 log 15 2x x         2 2 2 3log 2 15 x x    M1 Log laws     2 2 2 32 15 x x    M1 Definition of log     221 15 34 x x     2 21 15 9 64 x x x    23 24 21 0x x   2 8 7 0x x   M1   7 1 0x x   7x (NA) or 1x

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 2 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) 1x A1 Reject 7x [8] 6 (a)        2 2f 3 2 1 x xx x e e        M1   2 2 3 1xe x        2 2 7xe x  A1 (b) 2(3 ) xy x e   y = 0 2(3 ) 0 xx e  x = 3 2 0xe  M1 (3,0)   2d 2 7d xy e xx   x = 3 Gradient =   2(3) 2(3) 7e  M1 = 6e Gradient of normal = 6e M1 Equation of normal where the curve crosses x-axis Subst. (3,0) M1   6 3y e x  A1 o.e (c)     2f 2 7 xx e x    For 13 2x , 2 7 0x  B1 Since 2 7 0x  and 2 0xe  ,  f 0 x  . B1  f 0 x  with conclusion f is a decreasing function. [9] 7 (a)        1 14 3 sin 90 4 4 sin 202 2A      M1 Area of triangle 6cos 8sin 20    M1  cos sin 90    8sin 6cos 20    (shown) AG (b)  6cos 8sin cos R      6cos 8sin cos cos sin sinR R        6 cosR  and 8 sinR  2 2 2 6 8R   M1

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 2 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) 100 10R  8 4tan 6 3   M1 53.1301    20 10 cos 53.1A      A1 (c) Max A = 20 +10 =30 A1 When  cos 53.1301 1    53.1   A1 (d)  20 10 cos 53.1301 25     10 cos 53.1301 5      1cos 53.1301 2    M1 1 1cos 60 2         53.1301 60    113.1   Since 113.1   , A cannot be equal to 25 m2. R1 seen 113.1   [9] 8 (a) Period of 1 2y  B1 Amplitude of 2 3 2y  B1 B2 B2 1y : 1m for shape 1m for values as plotted (reasonably placed) 2y : 1m for shape 1m for values as plotted (reasonably placed) −1 1 2 3 x ߨ 4 ߨ 2 3ߨ 2 ߨ 2 1 2 − 1 2

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 2 Preliminary Examination 2022 Anglo-Chinese School (Barker Road) (c) 3tan 2 cos 2 12x x  2 tan 2 3cos 2 2x x  2 tan 2 3cos 2 2 0x x   M1 The intersections of the two graphs will be the x solutions of 2 tan 2 3cos 2 2 0x x   B1 With explanation [8] 9 (a) 3 2 2Let f ( ) ( ) 23 12x ax ax x x     2f(2) (1 4 )a a          3 2 2 32 2 2 23 2 12 4a a a a      M1 Subst. x = 2 3 24 4 7 30 0a a a    M1 Reduce to cubic equation 2( 2)(4 4 15) 0a a a    M1 Quadratic factor using long division o.e For 24 4 15 0a a      2 discriminant 4 4 4 15 224    M1 Use of discriminant to show 2 4 0b ac  Or use of quadratic formula to show no solution for 24 4 15 0a a   Since discriminant 0 , there is no real roots for 24 4 15 0a a   2a is the only real root. A1 (b) 212 (2 1) 4 4x k x kx     21 2 4 2 1 04 x kx x k        21 2 2 3 04 x k x k     M1 Reduce to quadratic equation discriminant =     2 12 4 2 3 4k k       