2022 HS AMath Prelims P2 Ans

Marking Scheme for A Maths Paper 2 Qn Working Marks Remarks 1(a) B1 B1 Shape of the curve y-intercept and horizontal asymptote x y O 2 3

(b) ( )( ) 2 2 2 3 e 2e 13 2ee 3e 1 2e 2e 3e 1 0 Let e 2 3 1 0 2 1 1 0 1 or 12 1e or e 12 1ln or ln12 0.693 or 0 xx x x xx xx x xx y yy yy yy xx xx −−= −= −= − + = = − + = − − = == == == =− = M1 M1 M1 A1

2(i) Since roots of f(x) = 0 are 1, k and k2 2f ( ) ( 1)( )( )x x x k x k=− − − − 2 2 23 32 (2) 7 (2 1)(2 )(2 ) 7 (2 )(2 ) 7 4 2 2 7 2 2 3 0 (shown) f kk kk k k k k k k =− − − − − =− − − = − − + = + + − = M1 M1 A1 (ii) 32 32 Let g( ) 2 2 3 g(3) 3 2(3) 2(3) 3 0 k k k k= − − − = − − − = Since g(3) = 0, k – 3 is a factor of g(k). 32 2 2 2 3 0 ( 3)( 1) 0 k k k k k k − − − = − + + = 2 22 3 or 1 0 4 1 4(1)( 1) 3 0 (no real roots) k k k b ac = + + = − = − =−  Therefore 32 2 2 3 0k k k− − − = has only 1 real root. M1 M1 M1, M1 M1 B1

3(i) 1 2 1 2 1 2 ( 2) 1 d1 ( 2)[ ( 1) ] 1(1)d2 1( 1) [ 1 1]2 3( 1) ( ) 2 3 21 y x x y x x xx x x x xx x x − − − = + − = + − + − = − + + − =− = − M1 M1 M1 A1 (ii) By Chain Rule, d d d d d d 3d2 d21 d 3(2)2d 2 2 1 2 units/s3 y y x t x t xx tx x t = = − = − = M1 A1

(iii) 55 22 5 2 23 d d 31 2 1 2 [( 2) 1]3 2 [7 4 4 1]3 2 (10)3 26 3 xx xx xx xx = −− = + − =− = =  M1 M1 A1

4(i) Let E be the point on AB such that BE is perpendicular to CE. F is the point on CE such that CF is perpendicular to FD. cos 4 4cos BE BE   = = sin 1 sin FD FD   = = 4cos sin (shown) AB BE FD  =+ =+ M1 for either one correct A1 (ii) 2241 17 R=+ = 1 1tan 4 1tan 4 14.036o   − = = = ( )17 cos 14.0AB  = −  M1 for finding R, M1 for finding  A1 (iii) Max 17 mAB= B1 Accept 4.12 m

For AB to be maximum, the value of ( )cos 14.036 − must be 1. ( )cos 14.036 1 14.036 0 14.036 14.0 (1 d.p.) o    −  = − = = = B1 (iv) ( )3 17 cos 14.036= −  ( ) 3cos 14.036 17 14.036 43.313 43.313 14.036 57.349 57.3 (1 d.p.)    −  = − =  = +  = =

5(a) 226y x x c= − + --------- (1) 28yx+= --------------- (2) Equating (1) & (2): 2 2 2 6 8 2 2 4 8 0 x x c x x x c − + = − − + − = The line is a tangent to the curve, 2 2 40 ( 4) 4(2)( 8) 0 16 8 64 0 10 b ac c c c −= − − − = − + = = M1 M1 A1 (b) 2 2 2 3( 5) 1 3 15 1 0 3 14 0 (3 7)( 2) 0 12 or 2 3 xx xx xx xx xx −  − − − +  − −  − +  −  M1 A1 A1

(c) ( )( ) 2 2 2 2, 1, For real roots, 4 0 1 4 2 0 1 8 0 1 8 Greatest value of integer 0 x x p a b c p b ac p p p p − + − =− = =− − − − −  −  = M1 M1 A1

6(i) ( ) ( ) ( ) 5 5 4 3 2 1 2 3 23 1 22 5 5 51 1 1 1 2 2 21 2 32 2 2 2 15 5 2032 8 x x x x x xx  −             = + − + − + − +                         = − + − + M1 A1 (ii) ( ) ( ) 5 2 2 2 3 2 2 2 2 2 11 3 2 2 151 3 5 20 ... 32 8 Considering , 5 3 135 8 32 2 5 3 13 5 8 32 2 1 2 4 ax x x xax x x x x x ax x x a a + + −  = + + − + − + − + = − + = =− M1 M1 M1 A1 (iii) ( ) ( ) ( ) ( ) ( ) 5 5 5 2 3 10.47 2 2 1 2 0.472 0.015 5 0.01510.47 5 0.015 20 0.015 ...32 8 0.02293 x x x =− −= = = − + − +  M1 M1 A1