CBSS Prelim 2024 4E AM P1 MS

4E Additional Mathematics Preliminary Examination 2024 Marking Scheme Qns Suggested Solutions Remarks 1 2 22 22 3 12 1 (3 1)( 1) 3 1 1 ( 1) 3 12 1 ( 1) (3 1)( 1) (3 1) x x A B C x x x x x x x A x B x x C x − + − = + ++ − + − − − + − = − + + − + + When 1x= , 3 12 1 (4) 2 C C − + − = = When 1 3x=− , 22 1 1 43 12 13 3 3 3 A A      − − + − − = −           =− When 1x= , 3A=− , 2C= 21 ( 3)( 1) (1)( 1) (2)(0 1) 0 B B − = − − + − + + = 2 22 3 12 1 3 2 (3 1)( 1) 3 1 ( 1) xx x x x x − + − =− ++ − + − [M1] [M1] [M1] [M1] [A1] 2(a)(i) 3cos x Amplitude = 3 Period = 360 [B1] 2(a)(ii) 1 4sin 2 x− Amplitude = 4 360Period 2 180 = = [B1] [B1]

2(b) [M1] Correct shape [M1] Smoothness [M1] Intersection 2(c) 4 solutions [B1] 3(a)(i) ( ) ( ) 72 657 2 772 2 2 2 ... 123 3 3 448 224128 ... 33 x x x xx         − = + − + − +                  = − + + [M1] [A1] 3(a)(ii) ( ) ( ) 7 2 2 2 2 2 2 2 2 22 448 2242 2 128 ...3 3 3 224 896 128 ...33 224 896 128 ...33 xp x p px x x x p x px x p p x    + − = + + − + +       = − + + = − + + 22 2 2 224 896 32 1283 3 3 256 896 384 0 2 7 3 0 ( 3)(2 1) 0 p p p pp pp pp − + =− − + = − + = − − = 3 1 (rejected)2 p p = = [M1] [M1] [M1] [A1] 3(b) 17 2 1 2x x  − 1 4sin 2yx=− 3cosyx=

( ) ( )( ) ( ) 172 1 34 2 34 3 17 1 2 17 111 2 17 11 2 rr r rr rr r r r Tx r x xr x xr − + − −  =−      =−         =−  Let 34 3 0rxx − = 34 3 0 34 3 1113 r r −= = = Since r is not an integer, there is no independent term. [M1] [M1] [M1] [A1] 4(i) 2 4 3 7 (1) 2 5 (2) y ax x a y ax = + + − =− Sub (2) into (1) 2 2 2 2 22 2 2 4 3 7 2 5 (4 2 ) 3 2 0 40 (4 2 ) 4 (3 2) 0 16 16 4 12 8 0 8 8 16 0 20 ( 2)( 1) 0 2, 1 ax x a ax ax a x a b ac a a a a a a a aa aa aa aa + + − = − + − + − = −= − − − = − + − + = − − + = + − = + − = =− = [M1] [M1] [M1] [A1] 4(ii) Since gradient is positive, a = 1. 2 4 4 (1) 2 5 (2) y x x yx = + − =− Sub (2) into (1)

2 2 2 4 4 2 5 2 1 0 ( 1) 0 1 x x x xx x x + − = − + + = += =− Sub 1x=− , 25 7 ( 1, 7) y y Q =− − =− −− [M1] [A1] 5(i) t (years) 5 10 15 20 25 V (in thousands $) 192 370 714 1374 2642 lgV 2.28 2.57 2.85 3.14 3.42 [G1] Axes [G1] Points plotted correctly [G1] Straight line passing through all plotted points.

5(ii) From graph 2 lg 2 10 100 V V V = = = The initial value of investment is $100 000. [M1] [A1] 5(iii) lg lg lg t o o V V k V t k V = =+ From graph, Gradient 3.08 2.4 17 19 7 300 −== − 17lg 300 1.13937 1.14 k k k = =  [M1] [M1] [A1] 5(iv) Increase by 50% → investment is at $150 000 150 lg 2.18 V V =  From graph, when lg 2.18V = , 3.25 yearst= [M1] [A1] 6(a) 2 2 2 ( 3) 3 5 2 6 3 5 2 3 0 ( 1)(2 3) 0 11 or 1 2 x x x x x x xx xx xx −  − −  − − −  + −  −  [M1] Expansion [M1] Factorisation [A1] 6(b) 2 (1) 62 (2) y k x yx x += =+ Sub (2) into (1)

22 2 622 124 4 12 3 12 0 x k xx x k xx x kx x x kx  + + = + + = + + = + + = 2 2 2 40 4(3)(12) 0 144 0 ( 12)( 12) 0 12 12 b ac k k kk k − − − + −  −   Therefore, the greatest value of k is 11. [M1] [M1] [M1] [A1] 7(a) ( ) 2 2 2 2 2 6 9 2 3 9 3929 24 392 22 y x x xx x x = − + = − + = − − + = − + Since 39,22   is a minimum point, it is not possible for the curve to have a value smaller than 9 2 . [M1] [A1] 7(b) 2 3 5 (1) 2 6 9 (2) yx y x x =+ = − + Sub (1) into (2) 2 2 2 6 9 3 5 2 9 4 0 ( 4)(2 1) 0 14, 2 x x x xx xx xx − + = + − + = − − = == [M1]

Sub x = 4 3(4) 5 17 (4,17) y=+ = Sub 1 2x= 135 2 13 2 1 13,22 y =+ =   Length PQ 22 1 134 1722 490 4 490 7 10 7 or or 102 3 2    = − + −       = = [M1] [M1] [A1] 8(a) ( ) ( ) ( ) 3 4 4 827 2 24 2 24 2 y x dy xdx x − =+ − =− − =− − Since 0dy dx  for all 2x , the curve does not have a stationary point. [M1] [M1] [A1] 8(b) Sub x = 0

( ) 3 827 2 26 (0, 26) y B =+ − = Sub y = 0 ( ) ( ) ( ) 3 3 3 3 80 27 2 827 2 27 2 8 8( 2) 27 22 3 113 11 ,03 x x x x x x A =+ − −= − − − = − =− − =− =   [B1] [B1] 8(c) Sub x = 0 ( ) 4 24 2 3 2 dy dx =− − =− Gradient of normal 2 3= 226 ( 0)3 2 263 yx yx − = − =+ [M1] [M1] [A1] 8(d) Sub y = 0

2 263 39 x x =− =− Area of triangle BOC 2 1 39 262 507 units =   = [M1] [A1] 9(i) (alternate segment Thm) 90 (angle in semi-circle) 90 (angle on a straight line) 90 is similar to (AA) PAB BCA ABC ABP ABP ABC BAP BCA  = = =  = =   [M1] [A1] 9(ii) Let , 90 (tangent perpendicular to radius) 90 (sum of angles in triangle) 90 (angle bisector) 90 (angle in semi-circle) 90 (angle on a straight line) 180 90 AKH x KAP APH x HPB x ABC ABP BHP = =  = −  = − = =  = − − ( )90 (sum of angles in triangle) (vertically opposite angles) Since , (isosceles triangle) x x AHK x AHK AKH AH AK − = =   = = [M1] [A1] 9(iii) Since is similar to ( from (ii)) (Proved) BHP AKP AK KP BH HP AH KP AK AHBH HP AH HP KP BH  =  = =   =  [M1] [A1]

10 2 22 2 22 2 2 2 1 2 (1 ) (1 ) 32 (1 ) (3 2 ) (1 ) x xx xx x ey x dy e x e dx x e xe x ex x = − −+= − −= − −= − Since 3 2x , ( ) ( ) ( ) 2 2 2 2 2 (1 ) 0 1 0 1 (3 2 ) 0 1 e (3 2 ) 0 1 0 x x x x x x x dy dx −  − −  − −  −  Since 0dy dx  , y is always decreasing. [M1] [M1] [M1] [M1] [A1] 11 2 3 2 3 3 3 36 36 36 3 12 xx xx xx xx dy eedx dy e e dxdx e e c e e c − − − − =− =− = + + = + +  Sub 14, 0dy xdx == 14 12 1 1 c c = + + = [M1] [M1] [M1]