FFMS 2024 A Math Prelim P2 Answer Key

FMS(S) Sec 4 Express Add. Math Preliminary Exam 2024 Paper 2 1 1 2 3 (rej. as -ve) 2 2ln 2 1.39 (3 s.f) Therefore, the equation has only one solution where =1.39 x e or x x =− == 10a 6cos 4 When 0, 6 Initial velocity of the particle is 6m/s. vt tv = == 2 57,66ab== 10b 0.870 (3 s.f)t= 3b 33 2 2 ln 2 1 ln 2 1d 22 xx xCx x x =− + + 10c 3 sin 42 When = 0, = 0, = 0 3 sin 42 0.432 Displacement = 0.432m (3 sf) s t c ts c st s =+ = =− − 4a 53 a−   4b 2 a=− 5a 1 4 5b 24sin 4sin− 5c 7 11,66  = 6i 8 2 15yx x=− − + 10d Total distance travelled = (1.5 2) 1.5 4.5m + = 6ii 2 or 2xx= =− 6iii 2 23 2 23 2 23 16 At 2, 16 202 Maximum point at 2 At 2, 16 16 20( 2) 8 Minimum point at 2 dy dx x x dy dx x x dy dx x =− = =− =−  = =− =− =− = −−  =− 11a 1 2 ( 6)Grad E 4 11 4 2 1Grad D 7 ( 1) 4 1Since Grad Grad 4 1 4 90 (Right angle in semi-circle) is the diameter of F F DF EF DF EF DFE DE C −−= =−−− −== −−  =−  =− ⊥  =   Centre of 1C = 1 7 6 4,22 + − +  = (4, 1)− 7a 18 15 2 129 m 36 ....x x m x+ + + 7bii 2 84 126+ = 294 8b 0.1500 tPe −= 8c ln100 4.6 When ln 4.6 16 (nearest year) P t = = = 11b 22 8 2 17 0x y x y+ − + − = 9b 9cos 4sin 97 cos( 24.0 )  + = −  11c 22( 8) ( 1) 34xy+ + + = 9c 76.427 76.4 =   11d 1(3 , 4) lies within C only 9d Maximum value of d = 97 and occurs when 24.0 =