2022 AMKSS AMath Prelims P1 Ans

Solutions to 2022 AMKSS Prelim AMP1 Qn Solution Marks 1i [2] Opp side = 21 p− 3rd quadrant 2 21sin 1 1 pAp −−= =− − M1 for opp side A1 1ii [2] 2 tan( ) tan tan 1 tan tan tan 0 1 tan (in 3rd quad) 1 A A A A A p p    − −= + −= = −−= M1 for use of formula A1 or B2 2 [4] 2 2 2 2 2 2 ( 6) 3 2 ( 6) 1 0 No real roots as curve does not touch line 40 ( 6) 4(1)(1) 0 12 36 4 0 12 32 0 ( 4)( 8) 0 48 x h x x h x b ac h hh hh hh h − + − − =− − − + = − − −  − + −  − +  − −   M1 form equation =0 M1 for < 0 M1 form correct discriminant A1 3 [4] 3 3 d 0.05d d 8d d8Sub 2, 1 d (2) d d d dt d d d0.05 1 d d1 0.05units/sd 20 y t y xx yx x y y x xt x t x ORt − =− =− −= = =− = − =−  = M1 correct differentiation M1 find value M1 correct relation only if -0.05 used A1 𝛼 p 1

4i [1] Sub 0t = , 1.2 0.015 $1.215millionV = + = B1 4ii [2] 2 2 2 1.3 1.2 0.015 0.1 0.015 Take ln on both sides 20ln ln3 20ln 2 lne3 20ln 2 3 0.949 (to 3sf) k k k e e e k k k =+ =  =  =  = = M1 for simplify to 2ke A1 4iii [2] 20ln 3 2 20ln 3 2 20ln 3 2 20ln 3 2 2 1.215 1.2 0.015 2.43 1.2 0.015 Take ln on both sides ln82=ln ln 82 ln 4.65 years (to 3sf) t t t t e e e e t         = + − = = = M1 to simplify to e expression A1

5i [2] 2 2 2 2 2 2 2 4 16 44 4 4 2 2 4 ( 2) 4 4( 2) 16 h t t tt tt t t =− + =− − =− − + − =− − − =− − + M1 for 22 A1 5ii [2] Max height = 16m Occurs when t = 2s B1 B1 5iii [3] Sub h = 2.5 2 2 2.5 4( 2) 16 13.5 ( 2)4 13.52 4 3.84 0.163(NA as ball going upwards) t t t t or =− − + − =−− = = Sub h = 0 4 ( 4) 0 04 tt t or t − + = == 3.84 4 t M1 to calculate t when h =2.5 M1 to calculate t when h = 0 A1

6i [5] ( ) 2 2 2 2 22 2 2 2 2 2 2 6 21 6 2 1 dd 22dd d 2 (2 1) 2( 6) d (2 1) 4 2 2 12 (2 1) 2 2 12 (2 1) d 2 2 12Sub 0, 0d (2 1) 2( 6) 0 ( 2)( 3) 0 2 3 (NA as 0) 10 5 Turning point 2, 2 xy x u x v x uv xxx y x x x xx x x x x xx x y x x xx xx xx x OR x x y += + = + = + == + − += + + − −= + +−= + +−== + + − = − + = = =−  = M1 for either correct differentiation M1 for correct quotient rule M1 for = 0 M1 for solve x A1 6ii [3] 2 2 d 2 2 12 d (2 1) dSub 0, 12 d 6 1Equation of normal: 6 12 y x x xx yx x y yx +−= + = =− = =+ M1 gradient M1 for y-intercept A1

7a [4] ( ) 2 22 6 4 8 6 22 2 22 2 2 2 2 2 Term in (4) (1) 2 =15(256) 284 =960 28 960 28 1212 9 3 (reject -3 since is positive) xx C C mx x mx x m x m m mm = + −  + + += = = M2 for correct term of each expression M1 equate to 1212 A1 7b [5] 1 1 2 2 12 22 2 2 2 (1 ) 1 (1) ( ) (1) ( ) ... ( 1)1 ... 2 Compare coefficient