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 of term in 27 27 (1) Compare coefficient of term in ( 1) 324 (2)2 Sub (1) into (2) ( 1)(729) 322 n n n n nkx C kx C kx nnnkx k x x nk k n x n n k nn n −−+ = + + + −= + + + =− −= −−− − = −−− − = 2 22 2 4 8( 1) 9 8 09 1 09 1 109 0 ( ) or 9 3 n n n n n n nn nn n NA n k −= − − = −=  −= == =− M1 for expansion M1 form 2 eqn by equating coeff M1 substitution mtd only if 1 linear and 1 quad eq A1 A1

8i [3] Let 180 (adjacent angles on a straight line) 180 (angles in opposite segment) =180 (180 ) So (common angle) Since two corresponding angles are equal, and CAB a DAB a DXC DAB a a DXC CAB ACB XCD ABC XDC =  = −  = − −− =  =  =  are similar (AA Similarity Test) M1 for reason M1 for reason M1 for reason Minus 1m if no proper conclusion 8ii [3] (angles in alternate segment, tangent YD) (vertically opposite angles) So From 8i, So (base angles of isosceles triangle ) Since is an isoscele CBY CAB CBY DBX DBX CAB CAB CXD BXD DBX BXD BDX BDX  =  =  =  = =  =  s triangle, .DB DX= M1 for reason M1 for reason M1 for reason base angles

9i [1] In , is midpoint of and is midpoint o f . Using midpoint theorem, is parallel to . ABC R AC Q BC RQ AB  B1 with correct quote of ONLY TWO midpoints and midpoint theorem. 9ii [2] 40Gradient 1 22 Equation of : Sub (1, 5) 51 6 6 RQ AB y x c c c yx −= =−−− =− + =− + = =− + M1 for gradient A1 9iii [2] :6 : 5 6 Solve simultaneous eq 6 5 6 12 4 3 9 ( 3,9) AB y x AC y x xx x x y A =− + =− − − + =− − =− =− = − M1 A1 9iv [2] 2 3 2 2 1 31 9 4 0 5 92 1 ( 12 10 9 18 8 15)2 16 units Area − − −= = − + + + − + = M1 for 5 coordinates with correct “shoe- lace” method A1 only if coord in formula in anti- clockwise direction

10i [2] Amplitude = 2 Period = 2 B1 B1 10ii [4] Each graph – B1 correct shape, B1 correct y-intercept B2 B2 Minus 1m if max point of both graphs not aligned at x = 10iii [4] 2 2 2 2 11 2cos sin 2 111 2 1 2sin sin 22 1Let sin 2 1 2(1 2 ) 1 2 4 4 1 0 1 1 4(4)( 1) 8 1 17 8 1 1 17 1 17sin 2 8 8 1(reject since negative as 0 ) 2 Basic angle = 0.695004 1 0.695004, 02 xx xx ux uu uu uu u x or x x   −=  − − =   = − − = − + = − − =  − −= = +−=  =− .695004 1.39, 4.89 (to 3sf)x= M1 for use of formula M1 form quad eq M1 for solving for trigo fn A1 for both values 10iv [1] 12For Graph above Graph 1.39 4.89 yy x B1

11i [2] 2 d d 96 Sub 1, 3 / va t t t a m s = =− == M1 for correct differentiation A1 11ii [3] 2 2 2 dFor max , 0 d 9 6 0 1.5 dCheck 6 0 (max )d max 9(1.5) 3(1.5) 6.75 / vv t t t v vt v ms = −= = =−  =− = M1 for t M1 for check max v A1 11iii [5] 2 23 23 2 23 d 9 3 d 93 23 Sub 0, 3 3 4.5 3 At rest, 0 9 3 0 3 (3 ) 0 03 Sub 3, 4.5(3) (3) 3 16.5 Distance 16.5 s v t t t t tt c ts c s t t v tt tt t or t t s m OP m = =− = − + == = = − + = −= −= == = = − + = =   M1 for integration M1 for c-value OR for definite integral M1 for set v = 0 M1 for t A1 No A1 if earlier c=0 and +3 in final step

12i [3] 2 1 d 9(2 1)d 9(2 1) ( 1)(2) Sub (2,12) 912 2(2 2 1) 912 6 13.5 91 132(2 1) 2 y xx xyc c c c y x − − =− −=+ − =+−  − =+− = =− + − M1 for correct integration M1 for sub point A1 12ii [2] 2 2 3 2 3 2 2 d 9(2 1)d d 18(2 1) (2)d 36 (2 1) dSub 0, 36 d y xx y xx x yx x − − =− =− − −= − == M1 for chain rule A1 12iii [2] At a turning point, 2 d9 0d (2 1) y xx== − Since there is no solution for x, there is no turning point. The conclusion is wrong. OR 2 d9Sub 0, 9 0 d (0 1) dySince at 0, 0, there is no stationary poin t.dx yx x x = = =  − = OR 2 dSince (2 1) 0, so 0 d dSince is always positive, there is no t urning point.d yx x y x −   M1 only if followed by explanation M1 M1 only if followed by explanation M1 M1 M1