2022 HS AMath Prelims P1 Ans

2022 Prelim A Maths Paper 1 Marking Scheme Qn Solution Marks Remarks 1 2 38y x x= + − ----- (1) 37 22yx=− ----- (2) (1) = (2) 2 38xx+− = 37 22x− ( )( ) 22 3 9 0 2 3 3 0 3 or 32 5 or 84 xx xx xx yy + − = − + = = =− =− =− Coord of A and B are ( )35, & 3, 824  − − − . Distance btw A and B ( ) ( ) 221.5 3 1.25 8 8.11 units = + + − + = [M1] [M1] [M1] [A1] Accept exact surd 1053 4 2 ( )( ) ( ) 22 78 2 1 12 1 1 1 x A B C xxx x x + = + ++−+ − − ( ) ( )( ) ( ) 27 8 1 2 1 1 2 1x A x B x x C x− = − + + − + + Let x = 1, 15 = 3C , C = 5 Let 1 2x=− , 293 22 A=−  , A = 2 Let x = 0, 28 2( 1) (1)( 1) 5(1)B= − + − + , B = ─1 ( )( ) ( ) 22 7 8 2 1 5 2 1 12 1 1 1 x xxx x x + = − ++−+ − − [M1] [M1] [A3]

3 ( ) 2 2 43 3 23 19 8 3 2 3 = 2 3 2 3 38 19 3 16 3 24 = 23 = 62 35 3 62, 35 ab ab − += + −−  +− −−+ − − = =− [M1] [M1] [A1] Accept alternative method ( )( ) ( ) 2 3 2 3 4 3ab+ + = − 4 ( ) 2 2 2 At ( 1, 4), 4 ( 1 2) 4 y a x k ak ak = + + − = − + + += Since a > 0 & k > 0 as curve lies above x-axis, let a = 1 , then k = 3 A possible equation for the curve is ( ) 223yx= + + [M1] [M1] [M1] [A1] (Any other suitable equation satisfying the relevant conditions accepted) 5(a) (b) (c) max min 1 ( 3) 1(shown)22r + + −= = =− [B1] Period of curve = 2 8 , 41 q q  == Amplitude = max min 1 ( 3) 222 − − −== , 2p=− [B1] [B1] Equation of the curve is 2sin 14 xy=− − [B1] 6(a) ( ) ( ) 2 f ' 6 2 32 x x dx x x c =+ = + +  At '1, f ( ) 11xx=− = ( ) 211 3 1 2( 1) 10 c c = − + − + = ( ) 2f ' 3 2 10x x x= + + [M1] [M1] [A1]

(b) ( ) ( ) 2 32 f 3 2 10 10 x x x dx x x x d = + + = + + +  At ( ) ( ) ( ) 321,10 , 10 1 1 10( 1) 20 d d − = − + − + − + = ( )f x = 32 10 20x x x+ + + [M1] [A1] (c) For y = f(x) to have stationary points, set ( )f ' 0x = . 2 22 3 2 10 0 4 2 4(3)(10) 116 0 xx b ac + + = − = − =−  ( )f ' 0x = has n o real solution, ie no stationary points. [M1] [M1] [A1] Solve equation to show no real roots accepted No marks if no real roots is not mentioned 7(a) ( ) ( ) 214.5, 4.5 18 4.53 729 8 xV   = = − = 729 8 81 98 dV dt  == cm/s ( or 110 8 cm/s) [M1] [M1.A1] Accept 10.125 31.8or cm/s (b) ( ) ( ) 23 2 183 36 33 V x x dV xxdx   =− =− ( ) ( )( ) 236 4.5 3 4.53 15 4.5,  3 4 Ax dV dx t   =− = = 81 8 135 4 Using 0.3 dV dV dx dt dx dt dx dt   = == The water level is rising at a rate of 0.3 cm/s [M1] [M1] [M1. A1]

8(a) ( )( ) 33 22 sin cos sin cos sin sin cos cos xx x x x x x x + = + − + [B1] (b) ( )( ) ( ) 33 22 22 sin cos sin cos sin cos sin sin cos cos sin cos 1sin 2sin cos cos2 11 sin 22 xxLHS xx x x x x x x xx x x x x x RHS += + + − + = + = − + = − = [M1] [A1, A1] No mark awarded if student do not show ( )1sin cos 2sin cos2x x x x= (c) 151 sin 224 11sin 224 1sin 2 2 x x x −= =− =− Basic angle, 30o = 2 210 ,330 ,570 ,690 105 ,165 , 285 ,345 o o o o o o o o x x = = [M1] [M1] [M1] [A1] 9(a) (given) (common) & are similar (AA Test) ACB ATC BAC CAT ABC ACT  =  =   [M1] [A1] Either statement (b) Since & are similarABC ACT , ( )( ) ( ) 2 2 22 () (shown) AB AC AC AT AC AB AT AT TB AT AT AT TB AC AT AT TB = = =+ = +  − =  [M1] [M1] [A1] (c) (alt. segment thm) & (given) BAX ACB ACB ATC BAX ATC  =  =  = By the alternate angle property, SC and XY are parallel. [M1] [M1] [A1]

10(a) 10 3 6 1 AB CB pm pm −= −= Since ABO = CBO, AB CBmm =− 10 (6 )3 10 18 3 28 4 (shown) p p pp p p − =− − − =− + − =− = [M1] [M1] [A1] Accept method involving tan  (b) 2 1 2 AB AD m m =− = Equation of line AD is ( ) ( ) 110 3 2 1 23 shown22 yx yx − = + =+ CD // AB, 2CDm =− Equation of line CD is ( )6 2 1 28 yx yx − =− − =− + At D, 1 23 22x+ = 28x−+ 57 22 7 7 54, 2 85 5 5 x xy =− =− =− − + = Coord of D is 7 54,55 − [M1] [A1] [M1] [M1] [A1] Accept ( -1.4, 10.8)

(c) Area of ABCD = ( ) ( ) 2 0 1 1.4 3 01 4 6 10.8 10 42 1 0 10.8 14 12 4 8.4 32.42 10.8 units −−= = + − − − − − = [M1] [A1] 11(a) ( )( )2 4 3 3 1x x x x+ + = + + By Factor Theorem, ( 3) 0 & ( 1) 0PP− = − = ( ) ( ) 3 2 3 2 5 3 ( 3) 3 0 9 132 (1) 5 1 ( 1) 1 0 4 (2) (1) (2) : 8 128 16, 12 ab ab ab ab a ab − + − + + = + = −−−− − + − + + = + = −−−− −= = =− [M1] [M1] [M1] [A1] *overall minus 1 mark if any expression is not set to 0 (b) ( )( )( )325 16 12 3 1x x x x x px q+ − − = + + + By comparison, 5 & 4pq= =− ( )( )( ) ( ) 0 3 1 5 4 0 43 or 1 or 5 Px x x x x = + + − = =− − [M1] [M1] [A1] (c) ( ) ( ) 322 2 250x a x x b+ − + = ---- (1) Let 2ux= , (1) becomes 3250 4From ( ), 3( ) 1( ) 5 u au u b b u rej or rej or + − + = =− − 2 4 5 22 or 55 x x = =− [M1] [A1] Accept 25 5

12(a) ( ) ( ) ( ) 1 2 2 7 7 6 161 7 6 1 (6) 42 61 yx x dy xdx x − − = = ++ =− + =− + At K, x = 1, 2 42 6 77 dy dx =− =− Equation of tangent at K is ( )611 7 6 13 77 yx yx − =− − =−− + Coord of B is 130, 7   [M1] [M1] [M1] [M1] [A1] (b) Area of the shaded region = Area under curve – Area of trapezium ( ) ( ) ( ) 1 0 1 0 2 7 1 13 116 1 2 7 1 107 ln 6 167 7 10ln 7 ln167 0.842 units dxx x = − + +  = + − = − − =  [M1,,M1] [M1] [M1] [A1] 13(a) ( ) ( ) ( )( ) 2 ln ln 2 2 3 3 2ln 3 ln 2ln 3 0 Let ln 2 3 0 3 1 0 3 or = 1 ln 3 or ln 1 1 or e x x xx ux uu uu uu xx x e x e e − += + − = = + − = + − = =− =− = = = = [M1] [M1] [M1] [A1] Accept x = 0.0498 or x = 2.72

(b) (i) ( ) ( ) 2 2 2 lg lg 22 22 24 1 2 4 4 12 p pqq p pqq p pq q p q q qp q  =+ =+ =+ −= = − [M1] [A1] (ii) Range of p is p > 0 2 2 4 012 Since 0 for lg 2 to be defined, 4 0 & 1 2 0 1 < 2 10 (shown)2 q q qq qq q q  −   −     [B1] [B1] (c) By observing the shape of the curve, a logarithmic function, ie equation (B) lny a x b=+ is a suitable model since the rate of growth of the head circumference gets much slower when the baby gets older over the months. [B1] [B1] -for correct model -for correct reasoning