AMKSS_Prelim_2024_AMath_P1_MS

2024 AMKSS 4E5N Prelim AM Paper 1 Solutions Comments: - Minus 1 mark overall for expressing coordinates in improper fraction (Q1 & Q12i) or missing unit such as degree (Q9i, Q10ii) Qn Solutions Marks 1 [5] 22 22 2 2 2 2 22 22 2 2 5 12 7 5 12 7 5 12 7 (1) 57 (2)2 Substitude (2)into (1) 575 12 7 2 25 70 495 12 7 4 5 3(25 70 49) 7 5 75 210 147 7 70 210 140 0 3 2 0 ( 2)( 1) xy y x xy xy xy xy xy xy xy xx xxx x x x x x x xx xx xx −−= −−= − =− − +=− +− =−  ++− =− − + + =− − − − =− + + = + + = ++ 0 2 or 1 1.5 or 1 ( 2, 1.5), ( 1,1) xx yy AB = =− =− =− = − − − M1 (substitution) M1 (expand & simplify to quadratic eqn) M1 (factorise/formula) A2 (coordinates should not be in improper fraction) 2(a) [2] 3 3 2 3 2 3 2 36 32 36 + −+ += − =+ M1 (conjugate) A1

2(b) [4] 22 2 22 2 2 2 2 2 2 30 3 6 2 2 4 30 (3 6) 4 4 16 30 (3 6) 4 30 9 6 6 6 4 45 6 6 4 4(45 6 6) 180 24 6 P P P P P P P     + +=            ++= + + = + + + = += =+ =+ M1 (Pythagoras theorem) M1 (attempt to form an eqn in P) A1, A1 3(i) [1] There is a fixed price of $100 million incurred, even when no submarines were assembled. B1 3(ii) [2] ( ) 2 2 22 2 2 2 5 20 1002 5 8 402 5 8 8 8 402 2 2 5 [( 4) 24]2 5 ( 4) 602 y x x y x x y x x yx yx = − + = − +     = − + − +         = − + = − + M1 A1 3(iii) [1] x = 4 occurs at the minimum point of the curve. Hence, the cost per submarine will be the lowest when we assemble 4 submarines B1

4(a) [4] ( )4 44 44 2 2 log 3log 2 3log 2 2 2log log 8 3 2 8 3 3 4 3 16 16 48 0 ( 4)( 12) 0 4 or 12 yy y y y y yy yy yy yy yy −−=  = − = − −= −= − + = − − = == M1 (quotient law of log) A1 (correct expression or equivalent) M1 (factorise/formula) A1 4(b) [3] 27 9 33 32 33 1 2 33 1 2 33 13 22 33 3 4 33 3 4 log log loglog log 3 log 3 log log 32 3log log 2 log log log log zy yz zy zy zy zy zy  = = = = = = = NOTE: If working end up with 3 2 2 33 44 or (Rej) zy z y z y = = =− M1 (change of base formula) M1 (power law) A1 Minus 1m if never show reject negative answer

5 [5] 32 2 32 2 32 32 2 22 2 2 6 1 ( 1)( 2) 2 6 1 ( 1)( 4 4) 2 6 1 34 Using long division, 92 ( 1)( 2) 9 ( 1)( 2) 1 2 ( 2) 9 ( 2) ( 1)( 2) ( 1) When 1; 9 9 1 When 2; 9 3 3 When 0; 9 4 xx xx xx x x x xx xx xx A B C x x x x x A x B x x C x xA A xC C x ++ −+ ++= − + + ++= +− + −+ = + +− + − + + = + + − + + − == = =− =− =− == 2 23 1 1 1 32 1 2 ( 2) B B x x x −+ =−  + − −− + + B1 2 () R Dx  + M1 (correct case) A2 (minus 1m for each incorrect ans) A1 6(a) [4] 17 22 Gradient of 2 Equation of : 2 6 2 6 (1) 5 6 (2) 5( 2 6) 6 10 30 6 9 36 4 2 (4, 2) yx AC AC y x yx xy xx xx x x y C =+ =− =− + =− + − + =− − + − + =− − + =− − =− = =−