Regent 4E5N AM Prelim Examination Paper 1 - Marking Scheme

Regent Secondary School Additional Mathematics Sec 4 Express Preliminary Examination 2020 Paper 1 (Setter: Ms Su RY) Marking Scheme Question Solution Marks Total Marker’s Report 1i cos 12 13 A =− B2 2 1ii sin( ) sin cos cos sin 5 4 12 3 13 5 13 5 56 65 AB A B A B + =+      = − − + − −           = M1 A1 2 2 Let 32 52 2 −− − xx x = )1)(3( 52 +− − xx x = A x - 3 + B x + 1 2x – 5 = A(x + 1) + B(x – 3) Let x = 3, 1 = 4A + 0 A = 1 4 Let x = –1, –7 = 0 – 4B B = 4 7 Hence 32 52 2 −− − xx x = )3(4 1 −x + )1(4 7 +x M1 M1 M1 M1 A1 5 3i B1 graph, B1 mark each 2

3ii )11,1( 11 1 1 121121 121)11( 3 2 2 = = = = = y x x xx xx M1 M1 A1 3 4 Length = 8 7 2 5 2 2 8 7 2 5 2 2 5 2 2 5 2 2 40 16 2 35 2 28 25 8 68 51 2 17 4 3 2 += − ++= −+ + + += − += + M1 M1 M1 A1 4 5i 25 7 8 (5 4)( 2) 0 4 or 2 (rejected)5 x x x xx xx + − = − + = = =− M1 M1 A1, A1 4 5ii 3 3 2 3 33 2 3 2 loglog 2 log 3 2log log 4 log 4 81 ba ab a b ab −= −= = = M1 M1 A1 3 6i 2 32 32 3 3 8 2 15 2 2 15 (2 ) 2 .2 15 Let 2 4 15 4 15 0 xx xx xx xu uu uu + + −= −= −= = −= − − = M1 M1 A1 3 6ii Let f(u) = 3 4 15 0uu− − = f(3) = 0 M1 M1

2 2 2 2 ( ) ( 3)( 3 5) consider 3 5 4 3 4(1)(5) 11 f u u u u uu b ac = − + + ++ − =− =− Since 2 40b ac− , the quadratic factor has no real roots. Hence u = 3 is the only real solution of this equation (shown) M1 A1 4 6iii f(u) = 0 u = 0 2x = 3 Taking lg on both sides lg 2 lg 3 lg 2 lg 3 1.58496... 1.58 x x x x = = = = M1 M1 A1 3 7i 𝑡 = 20 𝐴 = 48 × 12 = 576 𝑐𝑚2 𝐴 = 𝜋𝑟2 𝑟 = √576 𝜋 = 24 √𝜋 (shown) M1 M1 A1 3 7ii 𝑑𝐴 𝑑𝑡 = 48 𝑑𝐴 𝑑𝑟 = 2𝜋𝑟 𝑑𝐴 𝑑𝑡 = 𝑑𝐴 𝑑𝑟 × 𝑑𝑟 𝑑𝑡 48 = 2𝜋𝑟 × 𝑑𝑟 𝑑𝑡 𝑑𝑟 𝑑𝑡 = 48 2𝜋𝑟 = 48 2𝜋 (24 √𝜋) = 1 √𝜋 𝑐𝑚/𝑠 M1 M1 A1 √ based on 7i 3 8i 32 23 2 14 320023 23200 3 3200 2 3 r r h r h r hr r     + = =− =− M1 A1 2

8ii 2 2 2 22 2 1 422 3200 222 3 6400 42 3 6400 2 (shown)3 A r rh r r r r rr r rr      =  + = + −  = + − =+ M1 A1 √ based on 8i 2 8iii When A has a stationery value, 2 3 3 2 23 2 2 0 6400 4 03 1 1600 03 4800 11.51764.. 11.5 12800 4 3 When 11.5176..., 12.5665 (>0) dA dr rr r r r r cm dA dr r r dA dr     = − + = −= = = = =+ = = This value of A is a minimum M1 M1 M1 M1 A1 5 9i k = −1 as 2 is the mid-value of k and 5. B1 1 9ii 2 2 2 1 0 or 5 0 [( 1)( 5)] 0 [ 4 5] 0 ( 4 5) (2, 18) 18 (4 8 5) 18 ( 9) 2 when 2 2 8 10 8 10 xx a x x a x x y a x x sub a a a a y x x b c + = − = + − = − − =  = − − − − = − − − = − = = = − −  =− =− M1 A1 A1 A1 4 9iii 3 B1 1

10i 3 1 514 69 ofGradient = − −=PR 3 ofGradient −=QS     =    ++= 2 15 ,2 19 2 96 ,2 145 ofMidpoint PR Let the coordinates of S be )0 ,( Sx 12 363 2 5732 15 3 2 19 02 15 = = −= −= − − s s s s x x x x S = (12, 0) M1 M1 M1 M1 A1 5 10ii 12 3128 0 = −=− − k k M1 A1 2 10iii Area of PQRS = 0 12 6 5 12 8 9 14 0 12 2 1  726072481681082 1 −−−++= = 60 units2 M1 A1 2 11i 23 4sin 1 cos 234 2 3 2(1 cos 2 ) 2cos 2 1 2, 1 x x x x ab − −=−  = − − =+ == M1 A1, A1 3 11ii Greatest value of f(x) = 3 Least value of f(x) = -1 B1 B1 2 11iii Period of f(x) = )(accept 180  Amplitude of f(x) = 2 B1 B1 2

11iv Shape C1 Correct max and min values C1 2 11 cycles C1 3 12i 𝑑𝑦 𝑑𝑥 = (2𝑥 − 1) 1 2(3) − (3𝑥 + 4) 1 2 (2𝑥 − 1)−1 2(2) (2𝑥 − 1) = 6𝑥 − 3 − 3𝑥 − 4 (2𝑥 − 1) 3 2 = 3𝑥 − 7 √(2𝑥 − 1)3 M1 M1 A1 3 12ii y is increasing 3𝑥 − 7 √(2𝑥 − 1)3 > 0 3𝑥 − 7 > 0 𝑥 > 7 3 M1 A1 2