AMKSS Prelim 2019 4E5N AMath P2 Solution

AMath Prelim 2019 Paper 2 Answer Scheme Qn Answer Mark Allocation 1(i) 2 2 When 0; 50000 2; 100000 50000 2 2 ln 2 1 ln 22 k k tP te e k k = = = = = = = M1 M1 1(ii) 15( ln 2)2When 5; 50000 282842 280000 tP e P = = = ≈ M1 A1 1(iii) 1 ln 22 50000 450000 9 1 ln 2 ln 92 6.34 Year 2016 kt t e e t t = = = = = M1 M1 A1 2(a) 2 2 2 2 cos sin sin cos cos sin 2 cos 22 2 cos 2( cos sin ) 2 sin 2 cos 2 cos 2 sin 2 sin 0(shown) xx x xxx x x xx x xxx x dy e xe xdx dy e xe xe xe x dx ex d y dy ydxdx e x e x ex ex e xe xe xe x = + = − +++ = −+ = − ++ =−−+ = M1 M1 M1 M1

2(b) 2 2 2 2 2( 1) (2 16) ( 1) 18 ( 1) 2 2 182 ( 1) ( 1) 9 1 3 or 1 3 4 or 2(NA) dy x x dx x x dy dx dx dt dx dy dx dt dx dt dx dx dt dtx x xx xx −− += − −= − =− −= × −−= × − −= −= −= − = =− M1 M1 M1 M1 A1 3(a) 2 2 2 2 1 22 lg10lg 2 3 lg 2lg 3 lg 22 lg 3 lg 2(lg ) 2 3lg Let lg 2 3 20 (2 1)( 2) 0 1 or 22 1lg or lg 22 10 or 10 0.316 or 100 x x x x x x xx xy yy yy yy xx xx xx − −= −= −= −= = − −= + −= = −= = −= = = = = M1 M1 M1 M1 A1

3(b) 3 2 log log log 3log (1) log 2 log log (2) 2 3log2 6 log 2 2log 6 log log 2 26 32 6 42 6 2 3 mm mm m m m m m m mm x ya x ya xb xb bx b ya b ya aby xy b ab b ab ba ba += += − = = = − += += −= − −= − −+= −= −= M1 M1 M1 M1 A1 3(c) 21 3 2 21 42 84 2 22 2 32 2 1 22 4 2 3 3 81 33 3 42 24 2 4, 2 ab ab ab b a ab ab ab ba bb ba − − ×= ×= +=− −= − −= − = ×= += −+= = = M1 M1 M1 A1

4(a) 2 3 2 3 5 5 10 30 10 30 0 10 30 0 10 30( )(2 ) 1 96 dy xdx kx xkx kx k k −= + − += −+ = = = M1 M1 A1 4(b)(i) 3 20 2 3 x x +> >− B1 4(b)(ii) 13 23 2 3 2(3 2) dy dx x x =  + = + M1 A1 4(b)(iii) 0 0 ln 3 2 0 1 ln(3 2) 02 ln(3 2) 0 32 3 21 1 3 3 12[3 2] 3 3 2 Equation of normal: 2 3 12At( , 0),39 22 39 y x x x xe x x dy dx y xc c yx = += += += += += =− = ×− + = = −+ −= − = −− M1 A1 M1 M1 A1

5(i) 2 2 2 2 2 82 (2 1) 820 (2 1) 8 2(2 1) (2 1) 4 At( , ); (2 1) 4 21 2 o r 212 111 or (NA)22 5 dy dx x x x x pq p pp pp q = − − −= − =− −= −= −= −= − = =− = M1 M1 M1 A1 A1 5(ii) 2 23 2 2 3 32 (2 1) 1At ; 3 2 1 ,32 1 32At ; 012 (2 1) 2 1 , 3 is max point2 dy dx x xy dyx dx = − = −= − −− = −= < ×− − −− M1 A1 M1 A1 6(i) 221 6 15px p x−+ A1, A1 for both terms 6(ii) 2 22 22 2 2 22 (1 4 4 )(1 6 15 ) 15 24 4 Coefficient of 15 24 4 x x px p x p x px x xpp −+ − + = ++ = ++ M1 A1 6(iii) 2 2 2 64 ( 6 4) 15 24 4 2( 6 4) 15 24 8 8 0 15 36 12 0 (5 2)(3 6) 0 2 or 25 px x px pp p p pp pp pp pp −− = −− + +=− − + + += + += + += = −= − M1 M1 M1 A1

7(i) Midpoint of 2 63 2,22 (4, 2) 1Gradient of 2 Gradient of 2 2 At(4, 2);2 (2)(4) 6 26 AC AC BD y xc c c yx ++  = =− = = + = + =− = − M1 M1 M1 A1 7(ii) y = 0 2x – 6 =0 x = 3 B1 7(iii) 2 2 22 22 22 22 2 2 ( 3) 5 (4 3) 2 4[( 3) ] 25(5) 125( 3) (2 6) 4 4[( 6 9) 4 24 36] 125 5 30 13.75 0 5.5 or 0.5(NA) 5 (5.5, 5) xy xy xx xx x x xx x y D −+= −+ −+= −+−= −++ − + = −+ = = = M1 M1 A1 7(iv) Area 2 5.5 2 3 6 5.51 5 3 01 52 1 49.5 24.52 12.5units = = − = M1 A1 8(i) a = 3 b = 0.5 c = 1 B1 B1 B1 8(ii) Correct shape, amplitude 3 Correct turning points, end points sinyx=− M1 M1 B1

8(iii) k = 4 B1 9(i) ( 2sin 2 ) cos 2 2 sin 2 cos 2 x xx xx x −+ = −+ M1, M1 A1 9(ii) [ ] [ ] 2 0 2 0 2 0 2 2 0 0 2 0 2 0 ( 2 sin2 cos2 ) cos2 sin2( 2 sin2 ) cos22 sin( 2 sin2 ) 22 ( 2 sin2 3) 3 2 3 22 x x x dx x x xx x dx x x dx x x dx x π π π ππ π π π π ππ π ππ π −+ = − +=  − = −− − + = −+ = −+ = ∫ ∫ ∫ ∫ M1 M1 M1 M1 A1 10(a)(i) 2 2 2 2 2 30 40 4(2 3) 0 8 12 0 ( 2)( 6) 0 2 or 6 x px p b ac pp pp pp pp − + −= −> − −> −+> − −> <> M1 M1 M1 A1 10(a)(ii) y = 2(x – 2) p = 2 therefore, y = 2x – 4 is tangent to curve M1 A1

10(b) 22 2 2 2 22 4 ( 2) 4(1)( 2 ) 4 48 44 ( 2) 0 Since ( 2) 0, 4 0 roots are real for all value s of . b ac k k kk k kk k k b ac k − = −− − = − ++ =++ = + ≥ +≥ −≥ M1 M1 A1 11(i) sin 5 5sin cos 2 2cos 3 2cos 5sin 3 y y x x xy θ θ θ θ θθ = = = = += += M1 M1 11(ii) 22 sin 5 cos 2 25 29 5tan 2 68.1 29 cos( 68.2 ) R R R R α α α α θ = = = + = = = ° −° M1 M1 M1 A1 A B E C D θ 5 m 2 m 3 m θ

11(iii) 29 cos( 68.2 ) 3 56.14 68.2 56.14 12.1 Acute θ θ θ − °= ∠= ° − °=− ° = ° M1 A1