AM4 4049 BSSS 2023 P2 MS

Marking Scheme BSSS AM4 4049 P2 2023 1 By Factor Theorem, ( ) ( 2)(2 1)( ) By Remainder Theorem, (1) 6 (1 2)(2 1)(1 ) 6 ( 1)(3)(1 ) 6 12 1 fx x x x k f k k k k = − +− =− ⇒− + −= − ⇒ − −= − ⇒ −= ⇒= − M1 M1 A1 2a 3 2cos 2Period = 5min/10 = = 0.5min 4 rad/min d kt k k π π = − = M1 A1 2b >2, 3 2cos 4 2 1 cos 42 1Let cos 4 = , 2 , 4 lie in quad 1&43 4 ,2 ,2 ,4 ,4 ,. . . 5 7 15 17,,,,, . . .3(4 ) 3(4 ) 3(4 ) 3(4 ) 3(4 ) 1 5 7 15,,,, . . .12 12 12 12 from the sketch, for >2, 51 12 12 d t t t t t t t d t π π π παπ π απαπαπαπα πππππ πππππ −> > = =−+−+ = = = − 1 = 3 10In 5 min, for 10 cycles, time = min.3 M1 M1 M1 A1 3a 2ln ( 1) when 2, 2 2 ln( 1) 1 yx y xa a ae = − = = = − = + M1 A1 3b 2 22 2 00 2 2 0 1 2 2 0 2ln ( 1) 1 d 1 d 2 22 2 2 2 ln( 1) d = 2( +1) d = 2( +1) 2 = 2 (Show y y y e yx xe xy e y ye e e xx e xy ee + = − = + = + = +  = +− = −= − − ∫∫ ∫ ∫ n) M1 - x M1 - intg M1-intg A1 - ans M1 A1 4a 2 22 2 sec 1 1 sec cos 1 2 2cos 2 2 cos 2 1 2 2 (Shown)cos 2 3 x xx x x x =++ = + = ++ = + M1 M1 A1 4b 2 2 sec 2 4 1 sec 2 7 24 cos 4 3 7 14 cos 4 34 2cos 4 4 5 7 114 , , , ,...33 3 3 5 7 11, , , ,...12 12 12 12 for 2 2 , there will be 16 solutions. x x x x x x x x πππ π πππ π ππ =+ ⇒= + ⇒= + ⇒= ⇒= ⇒= − ≤≤ M1 M1 A1 A1

5a G1- Shape of Graph + y- Intercept G1-Labels 5b 22 22 22 d d Since 0 and 0, d 0, for all values of .d the function is an increasing function. xx xx xx y eex ee y ee xx − − − = + >> = +> ∴ M1 A1 5c 22 22 22 For equation of function, 11 22 when 0, 3, 3 11 322 xx xx xx yee ye e c xy c ye e − − − = + =−+ = = = ∴= − + ∫ M1 M1 A1 6a (Angle in alternate segment) = = (Given) DAG DCA ECF ECB ∠= ∠ ∠ ∠ M1 A1 6b is midpoint of AC (Given) F is midpoint of DG (Given) ACD is similar to ECF. (Midpoint Theore m) BCA= BDA (Angle in the same segment ADCB) BEC= AED (Angle in the same segment AB) BCE is similar t E ∴ ∠∠ ∠∠ ∴   o DCA. (AA) BCE is similar to FCE.∴   M1 M1 M1 A1 6c Since ACD is similar to ECF. 1 2 Since BCE is similar to ECF 1 2 1 = (Shown)2 AC CD DA EC CF FE BC CE EB EC CF FE EC BE BC FE BC DA AD BC = = = = = ∴×=× = × ×   M1 A1 Alternatively Since BCE is similar to ECF 1 (Midpoint Theorem)2 1 = (Shown)2 BC CE EB EC CF FE EC BE BC FE BC DA AD BC = = ∴×=× = × ×  M1 A1 7 ( ) ( ) ( ) 13 2 32 32 32 32 32 2 2 74 52 2(2 ) 4 7(2 ) 5(2 ) Let y = 2 , 2 47 5 2 7 5 40 Let ( ) 2 7 5 4, ( 1) 2( 1) 7( 1) 5( 1) 4 = 2754 = 0 By Factor Theorem, 1 is a factor of x xx x xx x y yy yyy fy y y y f y + += + += + += + − − += = − −+ − =− −− −−+ −−++ + 32 2 2 ( ). 2 7 5 4 ( 1)(2 4) When y =1, 2 7 5 4 (2)(2 4) 6 (2)( 6) 36 9 ( ) ( 1)(2 9 4) ( 1)(2 1)( 4) () 0 ( 1)(2 1)( 4) 0 11 or or 42 2 1 (Not admissible) or 2 2xx fy y y y y y by b b b b fy y y y y yy fy y yy y yy − − += + + + −−+= ++ −= + −−= =− =+ −+ = + −− = + − −= = −= = =−= 12 or 2 2 1 or 2 x xx − = = −= M1 A1 M1 M1 M1- factorise M1 A1 – rej A1 8a 2 2 2 22 2 2 47 6 16 6 16 3 16 At O, 23, 0, 23 3 16 23 For such that , 4 7 3 16 23 0 2 12 16 0 68 0 ( 2)( 4) 24 A B B B B B AB Vt t at V t dt V t tC Vt C Vt t t VV tt t t tt tt tt t =−+ = − = − =−+ = = = =−+ > −+> − + >−+ >−+ >− − << ∫ M1- - intg M1- C M1 M1 M1 A1

8b 2 32 32 2 32 32 32 32 32 22 4 7 1 273 At O, = 0, 0, 0 1 273 3 16 23 8 23 At O, = 0, 0, 0 8 23 For overtaking, 1 27 82 33 2 6 16 03 2( 9 A AA AA A B BB BB B AB s t t dt s t t tC ts C s ttt s t t dt s t t tC ts C st t t ss t t tt t t tt t tt t = −+ = − ++ = = =−+ = −+ = −++ = = = −+ = − +=− + −+= − ∫ ∫ 22 22 2 24) 0 0 or 9 24 0 Consider 9 24 0 Discriminant = ( 9) 4(1)(24) 0 the only solution to is 0. Hence, there will not be any overtaking after O. AB t tt tt ss t += = −+= −+= −− < ∴= = M1- s A M1 - s B M1 – Solve M1 – D<0 A1 9a 2 ln 2 ln 1d x xx x xxdx  − = +− B1 9bi At , = 0. 2 ln 0 0 or ln 0 0 or 1 1 ay xx xx xx a = = = = = ∴= M1 M1 A1 9bii . 1 1 1 For equation of line , 01gradient of 1 10 is 1 l l ly x −= =−− = −+ M1 A1 9biii ( ) 22 2 1 2 2 1 2 2 1 2 Area needed = 1 = 2 ln (1)(1) 2 1 = ln 1 2 1 1 = ln 22 1 = 2 ln 2 2 - 1 l2 x xdx d x x x x dxdx x xx x x +  − −+ + +−− +  −   ∫  ∫ 2 4 ln 2 11n1 1 22 11 = 22 2 4l n2=1    −  −+ ++ − M1 M1 M1 A1 10a 2 2 2 1 Comparing numerators, 2 ( 1) ( ) when 0, 2 when 1, 2 2 2 22 1 AB r r rr Ar Br r A r B B r r rr = +++ = ++ = = =− =− =− = −++ M1 M1 A1 10b 22 2 2 1 2 2 3 98 99 99 100 22 22 2 2 2 2 ...1 2 2 3 98 99 99 100 22 1 100 1.98 +++ +×× × ×      =−+−+ + − + −           = − =  M1 A1 10c [ ] [ ] 3 21 3 21 3 1 3 1 3 3 d 32 d2 32 2 d21 3 ln ln( 1) 3 ln 3 ln 4 ln 2 323ln 4 33ln 2 3ln 2 27ln 8 rrr rrr rrr rr + = + = − + = −+ = −+ ×=  =  =  =  ∫ ∫ ∫ M1 M1 M1 M1 A1 11a ln ln bty ae y a bt = = + G1-Eqn G1- Table of Values G1- Accurate Graph and Plotting G1- Labels

11b From the graph, 8 ln 8 = 2980.95 = 2980 (to 3 sig. fig.) 80 07 1.1428 1.14 (to 3 sig. fig.) a ae b = = −= − =− =− B1 B1 11c From the graph, When 3, ln 4.4 81.450 81.5 (to 3 sig. fig.) t y y y = = = = M1 A1 11d 3 1ln 3 2 t ey e yt = = − From the graph, M contains a higher level of active ingredients after t = 8 years. M1 G1 – Table of Values G1- Accurate Graph + Plotting A1