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14. (a) Use the substitution u = 3” to show that 2 : . ar =| a (1+3°) p (1+) 0 where k, p and qg are constants to be found. (3) (b) Hence, using algebraic integration, find the exact value of x42 | _#F 4, ; (1 2 3*) (5) 14(a) du 3193 Bl | 1b dx “3 l dx = | — du MI | 1.1b lay Ferree 2: Al | 21 hac (143) ding) (Tuy (3) (b) 2 3B i x 9 [ 1 dx =9 dx or du Bl 3.1 | (1+3") » (143°) In3 J, (I+u) rr [(l+u)° du..(1+u)” MI | Lb —F(l+u)? Al | 1b [-set+9 y (-25 (s1)") dM1 | 1.1b 2In3 2In3 i AL | 21 251n3 () 13(a) J xsin 3x dx = -...xeos3x+...feos3x dx MI 1lb Jxsin3x dx =—...xcos 3x +...sin 3x dM1 1.1b fxsin3x dy =—Txo0s3x+
) - - a ; wax? =p Magy? > [xte* dx = [5 du = = [ure du M1 3.1la dx ax z SS eee . MI om | sf e die e -F Jue dx Al L.lb = Miter See ie e" du M1 1.1lb _3 3 3 . =a wor ee 1 oe 2d = ne = ue a aye +e=agre = +° Ane I Al 2.1 = se" (x°-2x7+2)+e = (x x ) e (5) Alternative: eae dx = fare dx M1 3.1la 6.2.3 an We J 5x? M1 21 [x°x7e a Be —2] ae” dx Al lib [ave dx = =e - 2f xx*e” dx 1 l : : M1 1.1b =—x*%e* —2|—x%e* — [x°e" ax 3 3 1 6s Fox": 22 1s 6 3 =—x"e” ——x%e* +e" +ce=—e (x —2x +2)+e Al 2.1 3 3 3 3 (7 marks) Notes (a) Question Scheme Marks. AOs 3(a) h=0.1 Bl lila A= 22 {1.632 +1.930+2(1.711 +1.786 + 1.859)} MI 1.1b = 0.714 Al 1.1b (3) (b) a f (3f(x)+2) dx =3x"0.714"4... MI 1.1b os os J (3f (x) +2) dx =..4+2%0.4 Mi | 3.1a 05 oe f (3f (+)+2) dv =3="0.714"4 0.8 = 2.942 Alft 2.2a os (3) (6 marks) Notes (a) B1: States or uses fr = 0.1 M1: Correct attempt at the trapezium rule. Must be an attempt at the correct structure e.g. A fy. s+ Yoo +2( a5 + Yor + Yos } with brackets as shown unless they are implied by subsequent work Al: For awrt 0.714 (b) M1: For multiplying their answer to part (a) by 3 M1: For a correct strategy for the “+ 2” part of the integral. May see e.g. 20.4 or 2«(0.9 — 0.5) as or 2 dx =[2x]}? =2x0.9—2x0.5 o$ Alft: For awrt 2.94 or follow through 3 = their answer to part (a) + 0.8 14, (a) Use the substitution w= 1+ sin’x to show that 2 q I’ chic ax=| du o |+sin*x gu(2—w) where p and gq are constants to be found. 6) (b) Hence, using algebraic integration, show that iT (= dx =InA o 1+sin* x where 4 is a rational number to be found. (6) In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. In Harbour X, the depth of the water, H metres, at time ¢ hours after midnight is modelled by the equation at at H=A + Bsin() + ceos( =) oO=rf=< 24 where 4, 8 and C are constants. Given that the depth of the water is = 8.9m at midnight a 4.1m at 06:00 - 3.3m at 09:00 (a) find a complete equation for the model. In Harbour Y, the depth of the water, D metres, at time f hours after midnight is modelled by the equation “. wet awit D=6.8 + 3sin( =) +2c0s( 2) o=zr< 24 (b) Express D in the form 6.8 + Rsin( + a). where R > O and O0O 31) = 0.3085... Bl l.1lb awrt 0.309 (2) (by) (i) 0.0668 (1 — 0.0668)" Mil =0.050665... Al awrt 0.0507 (2) Gi) ¥~BCS5, 0.0668) M1 3.3 P(¥Y >1)=1-Pc(Y =1) Mil 3.4 = 1-— 0.9610... = 0.0390 awrt 0.039 Al l.1b (3) (c) A~NCe, 1.57) PCA > 42) = 0.0005 or P(H =< 42) = 0.9995 Mi Zz = 3.2905268 ... Bl awnrt 3.29 42— M1 2.1 z= [= 3.29 ... uw = 37.0642... Al 1.1b awnrt 37.1 °c (4) (11 marks) Notes: (a)Qi) Bl for 0 no working or justification required Gi) awrt 0.309 (by) M1 pCi —p)* Al awrt 0.0507 Gu) 1* M1 Binomial B(5, 0.0668) 2™ M11 use of correct Binomial to find P(¥ > 1) Al awrt 0.039 (c) Bl awrt 3.29 2" M1 correct standardised expression with = >2 Al awrt 37.1 1 12 Figure 1 Figure 1 shows a Venn diagram with two events, 4 and 8, and their associated probabilities. (a) Explain whether or not events 4 and & are independent. Show your working clearly. G) (b) Find P(8|.4") @) (c) Complete the tree diagrarn in Figure 2 by calculating the probabilities associated with each branch. Bs B B B Figure 2 (4) Turn over for a spare copy of Figure 2 if you need to redraw your tree diagram. 1. Find 1 29. [> (2x -— 5) ax 3 writing each term in simplest form. (4) Question Scheme Marks AOs 1 F a 3 us = i + 2(2x-5 2 2 x? (2x-S)=287 tux? or = (23 J at Ml 1.1b 3 3 , « 2x? 5x7 Al L.lb 3 zo ; 2 > 2x7 5X ae ty? (+c) dM1 1.1b 3 3 5 gf Big J en, Al L.lb 15 (4) (4 marks) Notes 
