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1 . The diagram above shows a cylindrical water tank. The diameter of a circular cross-section of the tank is 6 m. Water is flowing into the tank at a constant rate of 0.48π m min. At time t minutes, the depth of the water in the tank is h metres. There is a tap at a point T at the bottom of the tank. When the tap is open, water leaves the tank at a rate of 0.6π h m min. (a) Show that t minutes after the tap has been opened 75 ( 4 5h) (5) When t 0, h 0. Find the value of t when h 0.5 (6) (Total marks) 9x + 6. (a) Find, x > 0. x () Edexcel Internal Review

2 Given that y 8 at x, solve the differential equation y d (9x + 6) y x giving your answer in the form y g(x). (6) (Total 8 marks). (a) Express 4 y in partial fractions. () Hence obtain the solution of dy cot x (4 y ) for which y 0 at π x, giving your answer in the form sec x g(y). (8) (Total marks) 4. Liquid is pouring into a large vertical circular cylinder at a constant rate of 600 cm s and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is 4000 cm. (a) Show that at time t seconds, the height h cm of liquid in the cylinder satisfies the differential equation 0. 4 k h, where k is a positive constant. () Edexcel Internal Review

3 When h 5, water is leaking out of the hole at 400 cm s. Show that k 0.0 () (c) Separate the variables of the differential equation h, to show that the time taken to fill the cylinder from empty to a height of 00 cm is given by h () Using the substitution h (0 x), or otherwise, (d) find the exact value of 0 0. h (6) (e) Hence find the time taken to fill the cylinder from empty to a height of 00 cm, giving your answer in minutes and seconds to the nearest second. () (Total marks) 5. A population growth is modelled by the differential equation dp kp, where P is the population, t is the time measured in days and k is a positive constant. Given that the initial population is P 0, (a) solve the differential equation, giving P in terms of P 0, k and t. (4) Edexcel Internal Review

4 Given also that k.5, find the time taken, to the nearest minute, for the population to reach P 0. () In an improved model the differential equation is given as dp λp cos λt, where P is the population, t is the time measured in days and λ is a positive constant. Given, again, that the initial population is P 0 and that time is measured in days, (c) solve the second differential equation, giving P in terms of P 0, λ and t. (4) Given also that λ.5, (d) find the time taken, to the nearest minute, for the population to reach P 0 for the first time, using the improved model. () (Total 4 marks) x 6. (a) Express ( x )( x ) in partial fractions. () Given that x, find the general solution of the differential equation (x )(x ) dy (x )y. (5) (c) Hence find the particular solution of this differential equation that satisfies y 0 at x, giving your answer in the form y f(x). (4) (Total marks) Edexcel Internal Review 4

5 7. At time t seconds the length of the side of a cube is x cm, the surface area of the cube is S cm, and the volume of the cube is V cm. The surface area of the cube is increasing at a constant rate of 8 cm s. Show that (a) k, where k is a constant to be found, x V (4) (4) Given that V 8 when t 0, (c) solve the differential equation in part, and find the value of t when V 6. (7) (Total 5 marks) 8. The rate of decrease of the concentration of a drug in the blood stream is proportional to the concentration C of the drug, which is present at that time. The time t is measured in hours from the administration of the drug and C is measured in micrograms per litre. dc (a) Show that this process is described by the differential equation kc, explaining why k is a positive constant. () Find the general solution of the differential equation, in the form C f(t). () Edexcel Internal Review 5

6 After 4 hours, the concentration of the drug in the blood stream is reduced to 0% of its starting value C 0. (c) Find the exact value of k. (4) (Total 8 marks) 9. The volume of a spherical balloon of radius r cm is V cm, where V 4 πr. (a) Find dr () The volume of the balloon increases with time t seconds according to the formula 000, 0. (t + ) t dr Using the chain rule, or otherwise, find an expression in terms of r and t for. () 000 (c) Given that V 0 when t 0, solve the differential equation (t + ) in terms of t. (d) Hence, at time t 5,, to obtain V (4) (i) find the radius of the balloon, giving your answer to significant figures, () (ii) show that the rate of increase of the radius of the balloon is approximately.90 0 cm s. () (Total marks) Edexcel Internal Review 6

7 0. A spherical balloon is being inflated in such a way that the rate of increase of its volume, V cm, with respect to time t seconds is given by V k, where k is a positive constant. Given that the radius of the balloon is r cm, and that V 4 π r, (a) prove that r satisfies the differential equation dr B 5, where B is a constant. r (4) Find a general solution of the differential equation obtained in part (a). () When t 0 the radius of the balloon is 5 cm, and when t the radius is 6 cm. (c) Find the radius of the balloon when t 4. Give your answer to significant figures. (5) (Total marks). Liquid is pouring into a container at a constant rate of 0 cm s and is leaking out at a rate proportional to the volume of the liquid already in the container. (a) Explain why, at time t seconds, the volume, V cm, of liquid in the container satisfies the differential equation 0 kv, where k is a positive constant. () Edexcel Internal Review 7

8 The container is initially empty. By solving the differential equation, show that V A + Be kt, giving the values of A and B in terms of k. (6) Given also that 0 when t 5, (c) find the volume of liquid in the container at 0 s after the start. (5) (Total marks). (a) Use integration by parts to show that π x cosec π π x + x cot x + + ln 6 6 sin x + + c, 6 π π < x <. 6 () Solve the differential equation sin π x + 6 dy xy(y + ) to show that ln y y + π π xcot x + + ln sin x + + c. 6 6 (6) Given that y when x 0, π (c) find the exact value of y when x. (6) (Total 5 marks) Edexcel Internal Review 8

9 . A drop of oil is modelled as a circle of radius r. At time t r 4( e λt ), t > 0, where λ is a positive constant. (a) Show that the area A of the circle satisfies da π λ (e λt e λt ). (5) In an alternative model of the drop of oil its area A at time t satisfies da A, t > 0. t Given that the area of the drop is at t, find an expression for A in terms of t for this alternative model. (7) (c) Show that, in the alternative model, the value of A cannot exceed 4. () (Total marks) 4. (a) Express x (x )( x + ) in partial fractions. (4) Given that y 4 at x, use your answer to part (a) to find the solution of the differential equation dy y ( x), x >.5 (x )( x + ) Express your answer in the form y f(x). (7) (Total marks) Edexcel Internal Review 9

10 5. Fluid flows out of a cylindrical tank with constant cross section. At time t minutes, t 0, the volume of fluid remaining in the tank is V m. The rate at which the fluid flows, in m min, is proportional to the square root of V. (a) Show that the depth h metres of fluid in the tank satisfies the differential equation k h, where k is a positive constant. () Show that the general solution of the differential equation may be written as h (A Bt), where A and B are constants. (4) Given that at time t 0 the depth of fluid in the tank is m, and that 5 minutes later the depth of fluid has reduced to 0.5 m, (c) (d) find the time, T minutes, which it takes for the tank to empty. Find the depth of water in the tank at time 0.5T minutes. () () (Total marks) 6. Liquid is poured into a container at a constant rate of 0 cm s. At time t seconds liquid is leaking from the container at a rate of 5 V cm s, where V cm is the volume of liquid in the container at that time. (a) Show that 5 V 450. () Given that V 000 when t 0, find the solution of the differential equation, in the form V f(t). (7) (c) Find the limiting value of V as t. () (Total marks) Edexcel Internal Review 0

11 . (a) 0.48π 0.6πh M A V 9πh 9π B 9π 0.48π 0.6πh M Leading to h cso A 5 75 d h d t 4 5h separating variables M 5ln ( 4 5h) t ( + C) M A 5ln ( 4 5h) t + C When t 0, h 0. 5ln C M t 5ln 5ln 4 5h ( ) When h 0.5 t 5ln 5ln.5 5ln 5ln.5 awrt 0.4 M A Alternative for last marks t [ 5ln (4 5h) ] ln.5 + 5ln M M 5ln 5n awrt 0.4 A 6.5 [] 9x (a) 9 x + M x 9x + 6ln x (+C) A Edexcel Internal Review

12 9x + 6 dy Integral signs not necessary B y x y d y 9 x + 6 x y 9x + 6 ln x ( + C) ± ky their (a) M y 8, x y 9x + 6 ln x + ( C) ft their (a) Aft ln + C C y + ( 9x 6ln x ) ( ) y (6x + 4ln x ) 8( x + ln x ) M A A 6 [8]. (a) A B + 4 y ( y)( + y) ( y) ( + A( + y) + B( y) Let y, B(4) B y ) Let y, A(4) A giving + ( y ) ( + y) Forming this identity. NB: A & B are not assigned in this question Either one of A or B M A +, aef A cao ( y ) ( + y) (If no working seen, but candidate writes down correct partial fraction then award all three marks. If no working is seen but one of A or B is incorrect then M0A0A0.) Edexcel Internal Review

13 dy 4 y cot x + ( y) ( + y) tan x ln( y) + ln( + y) ln(sec x) + ( + c) π y 0, x ln + ln ln + c π ( ) cos {0 ln + c c ln } ln( y) + ln( + y) ln(sec x) ln + y sec x ln ln y + y sec x ln ln y + y sec x ln ln y + y sec x y y Hence, sec x y Separates variables as shown. Can be implied. Ignore the integral signs, and the. ln(secx)or ln(cosx) Either ±a ln(λ y) or ± b ln(λ + y) M; their x cot x d LHS correct with ft for their A and B and no error with the with or without +c B B Aft Use of y 0 and x π in an integrated equation containing c; M* Using either the quotient (or product) or power laws for logarithms CORRECTLY. Using the log laws correctly to obtain a single log term on both sides of the equation. M dm* 8 + 4y sec x A aef 8 y [] Edexcel Internal Review

14 4. (a) 600 c h or 600 k h, (V 4000 h ) c h 600 c h Either, 0. 4 k k h 600 k h or 0. 4 k h h Either of these statements 4000 or Convincing proof of 4000 M M A AG When h 5 water leaks out such that c h 400c c(5) c 80 c 80 From above; k as required 4000 Proof that k 0.0 B AG Aliter Way k h k k(0000) k Using 400, 4000 and h 5 or h 5. Proof that k 0.0 B AG (c) 0.4 k h 0.4 k h 00 time required h time required h 0 h d Edexcel Internal Review 4

15 Separates the variables with and 0.4 k h with integral signs not necessary. on either side M oe Correct proof A AG (d) h 0 h d with substitution h (0 x) 0 (0 x)( ) or (0 x) h (0 x) h 0 x x (0 x) 0 h x x 0 00 x 0 00 x 00(x 0 ln x) (+ c) h change limits: when h 0 then x 0 and when h 00 then x [ 00x 000 ln x] h or 0 0 h ( ln 0) ( ln 0) 000 ln ln ln 000 [ 00(0 h) 000 ln(0 h) ] 00 0 Correct use of limits, ie. putting them in the correct way round Either x 0 and x 0 or h 00 and h 0 ddm Combining logs to give ln 000 or 000 ln 000 A aef 6 (e) Time required 000 ln sec 86 seconds (nearest second) 6 minutes and 6 seconds (nearest second) 6 minutes, 6 seconds B [] Edexcel Internal Review 5

16 dp 5. (a) kp and t 0, P P 0 () dp k P ln P kt; (+c) When t 0, P P 0 ln P 0 c (or P Ae kt P 0 A) In P kt + ln P 0 e ln P kt+ ln P0 kt ln P0 e e.e Hence, P P 0 e kt 4 M dp Separates the variables with and k P on either side with integral signs not necessary. A Must see ln P and kt; Correct equation with/without + c. M Use of boundary condition () to attempt to find the constant of integration. A P P 0 e kt P P 0 e kt written down without the first M mark given scores all four marks. Aliter Way dp kp and t 0, P P 0 () dp kp ln P t; ( + c) k When t 0, P P 0 k ln P0 c (or P Ae kt P 0 A) ln P t + ln P0 ln P kt + ln P k k ln P kt+ ln P0 kt ln P0 e e e.e Hence, P P 0 e kt dp kp with integral signs not necessary. M Separates the variables with and d t A Must see P k ln M 0 on either side and t; Correct equation with/without + c. Use of boundary condition () to attempt to find the constant of integration. Edexcel Internal Review 6

17 A P P 0 e kt P P 0 e kt written down without the first M mark given scores all four marks. Aliter Way dp kp ln( kp) t; ( + c) k When t 0, P P 0 ln( kp0 ) c k (or kp Ae kt kp 0 A) ln( kp) t + ln( kp0 ) ln( kp) kt + ln( kp0 ) k k ln( kp) kt+ ln( kp0 ) kt ln( kp0 ) e e e. e kp e kt.(kp 0 ) kp kp 0 e kt (or kp kp 0 e kt ) Hence, P P 0 e kt dp kp M Separates the variables with and d t integral signs not necessary. on either side with A Must see ln( kp) and t; Correct equation with/without + c. k M Use of boundary condition () to attempt to find the constant of integration. P P 0 e kt written down without the first M mark given scores all four marks. P P 0 & k.5 P 0 P 0 e.5t e.5t ln e.5t ln or.5t ln...or e kt ln e kt ln or kt ln t ln days.5 t minutes t 99 min or t 6 hr 9 mins (to nearest minute) M Substitutes P P 0 into an expression involving P M A Eliminates P 0 and takes ln of both sides awrt t 99 or 6 hr 9 mins Edexcel Internal Review 7

18 (c) dp λp cos λt and t 0, P P 0 () dp λ cos λt P ln P sin λt; (+c) When t 0, P P 0 ln P 0 c (or P Ae sin λt P 0 A) ln P sin λt + ln P 0 e ln P sin λt+ ln P0 sin λt ln P0 e e.e Hence, P P 0 e sin λt 4 M dp Separates the variables with and λ cosλt on either side P with integral signs not necessary. A Must see ln P and sin λi; Correct equation with/without + c. M Use of boundary condition () to attempt to find the constant of integration. A sin kt P P 0 e P P 0 e sin kt written down without the first M mark given scores all four marks. Aliter Way dp λp cos λt and t 0, P P 0 () dp cos t P λ λ λ ln P sin λt; ( + c) λ λ When t 0, P P 0 λ ln P0 c (or P Ae sin λt P 0 A) ln P sin λt + ln P λ λ λ ln P sin λt+ ln P sin e e e sin λt Hence, P P 0 e 0 0 λt ln P0 ln P sin λt + ln P M Separates the variables with and cosλt. e dp λp with integral signs not necessary. 0 on either side A Must see ln P and sin λt ; Correct equation with/without + c. λ λ M Use of boundary condition () to attempt to find the constant of integration. Edexcel Internal Review 8

19 A sin kt P P 0 e P P 0 e sin kt written down without the first M mark given scores all four marks. Aliter Way dp λp cos λt and t 0, P P 0 () dp cos t P λ λ λ ln( λp) sin λt; ( + c) λ λ When t 0, P P 0 λ ln (λp0 ) c (or λp Ae sin λt λp 0 A) ln( λp) sin λt + ln P λ λ λ ln( λp) sin λt + ln( λp0 ) ln( λp) sin λt+ ln( λp ) e e e λp e sin λt.(λp 0 ) (or λp λp 0 e sin λt ) sin λt Hence, P P 0 e 0 sin λt 0. e ln( λp ) dp λp with integral signs not necessary. M Separates the variables with and cos t 0 λ on either side A Must see ln( λp) and sin λt ; Correct equation with/without + c. λ λ M Use of boundary condition () to attempt to find the constant of integration. A sin kt P P 0 e sin.5t (d) P P 0 & λ.5 P 0 P 0 e e sin.5t sin.5t ln...or... e λt sin λi ln t sin (ln ).5 t t minutes t 44 min or t 7 hr mins (to nearest minute) M Eliminates P 0 and makes sin λt or sin.5t the subject by taking ln s dm Then rearranges to make t the subject. (must use sin ) Edexcel Internal Review 9

20 A awrt t 44 or 7 hr mins [4] 6. (a) x A B + ( x )(x ) ( x ) (x ) x A(x ) + B(x ) Forming this identity. NB: A & B are not assigned in this question Let x, B( ) B 4 Let x, A( ) A 4 giving + ( x ) (x ) M either one of A or B 4. A both correct for their A, B. A & (c) dy y (x ) (x )( x ) 4 + ( x + ) (x ) ln y ln(x ) + ln(x ) + c Separates variables as shown Can be implied B Replaces RHS with their partial fraction to be integrated. Mft At least two terms in ln s M At least two ln terms correct Aft All three terms correct and + c A 5 y 0, x gives c ln 0 ln y ln(x ) + ln(x ) + ln 0 ln y ln(x ) + ln(x ) + ln 0 (x ) ln y ln + ln0 ( ) or x 0(x ) ln y ln ( x ) 0(x ) y ( x ) c ln0 B Edexcel Internal Review 0

21 Using the power law for logarithms Using the product and/or quotient laws for logarithms to obtain a single RHS logarithmic term with/without constant c. M M 0(x ) y or aef. isw Aaef 4 ( x ) Aliter Way dy y (x ) (x )( x ) Separates variables as shown Can be implied 4 + ( x ) (x ) Replaces RHS with their partial fraction to be integrated. ln y ln(x ) + ln(x ) + c At least two terms in ln s At least two ln terms correct All three terms correct and + c Decide to award B here!!! B Mft M Aft A B Note: The B mark (part (c)) should be awarded in the same place on epen as in the Way approach. ln y ln(x ) + ln(x ) + c Using the power law for logarithms ln y (x ) ln x + c Using the product and/or quotient laws for logarithms to obtain a single RHS logarithmic term with/without constant c. M M A(x ) ln y ln x ln y (x ) ln + c x or e e e A(x ) y ( x ) y 0, x gives A 0 0(x ) y ( x ) where c ln A (x ) ln x e c Edexcel Internal Review

22 A 0 for B award above 0(x ) y or aef & isw Aaef 5&4 ( x ) Aliter Way dy (x ) y (x )( x ) + ( x + ) ( x ) ln y ln(x ) + ln( x ) + c Separates variables as shown Can be implied B Replaces RHS with their partial fraction to be integrated. Mft At least two terms in ln s M At least two ln terms correct Aft All three terms correct and + c A 5 y 0, x gives c ln 0 ln ln 40 ln y ln(x ) + ln( x ) + ln 40 ln y ln(x ) + ln( x ) + ln 0 ( x ) ln y ln + ln 40 or ( x ) 40( x ) ln y ln ( x ) 40( x ) y ( x ) c ln 0 ln or c ln 40 B oe Using the power law for logarithms M Edexcel Internal Review

23 Using the product and/or quotient laws for logarithms to obtain a single RHS logarithmic term with/without constant c. 40( x ) y or aef. isw Aaef 4 ( x ) M Note: Please mark parts and (c) together for any of the three ways. [] 7. (a) d S From question, 8 B ds S 6x x B ds ds 8 ; ( k ) x x d S ds 8 Candidate s ; M; A oe 4 x V x x B d V x. ; x x d V Candidate s ;λx M;Aft d As x V, then V V AG A Use of x V, to give V (c) V B Separates the variables with or V V on one side and on the other side. integral signs not necessary. V V t ( + c ) Edexcel Internal Review

24 Attempts to integrate and must see V and t; M; Correct equation with/without + c. A ( 8) (0) + c c 6 Use of V 8 and t 0 in a changed M; equation containing c ; c 6 A Hence: V t + 6 ( 6 ) + t + 6 t 6 depm Having found their c candidate substitutes V 6 into an equation involving V, t and c. giving t Acao 7 [5] AliterWay x V & S 6 x S 6V Bft ds 4V or ds V 4 B ds 8. ; V ds 4V V AG M; A 4 ds Candidate s ; V ds Aliter Way B V Separates the variables with or V oe on one V side and on the other side. integral signs not necessary. (c) V ( )( ) V t ( + c ) Attempts to integrate and must see V and t; M; Edexcel Internal Review 4

25 Correct equation with / without + c. A ( 8) (0) + c c 4 Use of V 8 and t 0 in a changed equation containing c; c M*; A Hence: V t + 4 ( 6 ) + 4 t + 6 t depm * Having found their c candidate substitutes V 6 into an equation involving V, t and c. giving t. A cao 7 Aliter Way similar to way. V x x B d V ds x.8. ; x ds x ds Candidate s ;λx M;Aft ds d As x V, then V A 4 Use of x V, to give V Aliter Way V B Separates the variables with or V oe on one V side and on the other side. integral signs not necessary. (c) Edexcel Internal Review 5

26 V V 4 t ( + c) Attempts to integrate and must see V and 4 t; M; Correct equation with / without + c. A ( ) 4 (0) + c c 4 8 Use of V 8 and t 0 in a changed equation containing c; c 4 M;A 4 Hence: V t + 4 ( 4 6 ) 4 t t + 4 depm Having found their c candidate substitutes V 6 into an equation involving V, t and c. giving t. A cao 7 8. (a) dc kc ; rate of decrease/negative sign; k constant of proportionality/positive constant B dc k C d t ln C kt + lna M C Ae kt A M (c) At t 0 C C 0, A C 0 B and at t 4 C 0 C 0, 0 C 0 C o e 4k, B 0 e 4t and 4k ln 0, k 4 ln 0 M, A 4 [8] Edexcel Internal Review 6

27 9. (a) dr 4πr B Uses dr. dr in any form, 000 4πr (t + ) M, A (c) V 000 (t + l) and integrate to p(t + ), 500(t + ) (+c) M, A Using V 0 when t 0 to find c, (c 500, or equivalent) M V 500( ) t + (any form) A 4 (d) (i) Substitute t 5 to give V, then use r V 4 π to give r, 4.77 M, A (ii) Substitutes t 5 and r their value into their part M d r (.90 0 ) (cm/s) AG A [] 4 0. (a) As V πr, then 4πr M dr dr k Using chain rule ; M A 4 dr πr 4πr B (*) A 4 5 r 5 r dr B B 6 r 6 Bt + c (allow mark at this stage, does not need r ) M A Edexcel Internal Review 7

28 5 6 (c) Use r 5 at t 0 to give c 6 or 604 or 600 M Use r 6 at t to give B or 586 or 588 or 590 M Put t 4 to obtain r 6 (approx 78000) M A Then take sixth root to obtain r 6.5 (cm) A 5 []. (a) is the rate of increase of volume (with respect to time) B kv : k is constant of proportionality and the negative shows decrease (or loss) giving 0 kv (*) B These Bs are to be awarded independently 0 kv separating variables M k ln(0 kv) t (+C) M A Using V 0, t 0 to evaluate the constant of integration c k ln 0 M 0 t ln k 0 kv Obtaining answer in the form V A + B e kt M -kt V 0 e k k A 6 Accept 0 ( e kt ) k Alternative to Using printed answer and differentiating kb e kt M Substituting into differential equation kb e kt 0 ka kb e kt M 0 A k M A Using V 0, t 0 in printed answer to obtain A + B 0 M 0 B k A 6 Edexcel Internal Review 8

29 (c) 0, t 5 0 0e kt k ln M A 75 At t 0, V ln M A 5 awrt 08 []. (a) I x cosec π π (x + ) xd( cot(x + )) 6 6 M π π x cot(x + ) + cot(x + ) 6 6 A π π x cot( x + ) + ln(sin( x + ) / c (*) 6 6 Ac.s.o. π ) y( + y) 6 M LHS dy y + y M A y lny ln + y or ln (a) + y M M y π π ln x cot x + + ln sin x + + c (*) + y 6 6 Ac.s.o. 6 xcosec ( x + (c) y, x 0 ln ln sin 6 M c ln A M x y π ln. + ln ln + y A "c" y π (i.e. ln ) + y 6 π y + e 6 y (o.e.) π y( e 6 ) M Edexcel Internal Review 9

30 y π e 6 (o.e.) A 6 [5]. (a) A πr d r, 4λe λt B, B d A d r d A πr, π 4( e λt ) 4λe λt M, M d A πλ( e λt e λt ) Acso 5 A da t M Separation A t (+ c) M, A + c Use of (, ) M c A So A t + A Attempt A or A M t + t 4t i.e. A (or equivalent) A 7 ( + t) t (c) Because ( + t) < or t < ( + t) ( A < 4) B [] 4. (a) A B Uses + ( x ) ( x + ) M Considers x + A (x + ) + B (x ) and substitutes x M or x.5, or compares coefficients and solves simultaneous equations To obtain A 4 and B. A, A 4 Separates variables 4 dy d y x x + x ln y ln(x ) ln(x + ) + C Substitutes to give ln 4 ln ln + C and finds C (ln 08) ln y ln(x ) ln(x + ) (+ ln 08) M A, B ft M M Edexcel Internal Review 0

31 C(x ) ln ( x + ) y 08(x ) ( x + ) Or y e ln(x ) ln(x+)+ln08 special case M A A A cso 7 [] 5. (a) ± c V or V M As V Ah, A or V h M c c use chain rule to obtain V A A h k h A h k d t h A kt M< M A A kt h h (A Bt) (*) A 4 (c) t 0, h : A B t 5, h 0.5: 0.5 ( 5B) h 0, t A B B ( 0.5) 5 B ) 5 7. min B 0.5 B (d) h A 0.5 m M A 4 [] 6. (a) 0 V M A V, no wrong working seen A* d t Edexcel Internal Review

32 Separating the variables Integrating to obtain 5 M V ln V 450 t OR 5 Using limits correctly or finding c ( ln550 OR V 450 ln Rearranging to give V ln V 5 t dm A 5 ln775) M t, or equivalent A t 775e. dm A 7 (c) V 5 B [] Edexcel Internal Review

33 . Many found part (a) difficult and it was quite common to see candidates leave a blank space here and proceed to solve part, often correctly. A satisfactory proof requires summarising the information given in the question in an equation, such as 0.48π 0.6πh, but many could dt not do this or began with the incorrect 0.48π 0.6πh. Some also found difficulty in obtaining a correct expression for the volume of water in the tank and there was some confusion as to which was the variable in expressions for the volume. Sometimes expressions of the form V π rhwere differentiated with respect to r, which in this question is a constant. If they started appropriately, nearly all candidates could use the chain rule correctly to complete the proof. Part was often well done and many fully correct solutions were seen. As noted in the introduction above, some poor algebra was seen in rearranging the equation but, if that was done correctly, candidates were nearly always able to demonstrate a complete method of solution although, as expected, slips were made in the sign and the constants when integrating. Very few candidates completed the question using definite integration. Most used a constant of integration (arbitrary constant) and showed that they knew how to evaluate it and use it to complete the question.. Part (a) of this question proved awkward for many. The integral can be carried out simply by decomposition, using techniques available in module C. It was not unusual to see integration by parts attempted. This method will work if it is known how to integrate ln x, but this requires a further integration by parts and complicates the question unnecessarily. In part, most could separate the variables correctly but the integration of, again a C topic, was frequently y incorrect. Weakness in algebra sometimes caused those who could otherwise complete the question to lose the last mark as they could not proceed from y 6x + 4n x to y (6x + 4n x ). Incorrect answers, such as y 6x + 64 ln x 8, were common in otherwise correct solutions.. In part (a), many candidates realised that they needed to factorise the denominator to give two linear factors, and usually proceeded to give a fully correct solution. A few candidates, however, thought that 4 y was an example of a repeated linear factor and tried to split up their fraction up accordingly. Some candidates struggled with factorising 4 y giving answers such as (4 + y)(4 y) or (y + )(y ). The majority of candidates were able to write down the correct identity to find their constants, although a noticeable number of candidates, when solving 4A found A. Edexcel Internal Review

34 A significant minority of candidates who completed part (a) correctly made no attempt at part. About half of the candidates in part were able to separate out the variables correctly. Many of these candidates spotted the link with part (a). It was pleasing that candidates who progressed this far were able to correctly integrate tan x and correctly find the two ln terms by integrating their partial fraction. Common errors at this point were integrating tan x to give sec K x and the sign error involved when integrating. A significant number of candidates y at this point did not attempt to find a constant of integration. Other candidates substituted x π and y 0 into an integrated equation which did not contain a constant of integration. A majority of candidates who found the constant of integration struggled to simplify their equation down to an equation with a single ln term on each side. The most common error of these candidates was to believe that ln A + ln B ln C implies A + B C. Of all the 8 questions, this was the most demanding in terms of a need for accuracy. Fewer than 0% of candidates were able to score all marks in this question, although statistics show that about half of the candidates were able to score at least 5 marks. 4. This proved by far the most difficult question on the paper and discriminated well for those candidates who were above the grade Athreshold for this paper. Only a few candidates were able to score above 8 or 9 marks on this question. Many fudged answers were seen in part (a). A more rigorous approach using the chain rule of was required, with candidates being expected to state and (or its reciprocal). The constant of proportionality also proved to be a difficulty in this and the following part. Few convincing proofs were seen in part with a significant number of candidates not understanding how to represent 400 cm s algebraically. Only a minority of candidates were able to correctly separate the variables in part (c). Far too often, expressions including 0.0 h 0.4 were seen by examiners. There were a significant number of candidates who having written could not progress to 0.4 k h 50 the given answer by multiplying the integral on the left hand side by. 50 Edexcel Internal Review 4

35 Despite struggling with the previous three parts, a majority of candidates were able to attempt part (d), although only a few candidates were able to produce the correct final exact answer. A majority of candidates who attempted this part managed to correctly obtain x 40 and then use this and the given substitution to write down an integral in x. At this point a significant x 0 number of candidates were unable to manipulate the expression k into an expression x 0 of the form k. The converted limits x 0 and x 0,caused an added problem for x those candidates who progressed further, with a significant number of candidates incorrectly applying x 0 as their lower limit and x 0 as their upper limit. A time of 6 minutes 6 seconds was rarity in part (e). 5. Many candidates, who answered part (a), were able to separate the variables correctly and integrate both sides of their equation to obtain ln P kt. At this point a significant number of candidates either omitted the constant of integration or were unable to deal with the boundary conditions given in the question. Some candidates, for example, wrote down P e kt + c; and stated that c P 0 to give the common incorrect solution of P P 0 + e kt. Other candidates used P 0 instead of P in their attempts, and then struggled to find the constant of integration. Some candidates, who correctly evaluated the constant of integration, did not make P the subject of the equation but left their answer as ln P kt + ln P 0. Those candidates who had successfully answered part (a) were able to gain most of the marks available in part. A few of these candidates, however, struggled to convert the correct time in hours to the correct time in minutes. Those who did not progress well in part (a) may have gained only a method mark in part by replacing P in their part (a) equation with P 0. Those candidates who were successful in the first two parts of this question usually succeeded to score most of the marks available in parts (c) and (d). In part (c) some candidates incorrectly integrated λ cos λt. In part (d), a significant number of candidates found difficultly in solving the equation sin(.5t) ln. It was not uncommon for some of these candidates to write t ln. Also, in part (d), some candidates did not work in radians when evaluating sin(.5) t arcsin(ln ). There were a significant minority of candidates who tackled this question with ease scoring at least thirteen of the fourteen marks available. Edexcel Internal Review 5

36 6. In part (a), nearly all candidates were able to form the required partial fractions accurately and efficiently. Most candidates substituted x and x into the identity x A(x ) + B(x ), but the cover-up rule and the method of equating coefficients were also used. In part, some candidates failed to proceed much further, through not knowing how to separate the variables or not recognising that their solution to part (a) provided a clue as how to proceed to solve the differential equation. Many candidates were able to write the general solution as an equation involving three logarithmic terms and a constant of integration c. Some candidates, however, omitted this 4 constant, whilst other candidates incorrectly integrated to give either 4 ln x or x ln x. In part (c), a majority of candidates realised that they were required to find the constant of integration and were able to evaluate this constant correctly. Only a minority of candidates, however, were able to use the laws of logarithms in the correct order to give their particular solution in the form y f(x). There were a significant number of candidates who arrived at the (x ) incorrect particular solution of y + 9. ( x ) 7. A significant number of candidates found parts (a) and difficult although other candidates ds answered these two parts of the question with ease. Those candidates who used in ds part (a) and in part managed better than those candidates who worked with V and S or V, S and x. The most common error in these parts was for candidates to incorrectly quote the surface area S, as x or 4x instead of 6x. Part (c) was tackled better than the rest of the question with many candidates recognising the need to separate the variables, integrate, find the constant of integration and substitute for V. Many candidates were able to score full marks easily on this part. There was, nevertheless, plenty of scope for errors to occur at all stages in the solution. Those who separated out V 4 frequently simplified this to V. After integration incorrect expressions such as V, V and ln V all regularly appeared. A number of candidates did not use a constant of integration. Other candidates found difficultly in working with ( 6 ). 8. Part (a) specifically asked the candidates to explain why k was a positive constant. Although candidates linked the negative sign with the rate of decrease, they often did not then explain how this related to k. When explanations were given they tended to be verbose. Edexcel Internal Review 6

37 Candidates who used C as a constant of integration in part often confused themselves. It was not unusual to see no constant of integration or to see the correct statement ln C kt + A leading to the incorrect statement C e kt + e A. It seemed that candidates were familiar with an exponential decay result of the form C Ae kt but were not necessarily sure how to get to this result from a given differential equation. Errors in method were seen in part (c) when candidates had omitted the constant of integration in part and full marks were only given to a full and correct solution. Many candidates ignored the given starting value of C C 0 at time t 0 and instead used, for example, C 0 or C This was allowed provided the candidates went on to use C 0. their value of C 0 at time t 4. The most frequent error was the use of 0.9C 0 or its equivalent. This was another question where some candidates ignored the requirement for an exact value. 9. The fact that this question had so many parts, with a good degree of independence, did enable the majority of candidates to do quite well. All but the weakest candidates scored the first mark and the first were gained by most. The integration in part (c ) did cause problems: examples of the more usual mistakes were to write k or or or, or to omit the constant of integration or (t + ) t + t + t + (t + ) assume it equal to zero; two of the mistakes which came more into the howler category were ln000v or V ln000 and 000 d ( + ) t t 4t + 4t + 4t t Many candidates were able to gain the method marks in parts (d) and (e). 0. This was well done on the whole but there were some slips simplifying the algebra in part (a) and a significant number of candidates persevered with complicated constants, involving π, throughout the question. The separation of variables and solution of the differential equation was answered well. In c) most candidates had some success trying to find their constants, but a number found the arithmetic difficult and did not obtain a correct answer. Edexcel Internal Review 7

38 . This proved by far the most difficult question on the paper. When an explanation is asked for an equation, the candidate must make specific reference to the elements of that equation in their explanation. For example, d V, needed to be explicitly identified with the rate of change (or increase) of volume with respect to time. The loss of liquid due to leakage at a rate proportional to the volume needed to be identified with the term kv. Few could make any progress at all with part. The question was at a level of abstraction unexpected by most candidates and there was much confusion between the various constants, k, A and B, occurring in the question. After an attempt at part (a), it was not uncommon for candidates to simply give up. The majority were unable to separate the variables. When separation and integration were achieved the constant of integration, if it was recognised at all, tended to be confused with k or subsumed into A or B. Those who got an answer in the correct form in part, even if it was incorrect, could usually demonstrate a correct method in part (c) and obtain some credit.. Candidates found this question challenging; however those who read all the demands of the question carefully were able to score some marks, whilst quite an appreciable minority scored them all. In part (a), the crucial step involved keeping signs under control. Seeing x x x cot x + cot x or a correct equivalent, demonstrated to examiners a clear method. Sign confusions sometimes led to a solution differing from the printed answer. Part was the main source for the loss of marks in this question. It was disappointing that so many candidates rushed through with barely more than three lines of working between separating the variables and quoting the printed answer, losing the opportunity to demonstrate their skills in methods of integration. The majority separated the variables correctly. Very few made any attempt to include the critical partial fractions step, merely stating dy ln y( y + ) + y y as printed. Some did not recognise the right hand side of their integral related to part (a), producing copious amounts of working leading to nowhere. In part (c), the working to evaluate the constant c was often untidy and careless. Those who persevered to a stage of the form ln P Q + R generally were unable to move on to P e Q+R in a satisfactory manner, often writing P e Q + e R. Edexcel Internal Review 8

39 . There were two common approaches used in part (a); substituting for r to obtain a formula for A in terms of t or using the chain rule. The inevitable errors involving signs and λ were seen with both methods and the examiners were disappointed that some candidates did not seem to know the formula for the area of a circle: πr, πr and 4πr were common mistakes. Part proved more testing. Most could separate the variables but the integration of negative powers caused problems for some who tried to use the ln function. Many did solve the differential equation successfully though sometimes they ran into difficulties by trying to make A the subject before finding the value of their arbitrary constant. The final two marks were only scored by the algebraically dexterous. There was some poor work here and seeing A t + 4 A followed by + or t + was not uncommon. The final part eluded most. A t Those who had a correct answer to part sometimes looked at the effect on A of t but only a small minority argued that since t >0 then t < ( + t), and therefore A <4. 4. The partial fractions were found easily by most of the candidates, with very few errors. Some had difficulty separating the variables, but it was still possible to continue with the answer and to obtain some credit. The log integration was performed well this time and the majority of difficulties were finding the correct constant and using it correctly in conjunction with combining the logs. 5. The first part of this proved to be particularly troublesome. M was given quite often for kv, and a number recognised that the constant cross section implied that V was proportional to r. Many candidates had elements of the correct method, but they failed to link these together to form a convincing argument. Several candidates however were not able to make use of the information about the constant cross section, or to interpret the rate and the proportionality. In part most candidates did try to separate the variables. Integrating / h caused problems, with many candidates giving an answer involving ln h, and with better attempts losing the factor. The constant of integration was often missing. Several candidates failed to make a convincing link between their answer and the form given on the question paper. The standard of algebra and arithmetic in the last two parts was disappointing. Most candidates did make some attempt to substitute the given values, but few reached correct answers. Many found A, then when finding B expanded the expression into a quadratic which then they tried to solve using the formula. Others square rooted both sides instead of squaring to remove the square root. Many candidates substituted 0.5 rather than 0.5T in part (d), indicating lack of comprehension of the question. Edexcel Internal Review 9

40 6. This was found to be the most difficult question on the paper. Setting up the differential equation in part (a) was usually answered well, as the answer was printed. The separation of variables proved difficult and this meant that part was virtually inaccessible for many candidates. Some of those who correctly separated the variables had difficulties with signs, having presumably not noticed the boundary conditions. Another group did not deal correctly with the constant of integration, leaving it as part of the exponential. It was possible to obtain the answer to part (c) from the original question, even if part had proved impossible. Some (Further Mathematics) candidates did use the integrating factor method to answer part. Edexcel Internal Review 40