June 2005 - 6666 Core C4 - Mark scheme - Edexcel

1 June 2005. 6666 Core C4. Mark scheme Question scheme Marks Number 1. 9x 2. 4 9x . 1. 1. 2. 2 1 B1. 4 . 12 9 x 12 12 9 x 2 1. 12 32 9x . 3.. 2 1 . 2.. M1. 1 4 4 4 .. 9 81 2 729 3 . 2 1 x x x .. 8 128 1024 . 9 81 2 729 3. 2 x, x , x .. A1, A1, A1. 4 64 512. [5]. 1. 12 1. 12 23 .. 2 2 2 3. Note The M1 is gained for or Special Case 9 81 2 729 3 . If the candidate reaches 2 1 x x x .. and goes no further 8 128 1024 . allow A1 A0 A0. 6666 Core C4 1. June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics Question scheme Marks Number dy dy 2. 2x 2x 2 y 6 y 0 M1 (A1) A1. dx dx dy 0 x y 0 or equivalent M1. dx Eliminating either variable and solving for at least one value of x or y. M1. y 2 2 y 2 3 y 2 16 0 or the same equation in x y 2 or x 2 A1. 2, 2 , 2, 2 A1. [7]. dy x y Note.

2 Dx 3 y x Alternative 3 y 2 2 xy x 2 16 0. 2 x 16 x 2 192 . y . 6. dy 1 1 8x . dx 3 3 16 x 2 192 M1 A1 A1. dy 8x 0 1. 16 x 192 . dx 2 M1. 64 x 2 16 x 2 192. x 2 M1 A1. 2, 2 , 2, 2 A1. [7]. 6666 Core C4 2. June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics Question scheme Marks Number 5x 3 A B. (a) . 3. 2 x 3 x 2 2 x 3 x 2. 5 x 3 A x 2 B 2 x 3 . Substituting x 2 or x 32 and obtaining A or B; or equating coefficients and M1. solving a pair of simultaneous equations to obtain A or B. A 3, B 1 A1, A1. (3). If the cover-up rule is used, give M1 A1 for the first of A or B found, A1 for the second.. 5x 3 3. (b) dx ln 2 x 3 ln x 2 M1 A1ft 2 x 3 x 2 2.. 3 ln 9 ln 2. 6. M1 A1. 2 2. ln 54 cao A1 (5). [8]. 6666 Core C4 3. June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics Question scheme Marks Number.

3 1 1 dx dx cos d Use of x sin and cos . 1 x2 1 sin 2 . 4. 1. 2. 3. 2 d M1.. 1. d M1 A1. cos 2 . sec 2 d tan M1 A1.. Using the limits 0 and 6 to evaluate integral M1. 1 3 . tan 0.. 6. cao A1. 3 3 . [7]. Alternative for final M1 A1. 1. Returning to the variable x and using the limits 0 and 2 to evaluate integral M1. 1. x 2 1 3 . cao A1. 1 x 0 . 2. 3 3 . 6666 Core C4 4. June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics Question scheme Marks Number xe e 1 1. dx x e . 2x 2x 2x 5. (a) dx Attempting parts in the right direction M1 A1. 2 2. 1 1. x e2 x e2 x A1. 2 4. 1. 1 2x 1 2x 1 1 2. 2 x e 4 e 4 4 e M1 A1. 0. (5). (b) x y 22. x y 43 Both are required to 5 B1.. (1). I .. 1. (c) B1. 2.. 0+ 06+2 +.890 22+ 07+ 43 M1 A1ft ft their answers to (b). 22. cao A1 (4). [10]. 1 1 2.

4 Note e . 4 4. 6666 Core C4 5. June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics Question scheme Marks Number dx dy 6. (a) 2 cosec 2 t , 4sin t cos t both M1 A1. dt dt d y 2sin t cos t dx . cosec 2 t 2sin 3 t cos t M1 A1. (4). (b) At t 4 , x 2, y 1 both x and y B1. dy 1 . Substitutes t 4 into an attempt at to obtain gradient M1. dx 2 . Equation of tangent is y 1 . 1. 2. x 2 M1 A1. Accept x 2 y 4 or any correct equivalent (4). (c) Uses 1 cot 2 t cosec2 t , or equivalent, to eliminate t M1. 2. x 2. 1 correctly eliminates t A1. 2 y 8. y cao A1. 4 x2. The domain is x 0 B1 (4). [12]. An alternative in (c). 1 1. y 2 x x y 2. sin t ; cos t sin t . 2 2 2 2 . y x2 y sin 2 t cos 2 t 1 1 M1 A1. 2 4 2. 8. Leading to y A1. 4 x2. 6666 Core C4 6. June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics Question scheme Marks Number 7.

5 (a) k component 2 4 2 1 M1 A1. Note 2. Substituting their (or ) into equation of line and obtaining B M1. B: 2, 2, 2 Accept vector forms A1. (4). 1 1 .. (b) 1 18; 1 2 both B1. 4 0 .. 1 1 .. 1 1 1 1 0 2 B1. 4 0 .. 2 1. cos cao M1 A1. 18 2 3. (4). 2 . (c) AB i j 4k AB 18 or AB 18 ignore direction of vector M1. 2 . BC 3i 3 j BC 18 or BC 18 ignore direction of vector M1.. Hence AB = BC A1 (3).. (d) OD 6i 2 j 2k Allow first B1 for any two correct B1 B1. Accept column form or coordinates (2). [13]. 6666 Core C4 7. June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics Question scheme Marks Number dV. 8. (a) is the rate of increase of volume (with respect to time) B1. dt kV : k is constant of proportionality and the negative shows decrease (or loss). dV B1. giving 20 kV These Bs are to be awarded independently dt (2).

6 1. (b) dV 1dt separating variables M1. 20 kV. 1. ln 20 kV t C M1 A1. k Using V 0, t 0 to evaluate the constant of integration M1. 1. c ln 20. k 1 20 . t ln . k 20 kV . Obtaining answer in the form V A B e kt M1. V . 20 20 kt k e k Accept 20. k 1 e kt A1 (6). dV. (c) 20 e kt Can be implied M1. dt dV 1. 10, t 5 10 20 e kt k ln 2 M1 A1. dt 5. 75. At t 10, V awrt 108 M1 A1 (5). ln 2. [13]. Alternative to (b). dV. Using printed answer and differentiating kB e kt M1. dt Substituting into differential equation kB e kt 20 kA kB e kt M1. 20. A M1 A1. k Using V 0, t 0 in printed answer to obtain A B 0 M1. 20. B A1 (6). k 6666 Core C4 8. June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics