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, 2 A1. [7]. dy x y Note: . 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).

3 [8]. 6666 Core C4 3. June 2005 Advanced Subsidiary/Advanced Level in GCE Mathematics Question scheme Marks Number . 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).

4 (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. 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).

5 [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. (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).

6 [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).. 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).

7 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