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EDEXCEL CORE MATHEMATICS C4 (6666)

MOCK PAPER MARK SCHEME

 Question Scheme Marks number 1. Uses d u = 6x M1 d x To give    1 d u A1 u 2 3 1 Integrates to give  3u M1, A1 Uses correct limits 16 and 4 (or 2 and 0 for x) M1 To obtain  1 48  1 12  1 16 A1 (6) (6 marks) 2. Differentiates w.r.t. x to give 3x 2 ,  2x d d At (4, 3) y x + 2y, 4 + 3y 2 d y d x = 0 M1, B1, A1 48  (8y + 6)  4 + 27y = 0 M1  y =  38 = 2 A1 19 1 Gradient of normal is 2 M1 y  3 = 1 2 (x  4) M1 i.e. 2y  6 = x  4 x – 2y + 2 = 0 A1 (8) (8 marks)

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EDEXCEL CORE MATHEMATICS C4 (6666)

MOCK PAPER MARK SCHEME

Question

 3. (a) (b) (c)
 4. (a) (b) (c)

Mark Scheme

1
 14 x
A
B
and attempt A and or B
(1
x
)(1
2
x
)
1
 x
1
 2 x
A = 5, B = 4
5
4
d x
= [5ln 1  x  2 ln1 + 2x]
1 
x
1
2
x
2
5
=
(5 ln
 2 ln
)  (5 ln
5  2 ln
3
3
4 )
6
3
5
4
=
5 ln
+ 2 ln
4
5
5
125
=
3 ln
= ln
4
64
5(1  x) 1 – 4(1 + 2x) 1
( 
1)(
2)(2 ) 2
x
(
1)(
2)(
3)(2 x ) 3
= 5(1 + x + x 2 + x 3 )  4( 1  2x +
+
)
2
6
= 1 + 13x  11x 2 + 37x 3 ….
11 + 4 = 24 + 7
5
+ 2 = 4 + 
Give 2 of these equations and eliminate
variable to find  or , find other
6
+ 4 = 13 + 5
5
= 11 + 2
 = 3;  = 2
Check in 3rd equation
Use  = 3 or  = 2 to obtain (3, 1, 2)
4
   
7
2
1
4
5
50
cos  =
=
2
2
2
2
2
2
4
2
4
7
1
5
36
75
50
50
3
5
3
cos  = 6 5 3
=
=
90
9

Marks

 M1 A1, A1 (3) M1 A1 M1 M1 A1 (5) B1 ft M1 A1 M1 A1 (5)

(13 marks)

 M1 A1 A1 B1 (4) M1 A1 (2) M1 A1 M1 A1 (4)

(10 marks)

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EDEXCEL CORE MATHEMATICS C4 (6666)

MOCK PAPER MARK SCHEME

 Question Mark Scheme Marks d x =  sin t, d y = 2 cos 2t  d y = 2 cos 2 t 5. (a) d t d t d x  sin t M1 A1 A1 (3) (b) 2 cos 2t = 0 2t =  , 3 2  , 5 2  , 7  M1 2 2 So t =  , 3 4  , 5 4  , 7  A1 A1 (c)  1    2 ,1)         1 2 4  , 1)         1 2 ,1)          4 1 2 ,1)     M1 A1 (3) (2) (d) y = 2 sin t cos t M1 = 2 2 1  cos t cos t = 2x 2 1 x M1 A1 (3) (e) y = 2x 2 1 x B1 (1) (12 marks) 2  6. (a) R =     x 2 sin   1  2 x    dx = 2x 2 cos = 2x 2 cos = 2x 2 cos         1  2 1 2 1 2 x   x   x    +    4x cos   1  2 x    dx   + 8x sin  1  x     2 –    8 sin ( +   + 16 cos (  8x sin 1 2 x    1 2 1 2 ) x x ) M1 A1 M1 A1 A1 Use limits to obtain [8 2  16]  [8] M1 A1 (7) (b) Requires 11.567 B1 (1) (c) (i) Area =  , [9.8696 + 0 + 2  15.702] (B1 for  in (i) or  in (ii)) B1, M1 4 4 8 = 32.42 A1 (ii) Area =  [9.8696 + 0 + 2(14.247 + 15.702 + 11.567)] M1 8 = 36.48 A1 (5) (13 marks)

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EDEXCEL CORE MATHEMATICS C4 (6666)

MOCK PAPER MARK SCHEME

 Question Mark Scheme Marks d M 7. (a) = kM, where k > 0 M1 A1 (2) (b) d t M d ln(0.98)  10(0.98) t , =  0.02M d t =  10 d M =   k d . t B1, B1 (2) (c)   10 M 1   ln (10M  1) =  kt + c B1 M1 A1 At t = 0 M = 10  c = ln 99 M1 A1 At t = 10 M = 8.5  k = 1 ln 99 (= 0 .0164). M1 A1 10 84 Uses 10 M  1 = 99 e kt with values for k and t = 15 M1 To give 7.8 grams accept awrt 7.8 A1 (9) (13 marks)

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EDEXCEL CORE MATHEMATICS C4 (6666)

MOCK PAPER MARK SCHEME

 Qn Specifications Section Synoptic AO1 AO2 AO3 AO4 AO5 Totals Marks Total Q1 5.3 4 2 6 5 Q2 4.1 5 3 8 6 Q3 1, 3, 5.1, 5.4 5 6 2 13 8 Q4 6.1, 6.2, 6.3, 6.5, 6.6 4 5 1 10 4 Q5 2, 4.1, 5 6 1 12 10 Q6 5.1, 5.3, 5.6 4 4 5 13 8 Q7 4.3, 4.2, 5.5 3 2 1 5 2 13 4
TOTAL
30
28
5
5
7
75
45

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