Question 6:
--- 6(a) ---
6(a) | 1
dy 1 
2 8 x 2
dx 2 | B1
1

24x 2 0 | M1 | dy
Equating their two term , with at least one term
dx
correct, to 0.
[A is] 4, 8 or x4, y8 | A1
[B is] 16, 0 or x16, y0 | B1
4 | dy
Note: Correct answers without use of can be
dx
awarded 4/4.
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | 2x2 8 3
  x2 C
2 3
2 | B1 | Seen correct in unsimplified form or better.
x2 32x 2x322
 or C
3 12 | B1 | Seen correct in unsimplified form or better.
Attempt to integrate, defined by at least one correct power in each expression,
and then subtract. | M1 | Multiplying by 3 before integration scores M0.
   
 8 3  8 3 162 3216 42 324
162  .16242  .42    
 3   3   3   3 
 2   2  | M1 | Use of their x values, > 0, from (a) as limits in their
integrated expressions. Allow, for correct limits,
sight of
 256  80  256  112
      .
 3   3   3   3 
If incorrect limits are used, then clear substitution
must be seen.
Question | Answer | Marks | Guidance
6(b) | Alternative Method 1 for first 4 marks of Question 6(b)
 1  8 3
 2x8x2 dx x2  x2 C
3
 
2 | (B1) | Seen correct in unsimplified form or better.
[Area of triangle =] 48 | (B1)
Attempt to integrate, defined by at least one correct power, and then subtract
their triangle area. | (M1)
   
 8 3  8 3
162  .16242  .42
3 3
   
 2   2  | (M1) | Use of their x values, > 0, from (a) as limits in their
integrated expression. Allow sight of
 256  80
      .
 3   3 
If incorrect limits are used, then clear substitution
must be seen.
Question | Answer | Marks | Guidance
6(b) | Alternative Method 2 for first 4 marks of Question 6(b)
Subtract and then integrate, defined by at least two correct powers.
Condone functions being the wrong way round. | (M1) | If terms in x have not been combined use the first
scheme.
 
 4 8 3 32x
  x2  x2  
32 3 3
 
 2  | (B2,1,0) | B2 for 3 correct terms, B1 for any 2 correct terms.
   
 4 8 3 3216  4 8 3 324
  162  162   42  42  
 32 3 3   32 3 3 
 
 2   2  | (M1) | Use of their x values, >0, from (a) as limits in their
 32
integrated expression. Allow sight of   0 .
 3 
If incorrect limits are used, then clear substitution
must be seen.
32 2
, 10 or 10.7
3 3 | (B1) | AWRT
32 32 32
Allow  or  changed to + for this mark.
3 3 3
(5) | Condone the inclusion of π for the first 4 marks but
use of y2 scores a maximum of B1 for the triangle.
Question | Answer | Marks | Guidance