Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9231/43 Cambridge International AS & A Level – Mark Scheme May/June 2021
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2021 Page 5 of 13
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | 1
x 0x<1,

8
F(x)=

1  1  1
 8x− x2 − 1x8.
 
28 2  7 | M1
A1 | Integrate both parts (all powers of x increase by 1), allow M1 if
no constant term.
Both parts correct, with domains.
2 1 1
Any equivalent form, for example: x− x2 − .
7 56 7
F(x)= 0 for x < 0 and F(x) = 1 for x > 8. | A1
Alternative method for question 6(a)
1
x 0x<1,

8
F(x)=

 1− 1 ( 8−x )2 1x8.
 56 | M1
A1 | Integrate both parts (all powers of x increase by 1), allow M1 if
no constant term.
Both parts correct, with domains.
1 ( )
Any equivalent form, for example: 16x−x2 −8 .
56
F(x)= 0 for x < 0 and F(x) = 1 for x > 8. | A1
3
--- 6(b) ---
6(b) | 5 1  1  1 5
F(a) = leading to 8a− a2 − =

7 28 2  7 7 | *M1 | Correct method and attempt to simplify, ft their part (a).
a2 −16a+48=0
[ ]
a=4 a=12 | DM1 | Obtain quadratic and solve.
a = 4 | A1 | Correct single answer, WWW.
3
Question | Answer | Marks | Guidance
--- 6(c) ---
6(c) | 1
y3 0  y <1,

8
G(y)=

1 ( ) 1
 16y3 − y6 − 1 y 2.
56 7 | M1
A1 | Substitute x = y3 into both parts of their F(x) from part (a).
Correct functions in terms of y, any equivalent form.
0  y < 1 and 1  y  2seen in CDF or PDF | A1 | Condone  instead of <.
3
y2 0  y< 1,

8

 3 ( )
g(y)=  8y2 − y5 1  y<2,
28


0 otherwise.

 | M1
A1 | Differentiate their G(y), all powers of y decrease by 1.
Must include ‘otherwise’.
5
Question | Answer | Marks | Guidance
6(c) | Alternative method for question 6(c)
1 1 −2
y= x3, dy= x 3dx, dx=3y2dy
3 | M1
0x<1, f ( x ) dx= 1 dx= 3 y2dy
8 8 | M1
( )= 3
g y y2
8 | A1
1x8, f ( x ) dx= 1 ( 8−x ) dx= 1 ( 8−y3 ) 3y2dy
28 28
g ( y )= 3 ( 8y2 −y5 )
28 | A1
0  y < 1 and 1  y  2 seen in CDF or PDF | A1
5