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
9709/12 Cambridge International AS & A Level – Mark Scheme May/June 2023
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 2023 Page 5 of 22
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | 1 OA x
OA xcos or  or
2 sinπ2 sin
OA2  x2 x2 2x2cosπ2 or
x2 r2 x2 2rxcos or other valid method. | *B1 | 1
Correct expression containing OA, OA or OA2(allow p,
2
a or r for OA) containing only terms with x and  but not
just OA2xcos .
Do not condone sinπ2 until missing brackets
recovered or cos(1802) until it becomes cos2 etc.
OA = 2xcos leading to Arc length = 2xcos | DB1 | AG Complete correct method showing all necessary
working. Condone 2xcos .
2 | If B0 but www then SCB1 for OA = 2xcos leading to
Arc length = 2xcos.
--- 6(b) ---
6(b) | 1
2xcos2
Sector area = θ
2 | M1 | OE Using sector formula with a correct OA. Condone
cos2 for cos2 and missing brackets.
1 1
Triangle area = 2xcosxsin OR x2sinπ2
2 2 | M1 | Using a correct triangle formula for the correct triangle.
Condone missing brackets and 180 for π.
[Area APB =] Their sector area – their triangle area | M1 | Both expressions must be areas involving terms with x2 and
 only. Condone missing brackets and 180 for π for the
triangle. Condone calling the sector a segment.
1 1
[Area APB =] 2xcos2  x2sinπ2
2 2
1
[ x2(2cos2 sin2) or x2cos(2cos sin)]
2 | A1 | OE
A correct expression. Mark the first unsimplified result of
subtraction and ISW any incorrect ‘simplifications’.
4
Question | Answer | Marks | Guidance