Question 3:
--- 3(i) ---
3(i) | E.g. n = 1 : 23−13 =7, n = 2 : 3 3 − 2 3 = 1 9 , n = 3 : 4 3 − 3 3 = . . .
Or identifies counterexample directly. | M1
e.g. 6 3 − 5 3 = 9 1 = 7  1 3 so not true for n = 5, hence statement is not true. | A1
(2)
Notes for part (i)
M1: Shows evidence of trying to find a counter example for a positive integer (at
least one attempt).
23−13 is prime is sufficient.
A1: Gives a correct counter example with reason (shows factorisation) and
concludes e.g. “which is not prime”. Ignore any previous “incorrect” attempts e.g.
63− 53 = 91 which is prime.
Note n = 7 (169 = 13×13) and n = 8 (217 = 7×31) and n = 12 (469 = 7×67) are the
next few counter examples. (Bigger examples are not likely to be seen!)
Allow equivalent reasons for not being prime e.g. 169/13 = 13 or 169 is divisible by
13 (condone “can be divided by 13”)
Generally algebraic approaches score no marks unless they substitute numbers
as indicated above.
--- 3(ii) ---
3(ii) | The majority of methods here will follow ways 1, 2 or 3 below
In these cases the general guidance is as follows:
M1: Attempts to find
• the gradient of any relevant line, e.g. AC or BC or
• the length of any relevant line, e.g. AB/AB2 or BC/BC2 or AC/AC2 or
• the mid-point M of line AB
A1: Correct relevant calculation of
• gradients AC and BC
• lengths of lines AB/AB2, BC/BC2 and AC/AC2
• mid-point of line AB
dM1: Full attempt at combining all relevant information required to solve the prob-
lem
• attempts product of gradients or equivalent
• attempts to show Pythagoras AB2 = AC2 + BC2
• attempts to show MA2 = MC2
A1: Correct calculations or equivalent providing required evidence for the above
A1: Provides correct reason and conclusion with all previous marks scored.
Way 1 | − 6 − 0 − 6 − ( − 1 0 )
m = = . . . or m = = . . .
A C 7 − 1 B C 7 − 3 | M1
m = − 1 and m = 1
A C B C | A1
So m  m = − 1  1 = − 1
A C B C
or e.g. m is negative reciprocal of m
A C B C | dM1A1
So e.g. angle (at C) is a right angle hence AB is a diameter
(Or equivalent) | A1
(5)
M1: Attempts the gradients of AC or BC. Allow slips but score M0 if both attempts are clearly
incorrect.
A1: Correct gradients from correct formulae
dM1: Applies perpendicular condition. May be seen as shown but allow equivalent work.
A1: Correct calculations or equivalent
A1: Suitable explanation and conclusion given with no errors and all previous marks awarded with
no incorrect statements seen.
Way 2 | A B = ( 3 − 1 ) 2 + ( − 1 0 − 0 ) 2 = . . . o r A C = ( 7 − 1 2 ) + ( − 6 − 0 ) 2 = . . .
o r B C = ( 7 − 3 2 ) + ( − 6 + 1 0 ) 2 = . . . | M1
( ) ( ) ( )
A B = 1 0 4 2 2 6 , A C = 7 2 6 2 , B C = 3 2 4 2 | A1
AB2 =104=72+32= AC2 +BC2 | dM1A1
Hence ABC is a right-angle triangle with hypotenuse AB hence AB is a
diameter. (Or equivalent) | A1
(5)
M1: Attempts length of AB or AC or BC or their squares. Allow slips but score M0 if attempts
are clearly incorrect.
A1: Correct values for AB, AC and BC or their squares.
dM1: Applies Pythagoras’ theorem with their values. (May see cosine rule used.)
A1: All calculations correct for this approach.
A1: Suitable explanation and conclusion given with no errors and all previous marks awarded with
no incorrect statements seen.
Way 3 |  1 + 3 0 − 1 0 
If AB is diameter centre must be midpoint of AB ie M ,
2 2 | M1
= ( 2 , − 5 ) | A1
M A = ( 2 − 1 ) 2 + ( − 5 − 0 ) 2 = 2 6 , M C = ( 2 − 7 ) 2 + ( − 5 − ( − 6 ) ) 2 = 2 6 | dM1A1
MA= 26,MC = 26 so MA=MC(=MB)
As the length from M to each of A and C is the same M is the centre of the circle
hence AB is a diameter. (Or equivalent) | A1
(5)
M1: Attempts midpoint of AB. If no method is shown accept one correct coordinate as evidence.
A1: Correct midpoint
dM1: Attempts length of MC and at least one of MA or MB, or AB. As M is midpoint of AB there is no need to
find both MA and MB, these may be assumed to be the same. If they find AB then they must halve it to find
the radius.
A1: All required calculations correct for this approach.
A1: Suitable explanation made which may be in a preamble and conclusion given with no errors and all
previous marks awarded.
Way 4 |  1 + 3 0 − 1 0 
If AB is diameter centre must be midpoint of AB ie M ,
2 2 | M1
= ( 2 , − 5 ) | A1
M A = r = ( 2 − 1 ) 2 + ( − 5 − 0 ) 2 = 2 6
 ( x − 2 ) 2 + ( y + 5 ) 2 = 2 6
C ( 7 , − 6 )  ( 7 − 2 ) 2 + ( − 6 + 5 ) 2 = 5 2 + 1 2 = 2 6 | dM1A1
As C also satisfies the equation of the circle then AB must be the diameter
(or equivalent)
There must be some further justification as above rather than just “AB is a
diameter” which may be in a preamble e.g. If C lies on the circle… | A1
(5)
M1: Attempts midpoint of AB. If no method is shown accept one correct coordinate as evidence.
A1: Correct midpoint
dM1: Attempts length of MA or MB to find r or r2, forms equation of the circle and substitutes the
coordinates of C.
A1: All required calculations correct for this approach.
A1: Suitable explanation and conclusion given with no errors and all previous marks awarded.
Question | Scheme | Marks