Mathematics · May / June 2021 · Paper 41 · Question 9
YMA01
As printed in the paper
Show the answer and mark scheme
9(i)[3]
Attempts two of AB =
b
−
a , AC =
c
−
a and BC =
c
−
beither way around
(
)
(
)
Attempts
c
−
b =
2
×
b
−
a oe such as
c
−
a = 3
×
b
−
a
⇒
c
=
3
b
−
2
a *
9(ii)[5]
2n is a multiple of 3
Assume that there exists a number n that isn't a multiple of 3 yet
(
)
If n is not a multiple of 3 then
m
=
3
p
+ or
1
m
=
3
p
+
2
p∈ giving
2
2
(
3
1
)
2
9
6
1
m
=
p
+
=
p
+
p
+
(
)
2
(
)
2
2
2
Or
m
=
3
p
+
2
=
9
p
+
12
p
+
4
=
3 3
p
+
4
p
+
1
+
1
(
(
)
)
2
2
(
3
p
1
)
2
9
p
6
p
1
3 3
p
2
p
1
is one more than a multiple of 3
+
=
+
+
=
+
+
(
)
2
3
p
+
2
2
=
9
p
+
12
p
+
4
is not a multiple of 3 as 3 does not divide into 4 (exactly)
2n is a multiple of 3
Hence if n is a multiple of 3 then
(i)
M1: Attempts any two of
AB
,
AC
and BC
.
Condone the wrong way around but it must be subtraction.
Allow marked in the correct place on a diagram
dM1: Uses the given information.
1
2
Accept
,
BC
2
AB
,
etc condoning slips as in previous M1.
AB
=
AC
=
×
BC
=
×
AC
3
3
A1*: Fully correct work inc bracketing leading to the given answer
c
=
3
b
−
2
a
(
)
Expect to see the brackets multiplied out. So
c
−
b =
2
×
b
−
a
⇒
c
−
b = b
2
−
2
a
⇒
c
=
3
b
−
(ii)
B1: For setting up the contradiction.
Eg Assume that there exists a number n that isn't a multiple of 3, yet
2n is a multiple of 3
As a minimum accept something like '' define a number n such that n is not a multiple of 3 but n2 is''
M1: States that
3
+ or
1
3
2
and attempts to square.
m
=
p
m
=
p
+
Alternatives exist such as
m
=
3
p
+ or
1
m
=
3
p
−
1
Using modulo 3 arithmetic it would be 1
→ and 2
1
→
4
=
1
M1: States that
m
=
3
p
+ AND
1
m
=
3
p
+
2
and attempts to square o.e.
A1: Achieves forms that can be argued as to why they are NOT a multiple of 3
(
)
2
2
2
E.g.
m
(
3
p
1
)
2
3 3
p
2
p
+ or even
1
9
p
+
6
p
+
1
=
+
=
+
(
)
2
2
2
and
m
=
(
3
p
+
2
)
2
=
3 3
p
+
4
p
+
1
+ or even
1
9
p
+
12
p
+
4
A1: Correct proof which requires
Correct calculations2
Correct reasons. E.g.9
p
+
12
p
+
4
is not a multiple of 3 as 4 is not a multiple of 3
(
)
2
(
)
2
2
There are many ways to argue these. E.g
m
=
3
p
+
1
=
3 3
p
+
2
p
+ is
1
sufficient as long as followed (or preceded by) ''not a multiple of 3''
Minimal conclusion such as Π. Note that B0 M1 M1 M1 A1 is possiblePearson Education Limited. Registered company number 872828
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