Mathematics YMA01 Oct / Nov 2023 Paper 41 Question 4

4(a) 9 Assume that there exists a positive number k such that k + < 6 k 9 2 2 k + < 6 k + 9 < 6 k k − 6 k + < 9 0 k or 9 9 ( )( ) k + < 6 k + − 6 < 0 k ... k ... < 0 k k or 2 9 9 81 2 k + < 6 k + < 36 k + 18 + − 36 < 0 k k 2 k 2 2 ( ) 3 9 2 k − 3 < 0 or k − < 0 or k − < 0 k k 9 But numbers squared…

Mathematics · Oct / Nov 2023 · Paper 41 · Question 4

YMA01

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4(a)[4]
9
Assume that there exists a positive number k such that
k
+
<
6
k
9
2
2
k
+
<
6
k
+
9
<
6
k
k
−
6
k
+ <
9
0
k
or
9
9
(
)(
)
k
+
<
6
k
+
−
6
<
0
k
...
k
...
<
0
k
k
or
2
9
9
81
2
k
+
<
6
k
+
<
36
k
+
18
+
−
36
<
0
k
k
2
k
2
2
(
)
3
9
2
k
−
3
<
0
or
k
−
<
0
or
k
−
<
0
k
k
9
But numbers squared are
0
, hence
k
+
6
k
4(b)[1]
9
(
)
k = −
3,
E.g. When
−+
3
= −
6
which is not
6
−
3
9
9
(b) Alt
k
<
0
<
0
k
+
<
0
which is not
6
k
k
(a)
B1: For setting up the correct contradiction. It must include a word/words such as “assume” or “let” and
9
must be a strict inequality. As a minimum accept just “assume/let
6
”.
k
+
<
k
9
Condone e.g. Assume that for all positive numbers k,
k
+
<
6
k
9
9
9
M1: Starting from
k
+
<
6
or
k
+
6
or
k
+
=
6
, either
k
k
k
2...
2...
2... 0
multiplies by k and attempts to reach
k
<
0
or
k
0
or
k
=
9
collects terms to one side to reach
k
+
±
6
/
<=
/
0
and attempts to factorise to
k
(
)(
)
k
...
k
...
/
<=
/
0
2...
2...
2... 0
k
0
k
0
k
=
squares both sides and collects terms to reach
<
or
or
2
2
A1: Reaches (
)
3
9
2
k −
3
...0
or e.g.
k
−
...0
or e.g.
k
−
...0
where … is < or or =
k
k
Requires
correct calculations/algebra
a reason: “numbers squared are 0 “this is impossible”, “numbers squared are not negative”
9
a minimal conclusion. E.g. hence
k
+
6
, hence proven, QED, etc. but not just “contradiction”
k
Alternatives for A1A1*:
1. Using discriminant:
2
4
6
2
4 9
0
A1:
b
−
ac
=
−×=
so one root and ‘positive’ quadratic
A1*: For a fully correct proof.
Requires
correct calculations/algebra
2
a reason e.g. so
k
−
6
k
+
9
0
or curve is on or above the x/k axis
9
a minimal conclusion. E.g. hence
k
+
6
, hence proven, QED, etc. but not just “contradiction”
k
2. Via sketch:
2
A1: Sketches
y
=
k
−
6
k
+
9
:
A1*: For a fully correct proof.
Requires
correct calculations/algebra
2
a reason e.g. graph always on or above (or never below) the xaxis so
k
−
6
k
+
9
0
9
a minimal conclusion. E.g. hence
k
+
6
, hence proven, QED, etc. but not just “contradiction”
k
3. Via differentiation:
2
d
k
−
6
k
+
9
=
2
A1:
2
k
−
6
=
0
k
=
3
k
−
6
k
+
9
=
0
so minimum at (3, 0)
d
k
A1*: For a fully correct proof.
Requires
correct calculations/algebra
2
a reason e.g. so
k
−
6
k
+
9
0
or curve is on or above the x/k axis
9
a minimal conclusion. E.g. hence
k
+
6
, hence proven, QED, etc. but not just “contradiction”
k
If candidates use another variable for k, e.g. x or m then withhold the final mark
unless they revert to k in their conclusion.
This is also acceptable (working backwards):
9
Assume
k
+
<
6
B1
k
(
)
9
2
2
k
−
3
0
k
−
6
k
+
9
0
k
−+
6
0
M1
k
Condone > for
From (
)
2
k −
3
0
expands and divides by k
9
6
k
+
A1
k
Condone > for
9
This contradicts the assumption and so
k
+
6
A1*
k
Note: Attempts that use the contradiction: “Assume there are negative numbers for which
k
+
generally score no marks but use review if necessary.
(b)
9
B1: Chooses a negative number, shows the result of substituting into
k
+
and gives a
k
9
minimal conclusion e.g. It is not necessary to evaluate e.g.
−+ − is sufficient.
3
3
which is not
6
is < 6
(b)Alt:
9
B1: States that if k < 0 then
k
+
<
0
and gives a minimal conclusion e.g.
k
which is not
6
is < 6
There must be no incorrect work and no contradictory/incorrect statements.
Do not allow arguments that refer to k = 0 being not valid.
2

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