Mathematics · May / June 2024 · Paper 11 · Question 3

YMA01

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The question as text

3. y 2 –1 O 3 x Figure 1 Figure 1 shows a sketch of the curve with equation y = f(x). The curve passes through the points (–1, 0) and (0, 2) and touches the x‑axis at the point (3, 0). On separate diagrams, sketch the curve with equation

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Show the answer and mark scheme

3(a)[3]
+
3
3 3
+ + +
3
3
3
=
3
−
1
3 1
−
A1: For 3
+
2 3
with at least one intermediate step e.g.
3
+
3
3 3
+ + +
3
3
3
3
+
3
3
+
1
4 3
+
6
3
+
3
3
+
1
3 3
+ + +
3
3
3
=
or
×
=
or
×
=
(
)(
)
3 1
−
3 1
−
3 1
−
3
+
1
2
3 1
−
3
+
1
3 1
−
3
+
1
Allow 2 3
3.
It is for this expression and not for just p = 3, q = 2 unless 3
2 3
is seen.
+
+
Example of insufficient work:
(
)
3
+
3
x
3
−
3
=
x
+
3
x
3
−
x
=
3
+
3
x
3
−
1
=
3
+
3
x
=
3
−
1
3
+
3
3
+
1
=
×
=
3
+
2 3
3
−
1
3
+
1
Scores M1M1A0
Note that:
(
)
3
+
3
x
3
−
3
=
x
+
3
x
3
−
x
=
3
+
3
x
3
−
1
=
3
+
3
x
=
=
3
+
2 3
3
−
1
Scores M1M0A0
(
)
(ii)
x
3
−=
3
x
+
3
p
+
q
3
3
−=
3
p
+
q
3
+
3
p
3
+
3
q
−=
3
p
+
3
+
q
3
ALT 1
p
=
q
+
1, 3
q
−=
3
p
p
=
q
+
1, 3
q
−=
3
p
p
=
...,
q
=
...
x = +
3
2 3
Notes:
M1: Substitutes
, collects terms, compares coefficients and forms 2 equations in p and q
x
=
p
+
q
3
It is for obtaining 2 equations of the form
α
p
+
β
q
=
γ
,
αβγ
, ,
≠
0
M1: Solves simultaneously with evidence of algebra i.e. answers not just written down from a calculator.
Not formally dependent but requires the solution of 2 equations of the form
α
p
+
β
q
=
γ
,
αβγ
, ,
A1: For 3
+
2 3.
Allow 2 3
+
3.
It is for this expression and not for just p = 3, q = 2 unless 3
+
2 3
(ii)
x
3
−=
3
x
+
3
3
x
2
−
6 3
x
+
9
=
x
2
+
2 3
x
+
3
ALT 2
2
x
2
−
8 3
x
+
6
=
0
(
)
2
4 3
±
4 3
−××
4 1 3
x
2
−
4 3
x
+
3
=
0
x
=
2
4 3
±
36
±
x
=
x
=
2 3
3
2
x =
2 3
+
3
Notes:
M1: Squares both sides. Must obtain a 3 term quadratic expression on both sides and collect terms to one side.
M1: Solves a 3TQ of the form
2
3
0
by a correct noncalculator method.
px
+
q
x
+
r
=
A1: For 3
+
2 3.
Allow 2 3
+
3.
It is for this expression and not for just p = 3, q = 2 unless 3
+
2 3
If any other answers are offered and clearly not rejected score A0.
(
)(
) (
)(
)
(ii)
x
3
−=
3
x
+
3
x
3
−
3
x
−
3
=
x
+
3
x
−
3
ALT 3
(
)
3
2
6
3 3
2
3
3 1
2
6
3 3
3
0
x
−
x
+
=
x
−
−
x
−
x
+
+
=
(
)(
)
6
36
4
3 1 3 3
3
±
−
−
+
(
)
3 1
−
x
2
−
6
x
+
3 3
+
3
=
0
x
=
(
)
2
3 1
−
6
36
24
6
12
3
3
3
3
3
1
±
−
±
±
±
+
=
=
=
=
×
(
)
(
)
3 1
3 1
3
1
2
3 1
−
2
3 1
−
−
−
+
3 3
+ ±±
3
3
3
=
=
2 3
+
3,
3
2
x =
2 3
+
3
Notes:
M1: Multiples both sides by e.g.(
)
x −
3
and collects terms to one side.
Must obtain a 3 term quadratic expression on lhs and a 2 or 3 term quadratic expression on rhs.
M1: Solves a 3TQ of the form (
)
A
+
B
3
x
2
+
Cx
+
D
+
E
3
=
0,
A B C D E
, , ,
,
≠
0
by a correct noncalculator
method and then attempts to rationalise the denominator as in the main scheme.
A1: For 3
+
2 3.
Allow 2 3
+
3.
It is for this expression and not for just p = 3, q = 2 unless 3
+
2 3
If any other answers are offered and clearly not rejected score A0.
3(a)
(0, 0)
(−4, 0)
Notes:
If there is any ambiguity regarding coordinates, the sketch takes precedence.
B1: Same shape translated parallel to the xaxis with no evidence from the xintercepts (if labelled) that the
transformation is anything other than a translation. So the x intercepts (if labelled) must be 4 units
apart. Ignore any coordinates on or near the maximum.
B1: Minimum at the origin. The origin does not have to be labelled.
B1: Passes through (−4, 0) or at least “reaches” this point from above or below.
Allow just −4 marked in the correct place and condone (0, −4) as long as it is marked in the correct place.
Allow to score away from the sketch as long as their curve passes through (or reaches) this point but must
be identified correctly as (−4, 0) and correspond to the sketch.
See examples below:
B1: Translation (the maximum at (−3, 2) can be ignored)
B1: Minimum at the origin
B1: Passes through (−4, 0)
B0: The x intercepts are contradictory for a translation (not 4 units apart)
B0: No minimum at the origin
B1: Passes through (−4, 0)
B1: Translation
B1: Minimum at the origin (the intention is clear)
B1: Passes through (−4, 0)
B0: Incorrect shape
B1: Minimum at the origin
B1: Passes through (−4, 0)
(0, 2)
(−1, 0)
1 , 0
3
Notes:
If there is any ambiguity regarding coordinates, the sketch takes precedence.
B1: Reflection in the yaxis. This requires a minimum on the negative xaxis and a maximum to the left of the
yaxis in quadrant 2. If the maximum is on the yaxis score B0.
The curve must continue upwards to the left of the yaxis to reach a maximum in quadrant 2.
The curve must extend into quadrant 4 as shown.
B1: Minimum at (−1, 0). Allow just −1 marked in the correct place and condone (0, −1) as long as it is marked
in the correct place. Allow to score away from the sketch but must be identified correctly as (−1, 0) and
correspond to the sketch.
B1: Passes through (0, 2) and passes through or touches 1 , 0
. For both points, allow just the intercepts
3
1
marked in the correct place and condone (2, 0) and/or
0, 3
as long as they are marked in the correct
place. Allow to score away from the sketch but must be identified correctly as (0, 2) and 1 , 0
3
correspond to the sketch.
See examples below:
B0: Maximum in quadrant 2
B1: Minimum at (−1, 0)
B1: Passes through (0, 2) and 1 , 0
3
B0: Maximum on the yaxis (we would have tolerated the slight “pointy” maximum)
B0: Minimum at (−2, 0)
B0: Does not pass through (0, 2) and 1 , 0
3
B0: Not a reflection in the yaxis
B0: No minimum at (−1, 0)
B1: Passes through (0, 2) and touches 1 , 0
3
B0: Must be a curve – a reflection in the yaxis with the same shape.
B1: Minimum at (−1, 0)
B1: Passes through (0, 2) and 1 , 0
3
3(b)[3]
+
3
3 3
+ + +
3
3
3
=
3
−
1
3 1
−
A1: For 3
+
2 3
with at least one intermediate step e.g.
3
+
3
3 3
+ + +
3
3
3
3
+
3
3
+
1
4 3
+
6
3
+
3
3
+
1
3 3
+ + +
3
3
3
=
or
×
=
or
×
=
(
)(
)
3 1
−
3 1
−
3 1
−
3
+
1
2
3 1
−
3
+
1
3 1
−
3
+
1
Allow 2 3
3.
It is for this expression and not for just p = 3, q = 2 unless 3
2 3
is seen.
+
+
Example of insufficient work:
(
)
3
+
3
x
3
−
3
=
x
+
3
x
3
−
x
=
3
+
3
x
3
−
1
=
3
+
3
x
=
3
−
1
3
+
3
3
+
1
=
×
=
3
+
2 3
3
−
1
3
+
1
Scores M1M1A0
Note that:
(
)
3
+
3
x
3
−
3
=
x
+
3
x
3
−
x
=
3
+
3
x
3
−
1
=
3
+
3
x
=
=
3
+
2 3
3
−
1
Scores M1M0A0
(
)
(ii)
x
3
−=
3
x
+
3
p
+
q
3
3
−=
3
p
+
q
3
+
3
p
3
+
3
q
−=
3
p
+
3
+
q
3
ALT 1
p
=
q
+
1, 3
q
−=
3
p
p
=
q
+
1, 3
q
−=
3
p
p
=
...,
q
=
...
x = +
3
2 3
Notes:
M1: Substitutes
, collects terms, compares coefficients and forms 2 equations in p and q
x
=
p
+
q
3
It is for obtaining 2 equations of the form
α
p
+
β
q
=
γ
,
αβγ
, ,
≠
0
M1: Solves simultaneously with evidence of algebra i.e. answers not just written down from a calculator.
Not formally dependent but requires the solution of 2 equations of the form
α
p
+
β
q
=
γ
,
αβγ
, ,
A1: For 3
+
2 3.
Allow 2 3
+
3.
It is for this expression and not for just p = 3, q = 2 unless 3
+
2 3
(ii)
x
3
−=
3
x
+
3
3
x
2
−
6 3
x
+
9
=
x
2
+
2 3
x
+
3
ALT 2
2
x
2
−
8 3
x
+
6
=
0
(
)
2
4 3
±
4 3
−××
4 1 3
x
2
−
4 3
x
+
3
=
0
x
=
2
4 3
±
36
±
x
=
x
=
2 3
3
2
x =
2 3
+
3
Notes:
M1: Squares both sides. Must obtain a 3 term quadratic expression on both sides and collect terms to one side.
M1: Solves a 3TQ of the form
2
3
0
by a correct noncalculator method.
px
+
q
x
+
r
=
A1: For 3
+
2 3.
Allow 2 3
+
3.
It is for this expression and not for just p = 3, q = 2 unless 3
+
2 3
If any other answers are offered and clearly not rejected score A0.
(
)(
) (
)(
)
(ii)
x
3
−=
3
x
+
3
x
3
−
3
x
−
3
=
x
+
3
x
−
3
ALT 3
(
)
3
2
6
3 3
2
3
3 1
2
6
3 3
3
0
x
−
x
+
=
x
−
−
x
−
x
+
+
=
(
)(
)
6
36
4
3 1 3 3
3
±
−
−
+
(
)
3 1
−
x
2
−
6
x
+
3 3
+
3
=
0
x
=
(
)
2
3 1
−
6
36
24
6
12
3
3
3
3
3
1
±
−
±
±
±
+
=
=
=
=
×
(
)
(
)
3 1
3 1
3
1
2
3 1
−
2
3 1
−
−
−
+
3 3
+ ±±
3
3
3
=
=
2 3
+
3,
3
2
x =
2 3
+
3
Notes:
M1: Multiples both sides by e.g.(
)
x −
3
and collects terms to one side.
Must obtain a 3 term quadratic expression on lhs and a 2 or 3 term quadratic expression on rhs.
M1: Solves a 3TQ of the form (
)
A
+
B
3
x
2
+
Cx
+
D
+
E
3
=
0,
A B C D E
, , ,
,
≠
0
by a correct noncalculator
method and then attempts to rationalise the denominator as in the main scheme.
A1: For 3
+
2 3.
Allow 2 3
+
3.
It is for this expression and not for just p = 3, q = 2 unless 3
+
2 3
If any other answers are offered and clearly not rejected score A0.
3(a)
(0, 0)
(−4, 0)
Notes:
If there is any ambiguity regarding coordinates, the sketch takes precedence.
B1: Same shape translated parallel to the xaxis with no evidence from the xintercepts (if labelled) that the
transformation is anything other than a translation. So the x intercepts (if labelled) must be 4 units
apart. Ignore any coordinates on or near the maximum.
B1: Minimum at the origin. The origin does not have to be labelled.
B1: Passes through (−4, 0) or at least “reaches” this point from above or below.
Allow just −4 marked in the correct place and condone (0, −4) as long as it is marked in the correct place.
Allow to score away from the sketch as long as their curve passes through (or reaches) this point but must
be identified correctly as (−4, 0) and correspond to the sketch.
See examples below:
B1: Translation (the maximum at (−3, 2) can be ignored)
B1: Minimum at the origin
B1: Passes through (−4, 0)
B0: The x intercepts are contradictory for a translation (not 4 units apart)
B0: No minimum at the origin
B1: Passes through (−4, 0)
B1: Translation
B1: Minimum at the origin (the intention is clear)
B1: Passes through (−4, 0)
B0: Incorrect shape
B1: Minimum at the origin
B1: Passes through (−4, 0)
(0, 2)
(−1, 0)
1 , 0
3
Notes:
If there is any ambiguity regarding coordinates, the sketch takes precedence.
B1: Reflection in the yaxis. This requires a minimum on the negative xaxis and a maximum to the left of the
yaxis in quadrant 2. If the maximum is on the yaxis score B0.
The curve must continue upwards to the left of the yaxis to reach a maximum in quadrant 2.
The curve must extend into quadrant 4 as shown.
B1: Minimum at (−1, 0). Allow just −1 marked in the correct place and condone (0, −1) as long as it is marked
in the correct place. Allow to score away from the sketch but must be identified correctly as (−1, 0) and
correspond to the sketch.
B1: Passes through (0, 2) and passes through or touches 1 , 0
. For both points, allow just the intercepts
3
1
marked in the correct place and condone (2, 0) and/or
0, 3
as long as they are marked in the correct
place. Allow to score away from the sketch but must be identified correctly as (0, 2) and 1 , 0
3
correspond to the sketch.
See examples below:
B0: Maximum in quadrant 2
B1: Minimum at (−1, 0)
B1: Passes through (0, 2) and 1 , 0
3
B0: Maximum on the yaxis (we would have tolerated the slight “pointy” maximum)
B0: Minimum at (−2, 0)
B0: Does not pass through (0, 2) and 1 , 0
3
B0: Not a reflection in the yaxis
B0: No minimum at (−1, 0)
B1: Passes through (0, 2) and touches 1 , 0
3
B0: Must be a curve – a reflection in the yaxis with the same shape.
B1: Minimum at (−1, 0)
B1: Passes through (0, 2) and 1 , 0
3

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