Mathematics · Oct / Nov 2023 · Paper 11 · Question 8

YMA01

As printed in the paper

The question as text

8. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. The curve C1 has equation 15 xy x x ≠ 0 5 2 The curve C2 has equation 7 3 y x x 5 2

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

8(a)[2]
2
+
4
+
12 15
+
−
4
2
=
0
=
8,
=
4 ,
=
12 15
+
−
4
2
x
px
p
p
a
b
p c
p
p
(
)
(
)
4
2
−××
4 8 (12
+
15
−
4
2
)
>
0
(they have just written x instead of p which is fine)
x
x
x
3
2
−
10
−>
8
0
A1*: Achieves
p
p
with all stages of working shown:
Multiplying out (
p
−
2 )(
x x
+
p
)
Collecting terms on one side of an equation (=0 may be implied)
Finding the discriminant
At least one intermediate stage of working between the discriminant and the final
There must be no errors seen including invisible brackets but condone a missing trailing
bracket if it does not alter the processing. e.g. (
)
2
4
−××
4 8 (12 15
+
−
4
2
p
p
p
Condone the > 0 appearing for the first time in their final answer, provided an incorrect
inequality has not been used in earlier working.
Do not withhold this mark if they state e.g. “
2
4
0
” as part of their working.
b
−
ac
=
M1:
Attempts to find critical values by solving the given quadratic by either
(
−
4)(
+
2
)
factorising (do not accept
p
p
unless they have divided both sides of the
3
inequality/equation by 3 first)
5
2
(
)
completing the square
3
p
−
±
...
p
=
...
3
quadratic formula
Usual rules apply (see general marking principles for guidance). May be in another
variable. Allow this mark to be scored if seen in (a).
This mark cannot be awarded from directly using a calculator and stating the roots.
M1:
Attempts to find the outside region for their critical values. May use another variable e.g. x
2
May be implied by
or incorrect use of inequalities e.g.
p
"
−
",
p
"4"
4
<
p
<−
3
This is not dependent on the first method mark so if a calculator has been used then this
mark can still be scored.
2
A1:
or
(or equivalent). (may appear on separate lines). Isw once a correct
p <−
p >
4
3
answer is seen, provided there is no contradiction and no part of the range is rejected.
This can only be scored provided both previous method marks have been awarded.
Must be in terms of p
2
2 ,
2
✓
:
<−
∪
>
4
Accept e.g.
p
<−
p
>
4
,
p
<−
p
>
4
p p
p
or variations of
3
3
3
2
(
)
−∞−
,
,
4,
∞
these. Accept
3
2
2
✓
:
<−
∩
>
4
Do not accept e.g.
p <−
and
p >
4
p p
p
or variations of these
3
3
2
2 ,
2
,
, [4,
)
Do not accept
−∞−
∞
−
4
4
<
<−
p
p
p
3
3
3
7ai
1
4(4)
2
+
10
−
7(4)
2
15
f (4)
′
=
=
1
2
4(4)
2
ii
15
2
"
"
→−
2
15
2
+ = −
1
"
"(
−
4)
y
x
15
2
+
15
+
7
=
0
x
y
b
1
4
2
+
10
−
7
3
1
x
x
2
−
= ±
...
x
2
±
...
x
2
±
...
1
4
x
2
3
1
5
7
Two of the terms of
−
x
2
+
x
2
−
2
4
3
5
1
7
2
5
1
7
−
x
2
+
x
2
−
d
x
=
x
2
+
5
x
2
−
x
(
+
c
)
2
4
5
4
2
5
1
7
5(4)
1
...
2
+
2
−
+
c
= −
c
=
5
4
5
1
2
7
84
(
)
f ( )
x
=
x
2
+
5
x
2
−
x
−
5
4
5
Mark (a) and (b) together so do not be concerned with labelling of the parts
(a)
(i)
15
B1:
2 oe stated (as the gradient of f(x) at P).
(ii)
M1
Finds the negative reciprocal of their gradient in part (i). If they do not have a gradient in (i)
2
1
then only allow
−
or an attempt at
−′
15
f (4)
M1
Attempts to find the equation of the normal using a changed gradient to that found in (i) and
with the coordinates in the correct positions. If they do not have a gradient in (i) then
(4,
−
1)
15
allow any gradient
≠
. If they use y
=
mx
+ they must proceed as far as
c
c =
...
2
A1:
2
x
+
15
y
+
7
=
0
or any multiple of this where all the coefficients are integers and all terms
are on the same side of the equation. e.g. 30
+
14
+
4
=
0
scores A1
y
x
(b)
M1:
Splits into three separate terms with at least one term with the correct index. The index
1
2
−
does not need to be processed. e.g.
x
2
3
5
1
7
A1:
Two of
−
(unsimplified but the indices must be processed). May appear as a
x
2
+
x
2
−
2
4
list of terms on different lines. May be implied by a correctly integrated expression.
dM1: Attempts to increase the power by one on at least one term. It is dependent on the previous
method mark. The index does not need to be processed for this mark. It cannot be scored for
attempting to integrate individual terms on the numerator or denominator.
5
5
x
2
5
1
7
x
2
2
A1ft:
or exact unsimplified equivalent (indices processed). e.g.
+
5
−
+
x
2
x
2
x
5
1
5
4
2
2
5
1
3
1
Ax
2
Bx
2
−
Follow through on an expression of the form
+
+
→
+
+
where A,
Ax
2
Bx
2
C
Cx
5
1
2
2
B and C are all nonzero.
Do not be concerned with the presence or omission of the constant of integration.
Accept
1x for x.
Ignore any spurious notation including the presence of the integral sign.
4
ddM1: Attempts to substitute
x =
into their changed expression (condone one slip in substituting
in), sets equal to 1
− and attempts to find c.
This mark cannot be scored if they do not have a constant of integration.
It is dependent on both of the previous method marks.
Do not be concerned by the mechanics of the rearrangement and condone arithmetical slips,
but they must achieve a value.
The substitution may be implied by a correct value for c for their integrated expression if no
84
working is shown (you may need to check this), or allow for
±
5
5
1
2
5
1
7
84
(
)
A1:
f ( )
=
+
5
−
−
or any equivalent expression e.g.
0.4
x
2
+
5
x
2
−
1.75
x
x
x
2
x
2
x
5
4
5
Accept
1x for x. Condone equivalent fractions for the coefficients provided both numerator
5
x
2
1
7
84
and denominator are integers. e.g. (
f( )
=
)
+
5
−
−
is A0
x
x
2
x
2.5
4
5
Withhold the final mark if there is still integration notation around the answer or if it is set
equal to 0
8a
7
15
x x
3
−
x
−
5
=
−
5
x
2
2
2
4
7
2
15
0
*
x
−
x
−
=
(
)
b
(2
2
+
3)(
2
−
5)
=
0
2
=
...
x
x
x
(
) ( ) 5
x =
±
7
15
(
)
3
(
)
5
...
or
y =
"
±
5"
−
"
±
5"
−
5
y =
−
=
2
2 "
×±
5"
3
3
−
5,
−
5
−
5
5,
5
−
5
2
2
=
("2 5")
2
+
("3 5")
2
=
...
PQ
PQ =
65
(a)
M1:
Sets up a correct equation in x. Invisible brackets may be implied by further work.
7
15
e.g.
4
2
5
5
x
−
x
−
x
=
−
x
2
2
A1*: Multiplies out and rearranges the equation to achieve the given answer with no errors
seen including invisible brackets. Condone the implied = 0 in their working, provided it is
present in their final answer. There must be at least one stage of intermediate working
between their starting equation and proceeding to the given answer.
7
15
4
−
2
−
5
=
−
5
2
4
−
7
2
−
15
=
0
is M1A0*
x
x
x
x
x
x
2
2
(b) Work for (b) may be seen in (a) which can score
M1:
Attempts to solve the quadratic equation in
x by either
2
factorising
completing the square
quadratic formula
Usual rules apply. Do not be too concerned by the labelling e.g. it is acceptable to use a
=
2
different variable as well as condoning
y
x
and
=
2
x
x
This mark cannot be awarded from directly using a calculator and stating the roots.
B1:
At least one of
5 and
−
5
(and no others)
M1:
Note that if they do not square root their solution to the quadratic in
x then this mark
2
cannot be scored (and no further marks).
(which must be of the form k b
+
and k b
−
Attempts to substitute in
±
"5"
where k
may be 1) into either of the equations of the curves and proceeds to find irrational
expressions for the y coordinates of P and Q of the form 5
±±
"5"
n
(which may be the
same for both). They cannot be decimals.
−
3 5
−
10
3 5
−
10
May be seen as e.g.
−
5,
5,
dM1: A full attempt to find the length PQ. It is dependent on the previous method mark.
They must be using (
, 5
), (
, 5
)
k b
±±
n b
−
k b
±±
n b
or equivalent.
Look for an expression for PQ of the form e.g.
(
)
2
(
)
2
(
−−
)
+
( 5
±±
)
−±±
( 5
)
k b
k b
n b
n b
They must be attempting to find (
)
2
for their x coordinates which would lead to
x
−
x
2
1
±
2k b
but condone ambiguity in the location of negatives when calculating (
−
y
2
2
(
)
10
+
3 5
−
10
+
3 5
2
x = ±
5
e.g.
:
( 5
−−
5)
+ −
−
scores B1M1dM1
2
2
2
(
)
110 15 11
+
−
110 15 11
−
2
e.g.
(
−
11
−
11)
+
−
scores B0M1dM1
x = ±
11
22
22
(as their (
)
−
2
is correct for their
"5"
and we condone sign errors when finding
x
x
±
2
1
(
)
2
using their y coordinates of P and Q )
y
−
y
2
1
If the subtractions are not shown then they must achieve either of the following:
(2
k b
)
2
+
(2
n b
)
2
(n does not have to be an integer) or equivalent
o
(2
)
2
+
0
2
o
k b
(where the y coordinates were the same)
(They must have still found the y coordinates to score this mark)
Do not condone invisible brackets unless it does not affect the processing. This mark may be
implied by their answer.
A1:
65 cao (this cannot be scored if the first method mark has not been scored)
8(b)[5]
M1:
Attempts to find critical values by solving the given quadratic by either
(
−
4)(
+
2
)
factorising (do not accept
p
p
unless they have divided both sides of the
3
inequality/equation by 3 first)
5
2
(
)
completing the square
3
p
−
±
...
p
=
...
3
quadratic formula
Usual rules apply (see general marking principles for guidance). May be in another
variable. Allow this mark to be scored if seen in (a).
This mark cannot be awarded from directly using a calculator and stating the roots.
M1:
Attempts to find the outside region for their critical values. May use another variable e.g. x
2
May be implied by
or incorrect use of inequalities e.g.
p
"
−
",
p
"4"
4
<
p
<−
3
This is not dependent on the first method mark so if a calculator has been used then this
mark can still be scored.
2
A1:
or
(or equivalent). (may appear on separate lines). Isw once a correct
p <−
p >
4
3
answer is seen, provided there is no contradiction and no part of the range is rejected.
This can only be scored provided both previous method marks have been awarded.
Must be in terms of p
2
2 ,
2
✓
:
<−
∪
>
4
Accept e.g.
p
<−
p
>
4
,
p
<−
p
>
4
p p
p
or variations of
3
3
3
2
(
)
−∞−
,
,
4,
∞
these. Accept
3
2
2
✓
:
<−
∩
>
4
Do not accept e.g.
p <−
and
p >
4
p p
p
or variations of these
3
3
2
2 ,
2
,
, [4,
)
Do not accept
−∞−
∞
−
4
4
<
<−
p
p
p
3
3
3
7ai
1
4(4)
2
+
10
−
7(4)
2
15
f (4)
′
=
=
1
2
4(4)
2
ii
15
2
"
"
→−
2
15
2
+ = −
1
"
"(
−
4)
y
x
15
2
+
15
+
7
=
0
x
y
b
1
4
2
+
10
−
7
3
1
x
x
2
−
= ±
...
x
2
±
...
x
2
±
...
1
4
x
2
3
1
5
7
Two of the terms of
−
x
2
+
x
2
−
2
4
3
5
1
7
2
5
1
7
−
x
2
+
x
2
−
d
x
=
x
2
+
5
x
2
−
x
(
+
c
)
2
4
5
4
2
5
1
7
5(4)
1
...
2
+
2
−
+
c
= −
c
=
5
4
5
1
2
7
84
(
)
f ( )
x
=
x
2
+
5
x
2
−
x
−
5
4
5
Mark (a) and (b) together so do not be concerned with labelling of the parts
(a)
(i)
15
B1:
2 oe stated (as the gradient of f(x) at P).
(ii)
M1
Finds the negative reciprocal of their gradient in part (i). If they do not have a gradient in (i)
2
1
then only allow
−
or an attempt at
−′
15
f (4)
M1
Attempts to find the equation of the normal using a changed gradient to that found in (i) and
with the coordinates in the correct positions. If they do not have a gradient in (i) then
(4,
−
1)
15
allow any gradient
≠
. If they use y
=
mx
+ they must proceed as far as
c
c =
...
2
A1:
2
x
+
15
y
+
7
=
0
or any multiple of this where all the coefficients are integers and all terms
are on the same side of the equation. e.g. 30
+
14
+
4
=
0
scores A1
y
x
(b)
M1:
Splits into three separate terms with at least one term with the correct index. The index
1
2
−
does not need to be processed. e.g.
x
2
3
5
1
7
A1:
Two of
−
(unsimplified but the indices must be processed). May appear as a
x
2
+
x
2
−
2
4
list of terms on different lines. May be implied by a correctly integrated expression.
dM1: Attempts to increase the power by one on at least one term. It is dependent on the previous
method mark. The index does not need to be processed for this mark. It cannot be scored for
attempting to integrate individual terms on the numerator or denominator.
5
5
x
2
5
1
7
x
2
2
A1ft:
or exact unsimplified equivalent (indices processed). e.g.
+
5
−
+
x
2
x
2
x
5
1
5
4
2
2
5
1
3
1
Ax
2
Bx
2
−
Follow through on an expression of the form
+
+
→
+
+
where A,
Ax
2
Bx
2
C
Cx
5
1
2
2
B and C are all nonzero.
Do not be concerned with the presence or omission of the constant of integration.
Accept
1x for x.
Ignore any spurious notation including the presence of the integral sign.
4
ddM1: Attempts to substitute
x =
into their changed expression (condone one slip in substituting
in), sets equal to 1
− and attempts to find c.
This mark cannot be scored if they do not have a constant of integration.
It is dependent on both of the previous method marks.
Do not be concerned by the mechanics of the rearrangement and condone arithmetical slips,
but they must achieve a value.
The substitution may be implied by a correct value for c for their integrated expression if no
84
working is shown (you may need to check this), or allow for
±
5
5
1
2
5
1
7
84
(
)
A1:
f ( )
=
+
5
−
−
or any equivalent expression e.g.
0.4
x
2
+
5
x
2
−
1.75
x
x
x
2
x
2
x
5
4
5
Accept
1x for x. Condone equivalent fractions for the coefficients provided both numerator
5
x
2
1
7
84
and denominator are integers. e.g. (
f( )
=
)
+
5
−
−
is A0
x
x
2
x
2.5
4
5
Withhold the final mark if there is still integration notation around the answer or if it is set
equal to 0
8a
7
15
x x
3
−
x
−
5
=
−
5
x
2
2
2
4
7
2
15
0
*
x
−
x
−
=
(
)
b
(2
2
+
3)(
2
−
5)
=
0
2
=
...
x
x
x
(
) ( ) 5
x =
±
7
15
(
)
3
(
)
5
...
or
y =
"
±
5"
−
"
±
5"
−
5
y =
−
=
2
2 "
×±
5"
3
3
−
5,
−
5
−
5
5,
5
−
5
2
2
=
("2 5")
2
+
("3 5")
2
=
...
PQ
PQ =
65
(a)
M1:
Sets up a correct equation in x. Invisible brackets may be implied by further work.
7
15
e.g.
4
2
5
5
x
−
x
−
x
=
−
x
2
2
A1*: Multiplies out and rearranges the equation to achieve the given answer with no errors
seen including invisible brackets. Condone the implied = 0 in their working, provided it is
present in their final answer. There must be at least one stage of intermediate working
between their starting equation and proceeding to the given answer.
7
15
4
−
2
−
5
=
−
5
2
4
−
7
2
−
15
=
0
is M1A0*
x
x
x
x
x
x
2
2
(b) Work for (b) may be seen in (a) which can score
M1:
Attempts to solve the quadratic equation in
x by either
2
factorising
completing the square
quadratic formula
Usual rules apply. Do not be too concerned by the labelling e.g. it is acceptable to use a
=
2
different variable as well as condoning
y
x
and
=
2
x
x
This mark cannot be awarded from directly using a calculator and stating the roots.
B1:
At least one of
5 and
−
5
(and no others)
M1:
Note that if they do not square root their solution to the quadratic in
x then this mark
2
cannot be scored (and no further marks).
(which must be of the form k b
+
and k b
−
Attempts to substitute in
±
"5"
where k
may be 1) into either of the equations of the curves and proceeds to find irrational
expressions for the y coordinates of P and Q of the form 5
±±
"5"
n
(which may be the
same for both). They cannot be decimals.
−
3 5
−
10
3 5
−
10
May be seen as e.g.
−
5,
5,
dM1: A full attempt to find the length PQ. It is dependent on the previous method mark.
They must be using (
, 5
), (
, 5
)
k b
±±
n b
−
k b
±±
n b
or equivalent.
Look for an expression for PQ of the form e.g.
(
)
2
(
)
2
(
−−
)
+
( 5
±±
)
−±±
( 5
)
k b
k b
n b
n b
They must be attempting to find (
)
2
for their x coordinates which would lead to
x
−
x
2
1
±
2k b
but condone ambiguity in the location of negatives when calculating (
−
y
2
2
(
)
10
+
3 5
−
10
+
3 5
2
x = ±
5
e.g.
:
( 5
−−
5)
+ −
−
scores B1M1dM1
2
2
2
(
)
110 15 11
+
−
110 15 11
−
2
e.g.
(
−
11
−
11)
+
−
scores B0M1dM1
x = ±
11
22
22
(as their (
)
−
2
is correct for their
"5"
and we condone sign errors when finding
x
x
±
2
1
(
)
2
using their y coordinates of P and Q )
y
−
y
2
1
If the subtractions are not shown then they must achieve either of the following:
(2
k b
)
2
+
(2
n b
)
2
(n does not have to be an integer) or equivalent
o
(2
)
2
+
0
2
o
k b
(where the y coordinates were the same)
(They must have still found the y coordinates to score this mark)
Do not condone invisible brackets unless it does not affect the processing. This mark may be
implied by their answer.
A1:
65 cao (this cannot be scored if the first method mark has not been scored)

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