Mathematics · Oct / Nov 2022 · Paper 11 · Question 5

YMA01

As printed in the paper

The question as text

5. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. The curve C has equation y = f(x), x > 0 Given that x f′(x) = 12 + • – 4 x 3 • the point P(9, 8) lies on C

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

Show the answer and mark scheme

5(a)[5]
Eg5
0,
2 − M0dM0B1A0 (no method seen at all)
,
3
This response below can score full marks.
3
2
5
5
2
M1
y
3
=
" "
y
=
" "
(or one of their positive solutions from part (a)). The question
2
2
states a calculator cannot be used so do not award for an implied method if they have
decimal solutions or the exact answer without the method to deal with the fractional power
seen. No marks can be scored without a positive solution from (a) or they restart in (b).
2
3 5
Allow notation such as
y =
and you do not need to see
y =
...
2
5 10
or 1.25 10 or 1
A1
1
10
only and no other solutions cso
4
4
2
5
Eg1
y
3
=
" "
y
=
3.95...
is M0A0
2
2
5
5
Eg2
y
3
=
" "
y
=
10
is M0A0
2
4
2
5
125
5
Eg3
y
3
=
" "
y
2
=
y
=
10
can score M1A1 as they have shown a method to deal
2
8
4
with the factional power.
5(a)
1
x
−
f ( )
′
=
12
+
−
4
x
x
2
3
1
1
−→
One of
, 4
−→−
4x
→
2
x
2
x
2
,
x
x
1
1
2
x
x
−+
f( )x =
12
2
−
4d
=
24
2
+
−
4
+
x
x
x
x
c
3
6
1
2
8
24(9)
4(9)
...
=
2
+
−
+
c
c
=
6
1
2
83
x
(
)
f( )
=
24
2
+
−
4
−
x
x
x
6
2
(b)
12
9
(
)
f (9)
′
=
+
−
4
=
3
3
9
1
3
→−
3
1
y −= −
8
"
"(0
−
9)
3
(0, 11)
(a)
1
1
−→
M1
Integrates by raising the power on one of the terms (ie
,
−→−
4
4x
1
,
→
x
2
x
2
x
The index does not need to be processed.
1
x
2
A1
Two terms correct of
24
+
−
4
or unsimplified equivalent which may appear as a list.
x
2
x
6
The indices must be processed. Allow
1x
1
2
x
A1
24
x
2
+
−
4
x
(
+
c
)
or unsimplified equivalent. Condone the lack of +c for this mark.
6
Allow
1x and ignore any spurious notation including = 0
dM1
Attempts to substitute
x
=
9,
y
=
8
into their f(x) and proceeds to find c. It is dependent on
the previous method mark. If they have no + c then this mark cannot be scored.
1
2
83
x
(
)
A1
f( )
=
24
+
−
4
−
or simplified equivalent. Withhold this mark if they attempt to
x
x
2
x
6
2
make the coefficients integers. Do not accept rounded decimals such as
0.166x . Also
2
withhold this mark if spurious notation around the answer is seen such as the integral and
dx or an = 0
(b)
M1
Attempts to substitute
x =
9
to find a value for f (9)
′
dM1
Attempts to find the negative reciprocal to find the gradient of the normal. It is dependent on
the previous method mark.
It is for the attempt to find the normal line passing though (9,8) , using a changed gradient
M1
from f (9)
′
and substituting
x =
0
. Alternatively, they may just use their gradient of the
normal to determine where the line crosses the yaxis knowing that for every “3 units to the
left it is 1 up”. Condone a changed gradient from finding f (9)
′′
instead of from f (9)
′
for
this mark.
A1
(0, 11) or allow it to be written as
=
0,
=
11
x
y
5(b)[4]
3
2
5
5
2
M1
y
3
=
" "
y
=
" "
(or one of their positive solutions from part (a)). The question
2
2
states a calculator cannot be used so do not award for an implied method if they have
decimal solutions or the exact answer without the method to deal with the fractional power
seen. No marks can be scored without a positive solution from (a) or they restart in (b).
2
3 5
Allow notation such as
y =
and you do not need to see
y =
...
2
5 10
or 1.25 10 or 1
A1
1
10
only and no other solutions cso
4
4
2
5
Eg1
y
3
=
" "
y
=
3.95...
is M0A0
2
2
5
5
Eg2
y
3
=
" "
y
=
10
is M0A0
2
4
2
5
125
5
Eg3
y
3
=
" "
y
2
=
y
=
10
can score M1A1 as they have shown a method to deal
2
8
4
with the factional power.
5(a)
1
x
−
f ( )
′
=
12
+
−
4
x
x
2
3
1
1
−→
One of
, 4
−→−
4x
→
2
x
2
x
2
,
x
x
1
1
2
x
x
−+
f( )x =
12
2
−
4d
=
24
2
+
−
4
+
x
x
x
x
c
3
6
1
2
8
24(9)
4(9)
...
=
2
+
−
+
c
c
=
6
1
2
83
x
(
)
f( )
=
24
2
+
−
4
−
x
x
x
6
2
(b)
12
9
(
)
f (9)
′
=
+
−
4
=
3
3
9
1
3
→−
3
1
y −= −
8
"
"(0
−
9)
3
(0, 11)
(a)
1
1
−→
M1
Integrates by raising the power on one of the terms (ie
,
−→−
4
4x
1
,
→
x
2
x
2
x
The index does not need to be processed.
1
x
2
A1
Two terms correct of
24
+
−
4
or unsimplified equivalent which may appear as a list.
x
2
x
6
The indices must be processed. Allow
1x
1
2
x
A1
24
x
2
+
−
4
x
(
+
c
)
or unsimplified equivalent. Condone the lack of +c for this mark.
6
Allow
1x and ignore any spurious notation including = 0
dM1
Attempts to substitute
x
=
9,
y
=
8
into their f(x) and proceeds to find c. It is dependent on
the previous method mark. If they have no + c then this mark cannot be scored.
1
2
83
x
(
)
A1
f( )
=
24
+
−
4
−
or simplified equivalent. Withhold this mark if they attempt to
x
x
2
x
6
2
make the coefficients integers. Do not accept rounded decimals such as
0.166x . Also
2
withhold this mark if spurious notation around the answer is seen such as the integral and
dx or an = 0
(b)
M1
Attempts to substitute
x =
9
to find a value for f (9)
′
dM1
Attempts to find the negative reciprocal to find the gradient of the normal. It is dependent on
the previous method mark.
It is for the attempt to find the normal line passing though (9,8) , using a changed gradient
M1
from f (9)
′
and substituting
x =
0
. Alternatively, they may just use their gradient of the
normal to determine where the line crosses the yaxis knowing that for every “3 units to the
left it is 1 up”. Condone a changed gradient from finding f (9)
′′
instead of from f (9)
′
for
this mark.
A1
(0, 11) or allow it to be written as
=
0,
=
11
x
y

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