Mathematics · May / June 2023 · Paper 21 · Question 7
YMA01
As printed in the paper
The question as text
7. The height of a river above a fixed point on the riverbed was monitored over a 7‑day period. The height of the river, H metres, t days after monitoring began, was given by t 2) + 17 0 t 7 H = 20 (20 + 6t – t Given that H has a stationary value at t = α
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[5]
3
5
1
1
1
3
3
t
2
t
2
d
H
1
9
1
−
17
H
=
t
2
+
−
+
=
t
2
+
t
2
−
t
2
10
20
d
2
20
8
t
1
1
9
1
1
3
1
9
1
e.g.
−+
...
5
α
2
−
18
α
−
20
=
0
*
t
2
t
2
−
t
2
=
0
+
t
−
t
2
=
0
2
20
8
2
20
8
7(b)[1]
9
+
181
α
=
=
awrt 4.491
5
7(c)[2]
3
1
1
d
2
H
1
3
9
1
3
1
(
)
(
)
−
(
)
−
−
"4.49"
−
9 "4.49"
3 "4.49"
2
2
2
= −
t
2
+
t
2
−
t
2
−
+
−
d
2
4
40
16
t
4
40
16
d
2
H
(
)
*
= −
0.3...
<
0
max
d
2
t
(a) If they change t to αbefore differentiating then score maximum M1A1A0dM1A0
M1
Multiplies out and attempts to differentiate. Award for one correct index from correct working (may
3
1
1
− (condone if these
be implied if they multiply out and differentiate at the same time)
t
2
or
t
2
or
t
2
3
1
1
− or any other variable) The indices do not need to be processed for this mark.
are
α
or
α
or
α
2
2
2
5 1
−
e.g.
2t
Alternatively, they may attempt the product rule so look for a correct general expression such as
d
1
1
H
−
=
...
t
2
(20
+
6
t
−
t
2
)
+
...
t
2
(
A
+
Bt
)
(condone if in terms of αor another variable instead)
d
t
1
1
9
1
1
3
−
A1
One of
t
2
,
t
2
,
−
t
2
(simplified or unsimplified) via the main scheme
2
20
8
If they differentiate using the product rule then this mark can only be awarded for one of
1
1
1
1
1
1
−
−
(20
+
6
−
2
) ...
+
(
+
)
...
(20
+
6
−
2
)
+
(6
−
2 )
40 t
2
t
t
t
2
A
Bt
or
t
2
t
t
t
2
t
(unsimplified).
20
In both cases the indices must be processed. Condone if in terms of α
d
H
1
1
9
1
1
3
d
H
1
1
−
−
(
)
20 18
5
2
A1
=
t
2
+
t
2
−
t
2
or equivalent e.g.
=
t
2
+
t
−
t
which may be
d
t
2
20
8
d
t
40
unsimplified. Also allow
1
1
d
H
1
1
−
(20
6
2
)
(6
2 )
=
t
2
+
t
−
t
+
t
2
−
t
e.g.
(via the product rule)
d
40
20
t
1
1
3
−±
dM1
Sets their derivative of the form
...
2
...
2
±
...
2
=
0
(may be implied) and proceeds via a correct
t
t
t
method to a 3TQ in t or α(or any other variable) where the powers and the coefficients of the terms
are integers. Condone sign slips only. It is dependent on the previous method mark.
Proceeding directly to the given answer is M0, unless they state how they have proceeded
1
from the derivative directly to the given answer e.g. set t
=
α
and
×−
40t
2
can score this mark for
20 18
+
t
−
5
t
2
d
1
1
H
−
(
)
2
2
e.g.
=
t
2
"20 18
+
t
−
5 "
t
=
0
"5
t
−
18
t
−
20"
=
0
d
t
40
A1*
Achieves
with no errors seen and all previous marks scored. Must be in terms of
5
α
2
−
18
α
−
20
=
0
α(do not be concerned if it looks like a) . The
=
0
must have been seen before achieving the given
answer.
Do not be concerned with the labelling of the derivative if seen e.g. d
y
x instead
d
Do not penalise invisible brackets as long as they are recovered and the intention is clear through
their subsequent lines of working.
Alternative method: They may attempt to multiply by e.g. 20 or take out a factor and then differentiate.
This can score full marks, provided for A1* they explain at some point that this does not affect the given
answer. You will need to check their terms are correct for the first A mark and withhold the A marks if e.g.
they do not multiply all terms including the +17 by 20
(b) Mark (b) and (c) together
B1
α=awrt 4.491 only. Accept the exact answer. If both roots are stated then awrt 4.491 must be the
d
2
H
only one used within their
, or if both values are used then the other is rejected. If awrt 4.491 is
d
2
t
seen then allow for it to be subsequently rounded to 2 or 3sf when used as evidence of selecting awrt
4.491
(c)
Attempts to differentiate their d
3
H
−
M1
t to achieve a second derivative of the form
...
t
2
+
...
d
(may be in terms of α or another variable) and either attempts to substitute in their value of αwhich
must be between 0 and 7 or considers the sign of their second derivative. Condone their αto have
been subsequently rounded before substituted in
1
1
1
3
(
)
(
)
−
−
t
2
20 18
+
t
−
5
t
2
→−
t
2
20 18
−
t
+
15
t
2
May be seen as
40
80
If differentiation in part (a) is via the product rule then look for an attempt to apply the product rule
1
1
d
H
−
...
(20
6
2
)
...
(
)
on their
=
t
2
+
t
−
t
+
t
2
A
+
Bt
leading to
d
t
3
1
1
−
−
...
(20
+
6
−
2
)
+
...
(
+
)
+
...
t
2
t
t
t
2
C
Dt
t
2
Alternatively, attempts to show that the second derivative is always negative. e.g. attempts to
complete the square on
20 18
−
t
+
15
t
2
and considers its sign.
d
2
H
A1*
(
)
when α=awrt 4.491. If evaluated, it must be 0.3
−
or better
= −
0.3
<
0
max
d
t
2
( 0.3174...
−
). The second derivative must be correct but condone in terms of α
In the alternative method look for correct working and valid argument explaining that eg
1
3
(
)
−
20 18
−
+
15
2
is always greater than zero so
−
2
20 18
−
+
15
2
is always negative for any
t
t
80 t
t
t
value of t
Condone poor phrasing as part of the explanation e.g. “
t =
4.491
is a maximum” instead of
“
t =
4.491
is the value at which H is at a maximum”
d
2
H
Condone poor notation for
and condone e.g.
y
''
d
2
t
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