Mathematics · Oct / Nov 2024 · Paper 21 · Question 1
YMA01
As printed in the paper
The question as text
1. A continuous curve has equation y = f (x). A table of values of x and y for y = f (x) is shown below. 0.5 1.75 3 4.25 5.5 x y 3.479 6.101 7.448 6.823 5.182 Using the trapezium rule with all the values of y in the given table,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
1(a)[3]
h =
1.25
1 1.25 3.479
(
)
A ≈
×
+
5.182
+
2 6.101 7.448
+
+
6.823
2
= 30.9
1(b)[2]
5.5
(
( )
)
5.5
f
+
4
d
=
30.9
+
2
2
=
30.9
+ ×
2 5.5
2
−×
2 0.5
2
x
x
x
x
0.5
0.5
or
5.5
5.5
1
(
)
( )
( )
(
)
f
x
+
4
x
d
x
=
f
x
d
x
+
4 0.5
×
+ ×
4 5.5
×=
5
...
2
0.5
0.5
= 30.9 + 60 = 90.9
(a)
Correct strip width of 1.25. Allow equivalent numerical expressions e.g. 5.5
B1:
May be implied by sight of e.g.1.25
or 5
8 or 0.625 in front of the bracket. May also
2
be implied by a correct answer if no incorrect working seen.
h = −
1.25
is B0 unless
they recover and subsequently use
h = +
1.25
M1:
Correct application of the trapezium rule with their h (which may be 1).
(
This requires a correct inner bracket structure
3.479 5.182
+
+
2 6.101 7.448 6.823
+
multiplied by 1 .
h Condone slips copying values from the table or the omission of the
2
final bracket on the rhs e.g. 1
2 ×
"1.25"(3.479
+
5.182
+
2(6.101
+
7.448
+
6.823)
1
(
)
but
2 ×
"1.25" 3.479
×
+
5.182
+
2 6.101
+
7.448
+
6.823
is M0 unless the missing
brackets are recovered or implied by the correct answer (you may need to check).
Also allow for a correct method adding individual trapezia using their h condoning
copying errors but the brackets must be correct (or recovered or implied by later
work)
e.g. 1
1
1
"1.25"(3.479
6.101)
"1.25"(6.101 7.448)
"1.25"(7.448
×
+
+
×
+
+
×
2
2
2
1
2 ×
"1.25"(6.823
+
5.182)
A1:
For awrt 30.9. isw once a correct answer is seen. Correct answer with no working
scores B1M1A1 but if there is evidence of using
h = −
1.25
then maximum awarded
is B0M1A0. The A mark cannot be awarded without both B1M1 being awarded in
this question. The exact answer is 9881
320 and scores A0 unless followed by awrt 30.9
For reference, note that use of h = 1 (from using 5 strips) gives 24.7… or 9881
Scores B1M0A0 (BOD)
(b)
5.5
( )
M1:
Attempts their answer to (a) +
...
2
=
their ans
wer to
a
+
...5.5
2
−
.. 0.
.
x
0.5
their answer to
( )
a
2 5.5
2
2
0
.5
2
May be implied by e.g.
+ ×
−×
if the integration is not
seen explicitly.
Or
1 4 0.5
their answer to (a)
+
(
×
+ ×
4 5.5
)
×
5
(trapezium)
2
1
(
)
oe e.g. their answer to (a) (
)
(
)
+
4 0.5 5
×
×
+
5
×
5 5.5
×
−×
5 0.5
(rectangle + triangle)
2
May be implied by their answer to (a) + 60
Condone clear misreads of the 4 in the 4x but do not condone a misread of e.g. 4x for 4.
Condone clear miscopy/misread of limits as long as they are nonzero E.g. 5 for 5.5.
5.5
5.5
(
( )
)
Condone poor notation e.g.
f
+
4
d
=
30.9
+
2
2
d
=
30.9
+ ×
2 5.5
x
x
x
x
x
0.5
0.5
A1ft: Correct answer of awrt 90.9 or correct ft e.g. 60 + their answer to part (a)
Allow exact or exact ft answers e.g. 29081
320 for 90.9
Correct answer only scores no marks as the questions says “making your method
clear”
Attempts to use the trapezium rule again by adding 4x to the y values score M0 in (b)
(
)
Note that
30.9
+
4 0.5 1.75 3 4.25 5.5
+
+ +
+
=
90.9
is fortuitous and scores M0
1
5
(
)
But
30.9
2
22
2 7 12 17
(
)
90.9
scores M1A1
+
×
+
+
+
+
=
2
4
Condone an error with the strip width as in part (a) for this method but the ft is not available
in this case.
Examples of minimal acceptable working in (b) for both marks:
30.9 + 60 = 90.9
5.5
2
30.9
+
2
x
=
90.9
0.5
5.5
5.5
f
( )
d
4 d
30.9
60
90.9
x
x
+
x x
=
+
=
0.5
0.5
30.9
2 5.5
2
2 0.5
2
90.9
+ ×
−×
=
All score M1A1
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