Mathematics · May / June 2024 · Paper 21 · Question 6
YMA01
As printed in the paper
Show the answer and mark scheme
6(a)[3]
Shape or asymptote M1
Intercept B1
(
)
0,5
Fully correct A1
y =
4
O
h =
0.3
0.3 0
(
)
Area
≈
+
4.3137
+ ×
2
0.3246
+
0.8629 1.6643
+
+
2.7896
6(b)[3]
2
= awrt 2.34
3.5
(
)
3.5
(
)
3.5
2
2
x
2
d
2
x
2
4
d
2.34
2
(c) (i)
+
x
x
=
−
x
+
x
x
=
+
x
2
2
2
=
2.34 16.5
+
=
18.84
3.5
(
+ −
)
3.5
(
)
x
1
x
2
4
d
=
2 2
−
2
d
=
2 2.34
×
=
4.68
(ii)
x
x
x
x
2
2
(a)
M1:
For either the correct shape, an increasing curve in any position in quadrants 1 and 2
or for the correct asymptote (labelled in some way at height 4) but must have a curve
approaching as an asymptote, not just the line drawn. Be tolerant on “pen slips” approaching
the asymptote.
B1:
Correct intercept (touching or crossing). Accept 5 on axis, or stated as (0,5). Condone (5,0)
on the graph if it is in the right place.
A1:
Fully correct. Shape in quadrants 1 and 2 only with the y intercept at 5 and asymptote of
y =
4
clearly drawn on the graph and labelled or stated separately. Be tolerant on “pen slips”
approaching the asymptote. If there is conflicting information, what is on the graph takes
precedence.
(b)
This is implied by sight of 0.3
B1:
For
h =
0.3
2 in front of the bracket.
M1:
Applies the trapezium rule with correct bracket condoning slips in copying (or even
truncating) values. May use separate trapezia. Allow a missing final bracket, but otherwise
bracketing must be correct or recovered by correct answer.
A1:
awrt 2.34 Note that the calculator answer for this integral is 2.30
(c) (i)
3.5
(
)
3.5
(
)
x
x
2
M1:
For realising that
2
+
2
d
=
2
−
2
+
4
d
=
2.34
+
x
x
x
x
x
kx
2
2
be seen with an attempt to integrate (increase in power) made (though may be implied).
1.5 4 2
(
Could also use answer to (b) and add the area under the trapezium
2
be implied by correct answer after seeing the integral correctly split.
3.5
x x
3.5
Another alternative is to isolate the
2
d
=
x
2
+
"2.34"
and use this to find a value
2
2
for the integral required.
A1ft: ''2.34'' 16.5
+
=
18.84
but follow through on their 2.34 and allow awrt to 3s.f..
There must be a suitable method for this mark to be awarded.
Answer via repeated trapezium rule scores M0A0
(c)(ii)
B1 ft: For 2 their ''2.34''
×
=
4.68
((awrt to 3s.f.) must be evaluated). Follow through on their
answer to (b).
Note: Answer via repeated trapezium rule is permitted to score the B1 as long the answer is twice
their (b).
Some examples of graphs for (a):
M0B1A0 Not in quadrant 2, but does have correct intercept.
M1B1A0 Does not approach asymptote
M1B0A0 Incorrect intercept and asymptote
M1B1A0 Correct asymptote and intercept M1B1A0 asymptote not drawn.
2
+
2
+
8
−
10
=
29
6(c)[3]
Shape or asymptote M1
Intercept B1
(
)
0,5
Fully correct A1
y =
4
O
h =
0.3
0.3 0
(
)
Area
≈
+
4.3137
+ ×
2
0.3246
+
0.8629 1.6643
+
+
2.7896
2
= awrt 2.34
3.5
(
)
3.5
(
)
3.5
2
2
x
2
d
2
x
2
4
d
2.34
2
(c) (i)
+
x
x
=
−
x
+
x
x
=
+
x
2
2
2
=
2.34 16.5
+
=
18.84
3.5
(
+ −
)
3.5
(
)
x
1
x
2
4
d
=
2 2
−
2
d
=
2 2.34
×
=
4.68
(ii)
x
x
x
x
2
2
(a)
M1:
For either the correct shape, an increasing curve in any position in quadrants 1 and 2
or for the correct asymptote (labelled in some way at height 4) but must have a curve
approaching as an asymptote, not just the line drawn. Be tolerant on “pen slips” approaching
the asymptote.
B1:
Correct intercept (touching or crossing). Accept 5 on axis, or stated as (0,5). Condone (5,0)
on the graph if it is in the right place.
A1:
Fully correct. Shape in quadrants 1 and 2 only with the y intercept at 5 and asymptote of
y =
4
clearly drawn on the graph and labelled or stated separately. Be tolerant on “pen slips”
approaching the asymptote. If there is conflicting information, what is on the graph takes
precedence.
(b)
This is implied by sight of 0.3
B1:
For
h =
0.3
2 in front of the bracket.
M1:
Applies the trapezium rule with correct bracket condoning slips in copying (or even
truncating) values. May use separate trapezia. Allow a missing final bracket, but otherwise
bracketing must be correct or recovered by correct answer.
A1:
awrt 2.34 Note that the calculator answer for this integral is 2.30
(c) (i)
3.5
(
)
3.5
(
)
x
x
2
M1:
For realising that
2
+
2
d
=
2
−
2
+
4
d
=
2.34
+
x
x
x
x
x
kx
2
2
be seen with an attempt to integrate (increase in power) made (though may be implied).
1.5 4 2
(
Could also use answer to (b) and add the area under the trapezium
2
be implied by correct answer after seeing the integral correctly split.
3.5
x x
3.5
Another alternative is to isolate the
2
d
=
x
2
+
"2.34"
and use this to find a value
2
2
for the integral required.
A1ft: ''2.34'' 16.5
+
=
18.84
but follow through on their 2.34 and allow awrt to 3s.f..
There must be a suitable method for this mark to be awarded.
Answer via repeated trapezium rule scores M0A0
(c)(ii)
B1 ft: For 2 their ''2.34''
×
=
4.68
((awrt to 3s.f.) must be evaluated). Follow through on their
answer to (b).
Note: Answer via repeated trapezium rule is permitted to score the B1 as long the answer is twice
their (b).
Some examples of graphs for (a):
M0B1A0 Not in quadrant 2, but does have correct intercept.
M1B1A0 Does not approach asymptote
M1B0A0 Incorrect intercept and asymptote
M1B1A0 Correct asymptote and intercept M1B1A0 asymptote not drawn.
2
+
2
+
8
−
10
=
29
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