Mathematics · Oct / Nov 2025 · Paper 31 · Question 6
YMA01
As printed in the paper
The question as text
6. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. 2 + 5) find d y
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(i)[2]
d
y
4
x
=
d
x
2
x
2
+
5
6(ii)[5]
21
x
7
(
)
d
x
=
ln(3
x
2
+
k
)
+
c
3
x
2
+
k
2
k
7
7
3
k
2
+
k
3
k
2
+
k
2
+
<
<
<
ln(3
x
k
)
7ln8
ln
7ln8
64
+
+
2
2
3
k
3
k
1
3
k
2
+
k
( )
(
)
<
2
−
−
<
=
64
3
k
63
k
192
0
k
...
+
3
k
k =
23
awarded if subsequent working was correct.
(i)
...
M1:
. Allow with anything for the …
2
x +
2
5
d
y
4
x
A1:
=
o.e. and isw
2
+
d
x
2
x
5
(ii)
M1:
Attempts to integrate to an expression of the form
A
ln(3
x
2
+
k
)
or
. Allow for cancelling x’s e.g
(Allow A or B to be 1)
7 ln(3
A1:
x
2
+
k
)
o.e. with or without the constant of integration. Condone
2
missing brackets for this mark and condone if extra integral symbols are
left in.
M1:
Attempts to substitute in the limits into their changed expression (which
must involve ln), sets the expression less than7ln8 (condone an equation)
and uses correct log laws to remove lns – including correct bracketing
used. You may allow ln 8 to be written as a decimal but any ln or exponential
work must be correct. Note the form for their integral must be such that
they are able to use log laws to remove lns.
dM1: Rearranges their inequality (may be an equation), proceeds to a threeterm
quadratic in k and attempts to find a value for k (usual rules apply for solving
a quadratic – if using a calculator the value(s) must be correct for their
equation). It is dependent on the previous method mark.
A1:
23 cao
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