Mathematics · Oct / Nov 2024 · Paper 41 · Question 9
YMA01
As printed in the paper
Show the answer and mark scheme
9(a)[2]
1
A
B
1
(
2
1
)
...or
...
+
A
x
−
+
Bx
A
=
B
=
(
)
x
2
x
−
1
x
2
x
−
1
1
2
1
−
(
)
x
2
x
−
1
2
x
−
1
x
9(b)[6]
d
h
1
t
1
1
t
(
)
=
h
2
h
−
1 cos
d
h
=
cos
d
t
(
)
d
t
50
10
h
2
h
−
1
50
10
1
2
1
(
)
d
h
=
−
d
h
=
ln 2
h
−
1
−
ln
h
(
)
h
2
h
−
1
2
h
−
1
h
2
1
1
t
1
t
(
)
−
d
h
=
cos
d
t
ln 2
h
−
1
−
ln
h
=
sin
(
+
c
)
2
h
−
1
h
50
10
5
10
(
)
e.g.
ln 2 2.5 1
×
−
−
ln2.5
=
c
or
t
=
0,
h
=
2.5
1sin
t
2
h
−
1
1
t
2
h
−
1
e.g.
ln
=
sin
+
c
=
X
e
5
10
t
=
0,
h
=
2.5
X
=
1.6
h
5
10
h
1sin
1
t
2
h
−
1
t
(
)
ln 2
h
−
1
−
ln
h
=
sin
+
ln1.6
=
1.6e
5
10
5
10
h
1sin
t
2
h
−
1.6 e
h
5
10
=
1
h
=
...
5
h
=
1sin
t
10
−
8e
5
10
( )
9(c)[2]
t
t
π
5
π
9
π
sin
=
1
=
,
,
,...
t
=
...
10
10
2
2
2
t
=
45
π
=
141 (s)
(a)
1
A
B
M1: Correct attempt at partial fractions. Sets (
and uses a correct method to
=
+
)
x
2
x
−
1
x
2
x
−
1
A
B
identify at least one constant. Incorrect work e.g.
+
=
1
Ax
+
B
(2
x
−
1)
etc. is M0.
x
2
x
−
1
A1: Correct partial fractions. Note that this mark is not just for the correct constants, it is for the
correct written fractions but allow this mark to be scored if it is seen in part (b) (in terms of h
or x).
2
−
1
Condone
+
or written as e.g.
2(2
x
−
1)
−
1
−
x
−
1
2
x
−
1
x
(b) Note that they may work in x throughout, but the final answer must be
h =
...
M1: Attempts to separate the variables. Do not be concerned by slips with the constants so look
1
cos ...
(
)
for as a minimum
=
t
o.e. May omit an integral sign on one of the sides
(
)
h
2
h
−
1
or may be implied by their integrated expression.
A
B
+
d
h
=
α
ln
h
+
β
ln(2
h
−
1)
M1: For
h
2
h
−
1
" "
A
" "
B
d
ln
ln(2
1)
or e.g.
λ
+
h
=
α
h
+
β
h
−
(if they attempt some rearranging first)
h
2
h
−
1
Condone invisible brackets.
A1ft: Fully correct equation following through their A and B. The “+c” is not required here.
A
B
1
t
B
1
t
o.e.
+
d
h
=
cos
d
t
A
ln
h
+
ln(2
h
−
1)
=
sin
(
+
c
)
2
1
50
10
2
5
10
h
h
−
Condone invisible brackets.
M1: States or uses t = 0, h = 2.5 followed by
c =
...
is sufficient for M1. You do not need to check
their substitution/rearrangement but if there is just a value for the constant then you may need
to check they used t = 0, h = 2.5. May be implied by e.g. ln8
ln5
, ln1.6, awrt 0.47
−
following correct integration. Note that they may have removed lns first.
M1: To score this mark they must have integrated and achieved an expression of the form
t
±
p
ln
h
±
q
ln(2
h
−
1)
= ±
r
sin 10
+
c
o.e.
Rearranges using correct algebra to make h the subject including when combining logs and
when dealing with the constant. May occur before finding their constant but there must have
been a constant.
Expect to see terms in h collected on one side as part of their manipulation rather than
proceeding to the given answer once they have established the value of k. Alternatively, they
may cancel e.g. 2
h
1
1
−→
2
−
and rearrange to make h the subject using correct algebra.
h
h
Do not condone sign slips for this mark.
A1: Correct expression (including
h =
...)
in the required form. Requires
Correct integrationCorrect algebraDo not penalise brackets recovered in their work or working in x rather than h until the final
answer.
(c) Note that an answer with no working seen scores M0A0
t
=
M1: Solves sin
1
to find a value for t. Do not allow this mark to be implied by
10
t
π
e.g. 5 , 25 , 45
π
π
πbut allow for e.g.
= Must be working in radians.
=
t
..
10
2
Note that attempts that do not use part (b) and just use the differential equation proceeding to
t
=
t
= −
cos
0
score M0A0. (Note if their k is negative condone if they solve sin
1
10
10
A1: awrt 141 (but do not allow just 45π) Units not required.
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