Mathematics · Oct / Nov 2023 · Paper 31 · Question 9
YMA01
As printed in the paper
The question as text
9. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. y y = f (x) A O B x x = k Figure 2 Figure 2 shows a sketch of the curve with equation y = | 2 – 4 ln (x + 1) | x > k where k is a constant. Given that the curve • has an asymptote at x = k • cuts the y-axis at point A • meets the x-axis at point B as shown in Figure 2,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
9(a)[1]
k = −
1
(b)(i)
f(0)
=
2
−
4ln(0 1)
+
=
2
−
0
=
2
(ii)
1
0
=
2
−
4ln(
x
+
1)
ln(
x
+
1)
=
x
=
e p
+
q
2
1
x =
2e
−
1
9(b)[3]
k = −
1
(b)(i)
f(0)
=
2
−
4ln(0 1)
+
=
2
−
0
=
2
(ii)
1
0
=
2
−
4ln(
x
+
1)
ln(
x
+
1)
=
x
=
e p
+
q
2
1
x =
2e
−
1
2
−
4
ln
(
x
+
1)
=
3
l
n
(
x
+
1
)
=
.
.
.
or
−
2
+
4ln(
x
+
1)
=
3
ln(
x
+
1)
=
..
.
2
4ln(
1)
3
..
.
and
2
4ln(
1
)
3
..
.
−
x
+
=
x
=
−
+
x
+
=
x
=
1
5
−−
CVs
e
1, e
−
1
4
4
1
5
−
" 1"
−
<
x
<
e
4
−
1 or
x
>
e
4
−
1
Notes
(a)
B1: For (
k = −. Accept
) 1
x = −or even
1
x >−
1
(b) (i)
B1: For 2 seen as the y coordinate. May be identified as y or f(0) or (0, 2).
(ii)
M1: Sets 0
=
2
−
4ln(
x
+
1)
x =
...
, rearranges and proceeds to
of the correct form e
p
+
q q
be implied by awrt 0.65.
1
− or accept
2
A1:
x =
2e
1
e
− or even
1
4e
− May be seen as part of a coordinate pair.
1
(c) Note this is M1A1A1M1A1 on epen but is being marked as M1M1A1M1A1.
M1: Forms one valid equation with the modulus signs removed and attempts to solve at least as far as
ln(x+1) . Accept with > or < etc instead of = for the M.
dM1: Both equations attempted, allowing for “=” or any inequality used, to achieve values (may be
decimals) for each via ln(x + 1) =… and suitable attempt to undo ln (raises to a base power, allow
an error with the base). For reference the decimals are −0.221 and 2.49 to 3 s.f.
1
5
A1: Both
−−
Must be exact.
e
1, e
1
4
4
−
ddM1: Chooses the outside region for their values. Allow with strict or nonstrict inequalities. It is
dependent on both previous method marks and requires two distinct critical values. The left hand
)
1
5
−
bound −1 may be missing for this mark (i.e. allow for
Allow with
x
<
e
4
−
1 or
x
>
e
4
−
1
decimal values for this mark. ddM0 if the middle section is also included but isw if they choose
the correct outside region but later reject the left hand portion due to confusion with
Watch out for x > … for both inequalities, which is ddM0.
1
5
A1ft:
−
or any equivalent form following through on their negative k
" 1"
e
1 or
e
1
−
<
x
<
4
−
x
>
4
−
from (a). Must be exact (not decimals). Allow “and” instead of “or” – ie both regions are
identified as the solution. But use of ∩ in set notation is A0 as it is incorrect.
9(c)[5]
2
−
4
ln
(
x
+
1)
=
3
l
n
(
x
+
1
)
=
.
.
.
or
−
2
+
4ln(
x
+
1)
=
3
ln(
x
+
1)
=
..
.
2
4ln(
1)
3
..
.
and
2
4ln(
1
)
3
..
.
−
x
+
=
x
=
−
+
x
+
=
x
=
1
5
−−
CVs
e
1, e
−
1
4
4
1
5
−
" 1"
−
<
x
<
e
4
−
1 or
x
>
e
4
−
1
Notes
(a)
B1: For (
k = −. Accept
) 1
x = −or even
1
x >−
1
(b) (i)
B1: For 2 seen as the y coordinate. May be identified as y or f(0) or (0, 2).
(ii)
M1: Sets 0
=
2
−
4ln(
x
+
1)
x =
...
, rearranges and proceeds to
of the correct form e
p
+
q q
be implied by awrt 0.65.
1
− or accept
2
A1:
x =
2e
1
e
− or even
1
4e
− May be seen as part of a coordinate pair.
1
(c) Note this is M1A1A1M1A1 on epen but is being marked as M1M1A1M1A1.
M1: Forms one valid equation with the modulus signs removed and attempts to solve at least as far as
ln(x+1) . Accept with > or < etc instead of = for the M.
dM1: Both equations attempted, allowing for “=” or any inequality used, to achieve values (may be
decimals) for each via ln(x + 1) =… and suitable attempt to undo ln (raises to a base power, allow
an error with the base). For reference the decimals are −0.221 and 2.49 to 3 s.f.
1
5
A1: Both
−−
Must be exact.
e
1, e
1
4
4
−
ddM1: Chooses the outside region for their values. Allow with strict or nonstrict inequalities. It is
dependent on both previous method marks and requires two distinct critical values. The left hand
)
1
5
−
bound −1 may be missing for this mark (i.e. allow for
Allow with
x
<
e
4
−
1 or
x
>
e
4
−
1
decimal values for this mark. ddM0 if the middle section is also included but isw if they choose
the correct outside region but later reject the left hand portion due to confusion with
Watch out for x > … for both inequalities, which is ddM0.
1
5
A1ft:
−
or any equivalent form following through on their negative k
" 1"
e
1 or
e
1
−
<
x
<
4
−
x
>
4
−
from (a). Must be exact (not decimals). Allow “and” instead of “or” – ie both regions are
identified as the solution. But use of ∩ in set notation is A0 as it is incorrect.
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