Mathematics · May / June 2022 · Paper 31 · Question 3
YMA01
As printed in the paper
The question as text
3. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Given that k is a positive constant,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[2]
9
x
3
(
)
d
x
=
ln 3
x
2
+
k
+
C
3
x
2
+
k
2
3(b)[4]
3
3
ln(75
+
k
)
−
ln(12
+
k
)
=
ln8
2
2
3
75
+
k
=
ln
ln8
oe
2
12
+
k
75
+
k
4
...
=
k
=
12
+
k
(
) 9
k =
(a)
A
ln(3
x
2
+
k
)
M1
Integrates to
with or without the + C. Allow A = 1 for this mark. Condone invisible
brackets.
3 ln 3
3 ln
(
)
(
)
A1
x
2
+
k
+
C
or
B
3
x
2
+
k
oe Must include the constant of integration that is
2
2
3x
2
+
k
different from k. Must have brackets around
3
(
)
Also allow eg
ln
B
3
x
2
+
k
2
(b) *Be aware they can solve equations on the calculator which is not acceptable*
M1
Attempts to substitute in 5 and 2 into their changed expression which must include x and k,
subtracting either way round and sets equal to ln8 . The values embedded in an equation is
sufficient to score this mark. Condone slips if they attempted to evaluate when substituting
Ignore the + C .
dM1
Attempts to apply the subtraction (or addition) law of logarithms once to an equation of the
(
)
(
)
form
±
A
ln
B
+
k
±
A
ln
C
+
k
=
ln
D
or equivalent (allow A = 1) .
B
+
k
(
)
(
)
=
A
ln
B
+
k
−
A
ln
C
+
k
=
ln
D
Typically look for
A
ln
ln
D
or equivalent
C
+
k
Do not be concerned with arithmetical errors and condone slips dealing with their A
Condone invisible brackets.
A
B
+
k
=
1
B
+
k
=
ddM1 Solves their equation of the form
ln
ln
D
A
or ln
ln
D
C
+
k
C
+
k
A ≠, and the A is correctly dealt with by
1
where
removing lns correctly
rearranging to form a linear equation in k
proceeding to find a value for k.
2
"75"
+
k
=
" "
eg
"ln
ln8
"75"
+
k
=
"48" "4 "
+
k
k
=
...
3
"12"
+
k
It is dependent on both of the previous method marks.
Condone arithmetical slips in their rearrangement and condone invisible brackets as long as
intention is clear. Also condone this mark to be scored if k is negative.
A1
9 cao (with intermediate working seen)
They cannot proceed from 3
3
to (
) 9
ln(75
+
k
)
−
ln(12
+
k
)
=
ln8
k =
without some
2
2
intermediate working seen (eg an equation with the lns removed)
(
)
ln 3
x
2
+
k
(
+
C
)
in (a) will score a maximum of M1dM1ddM0A0 in (b)
Note an answer of
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.