Mathematics · Oct / Nov 2024 · Paper 21 · 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. Using the laws of logarithms, solve log4 (12 – 2x) = 2 + 2log4 (x + 1) (5)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
×
2
4
×
2
C
ax
2
6
6
6 5
ax
×
ax
Alternatively writes e.g. (
)
6
2
+
=
2
6
1
+
=
2
6
1
+
+
ax
ax
2
2
2
2
which can also score M1 for the expansion up to at least the second term with an
acceptable structure for either the 2nd or 3rd term. They do not have to multiply out
the brackets for this mark but the
2 cannot be omitted unless it is later recovered.
6
6 5
××
2
ax
Condone missing brackets e.g.
2
2
A1:
For any 2 correct simplified terms in any order (Allow a2x2 or (ax)2 for the third term
for this mark)
Allow terms to be listed.
Do not allow
64x for 64 or
0
192ax for 192ax
1
A1:
All correct and in any order. Allow terms to be listed. Must be a2x2 not (ax)2 for the
3rd term but allow (ax) for ax.
Apply isw if they divide through by e.g. 2 or 4 or 16 or set their expression = 0
Ignore any extra terms if attempted.
(b)
1
2
B1:
Correct expansion of
3
+
unsimplified or simplified.
x
Note that this may be implied by the omission of the “9” as 9 64
×
=
576
“9” is not required.
1
1
Note that
3
2
1 2 3
is correct.
+
+
(
)
2
x
3
x
2
2
1
6
1
6
Condone missing brackets if recovered e.g.
3
+
=
9
+
+
=
9
+
x
x
x
x
1
2
M1:
Uses their expansion in part (a) and their expansion of
3
+
to extract the constant
x
2
1
β
γ
term. This depends on having obtained
3
+
=
α
+
+
oe with
αβγ non
2
x
x
x
β
γ
zero and an attempt at
α×
"64"
+
×
"192
"
+
"240
2
2
"
oe
ax
a x
2
x
x
An expression of this form is sufficient i.e. with the x’s still included.
Note that the 9 64
×
may not be seen as this cancels with the 576 so this mark may be
implied.
dM1: Sets their constant term = 576 and proceeds to obtain a nonzero value for a.
You do not need to be concerned about the processing for this mark.
This may be implied by e.g.
6 192
×
a
+
240
a
2
=
0
a
=
...
as 9 64
×
=
576
May be implied by their value(s).
Condone x = … here.
Depends on the previous method mark.
48
A1:
Correct value. Allow equivalents e.g. −4.8,
−
10
The questions asks for the value of a so just look for the correct value e.g. “a = …” is
24
not required but
x = −
scores A0.
5
If a = 0 is also given and not rejected, score A0
6
(
)
(
)
2log
+
1
=
log
+
1
2
x
x
4
4
12
2
−
x
(
)
(
)
2
e.g.
log
12
−
2
−
log
+
1
=
log
x
x
4
4
4
(
)
2
x
+
1
12
−
2
x
=
16
(
)
+
1
2
x
8
2
+
17
+
2
=
0
=
...
x
x
x
1
oe e.g. −0.125
x = −
8
(
)
(
)
B1:
For
2log
+
1
=
log
+
1
2
seen or implied.
x
x
4
4
M1:
Correct attempt to combine two log terms.
(
)
2
(
)
2
(
)
2
E.g.
2
+
log
+
1
=
log 16
+
log
+
1
=
log 16
+
1
x
x
x
4
4
4
4
Condone if they make a slip before combining as long as they are combining two log
terms correctly.
e.g.
(
)
(
)
2
(
)
(
)
2
(
log
12
−
2
=
2
+
log
+
1
log
12
−
2
+
log
+
1
=
2
log
12
−
2
x
x
x
x
4
4
4
4
4
or e.g.
(
)
(
)
2
(
)
2
(
)
log
12
−
2
x
=
2
+
log
x
+
1
=
log 2
+
log
x
+
1
=
log 2
x
+
1
4
4
4
4
4
A1:
Obtains this equation in any form not involving logs from correct work so must
follow B1M1.
M1:
Solves a 3TQ by any acceptable method including a calculator to obtain at least one
real value for x. May be implied by their value(s) and may follow incorrect log work
but the B1 must have been scored. You may need to check.
Note that we are allowing the use of a calculator to solve the 3TQ here. This means
that correct work leading to e.g.:
12
−
2
x
16
16
2
34
4
0
=
x
+
x
+
=
(
)
2
+
1
x
Followed by
1
(
8
1
)(
2
)
0
x
+
x
+
=
x
= −
8
Is acceptable for full marks as long as it is preceded by correct log work.
A1:
This value only from correct work and all previous marks scored.
The other value (−2) must clearly be rejected or omitted.
Special Case: Beware incorrect log work which leads to the correct answer:
(
)
(
)
2
2log
x
+
1
=
log
x
+
1
4
4
(
)
log
12
−
2
x
(
)
(
)
2
log
12
−
2
x
−
log
x
+
1
=
4
4
4
(
)
2
log
x
+
1
4
12
−
2
x
=
16
(
)
2
+
1
x
8
x
2
+
17
x
+
2
=
0
x
=
...
1
x = −
8
This scores B1M0A0M1A0
Allow this SC if no combination of logs is shown at all, but the correct quadratic is
produced.
12
2
−
x
(
)
(
)
e.g.
log
12
−
2
=
2
+
2log
+
1
=
16
etc.
x
x
4
4
(
)
2
x
+
1
or e.g.
log
(
12
−
2
)
=
log 16
+
log
(
+
1
)
2
12
−
2
=
16
(
+
1
)
2
etc.
x
x
x
x
4
4
4
Could both score B1M0A0M1A0
Special Case: Beware incorrect work removing logs which leads to the correct
answer:
(
)
(
)
2
2log
x
+
1
=
log
x
+
1
4
4
12
−
2
x
log
(
12
2
)
log
(
1
)
2
log
2
−
x
−
x
+
=
=
4
4
4
(
)
2
+
1
x
12
−
2
x
=
2
4
=
16
(
)
2
x
+
1
8
2
+
17
+
2
=
0
=
...
x
x
x
1
x = −
8
This scores B1M1A0M1A0
Alternative:
log 12
(
2
)
2
2log
(
1
)
−
x
=
+
x
+
4
4
(
)
(
)
(
)
2
4
log
12 2
−
x
=
4
2 2log
+
x
+
1
=
4
2 log
+
x
+
1
4
4
4
2
4
2
4
log
(
x+
1
)
=
×
4
(
)
2
12
−
2
=
16
+
1
etc.
x
x
Score as:
B1:
2log
(
1
)
log
(
1
)
2
x
+
=
x
+
4
4
+ =
M1: Uses correct index law e.g. 4
a b
4
a
×
4
b
Then follow main scheme.
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