Mathematics · Oct / Nov 2024 · Paper 11 · Question 3

YMA01

As printed in the paper

The question as text

3. In this question you must show all stages of your working. x + 5 2 f(x) = ( ) x > 0 x

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

3(a)[4]
=
= −
3
+
for M1 A1
54
3
3
Notes: The B mark can be implied by working as long as it is seen before the final answer.
8
3
+
4
27
72
+
12
3
So, for example,
×
=
=
..
.
is sufficient to score the B mark.
2
7
27
2
7
An acceptable equivalent of rationalising the denominator of an expression for the first M is
rationalising the coefficient of a in a rearranged expression, e.g.
The second M mark requires an attempt to make a the subject and make progress towards the
form
, which means the denominator must have been reduced to a single term with
12
−
4
3
two terms in the numerator. So
a
is not sufficient progress. In the example above
=
9
+
3
3
the M is not awarded until the final line where the denominator is 54.
However, once the first B and M are scored (having seen evidence before the final answer)
reaching the correct answer will then be sufficient for the M (and A if all correct).
12
−
4
3
to 8
−
4 3
But from
with no further work shown is 2nd M0 (calculator use with
3
9
+
3
3
4 3
no further progress). However, if a correct attempt to rearrange this to
−
+
3
terms reversed) follows (or similar if there were errors) then the 2nd M becomes eligible as
there is further work to support the method.
3
1
1
5
3
1
2
20
−
2
2
2
2
2
2
f ( )
x
d
x
=
x
+
10
x
+
25
x
d
x
=
x
+
x
+
50
x
+
c
3. (a)
5
3
1
1
3
3
25
−
−
(b) (i)
f ( )
′
2
5
2
2
x
=
x
+
x
−
x
2
2
1
1
3
3
3
−
25
−
f ( )
′
x
=
0
x
2
+
5
x
2
−
x
2
=
0
AND
2
×
x
2
2
3
25
2
1
2
x
+
5
x
−
=
0
3
x
+
10
x
−
25
=
0
*
2
2
(
)
5
(ii)
x =
only
3
(a)
1
M1: For attempting to divide by
x and increasing a correct power by 1 (ie must have
2
expanded). Must be seen in part (a)
A1: For one correct term (allow un simplified)
A1: For two correct terms simplified.
5
3
1
2
20
2
2
50
2
A1:
x
+
x
+
x
+ or exact equivalent. Must include the constant of integration.
c
5
3
Condone if they write “y =” and ignore any spurious integral signs or extra dx
(b)(i)
1
M1: For attempting to divide by
x and decreasing a correct power by one. Must be seen or
2
used in part (b). For attempts at the quotient or product rule look for correct
(
)
2
A
(
x
+
5)
x
−
B x
+
5
x
denominator and one correct numerator term in
(
)
2
x
product rule)
1
1
3
3
25
−
−
A1: For
′
2
2
2
(oe e.g from quotient rule) which may be left un
f ( )
x
=
x
+
5
x
−
x
2
2
simplified.
1
1
3
3
−
−
2
2
2
2
dM1: Sets
α
x
+β
x
+
x
=
0
and attempts to multiply by
x seen in at least two terms or
explicitly shown in a step before the final answer.
1
This may be derived in stages, first multiplying by
x and then by x. This is permissible for
2
the M (and A) as evidence of the correct method has been shown by the stages. The dM
can be scored if the index work is correct in at least two indices, and A1 if all is correct.
1
1
3
3
25
3
25
3
25
−
−
So
x
+
5
x
−
x
=
0
x
+
5
−
x
−
1
=
0
x
2
+
5
x
−
=
0
is sufficient for the
2
2
2
2
2
2
2
2
2
dM1 as long as the middle step is seen with two correct powers (and the A1 if they then
multiply through be 2).
Note also that dividing by
is an equivalent to multiplying by
.
A1*: Reaches
2
showing all steps with no errors. Note that they must work
3
x
+
10
x
−
25
=
0
correctly with the equation, if they state
f ( )
′
x
=
3
x
2
+
10
x
−
25
before setting equal to 0
then allow the M (if suitable method shown) but it is A0.
(b)(ii)
B1: (
)
5
x =
only. The other root need not be seen, but if shown it must be subsequently
3
rejected in some way. Note that
(1.6 recurring) is acceptable as the answer for B1,
but 1.67 is B0.
States or implies that f( )
x
=
kx x
(
−
4)
3(b)[5]
+
1
(
)
12
−
4 3
9
−
3 3
12
−
4 3
=
9
+
3 3
a
a
=
×
for M1
9
+
3 3
9
−
3 3
144
−
72 3
4
8
=
= −
3
+
for M1 A1
54
3
3
Notes: The B mark can be implied by working as long as it is seen before the final answer.
8
3
+
4
27
72
+
12
3
So, for example,
×
=
=
..
.
is sufficient to score the B mark.
2
7
27
2
7
An acceptable equivalent of rationalising the denominator of an expression for the first M is
rationalising the coefficient of a in a rearranged expression, e.g.
The second M mark requires an attempt to make a the subject and make progress towards the
form
, which means the denominator must have been reduced to a single term with
12
−
4
3
two terms in the numerator. So
a
is not sufficient progress. In the example above
=
9
+
3
3
the M is not awarded until the final line where the denominator is 54.
However, once the first B and M are scored (having seen evidence before the final answer)
reaching the correct answer will then be sufficient for the M (and A if all correct).
12
−
4
3
to 8
−
4 3
But from
with no further work shown is 2nd M0 (calculator use with
3
9
+
3
3
4 3
no further progress). However, if a correct attempt to rearrange this to
−
+
3
terms reversed) follows (or similar if there were errors) then the 2nd M becomes eligible as
there is further work to support the method.
3
1
1
5
3
1
2
20
−
2
2
2
2
2
2
f ( )
x
d
x
=
x
+
10
x
+
25
x
d
x
=
x
+
x
+
50
x
+
c
3. (a)
5
3
1
1
3
3
25
−
−
(b) (i)
f ( )
′
2
5
2
2
x
=
x
+
x
−
x
2
2
1
1
3
3
3
−
25
−
f ( )
′
x
=
0
x
2
+
5
x
2
−
x
2
=
0
AND
2
×
x
2
2
3
25
2
1
2
x
+
5
x
−
=
0
3
x
+
10
x
−
25
=
0
*
2
2
(
)
5
(ii)
x =
only
3
(a)
1
M1: For attempting to divide by
x and increasing a correct power by 1 (ie must have
2
expanded). Must be seen in part (a)
A1: For one correct term (allow un simplified)
A1: For two correct terms simplified.
5
3
1
2
20
2
2
50
2
A1:
x
+
x
+
x
+ or exact equivalent. Must include the constant of integration.
c
5
3
Condone if they write “y =” and ignore any spurious integral signs or extra dx
(b)(i)
1
M1: For attempting to divide by
x and decreasing a correct power by one. Must be seen or
2
used in part (b). For attempts at the quotient or product rule look for correct
(
)
2
A
(
x
+
5)
x
−
B x
+
5
x
denominator and one correct numerator term in
(
)
2
x
product rule)
1
1
3
3
25
−
−
A1: For
′
2
2
2
(oe e.g from quotient rule) which may be left un
f ( )
x
=
x
+
5
x
−
x
2
2
simplified.
1
1
3
3
−
−
2
2
2
2
dM1: Sets
α
x
+β
x
+
x
=
0
and attempts to multiply by
x seen in at least two terms or
explicitly shown in a step before the final answer.
1
This may be derived in stages, first multiplying by
x and then by x. This is permissible for
2
the M (and A) as evidence of the correct method has been shown by the stages. The dM
can be scored if the index work is correct in at least two indices, and A1 if all is correct.
1
1
3
3
25
3
25
3
25
−
−
So
x
+
5
x
−
x
=
0
x
+
5
−
x
−
1
=
0
x
2
+
5
x
−
=
0
is sufficient for the
2
2
2
2
2
2
2
2
2
dM1 as long as the middle step is seen with two correct powers (and the A1 if they then
multiply through be 2).
Note also that dividing by
is an equivalent to multiplying by
.
A1*: Reaches
2
showing all steps with no errors. Note that they must work
3
x
+
10
x
−
25
=
0
correctly with the equation, if they state
f ( )
′
x
=
3
x
2
+
10
x
−
25
before setting equal to 0
then allow the M (if suitable method shown) but it is A0.
(b)(ii)
B1: (
)
5
x =
only. The other root need not be seen, but if shown it must be subsequently
3
rejected in some way. Note that
(1.6 recurring) is acceptable as the answer for B1,
but 1.67 is B0.
States or implies that f( )
x
=
kx x
(
−
4)

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