Mathematics · Oct / Nov 2023 · Paper 21 · Question 4
YMA01
As printed in the paper
The question as text
4. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. 3 + ax 2 – 29x + b f (x) = 4x where a and b are constants. Given that (2x + 1) is a factor of f (x),
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Show the answer and mark scheme
4(a)[2]
Sets
×−
+ ×−
a
−
×−
+
b
=
2
2
2
2
1
4 a
+
b
+
14
=
0
a
+
4
b
= −
56
*
Sets ( )
3
2
4(b)[2]
f 2
= −
25
→
4 2
×
+ ×
a
2
−
29 2
×+
b
= −
25
4
a
+
b
=
1
(i) Solves
+
4
= −
56
with their 4
1
...,
...
4(c)[5]
a
b
a
+
b
=
a
=
b
=
a
=
4,
b
= −
15
)(
)
(
(ii)
4
3
+
4
2
−
29
−
15
=
2
+
1
2
2
+
−
15
x
x
x
x
x
x
(
)(
)(
)
=
2
x
+
1 2
x
−
5
x
+
3
cso
(a)
1
M1: Attempts to set
f
−
=
0
→
equation in a and b Condone slips (e.g. on signs) but
f
2
Score when you see embedded values within the equation or two correct terms on the lhs of the equation.
A1*: Completes proof with at least one intermediate simplified and correct line such as
1
1
1
29
+ +
14
=
0
. Simplifying each term is sufficient
−
+
+
+
=
0
. Any incorrect lines written
4 a
b
a
b
2
4
2
after the start of their ‘proof’ should be scored A0*
.............................................................................................................................................................................
Solutions where the = 0 is implied but not stated should be awarded M1 A0*
1
1
3
1
2
1
1
. So 1
f
−
=
4
29
14
E.g.
×−
+ ×−
a
−
×−
+
b
=
a
+
b
+
4 a
+
b
= −
14
2
2
2
2
4
................................................................................................................................................................................
(b)
M1: Attempts to set ( )
→equation in a and b Condone slips, e.g sign slips or setting ( )
f 2
= −
25
Setting (
)
f
−
2
= −
25
is not a slip and should be marked M0.
1
1
0
A1: 4
a
+
b
= or other simplified equivalent such as
1
a
+
b
−
=
. The constant terms must be collected.
4
4
Note that this is not a given answer so allow recovery.
(c)(i)
+
4
= −
56
M1: Attempts to solve
a
b
(allowing for slips) with their 4
a
+
b
=
1
a
=
...,
b
=
...
Award as long as values appear for both a and b, so allow these just written down (say from a calculator)
There is no need their calculations
A1: Both values correct:
a
=
4,
b
= −
15
. Both just written down without working scores M1 A1 provided (b) is
(c)(ii)
M1: Attempt to divide or factorise out (
2
1
)
3
2
x +
from their
4
x
+
4
x
−
29
x
−
15
or multiple thereof
For factorisation look for correct first and last terms E.g.
3
2
(
4
x
+
'' ''
a x
−
29
x
+
'' ''
b
=
2
x
+
1
2
For division look for a quadratic quotient of the form
2x
+
px
+
q
with some ''correct'' work to find the
term in x for example. Condone sign slip when attempting to find p
2
A1: Correct quadratic factor
2
x
+
x
−
15
A1: cso (
)(
)(
)
2
+
1 2
−
5
+
3
. Note that M1, A1 must have been awarded and the full product must be seen. It
x
x
x
is not acceptable to see just the factors
..............................................................................................................................................................................
The question states ''Hence'' so there is an expectation that they use the factor of (
)
2
x +
1
which is given.
3
2
(
)(
)(
)
Solutions that just state
4
x
+
4
x
−
29
x
−
15
=
2
x
+
1
2
x
−
5
x
+
3
score M0 and therefore A0 A0
If they wish to start with another factor they must prove it is a factor first.
E.g. (
)
(
)
is a factor followed by division by (
f
−
3
=
4
×−
27
+ ×−
4 9
29
×−−
3 15
=
0
+
3
x +
x
............................................................................................................................................................................
Tabular or methods of division may be seen in (a) and (b). The question does not demand use of the factor or
remainder theorem so these methods should be rewarded.
The M1 would be scored for a full attempt and A1 for a correct answer or proof.
If you cannot see what they are doing and find it hard to award, then put in review please
Example of a tabular version for (a):
(
)
2
x +
1
is a factor of
f( )
=
4
3
+
2
−
29
+
x
x
ax
x
b
(
)
2
−
29
−
2b x
b
2x
2
3
(
)
2bx
2x
4x
2
−
29
−
2b x
(
)
2
−
29
−
2b x
1
2x
b
Then states that (
)
2
−
29
−
2
+
2
=
b
a
Example of a long division method for (b).
is divided by (
)
When
f( )
=
4
3
+
2
−
29
+
x −
2
the remainder is 25
−
x
x
ax
x
b
2
4
x
+
(
a
+
8)
x
+
(2
a
−
13)
3
2
x
−
2
4
x
+
ax
−
29
x
+
b
3
2
4
−
8
x
x
2
(
8)
29
a
+
x
−
x
2
(
)
(
a
+
8)
x
−
2
a
+
16
x
(2
a
−
13)
x
+
b
(
)
(2
a
−
13)
x
+ −
4
a
+
26
b
+
4
a
−
26
Remainder is 25
−
+
4
−
26
= −
25
+
4
=
1
so
b
a
b
a
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