Mathematics · Oct / Nov 2024 · Paper 21 · Question 3

YMA01

As printed in the paper

The question as text

3. f (x) = 2x 3 – x 2 + Ax + B where A and B are integers. Given that when f (x) is divided by (x + 3) the remainder is 55

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

Show the answer and mark scheme

3(a)[2]
(
)
3
(
)
2
(
)
2
−
3
−−
3
+
A
−
3
+
B
=
55
or e.g.
−
54 9 3
−−
+
=
55
A B
−
54 9 3
−−
A B
+
=
55
3
118*
A B
−
= −
(a)
M1:
Attempts f( 3)
The −3 embedded in the expression set = 55 is sufficient.
−
=
55.
(
)
Condone missing brackets e.g.
2 3
−
3
−−
3
2
+
−
3
+
=
55
A
B
Note that (
2
3
)
3
( )
3
2
(
3
)
55
is acceptable.
−
−
+
A
−
+
B
=
May be implied by further work. If −3 embedded in the expression is not seen,
condone slips in their evaluation provided there is still evidence that the intention was
to substitute in −3 and set = 55 but f(3)
is M0
=
55
A1*: Completes proof with at least one intermediate simplified and correct line that is not
the final line such as 54 9 3
−
−−
+
=
55
or 118
= −
3A B
+
. Any incorrect lines written
A B
after the start of their ‘proof’ should be scored A0*
Must be the given equation and not e.g.
B
=
3
A
+
118
unless the correct equation is
seen previously but condone
−+
3
= −
118
B
A
2
(
3
)
3
(
3
)
2
(
3
)
55
3
118
scores M1A0
−
−−
+
A
−
+
B
=
A
−
B
= −
Tabular or division methods may be seen. 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 proof.
If you cannot see what they are doing and find it hard to award, then use review.
e.g. long division:
2
2
−
7
+
+
21
x
x
A
3 2
3
2
x
+
x
−
x
+
Ax
+
B
2
3
6
2
x
+
x
−
7
x
2
+
Ax
+
B
7
2
21
−
x
−
x
(
21
)
A
+
x
+
B
(
)
(
)
+
21
+
3
+
21
A
x
A
−
3
−
63
=
55
B
A
Score M1 for obtaining a quotient
...
2
±
...
±
...
+
...
and a remainder ...
x
x
A
set = 55 where “…” are nonzero constants. Then A1* for a fully correct proof.
e.g. tabular method:
2x2
−7x
21 + A
x
2x3
−7x2
(21 + A)x
3
6x2
−21x
63 + 3A
So
−
(
63 3
+
)
=
55
etc.
B
A
Score M1 for obtaining a constant ...
...
where … are nonzero constants and then
A+
sets B – their constant = 55. Then A1* for a fully correct proof.
3(b)[3]
3
2
5
5
5
2
−
+
A
+
B
=
0
2
2
2
3
−
= −
118
, 5
+
2
= −
50
A B
A
B
A
=
...,
or
B
=
...
A = – 26, B = 40
(b)
Do not allow misreads of (2x – 5) in part (b).
5
The 5
=
M1:
Attempts
embedded in the expression set = 0 is sufficient.
f
0.
2
2
5
f
−
=
0
scores M0.
2
May be implied by further work. Again, tabular or division methods may be seen.
e.g. long division:
2
+
2
+
+
5
x
x
1
A
2
2
x
−
5 2
x
3
−
x
2
+
Ax
+
B
2
x
3
−
5
x
2
4
x
2
+
Ax
+
B
4
2
−
10
x
x
(
10
)
A
+
x
+
B
(
)
(
)
+
10
−
5
+
5
A
x
1
A
2
B
+
5
A
+
25
=
0
2
Score M1 for obtaining a quotient
...
2
±
...
±
...
+
...
and a remainder ...
x
x
A
set = 0 where “…” are nonzero constants.
e.g. tabular method:
A
x2
2x
5
+
2
(
10
)
2x
2x3
4x2
+
A x
5
A
−
25
−
−5
−5x2
−
10x
2
5
A
So
−−
25
−
=
0.
Score M1 for obtaining a constant ...
A+
...
where “…” are
B
2
nonzero constants, and then sets B – their constant = 0
M1:
Solves 3
−
= −
118
simultaneously with their equation in A and B from having
A B
5
attempted
or long division/tabular method using the correct factor of
f
±
=
0
2
(2x – 5) and obtains a value for A or B.
May be via a calculator and may be implied by their values. You do not need to check
but the above conditions must be satisfied.
5
For reference, use of
f
−
=
0
gives A = −161, B = −365
2
A1:
Both correct values
2
x
2
+
13
x
+
65
(c)
Do not allow misreads of (x – 7) in part (c).
M1: Uses any appropriate method e.g. long division/inspection/table with numeric non
zero A and B to obtain 2x2 + px + q where p and q are nonzero.
There is no requirement to find the remainder but assuming (x – 7) is a factor is M0
(see example below)
A1:
Must follow correct values of A and B.
Correct expression
2
2
+
13
+
65
which is clearly identified e.g. by
x
x
circling/underlining or e.g. Quotient =
2
2
2
x
+
13
x
+
65,
Q
=
2
x
+
13
x
+
65
Any remainder, correct or incorrect, can be ignored. (NB correct remainder is 495)
Long division for reference:
Tabular method for reference:
2
x
2
+
13
x
+
65
x
−
7 2
x
3
−
x
2
−
26
x
+
40
2x2
13x
2x3
13x2
x
2
x
3
−
14
x
2
−7
−14x2
−
91x
13
2
−
26
+
40
x
x
13
x
2
−
91
x
65
x
+
40
65
x
−
455
495
But note that just the above score M1A0 with no indication that the quotient is
2
2
+
13
+
65
x
x
Some candidates assume that (x – 7) is a factor of f(x) and try to find the quadratic
factor:
This scores M0 in part (c)
But work that doesn’t assume (x – 7) is a factor is fine e.g.:
Scores M1A1
3(c)[2]
(
)
3
(
)
2
(
)
2
−
3
−−
3
+
A
−
3
+
B
=
55
or e.g.
−
54 9 3
−−
+
=
55
A B
−
54 9 3
−−
A B
+
=
55
3
118*
A B
−
= −
(a)
M1:
Attempts f( 3)
The −3 embedded in the expression set = 55 is sufficient.
−
=
55.
(
)
Condone missing brackets e.g.
2 3
−
3
−−
3
2
+
−
3
+
=
55
A
B
Note that (
2
3
)
3
( )
3
2
(
3
)
55
is acceptable.
−
−
+
A
−
+
B
=
May be implied by further work. If −3 embedded in the expression is not seen,
condone slips in their evaluation provided there is still evidence that the intention was
to substitute in −3 and set = 55 but f(3)
is M0
=
55
A1*: Completes proof with at least one intermediate simplified and correct line that is not
the final line such as 54 9 3
−
−−
+
=
55
or 118
= −
3A B
+
. Any incorrect lines written
A B
after the start of their ‘proof’ should be scored A0*
Must be the given equation and not e.g.
B
=
3
A
+
118
unless the correct equation is
seen previously but condone
−+
3
= −
118
B
A
2
(
3
)
3
(
3
)
2
(
3
)
55
3
118
scores M1A0
−
−−
+
A
−
+
B
=
A
−
B
= −
Tabular or division methods may be seen. 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 proof.
If you cannot see what they are doing and find it hard to award, then use review.
e.g. long division:
2
2
−
7
+
+
21
x
x
A
3 2
3
2
x
+
x
−
x
+
Ax
+
B
2
3
6
2
x
+
x
−
7
x
2
+
Ax
+
B
7
2
21
−
x
−
x
(
21
)
A
+
x
+
B
(
)
(
)
+
21
+
3
+
21
A
x
A
−
3
−
63
=
55
B
A
Score M1 for obtaining a quotient
...
2
±
...
±
...
+
...
and a remainder ...
x
x
A
set = 55 where “…” are nonzero constants. Then A1* for a fully correct proof.
e.g. tabular method:
2x2
−7x
21 + A
x
2x3
−7x2
(21 + A)x
3
6x2
−21x
63 + 3A
So
−
(
63 3
+
)
=
55
etc.
B
A
Score M1 for obtaining a constant ...
...
where … are nonzero constants and then
A+
sets B – their constant = 55. Then A1* for a fully correct proof.
3
2
5
5
5
2
−
+
A
+
B
=
0
2
2
2
3
−
= −
118
, 5
+
2
= −
50
A B
A
B
A
=
...,
or
B
=
...
A = – 26, B = 40
(b)
Do not allow misreads of (2x – 5) in part (b).
5
The 5
=
M1:
Attempts
embedded in the expression set = 0 is sufficient.
f
0.
2
2
5
f
−
=
0
scores M0.
2
May be implied by further work. Again, tabular or division methods may be seen.
e.g. long division:
2
+
2
+
+
5
x
x
1
A
2
2
x
−
5 2
x
3
−
x
2
+
Ax
+
B
2
x
3
−
5
x
2
4
x
2
+
Ax
+
B
4
2
−
10
x
x
(
10
)
A
+
x
+
B
(
)
(
)
+
10
−
5
+
5
A
x
1
A
2
B
+
5
A
+
25
=
0
2
Score M1 for obtaining a quotient
...
2
±
...
±
...
+
...
and a remainder ...
x
x
A
set = 0 where “…” are nonzero constants.
e.g. tabular method:
A
x2
2x
5
+
2
(
10
)
2x
2x3
4x2
+
A x
5
A
−
25
−
−5
−5x2
−
10x
2
5
A
So
−−
25
−
=
0.
Score M1 for obtaining a constant ...
A+
...
where “…” are
B
2
nonzero constants, and then sets B – their constant = 0
M1:
Solves 3
−
= −
118
simultaneously with their equation in A and B from having
A B
5
attempted
or long division/tabular method using the correct factor of
f
±
=
0
2
(2x – 5) and obtains a value for A or B.
May be via a calculator and may be implied by their values. You do not need to check
but the above conditions must be satisfied.
5
For reference, use of
f
−
=
0
gives A = −161, B = −365
2
A1:
Both correct values
2
x
2
+
13
x
+
65
(c)
Do not allow misreads of (x – 7) in part (c).
M1: Uses any appropriate method e.g. long division/inspection/table with numeric non
zero A and B to obtain 2x2 + px + q where p and q are nonzero.
There is no requirement to find the remainder but assuming (x – 7) is a factor is M0
(see example below)
A1:
Must follow correct values of A and B.
Correct expression
2
2
+
13
+
65
which is clearly identified e.g. by
x
x
circling/underlining or e.g. Quotient =
2
2
2
x
+
13
x
+
65,
Q
=
2
x
+
13
x
+
65
Any remainder, correct or incorrect, can be ignored. (NB correct remainder is 495)
Long division for reference:
Tabular method for reference:
2
x
2
+
13
x
+
65
x
−
7 2
x
3
−
x
2
−
26
x
+
40
2x2
13x
2x3
13x2
x
2
x
3
−
14
x
2
−7
−14x2
−
91x
13
2
−
26
+
40
x
x
13
x
2
−
91
x
65
x
+
40
65
x
−
455
495
But note that just the above score M1A0 with no indication that the quotient is
2
2
+
13
+
65
x
x
Some candidates assume that (x – 7) is a factor of f(x) and try to find the quadratic
factor:
This scores M0 in part (c)
But work that doesn’t assume (x – 7) is a factor is fine e.g.:
Scores M1A1

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