Mathematics · Oct / Nov 2020 · Paper 21 · Question 3
YMA01
As printed in the paper
The question as text
3. f(x) = ax3 – x2 + bx + 4 where a and b are constants. When f(x) is divided by (x + 4), the remainder is –108
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[2]
(
)
(
)
(
)
a
−
4
3
−−
4
2
+
b
−
4
+
4
= −
108
Attempts to set f( 4)
108
to obtain an equation in a and b. Score when you see “– 4”
−
= −
embedded in the equation or 2 correct terms (excluding the “+ 4”) on lhs.
May be implied by e.g. 64
−
a
−
16
−
4
b
+
4
= −
108
Condone minor slips on the lhs e.g. one sign error between terms but must use − 108
As an alternative for the first mark we will condone an attempt at long division.
This requires a complete method to divide (ax3 – x2 + bx + 4) by (x + 4) to obtain a remainder
in terms of a and b which is then equated to −108
For reference, the quotient is ax2 – (1+4a)x + 16a + b + 4 and the remainder is –4b – 64a – 12
Correct equation obtained with no errors
and at least one line of intermediate
−
64
a
−
16
−
4
b
+
4
= −
108
working if starting with e.g.
⇒
16
a
+
b
=
24*
(
)
3
(
)
2
(
)
a
−
4
−−
4
+
b
−
4
+
4
= −
108
3(b)[3]
3
2
1
1
1
a
−
+
b
+
4
=
0
2
2
2
1
=
Attempts to set
f
0
to obtain an equation in a and b. Condone slips. Score when you
2
see “ 1
” embedded in the equation or 2 correct terms (excluding the “+ 4”) on lhs.
2
a
1
b
May be implied by e.g.
−
+
+
4
=
0
8
4
2
The “= 0” may be implied when they attempt to solve simultaneously below
An alternative for the first mark is to attempt long division.
This requires a complete method to divide (ax3 – x2 + bx + 4) by (2x – 1) to obtain a
remainder in a and b which is then equated to 0
a
a
1
b
1
a
and the remainder is 15
b
a
For reference, the quotient is
x
2
+
−
x
+
−
+
+
+
2
4
2
2
4
8
4
2
8
Attempts to solve 16
a
+
b
=
24
16
a
+
b
=
24
,
a
+
4
b
= −
30
simultaneously with their equation in a and
⇒
a
=
...,
b
=
...
b. This may be implied if values of a and b
are obtained (e.g. calculator)
a = 2, b = −8
Correct values
3(c)[1]
Correct derivative (follow through their a
( )
f
x
=
2
x
3
−
x
2
−
8
x
+
4
and b). Allow unsimplified and apply isw if
( )
⇒
f
′
x
=
6
x
2
−
2
x
−
8
necessary. Allow with the letters “a” and
“b” and a “made up” “a” and “b”.
3(d)[4]
Sets their f ′(x) = 0 (may be implied) and
6
x
2
−
2
x
−
8
=
0
solves a 3 term quadratic. Apply general
(
)(
)
⇒
3
x
−
4
x
+
1
=
0
guidance if necessary. You may need to
⇒
x
=
...
check if a calculator has been used.
Uses at least one of their x values to find a
value for y using their f (x) where x is from
4 ,
x
=
−⇒
1
y
=
...
an attempt to solve f ′(x) = 0. You may need
3
to check their y values if working is not
shown.
4
100
(
)
,
−
or
−
1, 9
3
27
4
100
x
=
,
y
= −
x
1,
y
9
Or e.g.
and
= −
=
3
27
One correct point. The fractional coordinates must be exact but allow 1.3 with a dot over the
3 and 3.703 with dots over the 7 and 3. Note that it is not necessary for the points to be
written as coordinates as long as the pairing is clear.
Depends on having scored both previous M marks.
4
100
(
)
,
−
and
−
1, 9
3
27
4
100
Or e.g.
x
=
,
y
= −
and
x
= −
1,
y
=
9
3
27
Both correct points. The fractional coordinates must be exact but allow 1.3 with a dot over the
3 and 3.703 with dots over the 7 and 3. Note that it is not necessary for the points to be
written as coordinates as long as the pairing is clear.
Depends on having scored both previous M marks.
Fully correct answers with no working scores 4/4 following a correct part (c) i.e.
( )
⇒
f
′
x
=
6
x
2
−
2
x
−
8
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