Mathematics · May / June 2022 · Paper 21 · Question 7
YMA01
As printed in the paper
The question as text
7. f(x) = Ax3 + 6x2 − 4x + B where A and B are constants. Given that • (x + 2) is a factor of f(x) 5 ∫ • x x = f ( ) d 176 3 find the value of A and the value of B. (7)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
(
x +
2)
a factor
f( 2)
−
=
0
−
8
A
+
24
+ +
8
B
=
0
A
f
( )d
x
x
=
x
4
+
2
x
3
−
2
x
2
+
B
x
4
5
5
A
f( )d
x
x
176
x
4
2x
3
2
x
2
Bx
176
=
+
−
+
=
4
3
3
A
A
5
4
+
2(5
3
)
−
2(5 )
2
+
5
B
−
3
4
+
2(3
3
)
−
2(3 )
2
+
3
B
=
176
4
4
8
A
−
B
=
32
✓
A
=
.
..,
B
=
..
.
136
A
+
2
B
=
1
2
1
,
B = −
28
A =
2
Notes:
M1: Uses the factor theorem or long division (see below) to find one linear equation relating A and
B. The “= 0” must be seen or implied by later work.
A1: Correct equation. Award once a correct equation is seen unsimplified or simplified.
So allow e.g.
A
(
−
2
)
3
+
6
(
−
2
)
2
−
4
(
−
2
)
+
B
=
0
but not e.g.
A
×−
2
3
+ ×−
6
2
2
−×−+
4
2
B
unless the indices are processed correctly subsequently to e.g. 8
−
A
+
24
+
8
+
B
=
0
The “= 0” must be seen or implied by later work.
M1: Attempts to integrate the given expression, look for at least one power increased by 1.
This can also be evidenced by B → Bx
6
x
2 1
+
A1: Correct integration. Allow unsimplified e.g.
for
2x
3
2 1
+
A
f
( )d
x
x
=
x
4
+
2
x
3
−
2
x
2
+
Bx
Condone poor notation e.g.
and ignore any “+ c”
4
dM1: Applies limits 5 and 3 either way round and subtracts and equates to 176 to obtain an equation
in A and B. Depends on previous M.
Condone poor use of brackets when subtracting e.g. allow
A
A
5
4
+
2(5
3
)
−
2(5 )
2
+
5
B
−
3
4
+
2(3
3
)
−
2(3 )
2
+
3
B
=
176
4
4
dM1: Solves their two equations in A and B to find values for both constants.
Depends on all previous M marks but allow this mark if the only error was to attempt
f (2) = 0 at the beginning.
A1: Correct values.
Note that some candidates eliminate A or B by rearranging the first equation and substituting into the integral.
This essentially follows the main scheme:
f( 2)
−
=
0
−
8
A
+
24
+ +
8
B
=
0
B
=
8
A
−
32
A
f( )d
x
x
=
x
4
+
2
x
3
−
2
x
2
+
8
Ax
−
32
x
4
A
A
5
4
2(5
3
)
2(5 )
2
40
A
160
3
4
2(
3
3
)
2(3 )
2
24
A
9
6
176
+
−
+
−
−
+
−
+
−
=
4
4
A
=
...,
B
=
...
1
,
B = −
28
A =
2
Long Division for reference:
Score the first M1 for a complete method to obtain a remainder in terms of A and B that is set = 0
and A1 for a correct equation e.g. B – 8A + 32 = 0
d
y
C
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