Mathematics · May / June 2024 · Paper 21 · Question 4
YMA01
As printed in the paper
The question as text
f (x) = (x – 2)(2x 2 + 5x + k) + 21 4. where k is a constant.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[1]
The answer is a drawing: see the picture of this mark scheme row.
4(b)[2]
1
Sets
f
0
equation in k
±
=
→
2
Eg. 1
1
1
−
2
2
×
+ ×
5
+
k
+
21
=
0
2
4
2
3
+
k
=
14
k
=
11
*
)(
)
(
(i)
f( )
=
−
2
2
2
+
5
+
11
+
21
=
2
3
+
2
+
−
1
4(c)[5]
x
x
x
x
x
x
x
)(
)
(
=
2
x
−
1
x
2
+
x
+
1
(ii) Attempts to find the number of roots of their (
)
2
+
+
1
x
x
States ()
x
2
+
x
+
1
has no roots with reason
1
Concludes f( )x =
0
has 1 root (at
x =
)
2
(a)
B1:
States 21
(b) If division is attempted in (b), mark (b) and (c) as a whole.
1
M1:
Sets
equation in k. The “= 0” may be implied. If division is used look
f
±
=
0
→
2
for setting the remainder after dividing by (2x 1) equal to zero to form an equation in
k.
A1*: Solves a correct equation in k showing at least one correct intermediate line (the “= 0”
may be implied but must have been a correct equation preceding the answer).
Condone missing brackets if clearly recovered.
3
(
)
May see
k =
11
embedded so
−
3 11
+
+
21
= −
21
+
21
=
0
Π is acceptable.
2
(c) (i) Only allow work carried out in (b) to count in (c) if carried forward into (c).
)(
)
Attempts to multiply out (
2
2
2
5
11
21
M1:
x
−
x
+
x
+
+
to achieve a cubic. If an incorrect
value for k is used, allow two M’s for equivalent work.
1
dM1: Attempts to divide by or factor out (
)
or
from their attempt at
2
x −
1
x
−
2
)(
)
expanding (
x
−
2
2
x
2
+
5
x
+
11
+
21
.
2
x
−
1 "2
Long division – look for the x2 + extra terms and first subtraction
2
x
Inspection/comparing coefficients – look for correct magnitudes for first and last term
)(
)
(
in the brackets.
2
−
1
2
±
...
±
1
x
x
)(
)
(
A1:
2
x
−
1
x
2
+
x
+
1
from correct work. Must be brought together in one expression.
1
(
)
1
(
)
Accept
2
−
2
+
+
1
but not
−
2
2
+
2
+
2
Condone if set equal to
x
x
x
x
x
x
2
2
zero.
Award A0 if they subsequently incorrectly further factorise the quadratic term.
(c)(ii)
M1:
Must have attempted to take the correct factor out of their cubic. Attempts to find the
number of roots of their (
)
x
2
+
x
+
1
not of
2
x
2
+
5
x
+
11
Allow for an attempt any of
the following:
2
an attempt at−
4
b
ac
an attempt at the formula or completing the squarecorrect (imaginary) roots from a calculator (must be seen)Do not accept “no roots” (oe) seen with no evidence for why. Do not accept “Math
Error” without an explanation of the error.
A1:
Requires all of the following (or equivalents)
)(
)
(
correct factorisationf( )
x
=
2
x
−
1
x
2
+
x
+
1
or correct separate factors
correct reason as to why ()
2
+
+
1
=
0
hasn't any roots. E.g
x
x
2
−
4
= −
1 4
(
= −
3
)
<
0
hence no roots. There must be a (correct) calculation,
b
ac
reference to the sign of the discriminant and deduction/implication of no roots, or
equivalent via other methods
concludes f( )x =
0
has one solution
Ignore erroneous extra statements after a correct reason has been given.
Some examples (following correct factorisation).
Scores M1A1:
1
x
2
+
x + →
1
Δ
= −
1 4
<
0
no real solutions therefore
x =
is only solution.
2
2
−
4
= −<
3
0
so no solutions to
2
+
+ hence f(x) has only one real solution (2x 1).
1
b
ac
x
x
We ignore any reference to (2x1) being the solution.
Scores M1A0:
(ii) one real solution as
2
+
+ has
1
2
−
4
= − so no real solution. No reference to < 0
3
x
x
b
ac
2
−
4
=
1
2
−
4( )
1 (
1)
= −
3
0
so there is only 1 real solution as
2
+
+
1 0
so the only real
b
ac
x
x
1
solution is (2x – 1) which is
x =
Needs <0 to be stated
2
−±
1
1
2
−
4( ) )
1 (1
1
no solution, 2x – 1,
x =
is the only real solution. Needs either complex
2
2
roots stated (accepting
− for 3i) or sign of discriminant considered.
3
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(
)
3
2
2
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