Mathematics · May / June 2023 · Paper 21 · Question 2
YMA01
As printed in the paper
The question as text
2. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. 3 – 8x 2 + 5x + a f(x) = 4x where a is a constant. Given that (2x – 3) is a factor of f(x),
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
2(a)[2]
3
2
3
3
3
3
f
=
4
−
8
+
5
+
=
0
=
...
a
a
2
2
2
2
27
15
−
18
+
+
=
0
= − *
3
a
a
2
2
2(b)[4]
Example where 2
x − is a linear factor:
3
2
2
−
+
1
x
x
2
3 4
3
8
2
5
3
x
−
x
−
x
+
x
−
4
3
−
6
2
x
x
−
2
x
2
+
5
x
−
2
2
+
3
x
x
+
2
−
3
x
+
2
−
3
x
0
3
no real roots so
is the only one real root *
( 1)
−
2
−
4
×
2 1
×= −<
7
0
x =
2
(a)
3
M1
Substitutes
x =
into the expression, sets equal to 0 (may be implied for this mark) and
2
rearranges to find a. Condone invisible brackets for this mark only and condone arithmetical
slips in their working.
3
Substituting
x = −
is M0 and attempts via long division score M0.
2
a = − achieved with no errors seen including the omission of brackets.
3
A1*
3
3
There must be sight of
x =
which may be embedded in the expression or
f
followed
2
2
by at least one intermediate stage of working/manipulation (it may just be the left hand side
of the equation), before proceeding to the given answer. The "
=
0"
must be seen somewhere
in their working before the given answer.
e.g.
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