Mathematics · Oct / Nov 2022 · Paper 11 · Question 4
YMA01
As printed in the paper
The question as text
4. In this question you must show detailed reasoning. Solutions relying on calculator technology are not acceptable. f(x) = x2(2x + 1) – 15x
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Show the answer and mark scheme
4(a)[4]
2
(2
+
1) 15
−
2
3
+
2
−
15
=
(2
2
+ −
15)
x
x
x
x
x
x
x
x
x
(2
−
5)(
+
3)
=
0
=
...
x
x
x
x
5
Two of
x =
0,
,
−
3
2
5
x =
0,
,
−
3
2
4(b)[2]
3
2
5
5
2
=
=
y
3
y
2
2
5 10
4
The question says “In this question you must show detailed reasoning. Solutions relying on
calculator technology are not acceptable”. Correct answers do not imply full marks.
(a)
M1
Multiplies out the bracket to achieve a cubic, takes out a linear factor or cancels the x
leading to a quadratic factor (usually
2
2
+
−
15
). Award for
... (...
2
±
...
±
...)
or
x
x
x
x
x
May be implied by eg (2
May see = 0 or may be implied.
... (...
x
x
±
...)(...
x
±
...)
x
x
−
5)(
x
+
3)
dM1
Attempts to solve their quadratic either by factorising or using the quadratic formula or
completing the square. They cannot just state the roots and the fully factorised version or
values of a, b and c must match their quadratic. It is dependent on the previous method
mark. The = 0 can be implied and the factorised quadratic may appear under their solutions.
5
B1
Two of
x =
0,
,
−
3
2
5
A1
x =
0,
,
− provided all previous marks have been scored. Check for
3
x =
0
in earlier work
2
5
Eg1
3
2
− M1dM1B1A1
2
x
+
x
−
15
x
=
x
(2
x
−
5)(
x
+
3)
0,
,
3
2
1
15
5
5
Eg2
2
3
+
2
−
15
=
0
3
+
2
−
=
0
−
(
+
3)
=
0
0,
,
−
3
M1dM1B1A1
x
x
x
x
x
x
x x
x
2
2
2
2
5
5
2
3
2
15
(
3)
0,
,
3
Eg3
x
+
x
−
x
=
x x
−
x
+
−
M1dM0B1A0 (the cubic and factorised version
2
2
are not equal to each other)
5
Eg4
2
3
+
2
−
15
=
(2
2
+
−
15)
0,
,
− M1dM0B1A0 (no method seen to solve the
3
x
x
x
x
x
x
2
quadratic)
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