Mathematics · Oct / Nov 2024 · Paper 31 · Question 3
YMA01
As printed in the paper
The question as text
3. y O x Figure 2 In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Figure 2 shows a sketch of the curve with equation y = f (x) , where f (x) = 2x 2 – 10x x ∈
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[3]
y =
48
Graphical interpretation of
(
)
f
48
x
=
One of 8,
−
8
Attempts to solve an appropriate equation
E.g.
2
2
10
48
2
5
24
0
(
8
)(
3
)
0
...
x
−
x
=
x
−
x
−
=
x
±
x
±
=
x
=
8,
− with no additional values
8
x =
3(b)[4]
Graphical interpretation of
5
f ( )
x
…
x
2
5
y
=
x
2
5
25
Attempts to solve
2
2
−
10
=
4
2
−
25
=
0
=
x
x
x
x
x
x
2
4
5
15
OR attempts to solve (o.e.)
10
−
2
2
=
4
2
−
15
=
0
=
x
x
x
x
x
x
2
4
5
25
2
2
Attempts to solve
2
x
−
10
x
=
x
4
x
−
25
x
=
0
x
=
2
4
5
15
2
AND attempts to solve (o.e.)
10
x
−
2
x
2
=
x
4
x
−
15
x
=
0
x
=
2
4
15
25
Achieves both critical values
x =
,
x =
4
4
15
25
Correct set of values
x „
or
x …
4
4
This question demands that some working should be shown
(a) This part is now being marked B1 M1 A1 instead of M1 A1 A1
B1: Correctly obtains 8 or
−
8
as a solution (for x) . Condone solution appearing as
x =
M1: Attempts to solve (
)
f
=
48
x
via an appropriate method including a calculator, factors or formula
So, look for a correct equation followed by an attempt to solve.
The equation must be given in the form of a correct 3TQ, but not necessarily collected on one side of
the = sign. Solving is allowed by all methods including via a calculator.
All the following score this M mark
Example 1:
2
2
10
48 o.e
8,
(
3
)
…only the 8 is required
x
−
x
=
x
=
−
Example 2:
2
2
(
)(
)
2
x
−
10
x
=
48 o.e
x
−
5
x
−
24
=
0
x
−
8
x
+
3
=
0
x
=
...
Example 3:
(
)
2
x
2
−
10
x
−
48
=
0 or
x
2
−
5
x
−
24
=
0
leading to
x
or
x =
8,
−
3
Example 4:
( )(
)
−
10
±
100
−
4 2
−
48
2
2
+
10
−
48
=
0
=
=
...
x
x
x
( )
2 2
Example 5:
2
( )
…only the −8 is required
2
x
+
10
x
=
48 o.e
x
= −
8, 3
A1
8,
− with no additional values unless they have been rejected
8
x =
(b)
M1: Attempts to solve a correct equation or inequality, leading to a nonzero value for x.
The method must be shown with the terms being collected as a minimum response
5
Look for
2
2
. Allow with the = as any inequality including <
2
x
−
10
x
=
x
ax
±
bx
=
0
x
=
c c
,
≠
0
2
5
2
OR
2
2
10
0
,
0
. Allow with the = as any inequality including <
−
x
+
x
=
x
ax
±
bx
=
x
=
c c
≠
2
5
2
You may see as an alternative to the 2nd equation/inequality
2
2
−
10
= −
±
x
x
x
ax
bx
2
5
5
25
It is acceptable to divide through by x, so
2
2
−
10
=
2
−
10
=
2
=
x
x
x
x
x
x
2
2
2
dM1: Attempts to solve BOTH correct equations or inequations leading to two nonzero values for x
Allow even if there are extra equations solved.
15
25
A1: Achieves both critical values
x =
and
x =
and no other values apart from 0 (and perhaps 5)
4
4
15
following both equations or inequalities. Allow with incorrect inequalities
and
x …
x „
4
looking for the critical values only
15
25
15
25
A1: Correct range in allowable form. E.g.
x „
or
x …
,
but condone
−∞
,
∪
,
∞
4
4
4
4
15
25
, 25
15
15
25
Do NOT incorrect forms such as
and
,
x „
x …
,
,
„
x
„
−∞
∩
∞
4
4
4
4
4
The two aspects must be brought together on a single line. Mark their final answer
Alt Method: You may see attempts via squaring. It can be marked using the above guidelines
(
)
5
2
(
)
2
2
2
2
2
(
)(
)
2
x
−
10
x
=
x
x
16
x
−
160
x
+
375
=
0
x
4
x
−
15
4
x
−
25
=
0
2
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