Mathematics · May / June 2024 · Paper 31 · Question 1
YMA01
As printed in the paper
The question as text
1. y y = f (x) P O x Figure 1 Figure 1 shows a sketch of the graph with equation y = f (x) where f (x) = 2| x – 5| + 10 The point P, shown in Figure 1, is the vertex of the graph.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
1(a)[2]
5
(
)
−
2
−
5
+
10...6
..." "
1(b)[2]
Attempts to solve e.g.
x
x
x
2
5
x <
o.e.
2
(
)
7, 30
1(c)[2]
Condone missing brackets around pairs of coordinates provided the meaning is clear
(a) If there is a contradiction between coordinates on the graph, by the question and in the main
body of the work then the work in the main body of the work takes precedence.
B1:
One correct coordinate. Allow as
=
5 or
=
10
Do not allow them to be the wrong way round
x
y
B1:
Both coordinates correct. Allow as
5,
10
x
=
y
=
(b)
Answers with no working scores 0 marks
M1:
Attempts to solve a correct equation or inequality with at least one intermediate stage of working.
Allow this mark even if there are additional equations/inequalities. The modulus signs must have
been removed.
Alternatively, they may attempt to square both sides and then attempt to solve the quadratic (usual
rules apply). If via a calculator the root(s) must be correct (you may need to check these)
Condone slips including when squaring and multiplying out
−
5 ...3
−
5
2
−
10
+
25...9
2
−
30
+
25
0...8
2
−
20
...
x
x
x
x
x
x
x
x
x
In either method condone slips in their rearrangement and do not be concerned by the
direction of the inequality sign for this mark.
5
x <
A1:
o.e. only. isw if they attempt to simplify their fraction incorrectly. Must be identified as their
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