Mathematics · May / June 2022 · Paper 31 · Question 5
YMA01
As printed in the paper
The question as text
5. y A y = f(x) O x B Figure 2 Figure 2 shows part of the graph with equation y = f(x), where f(x) = ½kx – 9½ – 2 x ∈ and k is a positive constant. The graph intersects the y‑axis at the point A and has a minimum point at B as shown.
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Show the answer and mark scheme
5(a)[2]
7
(i)
9
(ii)
k
5(b)[3]
... ...
kx −−
9
2
=
0
and
−
kx
+ −
9
2
=
0
,
k
k
CVs 7 11
,
k
k
7
11
x
<
<
k
k
5(c)[3]
9
" 2"
−
= −×
2 " " 3
+
k
=
...
k
(
) 3.6
k =
k >
3.6
Alt(b)
−<
2
kx
−
9
<
2
...
...
<
x
<
k
k
7
11
<
x
<
k
k
(a) Check Figure 2 and by the questions
(i)
B1
7 Condone (0, 7) but not (7, 0)
(ii)
9
oe Condone 9 ,
9
B1
−
2
but not
−
2, k
k
k
(b)
M1
A correct attempt to find the two critical values. Eg
Forms two correct equations (or linear inequalities) with the modulus signs removed and solves to
find the values of x where the graph crosses the xaxis (see main scheme) or the values of x such
that
kx −
9
−
2
<
0
Solves the given inequality to isolate x in the middle (Alt(b)) by:
kx
−
9
−
2
<
0
kx
−
9
<
2
−<
2
kx
−
9
<
2
...
<
x
<
...
Solves the given inequality by squaring both sides and solving:
2
18
77
9
4
(
kx
−
9)
2
<
4
k x
2
2
−
18
kx
+
77
<
0
x
2
−
x
+
<
0
x
−
<
...
<
k
k
2
k
k
2
Do not be too concerned by their use of ,
=
,
, ,
><
throughout their working.
Correct critical values 7 11
A1
,
cao which may be seen in their final answer.
k
k
7
11
7
11
7
11
✓
A1
<
x
<
or simplified equivalent eg “
x
>
∩
x
<
” or “
x
∈
:
<
x
<
” or
k
k
k
k
k
k
7 11
x
∈
,
Must be in terms of x
k
k
There must be some attempt to bring the two inequalities together if separate. The mark cannot be
7
11
7
11
7 11
x
>
x
<
x
>
or
x
<
x
∈
,
scored for
DO NOT ACCEPT eg “
”
k
k
k
k
k
k
(c)
M1
Uses a correct method to find a value for k. For example:
9
"
", " 2"
−
3
2
Substitutes in
into
y
=
−
x
and attempts to find a value for k. They may solve
k
9
" 2"
−
= −
3
2
x
x
=
"2.5"
" "
<
"2.5"
k
>
...
k
Finds when
y
=
3
−
2
x
intersects with AB (
y
= −
kx
+
7
) and then when
y
=
3
−
2
x
intersects with B
4
9
<
"
"
and forms the inequality
or similar.
k
−
2
k
4
Finds the value of x when
y
=
3
−
2
x
intersects with AB,
y
= −
kx
+
7
,
=
and then when
k
−
2
14
y
=
3
−
2
x
intersects
y
=
kx
−
11
=
and sets equal to each other to find the value for k (or
k
+
2
9
solves simultaneously to find
x
=
"2.5"
" "
<
"2.5"
k
>
..
)
k
" 2" 3
−
−
2
Finds the gradient between (0,3) and B and deduces that
−>
oe leading k = …
9
"
" 0
−
k
Sets f( )
x
=
3
−
2
x
and attempts the discriminant leading to a value for k:
3
−
2
x
=
kx
−
9
−
2
(5
−
2 )
x
2
=
(
kx
−
9)
2
(4
−
k
2
)
x
2
+
(18
k
−
20)
x
−
56
=
0
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