Mathematics · Oct / Nov 2021 · Paper 31 · Question 2
YMA01
As printed in the paper
The question as text
2. y P O x Figure 1 Figure 1 shows a sketch of part of the graph with equation y = f(x), where f(x) = ½3x – 13½ + 5 x ∈ The vertex of the graph is at point P, as shown in Figure 1.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
2(a)[2]
13,5
3
(b)(i)
f( )
x …
5
(ii)
10
2(b)[2]
13,5
3
(b)(i)
f( )
x …
5
(ii)
10
Attempts to solve either 16
−
2 ...3
x
x
−
13
+ ⇒ a value or inequality in x
5
Or 16
−
2 ... 3
x
−
x
+
13
+
5
⇒ a value or inequality in x
24
2, 5
Both correct critical values
Selects inside region for their critical values
24
2
<
x
<
5
1
a
=
4,
b
=
3
Alt(c)
(11
−
2 ) ...(3
x
2
x
−
13)
2
⇒
121
−
44
x
+
4
x
2
...9
x
2
−
78
x
+
169
5
x
2
−
34
x
+
48...0
⇒ a value or inequality in x
24
2, 5
Selects inside region for their critical values
24
2
<
x
<
5
Note – Check the diagram and next to the questions for answers
(a)
13,
x
=
y
=
5
B1
One correct coordinate. Allow as
or exact equivalent and condone
3
missing brackets.
13,
5
B1
Both coordinates correct. Allow as
x
=
y
=
or exact equivalent and condone
3
missing brackets.
(b)(i)
[
)
B1ft
f( )
x … . Allow equivalent correct answers. E.g.
5
y … ,
y ∈
5,
∞ , f
… or using
5
5
set notation. Follow through on their y coordinate in (a). Do not allow “range
… ”
(b)(ii)
B1
10
(c)
M1
Attempts to solve a correct equation or inequality. Look for the equation or
inequality (ignore the direction) after the modulus signs have been removed allowing
slips in their rearrangement.
24
A1
Correct critical values for
x
...2, 5
or exact equivalent which may be part of an
incorrect inequality
dM1
Selects inside region for their critical values or both correct inequalities seen for
their values.
Allow this to be written separately and allow the critical values to be included within
their region.
24
48
2
<
x
<
2
<
x
<
A1
Allow other exact equivalent forms such as
or
5
10
24
or 24
>
x
>
2
x
∈
2, 5
5
The inequalities must be presented together on one line for this mark.
24
Accept two separate statements such as “
x
>
2 AND
x
<
” but not “
5
24
24
x
2 or
x
” or “
x
2,
x
”
>
<
>
<
5
5
(d)
1
a
=
4 or
b
=
B1ft
One correct value. Either
but ft on their part (a) so accept
3
20
13
a
=
or
b
=
''
'' 4
−
''5''
3
1
a
=
4,
b
=
B1
which may be embedded within
y
=
a
f (
x
+
b
)
3
2(c)[4]
Attempts to solve either 16
−
2 ...3
x
x
−
13
+ ⇒ a value or inequality in x
5
Or 16
−
2 ... 3
x
−
x
+
13
+
5
⇒ a value or inequality in x
24
2, 5
Both correct critical values
Selects inside region for their critical values
24
2
<
x
<
5
2(d)[2]
1
a
=
4,
b
=
3
Alt(c)
(11
−
2 ) ...(3
x
2
x
−
13)
2
⇒
121
−
44
x
+
4
x
2
...9
x
2
−
78
x
+
169
5
x
2
−
34
x
+
48...0
⇒ a value or inequality in x
24
2, 5
Selects inside region for their critical values
24
2
<
x
<
5
Note – Check the diagram and next to the questions for answers
(a)
13,
x
=
y
=
5
B1
One correct coordinate. Allow as
or exact equivalent and condone
3
missing brackets.
13,
5
B1
Both coordinates correct. Allow as
x
=
y
=
or exact equivalent and condone
3
missing brackets.
(b)(i)
[
)
B1ft
f( )
x … . Allow equivalent correct answers. E.g.
5
y … ,
y ∈
5,
∞ , f
… or using
5
5
set notation. Follow through on their y coordinate in (a). Do not allow “range
… ”
(b)(ii)
B1
10
(c)
M1
Attempts to solve a correct equation or inequality. Look for the equation or
inequality (ignore the direction) after the modulus signs have been removed allowing
slips in their rearrangement.
24
A1
Correct critical values for
x
...2, 5
or exact equivalent which may be part of an
incorrect inequality
dM1
Selects inside region for their critical values or both correct inequalities seen for
their values.
Allow this to be written separately and allow the critical values to be included within
their region.
24
48
2
<
x
<
2
<
x
<
A1
Allow other exact equivalent forms such as
or
5
10
24
or 24
>
x
>
2
x
∈
2, 5
5
The inequalities must be presented together on one line for this mark.
24
Accept two separate statements such as “
x
>
2 AND
x
<
” but not “
5
24
24
x
2 or
x
” or “
x
2,
x
”
>
<
>
<
5
5
(d)
1
a
=
4 or
b
=
B1ft
One correct value. Either
but ft on their part (a) so accept
3
20
13
a
=
or
b
=
''
'' 4
−
''5''
3
1
a
=
4,
b
=
B1
which may be embedded within
y
=
a
f (
x
+
b
)
3
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