Mathematics · Oct / Nov 2020 · Paper 31 · Question 4
YMA01
As printed in the paper
The question as text
4. y y = f (x) O x Figure 2 Figure 2 shows a sketch of part of the graph with equation y = f (x) where f (x) = 21 − 2 | 2 − x | x 0
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[2]
ff(6)
=
f(13)
= −
1
Attempts 21
+
2(2
−
x
)
=
5
x
⇒
x
=
...
or 21 2(
−
x
−
2)
=
5
x
⇒
x
=
...
4(b)[2]
25
x =
only
7
k
<
21or
k
17
4(c)[2]
Either
17
k <
21
1
4(d)[2]
a =
b =
4
7
(a)
(
)
M1
For attempting f ''twice''. Award for sight of f(13) or
21
−
2 2
−
21
−
2 2
−
x
followed by a value.
A1
For 1
−
(b)
M1
Attempts 21
+
2(2
−
x
)
=
5
x
⇒
x
=
...
25
A1
For
x =
only. Do not isw if students go on to find additional solutions unless they identify this as the only
7
answer by rejecting others.
(c)
M1
For either
k
<
21or
k
17
Condone for this mark
k
21or
k >
17
Alt: Allow for both k = 17 and k = 21 identified as critical values even if no inequalities are given.
17
k <
21
A1
May be written as separate inequalities. Accept alternative notations for the range such as
1
7
f(
)x <
2
1
Do not accept
(d)
1
B1
Either
a =
or
b =
4
7
1
1
a =
b =
4
y
=
f(
x
−
4)
B1
Both
and
Allow both marks if
is given.
7
7
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