Mathematics · Oct / Nov 2022 · Paper 21 · Question 2

YMA01

As printed in the paper

The question as text

2. A curve C has equation y = f(x) where f(x) = (2 – kx)5 and k is a constant. Given that when f(x) is divided by (4x – 5) the remainder is 243 32

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2(a)[2]
243
243
f( )
(2
k
)
5
(2
k
)
k
...
5
=
−×
5
=
−×
5
=
5
=
4
4
4
32
32
3
5
k
1
2
k
*
=
=
2
4
2
5
2(b)[3]
2
2
2
5
4
or
5
C
2
3
x
±
C
×
2
×±
x
±
×
×±
1
2
5
5
64
32
−
32
x
+
x
2
5
2(c)[2]
128
f ( )
′
x
= −
32
+
x
+
...
f (0)
′
=
...
5
f (0)
′
32
= −
(a)
5
into f( )x , equates to 243
M1: Substitutes
x =
32 and attempts to make k the subject by taking the
4
5th root of both sides.
2
A1*:
k =
with no errors and sufficient working shown.
5
243
3
2
Accept as a minimum
(2
−×
k
)
5
=
(2
−×
k
)
=
k
=
5
5
4
4
32
2
5
243
2
Note that just
(2
−×
k
5
)
5
=
k
=
scores no marks as the minimum for the M mark
4
32
5
requires taking the 5th root of both sides.
Alternative by verification:
5
2
5
3
243
,
(2
)
5
M1:
k
=
x
=
−×
2
5
=
=
5
4
5
4
2
32
2
5
Substitutes
k
=
and
x
=
and attempts to raise an evaluated 2 – kx to the power of 5
5
4
2
A1: Hence
k =
5
Fully correct work and makes a (minimal) conclusion e.g. Hence proven, QED, Therefore true,
etc.
243
243
Note that just
(2
−×
2
5
)
5
=
or
(2
−×
2
5
)
5
=
(2
−
1
)
5
=
scores no marks as the
2−×
5
4
32
5
4
2
32
or
2−
1
must be evaluated for the M mark.
2
(b)
M1: Attempts the binomial expansion to obtain the correct structure for the x or
2x term i.e. the
2 .
correct binomial coefficient with the correct power of 2 and the correct power of
±
5 x
The binomial coefficients do not have to be evaluated but must be correct if they are.
If awarding this mark for the x2 term you can condone missing brackets e.g.
±
5
C
×
2
3
×±
2
64
A1: For the correct simplified x or x2 term i.e.
−
32 or
x
+
x
2
5
64
which may be written as a list. Allow equivalents for 64
A1: For
32
−
32
x
+
x
2
e.g. 12.8
5
5
64
(
)
Condone
32
+ −
32
x
+
x
2
5
Ignore any extra terms but do not isw – mark their final answer.
If they don’t simplify in (b) do not allow simplified terms in (c) as recovery.
(b) Alternative takes out a power of 2:
5
5
2
2
1
1
5 4 1
×
5
5
2
−
x
=
2
1
−
x
=
2
1 5
−×
x
+
x
+
...
5
5
5
2
5
1
5 4
1
2
1
×
Score M1 for
2
5
...
±×±
5
5 x
+
...
or
2
5
...
±
±
5 x
+
...
condoning
±
5 x
2
as above
2
Then A marks as above.
(c)
M1: Attempts to differentiate their expansion and substitutes
x =
0
which may be implied.
For the differentiation, look for
x
n
→
x −
n
1
at least once including k →0 or kx →k
A1ft: 32
−
following correct differentiation
Or follow through on their q provided
the expansion in (b) was of the form
p
+
qx
+
rx
2,
p q r
, ,
≠
0
the differentiation is correct for their
p
+
qx
+
rx
2
i.e.
q
+
2
rx
In part (c) you can also ignore any extra terms e.g. do not penalise if they have differentiated any of
the extra terms incorrectly in an otherwise correct solution.

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