Mathematics · May / June 2023 · Paper 21 · Question 10
YMA01
As printed in the paper
The question as text
10. The curve C has equation 2 y = ( x − k ) x > 0 x where k is a positive constant.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
10(a)[5]
(
)
2
2
2
2
2
2
2
3
1
1
x
−
k
x
−
kx
+
k
x
kx
k
−
=
=
−
+
=
−
2
+
2
x
2
kx
2
k x
2
1
1
1
1
x
x
2
x
2
x
2
x
2
16
3
1
1
2
5
4
3
1
16
−
−
2
+
2
d
=
−
+
2
2
x
2
kx
2
k x
2
x
x
2
kx
2
k x
2
5
3
1
1
2
5
4
3
1
2
4
2046
2
2
2
2
6
2
84
2
−
k
2
+
k
2
−
+
k
−
k
=
k
−
k
+
5
3
5
3
5
10(b)[2]
(1
−
)
2
k
9
=
9
=
(1
−
)
2
k
1
k =
4
only *
36
10(c)[3]
q =
1
2046
1683
(9
+
"36") 15
×
−
"6"(4)
2
−
"84"(4)
+
=
2
5
10
(a)
M1
Attempts to multiply out the numerator and split the fraction into separate terms. Score for one
term with a correct index (which must be processed). Any attempts via substitution e.g. setting
u
=
x
send to review. Condone poor attempts at multiplying out the brackets such that
(
)
2
2
2
x
−
k
x
±
k
M1
Attempts to raise the power (which must be a fraction or decimal) by one on one of their terms.
(The index does not need to be processed).
5
3
1
2
4
A1
or unsimplified equivalent. Indices must be processed. Ignore any spurious
x
2
−
kx
2
+
2
k x
2
2
5
3
notation around the expression. May be with or without
+
c
dM1
Substitutes in 16 and 1 for x and subtracts either way round. The values embedded are sufficient
and condone invisible brackets. It is dependent on the previous method mark.
2046
A1
6
2
−
84
+
o.e. e.g.
6
2
−
84
+
409.2
isw after a correct answer is seen, but
k
k
k
k
5
withhold this mark if there is spurious notation such as or dx at either the start or end of
the answer.
(b)
M1
Substitutes the coordinates of A into the equation for C or, alternatively, substitutes in the
4
coordinates of A and
k =
into the equation for C.
A1*
Solves either:
9
=
(1
−
k
)
2
k
=
...
k
2
−
2
k
−
8
=
0
with terms collected to a 3TQ via any allowable method including via a
calculator.
With either method they must proceed to
k =
4
only with no errors seen, including invisible brackets. If
− is found it must be rejected or 4 clearly selected. If they use
2
k =
4
in their solution, then there must be
a minimal conclusion.
(c)
B1
q =
36
seen (may be in (b) or on the diagram) or implied for example by a correct area of a
triangle or trapezium.
M1
Attempts to find the area of R using their q and their answer to part (a). Condone arithmetical
slips and do not be concerned with poor notation for this mark. Must proceed to find a value for
the area but condone subtracting areas the wrong way round which may result in a negative area.
Possible methods include:
16
9
36 d
4
e.g. Line – curve:
x
+
x
− part (a) with
k =
5
5
1
e.g. Area of trapezium (
=
337.5
) – part (a) with
k =
4
Note: the correct values if the trapezium is split into a rectangle and a triangle are 135 and
202.5, respectively.
For evidence of use of part (a), look for
k =
4
to be substituted into their part (a) answer, or
2046
may be implied by a correct value for their
6
k
2
−
84
k
+
with
k =
4
5
16
(
x
−
k
) d
2
Condone
=
169.2
provided their part (a) is correct
x
x
1
1683
A1
10 oe (ignore units) Condone spurious notation.
Withhold this mark if their method results in a negative area but they just omit the negative
1683
sign. They would need to indicate area is positive for e.g.
A =
.... ....
−
= −
so
A =
10
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