Mathematics · May / June 2024 · Paper 41 · Question 5

YMA01

As printed in the paper

The question as text

5. y t = b t = a 1 R O x Figure 2 Figure 2 shows a sketch of the curve defined by the parametric equations 2 2   x t 2 t y a t b t ( 3 t ) where a and b are constants. The ends of the curve lie on the line with equation y = 1

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Show the answer and mark scheme

5(a)[2]
y
1
2
3
t
t
2
t
...
=
=
=
−
=
(3
)
t
−
t
(
)
(
)(
)
2
3
2
1
2
0
1,
2
t
−
t
+
=
t
−
t
−
=
a
=
b
=
t
"2"
2
2
=
5(b)[5]
d
x
2
Area under curve
=
y x
d
=
y
d
t
=
×
(2
t
+
2)d
t
d
t
t
(3
−
t
)
t
=
"1"
1
1
t
=
1
x
=
3,
t
=
2
x
=
8
2
(
)
2
So area of R =
1
×
8
−
3
−
×
(2
t
+
2)d
t
t
(3
−
t
)
1
2
t
+
1
=
5
−
4
d
t
t
(3
t
)
−
1
t
+
1
A
B
(c)(i)
=
+
−
t
(3
−
t
)
t
3
t
+ =
t
1
A
(3
−
t
)
+
Bt
1
4
E.g.
t
=
0
=
1
3
A
A
=
;
t
=
3
4
=
3
B
B
=
3
3
t
+
1
1
4
=
+
t
(3
−
t
)
3
t
3(3
−
t
)
(ii)
t
+
1
1
4
1
4
(
)
d
t
=
+
d
t
=
ln
t
−
ln
3
−
t
t
(3
−
t
)
3
t
3
(3
−
t
)
3
3
2
1
4
(
)
1
4
1
4
Area
=
5
−
4
ln
t
−
ln 3
−
t
=
5
−
4
ln 2
−
ln1
−
ln1
+
ln 2
3
3
3
3
3
3
1
1
4
=
5
−
4
ln 2
+
ln 2
3
3
20
=
5
−
ln 2
3
(a)
M1: Sets the y equation equal to 1 (condoning slips) and then attempts to solve the resulting quadratic
equation (usual rules). May be implied by correct values if a calculator is used.
A1: Correct values. That is
a = and
1
b =
2
. Condone (t =) 1 and 2 due to the demand of the question.
Correct answers with no (incorrect) working scores both marks but must be seen in part (a).
If they do state
a = and
3
b =
8
following
t = and
1
t =
2
score A0
(b) M1: Attempts d
x
t and multiplies by y condoning slips (e.g. missing brackets).
d
No requirement to see any integral.
''2''
2
A1: Correct unsimplified expression for the area under the curve
t ×
(2
t
+
2)d
t
(3
)
t
−
''1''
Allow with their a and b or even with a and b (as letters). May be seen within an expression.
Condone the omission of the dt for this mark. Allow missing brackets to be recovered.
B1: Sight of correct (x) values 3 and 8 at the ends of the given curve. May be awarded from work done
in (a). Award for sight of 5 as the width of the rectangle or for 5−as their attempt at the area of R
M1: Full attempt at finding the area of region R. Be aware that the form is given.
Look for the area of the rectangle subtract the area under the curve. dt’s not required here
It is dependent upon
ax and
bx or use of their t values from (a) to find
ax and
bx which may be implied by
Use of the correct
the width of the rectangle
''2''
−
d
x
(
)
Use of
1
×
x
−
x
y
d
t
to find area of R.
b
a
d
t
''1''
(
)
t
=
"2"
2
2
4
t
+
1
d
x
(
)
(
)
(
)
2
May be achieved by
1
−
y
d
x
=
1
−
y
d
t
=
2
t
+
2
−
d
t
=
t
2
+
2
t
−
d
t
t
(3
−
t
)
1
t
=
"1"
1
1
A1: Correct answer achieved. All aspects including the dt must be present at least once before the final
solution
(c)(i)
M1: Correct form for partial fractions which may be implied
M1: Correct method to find at least one of the constants.
If they use cover up rule etc it may be implied by one correct constant or one correct pf
A1: Correct partial fractions, not just correct constants. May be awarded in (c)(ii)
Allow fractions like 1
4
3
3
. Ignore any … which may be set equal to this.
+
(3
)
t
−
t
(c)(ii)
M1: Integrates to correct form
p
ln
t
+
q
ln(3
−
t
)
o.e.
Be aware that
p
ln
ct
+
q
ln (3
k
−
t
)
where c and k are constants is also correct
dM1: Substitutes their t limits (either way around) and uses the answer from (b) to find the area.
There must be some correct ln work seen. E.g ln1
=
0
or attempts to combine terms
To score this there must be nonzero values for M, k, a and b.
K
1
Condone with M and k leading to a correct expression in M and k. E.g.
M
−
ln 2
3
K
1
4
It is dependent upon the previous M1 and must be of the form
M
−
''
ln 2
+
ln 2 ''
3
3
A1: Correct answer or equivalent simplest form. E.g. 15
−
20ln 2.
ISW after a correct answer.
3
5
4
Condone other simple equivalents such as
5
−
ln16 or
5
−
ln32
3
3
2
(
)
(
)
(
)
2
2
2
5(c)[6]
y
1
2
3
t
t
2
t
...
=
=
=
−
=
(3
)
t
−
t
(
)
(
)(
)
2
3
2
1
2
0
1,
2
t
−
t
+
=
t
−
t
−
=
a
=
b
=
t
"2"
2
2
=
d
x
2
Area under curve
=
y x
d
=
y
d
t
=
×
(2
t
+
2)d
t
d
t
t
(3
−
t
)
t
=
"1"
1
1
t
=
1
x
=
3,
t
=
2
x
=
8
2
(
)
2
So area of R =
1
×
8
−
3
−
×
(2
t
+
2)d
t
t
(3
−
t
)
1
2
t
+
1
=
5
−
4
d
t
t
(3
t
)
−
1
t
+
1
A
B
(c)(i)
=
+
−
t
(3
−
t
)
t
3
t
+ =
t
1
A
(3
−
t
)
+
Bt
1
4
E.g.
t
=
0
=
1
3
A
A
=
;
t
=
3
4
=
3
B
B
=
3
3
t
+
1
1
4
=
+
t
(3
−
t
)
3
t
3(3
−
t
)
(ii)
t
+
1
1
4
1
4
(
)
d
t
=
+
d
t
=
ln
t
−
ln
3
−
t
t
(3
−
t
)
3
t
3
(3
−
t
)
3
3
2
1
4
(
)
1
4
1
4
Area
=
5
−
4
ln
t
−
ln 3
−
t
=
5
−
4
ln 2
−
ln1
−
ln1
+
ln 2
3
3
3
3
3
3
1
1
4
=
5
−
4
ln 2
+
ln 2
3
3
20
=
5
−
ln 2
3
(a)
M1: Sets the y equation equal to 1 (condoning slips) and then attempts to solve the resulting quadratic
equation (usual rules). May be implied by correct values if a calculator is used.
A1: Correct values. That is
a = and
1
b =
2
. Condone (t =) 1 and 2 due to the demand of the question.
Correct answers with no (incorrect) working scores both marks but must be seen in part (a).
If they do state
a = and
3
b =
8
following
t = and
1
t =
2
score A0
(b) M1: Attempts d
x
t and multiplies by y condoning slips (e.g. missing brackets).
d
No requirement to see any integral.
''2''
2
A1: Correct unsimplified expression for the area under the curve
t ×
(2
t
+
2)d
t
(3
)
t
−
''1''
Allow with their a and b or even with a and b (as letters). May be seen within an expression.
Condone the omission of the dt for this mark. Allow missing brackets to be recovered.
B1: Sight of correct (x) values 3 and 8 at the ends of the given curve. May be awarded from work done
in (a). Award for sight of 5 as the width of the rectangle or for 5−as their attempt at the area of R
M1: Full attempt at finding the area of region R. Be aware that the form is given.
Look for the area of the rectangle subtract the area under the curve. dt’s not required here
It is dependent upon
ax and
bx or use of their t values from (a) to find
ax and
bx which may be implied by
Use of the correct
the width of the rectangle
''2''
−
d
x
(
)
Use of
1
×
x
−
x
y
d
t
to find area of R.
b
a
d
t
''1''
(
)
t
=
"2"
2
2
4
t
+
1
d
x
(
)
(
)
(
)
2
May be achieved by
1
−
y
d
x
=
1
−
y
d
t
=
2
t
+
2
−
d
t
=
t
2
+
2
t
−
d
t
t
(3
−
t
)
1
t
=
"1"
1
1
A1: Correct answer achieved. All aspects including the dt must be present at least once before the final
solution
(c)(i)
M1: Correct form for partial fractions which may be implied
M1: Correct method to find at least one of the constants.
If they use cover up rule etc it may be implied by one correct constant or one correct pf
A1: Correct partial fractions, not just correct constants. May be awarded in (c)(ii)
Allow fractions like 1
4
3
3
. Ignore any … which may be set equal to this.
+
(3
)
t
−
t
(c)(ii)
M1: Integrates to correct form
p
ln
t
+
q
ln(3
−
t
)
o.e.
Be aware that
p
ln
ct
+
q
ln (3
k
−
t
)
where c and k are constants is also correct
dM1: Substitutes their t limits (either way around) and uses the answer from (b) to find the area.
There must be some correct ln work seen. E.g ln1
=
0
or attempts to combine terms
To score this there must be nonzero values for M, k, a and b.
K
1
Condone with M and k leading to a correct expression in M and k. E.g.
M
−
ln 2
3
K
1
4
It is dependent upon the previous M1 and must be of the form
M
−
''
ln 2
+
ln 2 ''
3
3
A1: Correct answer or equivalent simplest form. E.g. 15
−
20ln 2.
ISW after a correct answer.
3
5
4
Condone other simple equivalents such as
5
−
ln16 or
5
−
ln32
3
3
2
(
)
(
)
(
)
2
2
2

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