Mathematics · May / June 2023 · Paper 41 · Question 8
YMA01
As printed in the paper
The question as text
8. y C l P R O Q x Figure 2 Figure 2 shows a sketch of part of the curve C with parametric equations 1 1 x t y t t > 0 7 . t t The curve C intersects the x-axis at the point Q.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
8(a)[1]
x = 2
(
)
8(b)[1]
2.5, 1.5
1
8(c)[3]
1
+
d
y
t
2
=
1
d
x
1
−
t
2
d
y
5
3
(
)
t
=
2
=
y
−
1.5
= −
x
−
2.5
d
x
3
5
3
x
5
y
15*
+
=
l crosses xaxis at x = 5
8(d)[7]
1
15
1.5
2
(
5
2.5
)
π
×
−
=
π
3
8
2
d
x
1
1
2
2
V
=
π
y
d
x
=
π
y
d
t
=
π
t
−
1
−
d
t
d
t
t
t
2
1
1
3
1
π
t
2
−
2
+
1
−
d
t
=
π
t
2
+
−
−
3 d
t
=
...
t
2
t
2
t
2
t
4
π
t
3
3
1
−
+
−
3
t
3
t
3
t
3
2
t
3
3
1
8
3
1
1
1
3
t
6
3
3
π
−
+
−
=
π
−
+
−
−
−+
−
3
t
3
t
3
3
2
24
3
3
1
13
15
29
V
Cao
=
π
+
π
=
π
24
8
12
(a)
B1: Correct value. Just look for 2 and condone Q = 2
(b)
x
=
2.5,
y
=
1.5
B1: Correct coordinates. Allow these to be given separately
(c)
1
1
±
M1: Attempts to differentiate both parameters and use d
y
d
y
d
x
t
2
=
÷
to achieve
or equivalent via
d
x
d
t
d
t
1
1
±
t
2
quotient rule
dM1: Uses t = 2 in their d
y
x and uses the negative reciprocal gradient with their point P to form equation
d
of normal.It is dependent upon the previous M
A1*: Obtains the printed answer with no errors and sufficient working shown.
What is shown in the MS is sufficient
(d)
B1: Correct x intercept for l. This may appear on the xaxis on Figure 2.
It may be implied by the value appearing in the volume of the cone but not the volume under the
curve
1
M1: Fully correct method for the cone volume
(
)
π
×
''1.5''
2
''5'' ''2.5''
−
=
3
5
5
(
)
3
(
)
3
π
3
−
0.6
x
π
3
−
0.6 2.5
×
May be awarded for
(
3
0.6
x
)
2
d
x
π
−
=
=
−
1.8
1.8
2.5
2.5
Condone slips. If they expand (
3
0.6x
)
2
look for a quadratic expression integrating to a cubic of the
−
correct form followed by use of the limits 5 and their 2.5
M1: Correct strategy for the volume generated by rotating the area under the curve
2 d
x
2
1
k
2
V
=
π
y
d
x
=
π
y
d
t
Look for
leading to
π
t
−
1
±
d
t
, where k is a constant
d
t
t
t
2
1
Condone a sign slip on d
x
1
±
. The πmay be added later in the solution.
so allow
d
t
t
2
The dt may be omitted
2
1
k
M1: Attempt to expand
t
−
1
±
and integrate term by term.
t
t
2
2
1
k
Look for
polynomial in t containing positive and negative indices followed by
t
−
1
±
→
t
t
2
some correct integration for both positive and negative indices
2
1
1
t
3
3
1
A1: Correct integration of
3
which does not need to be simplified
t
−
1
−
→
−
+
−
t
t
t
2
3
t
3
t
3
1
2
k
M1: Applies the t limits “1” and 2 to an attempt at the integral of ( )
π
t
−
1
±
d
t
t
t
2
The π does not need to be included.
A1: 29
12 π or exact equivalent.
...............................................................................................................................................................
Part (c) using a calculator
Whilst it is possible to do the integration for the area under the curve using a calculator,
it will not score all of the marks.
2
1
2
1
E.g.
π
t
−
1
−
d
t
=
0.5416
π
t
t
2
1
The marks in (c) available to such a candidate are:
B1(for x = 5),
M1 (for volume of cone),
M1 (Sight of correct integral),
M0 (no expansion),
A0 ,
M1(sight of limits),
A0
..................................................................................................................................................................................
See below for Cartesian approach for (c) and (d)
(c)
1
1
d
y
x
= +
t
,
y
= −
t
x
2
−
y
2
=
4
2
x
−
2
y
=
0
t
t
d
x
or
1
d
y
1
(
)
(
)
−
y
2
=
x
2
−
4
y
=
x
2
−
4
=
x x
2
−
4
2
2
d
x
Attempts to differentiate a Cartesian equation of the form
to obtain
x
2
−
y
2
=
k
d
y
1
d
y
(
)
either
x
y x
0
or
−
α
+
α
=
=
kx x
2
−
4
2
d
d
x
1
t
+
d
y
1
(
)
−
d
y
x
2.5
2
t
or
=
2.5 2.5
−
4
2
=
=
=
d
x
d
x
y
1
1.5
t
−
t
3
(
)
y
−
1.5
= −
x
−
2.5
5
Uses t = 2 in their d
y
and uses the negative reciprocal gradient with their point P.
d
x
If using explicit differentiation, must attempt x using t = 2 first
3
x
5
y
15*
+
=
Obtains the printed answer with no errors and sufficient working shown.
l crosses xaxis at x = 5
(d)
1
15
(
)
π
×
1.5
2
5
−
2.5
=
π
3
8
Fully correct method for the cone volume
(
)
V
=
π
y
2
d
x
=
π
x
2
−
4 d
x
Correct strategy for the other volume
(
)
Condone for this
V
=
π
y
2
d
x
=
π
x
2
−
k
d
x
π
x
3
(
)
x
2
4 d
x
4
x
π
−
=
−
3
M1: Attempt to integrate (
)
x
2
−
k
d
x
A1: Correct integration of (
)
x
2
−
4 d
x
2.5
π
x
3
−
4
x
3
2
Applies the limits “2” and their x from t = 2 to an attempted integral of
( ) (
)
π
x
2
−
k
d
x
13
15
29
V
=
π
+
π
=
π
24
8
12
Cao
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.