Mathematics · Oct / Nov 2020 · Paper 41 · Question 4
YMA01
As printed in the paper
The question as text
4. y B A O x Figure 2 Figure 2 shows a sketch of part of the curve with parametric equations x = 2t2 − 6t, y = t 3 − 4t, t ∈ The curve cuts the x-axis at the origin and at the points A and B, as shown in Figure 2.
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Show the answer and mark scheme
4(a)[3]
A and B are where
t
3
4
t
0
t t
(
2
4)
0
t
2 or
2
−
=
⇒
−
=
⇒=
−
Substitutes
t
=
2,
x
=
2 4
×
−×
6 2
= − Hence
4
A = −
( 4,0)
When
(
)
Hence
B =
(20,0)
*
t
= −
2,
x
=
2 4
×
−×−=
6
2
20,
y
=
0
d
y
d
y
3
t
2
4
d
t
−
4(b)[5]
=
=
d
x
d
x
4
t
−
6
d
t
gradient = (
)
d
y
3
t
2
−
4
4
Sub
t = −into
2
=
⇒
−
d
x
4
t
−
6
7
Uses their (
)
4
−
and (20,0) to produce eqn of tangent
⇒
7
y
+
4
x
−
80
=
0
* M1 A1*
7
4(c)[4]
Substitutes
x
=
2
t
2
−
6 ,
t y
=
t
3
−
4 ,
t
into 7
y
+
4
x
−
80
=
0
(
)
(
)
3
2
⇒
7
t
−
4
t
+
4 2
t
−
6
t
−
80
=
0
7
t
3
8
t
2
52
t
80
0
⇒
+
−
−
=
⇒
(
t
+
2) (7
2
t
−
20)
=
0
20
t
=
"
"
⇒
x
=
....
7
40
x = −
49
(a)
M1: Sets
3
4
0
..
. to reach t = 2 or 2
−
t
−
t
=
⇒=
t
A1: Substitutes
t
=
2
⇒
x
=
2 4
×
−×
6 2
= − and states coordinates of
4
A = −
( 4,0)
Accept
t =
2
⇒
A
= −
( 4,0)
provided the M1 has been awarded
B1*: Show that the coordinates of
B =
(20,0)
Possible ways of this are
1) Solves
t
3
−
4
t
=
0
⇒= − and substitutes into
t
2
2
t
2
−
6
t
⇒
x
=
20
⇒
B
=
(20,0)
.
This is a show that and there must be some evidence for the ''20'', not just sight of
t = −
Look for
2
(
2
)
2
6
(
2
)
x
20
or 8 12
x
20
× −
−× −
⇒
=
+
⇒
=
This is a show that and there must be some evidence for the ''0'', not just sight of
t = −
(b)
d
y
y t and calculates d
y
d
t
M1: Attempts to differentiate ( ) and
x t
( )
x by using
in part (b)
d
d d
x t
d
y
3
t
2
−
4
A1:
=
d
x
4
t
−
6
t = − into their d
y
M1: Sub ‘their’
2
x to find the gradient of the tangent at B.
d
M1: Uses (20,0) and their d
y
x to find an equation of the tangent.
d
If they use y
=
mx
+ they need to proceed as far as c = ...
c
A1*: Achieves given answer. cso
7
y
4
x
80
0
The = 0 must be seen
⇒
+
−
=
ISW after sight of this
(c) The demand here is to use algebra which is satisfied by the setting up of the cubic equation.
M1: Attempts to substitute
x
=
2
t
2
−
6 ,
t y
=
t
3
−
4
t
into7
y
+
4
x
−
80
=
0
A1: Correct simplified equation in t:
7
t
3
+
8
t
2
−
52
t
−
80
=
0
The = 0 may be implied by further work
dM1: Uses a correct method to find x
t ≠− and substituting
2
Using a calculator this can be scored for finding a value ofthis in x (t).You may have to check this on a calculator if their equation is incorrect.
An accuracy of 2 sf is required for non exact solutions. Note that the equation may be in
20
different forms under this method so
7
t
3
+
8
t
2
−
52
t
=
80
⇒=
t
is fine
7
20
40
Note: (
)
(
)
3
2
7
t
−
4
t
+
4 2
t
−
6
t
−
80
=
0
⇒=
t
⇒
x
= −
would score M1 A0 dM1 A0
7
49
Via factorisation look for factors where the ends work,E.g.
''7
t
3
+
8
t
2
−
52
t
−
80''
=
(
Pt
+
a
)(
Qt
+
b
)(
Rt
+
c
)
with
PQR =
''7''
and
followed by substitution of a value of
t ≠− in x (t)
2
40
A1: CSO
x = −
Do not accept decimals here but remember to isw following a correct answer.
49
Ignore any references to the y coordinate. Penalise extra solutions, incorrect factorisation etc on this mark.
∫
∫
∫
ln
x
1
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