Mathematics · May / June 2022 · Paper 11 · Question 10
YMA01
As printed in the paper
The question as text
10. y C O x A B l Figure 5 Figure 5 shows a sketch of the curve C with equation y = 2 7 x3 + 1 7 x2 − 5 2 x + k where k is a constant.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
10(a)[2]
x =
x
2
+
x
−
d
7
7
2
7
d
y
6
7
2
2
7
5
10(b)[4]
(
)
At
x = −
,
=
−
+
−
−
=
...
=
7
2
d
x
7
2
7
2
2
So at B we know d
y
1
x = −
''
''
d
7
6
2
5
1
hence
x
2
x
+
−
= −
7
7
2
7
12
x
2
4
x
35
2
1
2
x
2
4
x
33
0
*
+
−
= −
+
−
=
(
)(
)
E.g.
12
x
2
+
4
x
−
33
=
0
2
x
−
3
6
x
+
11
=
0
x
=
...
10(c)[2]
From graph we can see the x coordinate is positive, so x = 3
at B
2
1 ''
10(d)[4]
''
1
Equation of l is
y
= −
x
−
7
1
7
1
7
Finds coordinates of A
x = −
y
=
''
−
''
×
−
−= −
1
2
7
2
2
7
1
2
1
5
Substitutes
x
= −
,
y
= −
''
''
into
y
=
x
3
+
x
2
−
x
+
k
k
=
...
2
2
7
7
2
5
k =
CSO
4
(a)
M1: Finds d
y
, look for at least two terms correct. They do not need to be simplified.
d
x
A1: Correct derivative, need not be simplified. ISW after a correct answer
(b) Marks cannot be retrospectively awarded from work in (d)
7
into their d
y
M1: Substitutes
−
to find the gradient of C at A
2
d
x
M1: Applies perpendicular condition to their gradient to find gradient at B
dM1: Equates d
y
to the gradient of the normal at B, depends on first M mark and a changed gradient.
d
x
6
2
5
A1*: Reaches the given equation with any correct intermediate line shown following
2
x
+
x
−
7
7
2
(c)
M1: Any valid method to solve the quadratic, factorisation, completing square, formula or calculator may be
used (implied by one correct answer). This may be awarded for work in (b)
3
A1: Correct coordinate
x =
given with reason. See scheme. The reason should reference the sketch, e.g.
B
2
11
E.g. cannot be
−
as that is negative, condone reasons like ''because B is positive''
6
y
= −
''
(d) Marks cannot be awarded from work in (b), but allow the transfer of answers. E.g.
Explanation of Main method:
Find equation for l, then find coordinates for A or B, then sub coordinates into equation for C to find k
M1: Uses their gradient of l and intercept −1 to form the equation of l.
The gradient must a result of a changed d
y
7
at
x = −
. It cannot be just made up
d
x
2
7
3
dM1: Finds the coordinates of either A or B using the equation for l and either
x = −
or
x =
A
2
B
2
7
1
7
1
3
1
3
17
FYI
x = −
y
=
''
−
''
×
−
−= −
1
and
x =
y
= −
'' 7
''
×
−=
1
−
2
7
2
2
2
2
1
4
ddM1: A full method to solve for k. This involves substituting the coordinates of A or B in the equation for
3
17 ''
2
1
5
x =
y
=
'' 1
−
curve C. E.g. See scheme but can also use
into
y
=
x
3
+
x
2
−
x
+
k
2
4
7
7
2
5
A1: CSO
k =
4
Explanation of Alt method:
Use the x coordinate for A or B in the equation for C to find the y coordinate for A or B in terms of k.
Then use the gradient and point A or B to form an equation for l in terms of k. Use the fact that the
intercept of l is 1 to form and solve an equation in k
Alt I
3
2
2
7
1
7
5
7
7
At A
y
=
−
+
−
−
−
+
k
=
...
=
k
−
(d)
7
2
7
2
2
2
4
7
1
7
"
" " "
"k
"
1
7
7
Equation of l is
y
−
k
−
= −
x
+
or y intercept is
−
×
+
−
4
7
2
7
2
4
1
9
9
y
= −
x
+
k
−
k
−
= −
1
k
=
...
7
4
4
5
k =
4
7
3
M1: Substitutes
x = −
or
x =
into the equation for C and finds the y coordinate in terms of k.
2
2
This cannot be scored if they substitute any value for y (except for the correct value which would mean that
we would be using the main method).
69
FYI at B the y coordinate is
k −
28
dM1: Uses their gradient for normal at A (or tangent at B) and their y coordinate to find an equation of the
69
1
3
line l or to find an expression for the intercept. For use of B expect
y
−
k
−
= −
x
−
28
7
2
ddM1: Sets their intercept to −1 and solves for k.
5
k =
A1:
4
Explanation of Alt II (d)
Find the equation for l (as in main method) but then equate with the equation for C. Use the fact that the
equation formed has a root of either 3/2 or 7/2 to set up and solve an equation in k.
1 ''
M1: For an attempt at the equation for l. Score for
y
= −
''
x
−
1
7
dM1: Equate equation for l with equation for C and use the fact that a root of this equation is known.
1
2
1
5
'
'
''
x
1
x
3
x
2
x
k
4
x
3
2
x
2
33
x
14
k
14
0
For example
−
−=
+
−
+
+
−
+
+
=
7
7
7
2
3
7
Set
g
±
=
0
or
g
±
=
0
where
g( )
x
=
4
x
3
+
2
x
2
−
33
x
+
14
k
+
14
to form an equation in k
2
2
3
=
7
g
0
k
...
g
0
k
...
ddM1: As above but sets
=
−
=
=
which must lead to a value for k
2
2
5
k =
A1:
4
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