Mathematics · May / June 2024 · Paper 11 · Question 7
YMA01
As printed in the paper
The question as text
7. The curve C has equation y = f(x) where 3 – kx 2 + 14x + 24 f(x) = 2x and k is a constant.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[3]
( )
( )
f
x
=
2
x
3
−
kx
2
+
14
x
+
24
f
′
x
=
6
x
2
−
2
kx
+
14
(
)
f
′′
( )
x
12
x
2
k
=
−
(ii)
( )
2
( )
( )
6
x
2
−
2
kx
+
14
=
12
x
−
2
k
6 5
−
2
k
5
+
14
=
12 5
−
2
k
k
=
...
k =
13
k
=
13
6
x
2
−
38
x
+
40
=
0
x
=
...
2
4
4
4
4
x
=
"
"
y
=
12 "
"
−×
2 13
or
y
=
6 "
"
−×
2 13
×
"
"
+
14
3
3
3
3
4 ,
x
=
y
= −
10
3
Notes:
In part (a), penalise the inclusion of “+ c” on either derivative only once on the first occurrence.
(a)(i)
M1: For x3 →…x2 or x2 → …x or 14x → 14
( )
A1: Correct simplified first derivative. The
"f
′
x
= is not required.
"
Do not allow x1 for x.
Isw can be applied once a correct simplied derivative is seen.
(ii)
A1ft: Correct simplified second derivative. Follow through their first derivative
( )
The
"f
′′
x
= is not required.
"
Isw can be applied once a correct simplied second derivative is seen. Must follow M1
(b)
M1: Sets f ′(x) = f ′′(x), substitutes x = 5 and solves a linear equation in k to find a value for k.
May substitute x = 5 into f ′(x) and f ′′(x) first, then equate and solve for k.
A1: k = 13
(c)
M1: Substitutes their k into the equated f ′(x) = f ′′(x) and solves the resulting 3TQ to obtain at least one value
for x by any means including a calculator.
M1: Substitutes their x (not 5) and their k into f ′(x) or f ′′(x) to find a value for y.
May be implied by their values so you may need to check.
A1: Correct coordinates. May be seen as x = .., y = … or as a coordinate pair but condone missing brackets
e.g. 4 ,
3 −
10.
Award this mark as soon as the correct values are seen.
Allow a “made up” k in (c).
7(b)[2]
( )
2
( )
( )
6
x
2
−
2
kx
+
14
=
12
x
−
2
k
6 5
−
2
k
5
+
14
=
12 5
−
2
k
k
=
...
k =
13
7(c)[3]
k
=
13
6
x
2
−
38
x
+
40
=
0
x
=
...
2
4
4
4
4
x
=
"
"
y
=
12 "
"
−×
2 13
or
y
=
6 "
"
−×
2 13
×
"
"
+
14
3
3
3
3
4 ,
x
=
y
= −
10
3
Notes:
In part (a), penalise the inclusion of “+ c” on either derivative only once on the first occurrence.
(a)(i)
M1: For x3 →…x2 or x2 → …x or 14x → 14
( )
A1: Correct simplified first derivative. The
"f
′
x
= is not required.
"
Do not allow x1 for x.
Isw can be applied once a correct simplied derivative is seen.
(ii)
A1ft: Correct simplified second derivative. Follow through their first derivative
( )
The
"f
′′
x
= is not required.
"
Isw can be applied once a correct simplied second derivative is seen. Must follow M1
(b)
M1: Sets f ′(x) = f ′′(x), substitutes x = 5 and solves a linear equation in k to find a value for k.
May substitute x = 5 into f ′(x) and f ′′(x) first, then equate and solve for k.
A1: k = 13
(c)
M1: Substitutes their k into the equated f ′(x) = f ′′(x) and solves the resulting 3TQ to obtain at least one value
for x by any means including a calculator.
M1: Substitutes their x (not 5) and their k into f ′(x) or f ′′(x) to find a value for y.
May be implied by their values so you may need to check.
A1: Correct coordinates. May be seen as x = .., y = … or as a coordinate pair but condone missing brackets
e.g. 4 ,
3 −
10.
Award this mark as soon as the correct values are seen.
Allow a “made up” k in (c).
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