Mathematics · Oct / Nov 2024 · Paper 11 · Question 8
YMA01
As printed in the paper
The question as text
8. A curve C has equation y = f(x). The point P with x coordinate 3 lies on C Given • f′(x) = 4x 2 + kx + 3 where k is a constant 1 the normal to C at P has equation y x • 5 24
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
8(a)[3]
2
Solves f (3)
′
=
24
to find k. E.g
4 3
×
+
k
×
3
+
3
=
24
k
=
..
.
24
−
39
3
k
+
39
=
24
k
=
= − *
5
3
4
5
2
3
2
f ( )
′
x
=
4
x
−
5
x
+
3
f( )
x
=
x
−
x
+
3
x
+
c
8(b)[4]
3
2
3
Substitutes
x
=
3,
y
= −
+
5
into
y
=
f( )
x
to find ''c'' dM1
24
4
5
141
3
2
f( )
x
=
x
−
x
+
3
x
−
3
2
8
Mark this question as a whole, some may try finding f(x) first in part (a), so award positively.
(a)
B1: States or implies that gradient of tangent (to the curve at P) is 24. This may occur
1
within the use of the perpendicular condition, e.g.
f (3)
′
= − implies B1.
1
×−
24
1
M1: Attempts to solve
f (3)
′
= ±
24
or 2
(or equivalent equations) to find k. Accept as
4
minimum at least one unsimplified equation in k before the given answer is stated.
Alternatively, allow for substitution of k = 5
− and x = 3 into f’(x) and simplifying to a
value to compare with the gradient (verification method).
A1*: Shows that
k = − with at least one correct equation with simplified terms (not
5
necessarily all gathered) and no incorrect work shown before stating the printed answer.
By verification all steps must be correct with conclusion given.
(b)
3
2
M1: Integrates to a form
px
+
qx
+
rx
, with at least
,p q ≠
0
with no requirement for + c.
Allow with k used in place of −5 for this mark.
4
5
3
2
A1:
f( )
x
=
x
−
x
+
3
x
with no requirement for + c
3
2
3
dM1: Substitutes
x
=
3,
y
= −
"
+
5"
into
y
=
f( )
x
and finds a value for ''c’'. Must have had
24
a constant of integration.
A1: (
)
4
5
141
3
2
f( )
x
=
x
−
x
+
3
x
−
Accept with y =… instead of f(x), or even with no label
3
2
8
at all.
x
…
−
5
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