Mathematics · Oct / Nov 2021 · Paper 11 · Question 10
YMA01
As printed in the paper
The question as text
10. A curve has equation y = f(x), x > 0 Given that 1 3 , where a is a constant • fʹ(x) = ax – 12x • fʹʹ(x) = 0 when x = 27 • the curve passes through the point (1, –8)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
10(a)[3]
1
2
−
f ( )
′
12
3
f ( )
′′
4
3
x
=
ax
−
x
⇒
x
=
a
−
x
1
4
Sets f (27)
′′
=
0
⇒
0
=
a
−×
4
⇒
a
=
9
9
10(b)[4]
1
4
1
(
)
2
f ( )
′
x
=
ax
−
12
x
3
⇒
f ( )
x
=
ax
−
9
x
3
+
c
2
Substitutes
x
=
1,f ( )
x
= −⇒
8
c
=
...
4
2
7
(
)
2
3
f ( )
x
=
x
−
9
x
+
9
9
Mark (a) and (b) together
Do not be too concerned with notation e.g. in (a) when they differentiate they may call it
f′(x) or e.g. d
y
x
d
(a)
2
−
B1
States or uses
f ( )
′′
x
=
a
−
4
x
3
which may be unsimplified
′′
M1
Sets their f (27)
=
0
and proceeds to a value for a.
2
−
3
It is dependent upon one correct term in f ( )
′′
x
e.g. a or
−
4x
(oe)
4
A1
a =
9
(b)
4
1
1
12
x
3
3
2
M1
Integrates
ax
−
12
x
with one term correct e.g.
ax
or
−
with the indices
2
4
3
processed.
1
4
1
(
)
2
f ( )
′
x
=
ax
−
12
x
3
⇒
f ( )
x
=
ax
−
9
x
3
+
c
A1ft
follow through on a or a numerical a
2
or a “made up” a but must include + c
4
4
2
x
3
12
x
9
−
+
c
Allow simplified or unsimplified so allow e.g.
2
4
3
dM1
Substitutes
x
=
1 and f ( )
x
= − to obtain a value for c. Must have numerical a now.
8
Depends on the first M mark.
4
2
7
Allow equivalent correct fractions for 2
7
(
)
2
3
A1
f ( )
x
=
x
−
9
x
+
.
,
9 or recurring
9
9
9
decimals
.
.
e.g. 0.2, 0.7
with clear dots over the 2 and 7.
4
3
4
3
Allow e.g.
x
for
x
Look for a correct expression so no need to see f(x) = … and isw if necessary.
4
11
Note that a fairly common error is to obtain
a = −
in part (a) leading to
c =
9
9
part (b)
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