Mathematics · Oct / Nov 2020 · Paper 11 · Question 9
YMA01
As printed in the paper
The question as text
9. A curve with equation y = f(x) passes through the point (9, 10). Given that 3x 2 – 21 − 5 x f′(x) = 27x x > 0 2 x find f(x), fully simplifying each term. (6) TOTAL FOR PAPER IS 75 MARKS END
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21
x
3
−
5
x
21
x
3
−
5
x
5
...
or
...
1
=
α
x
+
=
+
β
x
2
2
2
x
2
x
27
21
2
5
2
5
f
( )
3
7
3
(
)
9
3
3
7
3
(
)
x
=
x
−
×
x
+
×
x
+
c
=
x
−
x
+
x
+
c
2
2
2
2
3
2
7
2
3
3
5
( )
⇒( )
3
( )
7
( )
3
f 9
=
10
9 9
−
3 9
+
9
+
c
=
10
⇒
c
=
...
2
2
3
5
(
( )
)
f
x
=
9
x
3
−
3
x
7
+
x
3
−
35
2
2
3
Notes
On EPEN it is B1M1A1A1dM1A1. We are marking this as M1M1A1A1dM1A1
M1 Uses correct index laws to obtain at least one correct index from splitting the fraction.
Award for
α
x
5
+
...
or
...
+
β
x
1
2
2
x
n
→
x +
n
1
M1
correctly seen on one term (usually the
27
x
2
→
...
x
3
). The indices do not need to be
processed. This mark can also be awarded for integrating terms from incorrect attempts to split
21
x
3
−
5
x
7
3
7
9
3
5
the fraction. Eg
=
42
x
2
−
10
x
2
42
x
2
→
...
x
2
or
−
10
x
2
→
...
x
2
1
2
x
2
This mark cannot be awarded for only seeing:
21
x
3
−
5
x
...
x
4
±
...
x
2
→
1
3
2
...
x
2
x
2
A1
Two correct terms simplified or unsimplified, but the indices must have been processed.
A1
All correct simplified or unsimplified (+ c not required)
dM1 Uses f (9) = 10 and attempts to find c. Do not be too concerned by the mechanics of their
arrangement. It is dependent on the previous method mark.
5
3
7
3
A1
9
x
−
3
x
+
x
−
35
All correct and simplified. Accept other simplified equivalent expressions
2
2
3
1 (27
7
3
for f ( )x such as
x
3
−
9
x
+
5
x
−
105)
2
2
3
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