Mathematics · Oct / Nov 2021 · Paper 11 · Question 2
YMA01
As printed in the paper
The question as text
2. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. A curve has equation 5 + 4x 3 – x + 5 y = 3x The points P and Q lie on the curve. The gradient of the curve at both point P and point Q is 2 Find the x coordinates of P and Q. (5)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
d
y
y
=
3
x
5
+
4
x
3
−
x
+
5
⇒
=
15
x
4
+
12
x
2
−
1
d
x
15
x
4
+
12
x
2
−=
1
2
⇒
15
x
4
+
12
x
2
−
3
=
0
2
2
⇒
3(5
x
−
1)(
x
+
1)
=
0
o.e
1
⇒
x
= ±
o.e.
5
M1
Attempts to differentiate with
x
n
→
x
n
−
1
for one correct power
x
5
x
4
or
x
3
x
2
or
x
1
Allow for
→
→
→
d
y
A1
=
15
x
4
+
12
x
2
−
1
which may be left unsimplified. Just look for a correct
d
x
expression
i.e. no need to see d
y
x =
...
d
Sets their d
y
dM1
x =
2
and collects terms to one side to obtain a 3TQ in
2x
d
Depends on first method mark.
ddM1 Correct attempt to solve 3TQ in
2x . This may be by factorising, using the quadratic
formula, or completing the square (see general guidance). The attempt to factorise must
be consistent with their 3TQ.
The correct quadratic with the correct answers just written down scores M0
Must be solving for x2 not x to obtain at least one value for x2.
Depends on both previous method marks.
1
5
3
A1
x = ±
or exact equivalent such as
±
,
±
and isw once the correct answers
5
5
15
are seen.
1
Must see both values so
x =
is A0.
5
Ignore any attempts to find the y coordinates.
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