Mathematics · May / June 2025 · Paper 31 · Question 9
YMA01
As printed in the paper
The question as text
9. y C B A x O Figure 2 Figure 2 shows a sketch of the curve C with equation π 2 sin 3 π < y < 3 π x y 3 4 12 4 The curve intersects the y-axis at the points A and B as shown.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
9(a)[3]
2
π
π
x
=
0
sin 3
y
+
=
0
3
y
+
=
π
, 2 ,...
π
y
=
...
3
4
4
π
7
π
y
=
or
y
=
4
12
π
7
π
y
=
and
y
=
4
12
Condone any confusion about which point is which and just look for the correct values.
Look for answers given in the body of the question or on Figure 2
M1: Correct strategy to find y at A or B.
π
π
Requires an attempt to solve either 3
y
+
=
π
or 3
y
+
=
2
π
4
4
π
May be implied by at least one correct value and may be seen embedded e.g.
0,
, 0,
4
π
7
π
Condone e.g.
, 0 ,
, 0
4
12
π
π
π
π
π
2
π
Allow equivalent work e.g.
3
y
0
y
y
,
+
=
= −
= −
+
−
+
4
12
12
3
12
3
May be implied by decimal answers e.g. awrt 0.79 or awrt 1.8
π
A1:
0,
One correct value. Allow simplified or unsimplified and may be seen embedded e.g.
4
1
π
π or e.g.
2
π
π
for 7
π
Condone unsimplified for this mark e.g.
y
=
π
−
for 4
y
=
−
3
4
3
12
12
π
7
π
0,
, 0,
A1:
Both correct and simplified and may be seen embedded e.g.
4
12
π
7
π
and condone unsimplified as long as they are single fractions e.g. 2
, 0 ,
, 0
Condone e.g.
4
12
Alternative:
2
π
2
π
2
π
3
π
7
π
sin 3
y
+
=
0
sin3 cos
y
+
cos3 sin
y
=
0
tan3
y
= −
1
3
y
=
,
3
4
3
4
3
4
4
4
3
π
7
π
and then A1A1 as above.
Score M1 for an attempt to solve either
3
y
=
or
3
y
=
4
4
9(b)[4]
2
π
d
x
2
π
π
x
=
sin 3
y
+
=
×
3cos 3
y
+
=
2cos 3
y
+
3
4
d
y
3
4
4
2
d
y
1
d
y
1
1
=
=
=
d
x
π
d
x
π
π
2cos 3
y
+
4cos
2
3
y
+
4 1 sin
−
2
3
y
+
4
4
4
1
=
2
3
x
4 1
−
2
1
=
4
−
9x
2
Correct d
y or d
x
y
B1:
x in any form.
d
d
π
Do not condone missing brackets e.g. 2cos3
y
+
unless recovered or implied by subsequent work.
4
M1: For any 2 of the following applied to their d
x
:
d
y
finds reciprocal (may have already happened if they find dy
x )
d
squaresπ
π
appliescos
2
3
y
1 sin
2
3
y
+
= −
+
4
4
Note that this mark (and the next one) may be implied by equivalent work e.g.
2
π
π
3
x
π
4
−
9
x
2
x
=
sin 3
y
+
sin 3
y
+
=
cos 3
y
+
=
3
4
4
2
4
2
d
x
π
d
y
2
1
=
2cos 3
y
+
=
4
−
9
x
2
=
d
y
4
d
x
4
−
9
x
2
d
2
y
dM1: Uses all 3 of the above correctly to obtain
in terms of x.
d
x
A1:
Correct answer.
1
You can condone e.g. missing trailing brackets within their working e.g.
π
4(1 sin
−
2
3
y
+
4
Special case: allow this mark if the only error in their solution is to obtain d
x
2
3cos 3
= −
×
d
y
3
(b) Way 2
2
π
π
3
x
x
=
sin 3
y
+
3
y
+
=
sin
−
1
3
4
4
2
π
3
x
d
y
1
3
3
y
+
=
sin
−
1
3
=
×
4
2
d
x
2
9
x
2
1
−
4
2
2
d
y
1
3
d
y
1
3 d
=
×
=
x
9
x
2
2
d
x
9
x
2
1
−
2 1
−
4
4
1
=
4
−
9x
2
Way 2:
1 3
x
B1:
Correct equation in any form connecting y and
sin
−
.
2
d
y
k
or equivalent. Condone missing brackets around 3
M1: Differentiates to obtain
=
d
x
3
x
2
1
−
2
2
d
y
dM1: Squares and makes
the subject.
d
x
A1:
Correct answer.
9(c)[4]
1
Identifies that one of the gradients is "4" or that one of the gradients is
"4"
or
Identifies that one of the gradients is 2 or that one of the gradients is 1
2
π
7
π
1
At A:
y
−
''
''
=
''2''
x
and At B:
y
−
"
"
=
''
''
x
4
12
2
π
1
7
π
''2''
x
+
''
''
=
''
''
x
+
''
''
x
=
...
4
2
12
2
π
x =
9
1
Note that in this part the question says to use the answer to part (b) i.e. their
p
−
qx
1
If they use their
then the ft is available as shown.
p
−
qx
2
If candidates “start again” or e.g. use their d
x
or d
y
in terms of y with their y values from part (a) then
d
y
d
x
the value(s) must be correct.
B1ft: Using the answer to part (b):
p or 1
Award this mark for one gradient identified as
where p is their value from part (b).
p
or not using the answer to part (b):
One gradient identified as 2 or 1
2
M1: Correct straight line method for the equation of the normal at A and the tangent at B.
They must have
2 values for y from part (a) (ignore how these were labelled)
both gradients positive
For the normal at A this requires a gradient of
p or 2 with x = 0 and their smaller value of y from part (a)
If they use y = mx + c they must proceed as far as c = …
1
or 1
For the tangent at B this requires a gradient of
with x = 0 and their larger value of y from part (a)
p
2
If they use y = mx + c they must proceed as far as c = …
dM1: Correct method for both lines as defined above and solves simultaneously (may be via a calculator) to
find the x coordinate of D.
Depends on the previous mark.
2
π
A1:
Cso. Correct value
x
=
.
9
Ignore any attempts to find the y coordinate.
If they use part (b) then full marks can be awarded following an incorrect value for q provided all the
other work is correct.
Beware: It is possible to obtain the correct answer in (c) from incorrect values in (a) e.g.
This scores B1M1dM1A0
Diagram for reference:
π
y
=
2
x
+
4
1
7
π
y
=
x
+
2
12
D
B
A
2
π
9
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