Mathematics · May / June 2022 · Paper 31 · Question 7
YMA01
As printed in the paper
The question as text
7. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[4]
13
(
)
3cot 2
2
θ
−
7
=
2sin
θ
3(cosec 2
2
θ
−
1)
−
7
=
13sec
θ
or
oe
2sin
θ
cos
θ
13
2
3(cosec 2
θ
−
1)
−
7
=
2sin
θ
cos
θ
3cosec 2
2
θ
−
10
=
13cosec2
θ
3cosec 2
2
θ
−
13cosec
2
θ
−
10
=
0
*
7(b)[4]
3cosec 2
2
θ
−
13cosec2
θ
−
10
=
0
1
(
)
(
)
(3cosec2
θ
+
2)(cosec2
θ
−
5)
=
0
cosec2
θ
=
5
sin 2
θ
=
5
−
1
sin
1
" "
5
θ
=
=
...
2
(
) awrt 0.101, 1.47
θ=
Alt(a)
3(1 sin 2 )
−
2
θ
2sin
7
13sec
θ
−
=
θ
2
sin 2
θ
3(1 sin 2 )
−
2
θ
13
−
7
=
sin 2
2
θ
2sin
θ
cos
θ
3
13
−
10
=
sin 2
2
θ
sin 2
θ
3cosec 2
2
13cosec2
10
0
θ
−
θ
−
=
*
(a)
M1
Attempts EITHER
1
1
to divide both sides by ( )
2 sinθand use
sec
θ
=
cosec
θ
=
(or
) to achieve a term of
cos
sin
θ
θ
A
...
the form
oe (implied by
or …cosec sec
θ
θ).
B
sin
θ
cos
θ
...sin 2θ
sin2 (3cot 2
θ
2
θ−
7)
=
13
does not score this mark. They have not divided by (2)sinθ
Note
1 cot 2
2
cosec 2
2
cot 2θ term in the equation or in the alternative
2
to use
±±
θ
= ±
θ
to replace the
cos 2
2
θ
cot 2
2
θ
=
±±
1 sin 2
2
θ
= ±
cos 2
2
θ
method it is for using
and using
to replace the
sin 2
2
θ
Condone arithmetical slips and invisible brackets.
dM1
Attempts BOTH
1
1
to divide both sides by ( )
2 sinθand use
sec
cosec
θ
=
(or
θ
=
) to achieve a term of the
cos
θ
sin
θ
A
...
form
oe (implied by
or …cosec sec
θ
θ). May be implied
B
sin
θ
cos
θ
...sin 2θ
to use
±±
1 cot 2
2
θ
= ±
cosec 2
2
θ
to replace the
cot 2θ term in the equation or in the alternative
2
cos 2
2
θ
method it is for using
cot 2
2
θ
=
and using
±±
1 sin 2
2
θ
= ±
cos 2
2
θ
to replace the
sin 2
2
θ
Condone arithmetical slips and invisible brackets. It is dependent on the first mark.
ddM1 A full attempt to achieve a 3TQ in cosec 2θ only. It is dependent on both of the previous method
marks. They must
1
use the identitysin 2
θ
=
2sin
θ
cos
θ
(or
cosec2
θ
=
cosec sec
θ
θ
) and proceed to a cosec2θ
2
term
multiply out the brackets and collect terms on one side to form a 3TQ in cosec 2θ only.
Do not be concerned with the mechanics of their arrangement and they may proceed to the printed
answer at this stage. Condone arithmetical slips and invisible brackets.
A1*
Achieves
3cosec 2
2
θ
−
13cosec2
θ
−
10
=
0
with no arithmetical errors in the main body of
their solution. You must see either sin
cos
θ or sin 2θ within their working.
θ
Allow recovery of invisible brackets in their solution.
Withhold the final mark for poor notation such as
cosec 2θ being written as
2
cosec2θ or if
2
2 is missing from 2θ
They may work in a different variable, but the final answer must be in terms of 2θ
*Be aware that answers with no intermediate working seen is not acceptable*
(b)
M1
Attempts to solve the quadratic and finds the reciprocal of one of their roots (usual rules apply for
solving a quadratic). Do not be concerned by their labelling.
1
3
2
Look for eg
....
=
5
→
or eg
....
= −
→−
5
2
3
) 1
3
) 1
(
A1
sin 2
θ=
(Ignore any reference to
−
on its own is sufficient. Condone eg
5
2
5
1
1
1
sin 2
2
sin
X =
sin
θ=
or
or
θ=
5
5
5
Correct method to find a value for θby finding arcsin(their 1
dM1
) and dividing by 2.
5
May be implied by awrt 0.10 or awrt 1.5 (or in degrees awrt 5.8 or awrt 84).
It is dependent on seeing either the quadratic solved or their root from the quadratic (either
1 or
−
1
) or the reciprocal of the root being stated first.
It cannot be scored for solving eg
sin 2
2
θ=
...
(
) awrt 0.101, awrt1.47
θ=
A1
and no other angles within the given interval.
Some intermediate working must be shown.
Examples of minimum working shown in (b):
d
v
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