Mathematics · Oct / Nov 2023 · Paper 21 · Question 3
YMA01
As printed in the paper
The question as text
3. 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
3(a)[5]
States or uses
tan
θ
=
→
2
+
3sin
θ
=
0
cos
cos
θ
θ
(
)
sin
θ
2
+
3cos
θ
=
0
2
+
3cos
θ
=
0
2
cos
=
= awrt 132 or awrt 228
θ
−
θ
°
°
3
2
cos
=
= awrt 131 8 and awrt 228 2
.
.
A1
θ
−
θ
°
°
3
sin
θ
=
0
θ
=
180 360
°
,
°
3(b)[2]
Sets 2
x
+
40
°=
their131.8
°
x
=
...
x =
awrt 45.9
°
(a) Condone a lack of a degrees symbol throughout.
Condone solutions that drop θ’s in parts of the working
sin
sin
θ
θ
M1: States
tan
θ
=
or attempts to use
tan
θ
=
in the given equation. Condone slips in coefficients.
cos
θ
cos
θ
sin
θ
dM1: Attempts to use
tan
θ
=
o.e., cross multiplies and either factorises or cancels to form a linear
cos
θ
equation in cosθ
2
A1: Achieves
cos
θ
=
−
and finds one correct θ value to the nearest degree.
3
Condone radian solutions for this mark so allow awrt 2.3(0) or awrt 3.98 following
cos
2
A1: Achieves
cos
θ
=
−
and finds both θ values in degrees correct to one dp.
3
2
Any extra values in the range (for
cos
θ
=
−
) is A0
3
B1: Achieves sin
θ
=
0
followed by
θ
=
180 360
°
,
°
Condone 0 appearing as a solution
(b) Condone a lack of a degrees symbol
M1: Sets 2
x +
40
°=
their131.8
°and proceeds to a value for x.
''131.8 '' 40
°±
°
Look for correct order of operations so
2
+
40
°=
''131.8 ''
°
=
x
x
2
There may be multiple attempts, which can be ignored for this mark
Follow through on their smallest value (must be greater than 40°), from part (a)
A1:
x =
awrt 45.9
°following a correct equation. Do not accept multiple solutions here
........................................................................................................................................................................
Note that there are many parallel solutions in part (a) that can be marked similarly.
Alt I
Using sin
θ
=
tan cos
θ
θ o.e
(
)
2tan
3tan cos
0
tan
2
3cos
0
θ
+
θ
θ
=
θ
+
θ
=
In this case the B1 mark is for achieving tan
θ
=
0
θ
=
180 360
°
,
°
Condone 0 appearing as a solution.
Alt II
Multiplying throughout by sinθ and using
2
2
eventually leads to a cubic cosθwhich has
sin
θ
+
cos
θ
=
1
2
roots of
and cos
θ
=1
. Allow use of a calculator to solve cubic equations, etc.
cos
θ
=
−
3
(
)
1
cos
2
θ
sin
sin
sin
2
θ
×
θ
θ
(
)
2
+
3sin
θ
=
0
→
2
+
3sin
2
θ
=
0
→
2
+
3 1
cos
2
θ
=
0
cos
θ
cos
θ
cos
θ
2
→
3cos
3
θ
+
2cos
2
θ
−
3cos
θ
−
2
=
0
→
cos
θ
= −
,
cos
θ
= ±
1
3
sin
θ
M1: For
tan
θ
=
cos
θ
dM1: for cos
θ
=
,
<
1
,
≠
0
k
k
k
2
A1: For
cos
θ
= −
and a solution as in the main scheme
3
2
A1: : For
cos
θ
= −
and both solutions as in the main scheme
3
B1 Achieves cos
θ= ±followed by
180 360
,
. Condone 0 appearing as a solution
1
θ
=
°
°
Alt III
Multiplying throughout by cosθ and using
sin
2
θ
+
cos
2
θ
=
1
eventually leads to a quartic in sinθwhich has
5
roots
sin
θ
= ±
and sin
θ
=
0
3
SQUARING
2sin
θ
= −
3sin cos
θ
θ
→
4sin
2
θ
=
9sin
2
θ
cos
2
θ
(
)
4sin
2
9sin
2
1 sin
2
9sin
4
5sin
2
0
→
θ
=
θ
−
θ
→
θ
−
θ
=
5
→
sin
2
θ
=
,
sin
θ
=
0
9
sin
θ
M1: For
tan
θ
=
cos
θ
dM1: For
sin
2
θ
=
or sin
θ
=
,
<
1
,
≠
0
k
k
k
k
5
5
5
A1: For
sin
θ
=
,
sin
θ
= −
or
sin
2
θ
=
and a correct solution as in the main scheme
3
3
9
5
5
5
2
A1: For
sin
θ
=
,
sin
θ
= −
or
sin
θ
=
and both solutions (only) in the main scheme
3
3
9
B1: Achieves sin
θ
=
0
followed by
θ
=
180 360
°
,
°
Condone 0 appearing as a solution
............................................................................................................................................................................
The solution must be obtained from correct working.
Calculator solutions like 2tan
θ
+
3sin
θ
=
0
θ
=131 8 and 228 2
.
°
.
°
score no marks in (a)
Even the solution
180 360
,
cannot score the B1 mark if it is not preceded by sin
θ
=
0
or tan
θ
=
°
°
(b) is hence or otherwise so must follow their solution in (a) or be a restart.
3
2
f( )
x
=
4
x
+
ax
−
29
x
+
b
1
1
3
1
2
1
f
−
=
0
→
4
29
0
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