Mathematics · Oct / Nov 2021 · Paper 21 · Question 10
YMA01
As printed in the paper
The question as text
10. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. π < x < 2 π
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
10(i)[5]
4
4
3
3
π
⇒
x
=
...
−
÷
2
4
π
11
π
5
π
7
π
⇒
x
=
,
−
;
,
−
24
24
24
24
(
)
2
10(ii)[5]
2sin
θ
−
cos
θ
= ⇒
1
4sin
2
θ
−
4sin
θ
cos +cos
θ
2
θ
=1
sin
2
θ
+cos
2
θ
=1
⇒
4sin
2
θ
−
4sin
θ
cos
θ
=sin
2
θ
Attempts to use
(
)
⇒
sin
θ
3sin
θ
−
4cos
θ
=0
4
tan
θ
=
3
θ
=
53.1 ,233.1 ,180
°
°
°
(i)
M1: Correct order of operations. Takes square root followed by arctan.
π
π , 2
π
π
Implied by 2
x +
=
x +
=
1.047
. Condone for this mark only 2
x +
=
4
3
4
4
π
Allow if they use θ for 2
x
+
4
π
sin
2
2
x
+
4
π
π
=
3
⇒
sin
2
2
x
+
=
3cos
2
2
x
+
A longer method is to do
π
4
4
2
cos
2
x
+
4
sin
2
θ
+cos
2
θ
=1
use
to produce an equation in either sin or cos, before taking the
inverse. They must have a correct method up to slips in rearranging, and reach the stage
of taking arcsin or arccos in order to score the M.
dM1: Complete attempt to find one value for x.
π before dividing by 2. Condone
This would involve an attempt to move the 4
π
x
1.047
2
=
±
÷
4
π
11
π
A1:
One value of
,
−
. Condone decimals here awrt 0.13, −1.44
24
24
5
π
7
π
A1:
One value of
,
−
. Condone decimals here awrt 0.65, −0.92
24
24
π
5
π
7
π
11
π and no other values in the range
A1:
,
,
−
,
−
24 24
24
24
(
)
2
2sin
θ
−
cos
θ
=
1
⇒
2sin
θ −
cos
θ = ±
1
10(ii)
⇒
2sin
θ =
cos
θ ±
1
Alt
2
2
⇒
4sin
θ =
cos
θ ±
2cos
θ +
1
2
2
2
then attempts to use
sin
θ
+cos
θ
=1
⇒
4
−
4cos
2
θ =
co
s
±
2c
os
θ +
1
2
⇒
5cos θ ±
2cos
θ −
3
=
0
⇒
cos
θ =
...
3
3
cos
θ
=
, 1 or
−
co
s
θ = −
,
1
5
5
θ
=
53.1 ,233.1 ,180
°
°
°
M1: Attempts at least one of
2sinθ
−
cos
θ
= ± , rearranges and squares then attempts to use
1
sin
2
θ
+cos
2
θ
=1
to produce an equation in just cosθ or just sinθ . FYI in sinθ it is
2
2
4s n
i
θ
±
4sin
θ
+ = −
1
1 s
i
n
θ
dM1: Solves their quadratic in cosθ or sinθ to obtain at least one value for cosθ or sinθ
4
4
One correct pair of solutions. FYI in sinθ it is
sin
θ
=
0,
or s
in
θ
=
0,
−
A1:
5
5
θ
=
53.1 ,233.1
°
°
A1:
and no others in the range. (Note via this method some extra solutions
in the range are likely)
B1:
θ
=
180
°
Award when seen and allow however it arises.
It is possible someone could have studied WMA13.
(
)
M1: Attempts to write 2sin
θ
−
cos
θ in the form
R
sin
θ
−α
o.e. FYI
(
)
2sin
θ
−
cos = 5 sin
θ
θ
−
26.6
°
1
(
)
2
(
)
sin
θ
−α = ±
dM1: Proceeds from
R
sin
2
θ
−α =
1
to
R
1
(
)
sin
θ
−
26.6
° = ±
A1:
5
A1:
θ
=
53.1 ,233.1
°
°
and no others in the range
(Note: 0 and 360° are outside the range so ignore if given as solutions.)
B1:
θ
=
180
°
Award when seen and allow however it arises.
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