Mathematics · May / June 2021 · Paper 31 · Question 9
YMA01
As printed in the paper
Show the answer and mark scheme
9(a)[3]
R =
13
5
tan
α
=
⇒
α
=
awrt 0.395
12
9(b)[3]
π
g( )
θ
=
10 13sin 2
+
θ
−
−
0.395
6
(i)
(i) Minimum value is 3
−
(ii)
π
3
π
2
θ
−
−
0.395
=
⇒
θ
=
awrt 2.82
6
2
9(c)[2]
(
)
h( )
β
=
10 169sin
−
2
β
−
0.395
−
159
h
10
(a)
B1:
R =
13
(
R = ±
13
is B0)
5
12
M1:
tan
α
= ±
, tan
α
= ±
⇒
α
=
...
12
5
5
12
If R is used to find α accept
sin
α
= ±
or cos
α
= ±
⇒
α
=
...
R
R
A1:
α =
awrt 0.395
Note that the degree equivalent
α =
awrt 22.6
° is A0
(b)(i)
B1ft: States the value of 10
−
R
following through their R.
(b)(ii)
π
3
π
M1: Attempts to solve
2
"0.395"
...
θ
−
±
=
⇒
θ
=
6
2
A1:
. No other values should be given
θ =
awrt 2.82
(c)
M1: Achieves one of the end values, either 159
(or 10 – (their R)2 evaluated) or 10
−
A1: Fully correct range 159
( )
( )
−
h
10,
−
159
h
β
10,
−
159
range
10,
−
159
h
x
or equivalent correct ranges.
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