Mathematics · May / June 2023 · Paper 31 · Question 5
YMA01
As printed in the paper
The question as text
5. 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
5(i)[3]
3
3sec
x +
2
=
0
cos
x
= −
x
=
...
2
x = π
5
6
cos2
1 2sin
2
5(ii)[4]
Attempts to use
θ =
−
θ
2
6sin
θ+
10sin
θ −
3
=
0
θ = −
5
±
43
s
in
(
1.926...,0
.2595.
.
.
)
arcsi (..
n
.
=
−
θ=
6
=15.0 , 165
θ
°
°
Notes
(i)
B1
States
x =
2
. May be seen anywhere in (i) and don’t be concerned where it comes from.
1
3
M1
For a correct process to solve
3sec
x +
2
=
0
E.g.
sec
x
=
cos
x
= −
cos
x
2
slips in rearranging but must attempt to solve
cos
x
=
k
, |
k
| 1
< or sec
x
=
k
, |
k
| 1
150°) following a correct equation implies the M mark. Note some may use
sec
form a quadratic in tan x. These will need a correct identity, correct method to solve a quadratic
(which may be by calculator) and attempt to solve
tan
x
=
k k
,
≠
0
x = π
5
A1
and no other extra solutions in the range. Accept awrt 2.62 (and isw).
6
5
π
Note that
3
sec
x +
2
=
0
→
x
=
can score M1A1 as no incorrect work is seen, method implied.
6
x = π
5
Question required working to be shown
without seeing at least
3sec
x +
2
=
0
6
M0A0.
(ii)
2
M1
Attempts to use
cos2
θ =
±±
1
2sin
θ
to form a quadratic equation in sinθ. If using alternative
forms for the identity, must also use
c
os
2
θ
= −
1
s
in
2
θ
before gaining this mark.
2
A1
Correct 3 term quadratic equation
6sin
θ+
10sin
θ −=
3
0
or a multiple of this. Alternatively may
2
be scored for
6sin
θ+
10sin
θ =
3
if followed by completing the square on LHS to solve.
M1
Full attempt to find one value for θ from a quadratic in sinθ. Must involved
correct method to solve the quadratic in sinθ (usual rules, may use calculator) to produce avalue for sinθ
use of arcsin(..) to reach the value for θ (you may need to check the values if arcsin(…) is notshown). Radian answers can imply the mark (awrt 0.263, 2.88 if correct).
May be scored from an incorrect identity as long as a quadratic is achieved. Accept arcsin
expression for the M
A1
θ
=awrt15.0 , 165
°
°and no other solutions in the range. Accept just 15° for 15.0° (but not awrt
15° if it does not round to 15.0°)
Condone a different variable used than θ throughout.
(
)
2,
−
10
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