Mathematics · May / June 2024 · Paper 21 · Question 8
YMA01
As printed in the paper
The question as text
8. 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
8(i)[5]
cos
x
cos
x
(
)
5sin
2
13cos
cos
2
5 1 cos
2
13cos
cos
2
x
+
x
=
x
−
x
+
x
=
x
6cos
2
13cos
5
0
x
−
x
−
=
1
(
)(
)
3cos
x
+
1
2cos
x
−
5
=
0
cos
x
= −
3
=
1.91
x
5
20
=
10 12sin(6
+
+
18)
°
sin(6
+
18)
°=
8(ii)(a)[4]
k
k
6
(6
k
+
18)
=
56.4,123.6
=
6.41,17.59
k
(b) 22 C
°
Time of day = 11:14
8(ii)(b)[1]
cos
x
cos
x
(
)
5sin
2
13cos
cos
2
5 1 cos
2
13cos
cos
2
x
+
x
=
x
−
x
+
x
=
x
6cos
2
13cos
5
0
x
−
x
−
=
1
(
)(
)
3cos
x
+
1
2cos
x
−
5
=
0
cos
x
= −
3
=
1.91
x
5
20
=
10 12sin(6
+
+
18)
°
sin(6
+
18)
°=
k
k
6
(6
k
+
18)
=
56.4,123.6
=
6.41,17.59
k
(b) 22 C
°
Time of day = 11:14
''6.41''
t
+
18
=
90
=
t
11.23
(i)
sin
sin
x
x
B1: States or uses
tan
x
=
E.g.
5sin
x
×
+
13
=
cos
x
cos
cos
x
x
sin
x
M1: Attempts to use
tan
=
,
sin
2
+
cos
2
= and multiply bycos x to form a quadratic equation in
1
x
x
x
cos
x
cos x (allow if there are slips in coefficients but the trig terms must be correct)
2
(
)
A1: Correct simplified quadratic
6cos
x
−
13cos
x
−
5
=
0
The “=0” may be implied.
M1: Solves a 3TQ in cos x leading to at least one value for cos x
1
A1: awrt
x =
1.91
(following
cos
x = −
) and no other values in the range.
3
(
)
Note: Answers only with no working score no marks. From
6cos
2
−
13cos
−
5
=
0
to 1.91 directly, score
x
x
final M0A0.
(ii) (a)
M1: Attempts to use the given information and proceeds to sin(6
+
18)
°= , or may be implied by
k
c
B
sin(6
18
)
6
1
8
arcsin
(evaluated) (which scores M1dM1)
A
k
+
°=
B →
k
+
=
A
dM1: Takes arcsin leading to a value for 6
k +
18
. Accept radian values seen for this mark. (0.985, 2.16).
Allow awrt 2 s.f. answers for evidence.
A1: One value for k: 6.41or17.59 Must be in degrees.
SC Allow for 6.4 and 17.6 both given if no more accurate answers are stated.
A1: awrt
6.41,17.59
and no other values
k =
Note answers only scores no marks. – the requirements of the M marks must be seen to be able to score them.
(ii)(b)
B1: cao 22 C
° Condone just 22.
(ii)(c)
M1: Sets their ''6.41''
18
90
...
May be implied by 11.23.
t
+
=
=
t
= π is M0.
Note ''6.41''
t +
18
2
A1: cao Time of day = 11:14 o.e. (e.g. 11h 14m is acceptable).
3
3
5
8(ii)(c)[2]
''6.41''
t
+
18
=
90
=
t
11.23
(i)
sin
sin
x
x
B1: States or uses
tan
x
=
E.g.
5sin
x
×
+
13
=
cos
x
cos
cos
x
x
sin
x
M1: Attempts to use
tan
=
,
sin
2
+
cos
2
= and multiply bycos x to form a quadratic equation in
1
x
x
x
cos
x
cos x (allow if there are slips in coefficients but the trig terms must be correct)
2
(
)
A1: Correct simplified quadratic
6cos
x
−
13cos
x
−
5
=
0
The “=0” may be implied.
M1: Solves a 3TQ in cos x leading to at least one value for cos x
1
A1: awrt
x =
1.91
(following
cos
x = −
) and no other values in the range.
3
(
)
Note: Answers only with no working score no marks. From
6cos
2
−
13cos
−
5
=
0
to 1.91 directly, score
x
x
final M0A0.
(ii) (a)
M1: Attempts to use the given information and proceeds to sin(6
+
18)
°= , or may be implied by
k
c
B
sin(6
18
)
6
1
8
arcsin
(evaluated) (which scores M1dM1)
A
k
+
°=
B →
k
+
=
A
dM1: Takes arcsin leading to a value for 6
k +
18
. Accept radian values seen for this mark. (0.985, 2.16).
Allow awrt 2 s.f. answers for evidence.
A1: One value for k: 6.41or17.59 Must be in degrees.
SC Allow for 6.4 and 17.6 both given if no more accurate answers are stated.
A1: awrt
6.41,17.59
and no other values
k =
Note answers only scores no marks. – the requirements of the M marks must be seen to be able to score them.
(ii)(b)
B1: cao 22 C
° Condone just 22.
(ii)(c)
M1: Sets their ''6.41''
18
90
...
May be implied by 11.23.
t
+
=
=
t
= π is M0.
Note ''6.41''
t +
18
2
A1: cao Time of day = 11:14 o.e. (e.g. 11h 14m is acceptable).
3
3
5
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