Mathematics YMA01 Oct / Nov 2023 Paper 21 Question 10

10(i) 3 + 2x 6 5 ( ) 1 6 4 ( ) 2 6 2 ( ) 4 Any correct form for any term of C 3 2x or C 3 2x or C 3 2x M1 1 2 4 5 4 2 Any two correct: t(2) = 6 3 ( 2x ) 1 , t(3) = 15 3 ( 2x ) 2 t(5) = 15 3 ( 2x ) 4 A1 × × × × × × All 3 terms correct 2nd term = 2916x, 3rd term= 4860x2 , 5th term=2160x4 A1 4 2 2160…

Mathematics · Oct / Nov 2023 · Paper 21 · Question 10

YMA01

As printed in the paper

Show the answer and mark scheme

10(i)[6]
3
+
2x
6
5
(
)
1
6
4
(
)
2
6
2
(
)
4
Any correct form for any term of
C 3
2x or
C 3
2x
or
C 3
2x
M1
1
2
4
5
4
2
Any two correct: t(2) =
6 3
(
2x
)
1
, t(3) =
15 3
(
2x
)
2
t(5) =
15 3
(
2x
)
4
A1
×
×
×
×
×
×
All 3 terms correct 2nd term = 2916x, 3rd term= 4860x2 , 5th term=2160x4 A1
4
2
2160
4860
x
x
Method using consecutive terms of a GP e.g
=
2
2916
x
4860
x
...
=
...
x
15
x
=
4
2
a
8
sin
θ
(ii) (a) Attempts to use
∞=
=
θ
S
1
5
1 2cos
−
r
−
8
1 cos
−
2
θ
= −
θ
5
1 2cos
2
−
8 16cos
θ
= −
5 5cos
θ
5cos
2
θ−
16cos
θ
+ =
3
0
1
(b)
Solves
5cos
2
θ−
16cos
θ
+
3
=
0
cos
θ
=
5
1
2
1
Attempts
2
ar
=
sin
θ
×
2cos
θ
=
1
−
××
2
5
5
48
=
125
(i) (a)
M1: For an attempt at using the binomial expansion to find one of the required terms.
Condone the omission of the brackets on the 2x terms. Allow using combination notation.
Don't be concerned over any labelling (2nd term). May be awarded from a full expansion
6
A1: Any two correct terms, simplified or not. The
C must now be numerical.
2
Don't be concerned by labelling and accept as part of a large list or expansion
A1: All three terms correct and simplified.
They may be part of a larger or complete expansion, which is fine. Look for the 2nd , 3rd and 5th terms
2
4
Accept the 3 terms with +'s between. So accept for example
2916
x
+
4860
x
+
2160
x
2
6
Note that the expansion can be done by (
)
6
6
3
+
2
x
=
3
1
+
x
The unsimplified versions of the terms are:
3
2
4
2
6 5 2
×
6 5 4 3 2
×××
6
6
6
t (2) =
3
×
6 3 x
t(3) =
3
×
and t(5) =
3
×
3 x
4 3 2 1 3 x
2
×××
(i)(b)
M1: Uses the fact that the 2nd, 3rd and 5th terms of their binomial expansion in (a) are consecutive terms of a
GP to set up an equation in x.
There are a few ways in which the equation could be set up so look carefully at what is written.
4
2
2
''2160
x
''
''4860
x
''
''4860
x
''
''2916 ''
x
The most common ways are
=
,
=
and
2 '
2
''2916 ''
4
''4 60
8
''
x
''2
160
'
'
'
' 8 0
4
6
'
x
x
x
2
''4860
x
''
4
2
''2160
''
=
×
''4860
''
. Condone slips e.g. 2196 for 2916
x
x
''2916 ''
x
The candidates must be using what they believe are the 2nd, 3rd and 5th terms of the binomial expansion
and not the 2nd, 3rd or 4th or any other combination.
If a list is given it must be the 2nd, 3rd and 5th terms in
+
+
2
+
3
+
4
+
....
where , , ,
a
bx
cx
dx
ex
dM1: It is dependent upon the first M and having the correct index on the powers of x.
It is for proceeding to an equation of the form ...
x =
...
15
A1:
x =
or exact equivalent. Condone an extra
x =
0
4
.........................................................................................................................................................................
If there are unlabelled/mislabelled terms in (a) that are used in (b) and you feel that they deserve more credit
than that afforded by the scheme, please use review. A common error is described below.
2
3
5x in (i) (a) and (b). We will treat as a SC only
It seems that some candidates are finding the terms in
x
,
x and
when appearing in both parts. FYI: These are
4860
2
, 4320
x and
3
576x . These values would score M1 in
5
x
(i)(a) for the
2x term and potentially M1, dM1 (when proceeding to ...x =...) in (i)(b)
...........................................................................................................................................................................
(ii)(a) Allow use of a different parameter, e.g.
θ
x
2
a
8
sin
θ
M1: Attempts to use
∞=
=
θ Condone slips only. The formula must be correct
S
1
5
1 2cos
−
r
−
2
2
dM1: Uses the identity
sin
θ
= −
1 cos
θto form an equation in just cosθ
A1*: Completes proof with sufficient working.
2
2
Condone one slip in notation. E.g.
sin
θ
sin
θ cos
cos
θbut the final line must be correct and
include the = 0
(ii)(b)
1
B1: States or uses
cos
θ
=
5
1
1
2
2
M1: Attempts
=
sin
θ
×
2cos
θ
=
1 ''
−
''
××
2 ''
''
ar
5
5
1
Alternatively finds the value of θ from their
cos
and uses this in
sin
2
2cos
θ
θ
=
ar =
θ
×
5
A1: 48
125 or exact equivalent. Eg. 0.384.
It cannot be scored by rounding a non exact answer or from a decimal value of θ
So, for example, answers of 48
125 achieved using a value of
θ
=
78.46….. don't score the final A1
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
10(ii)(a)[3]
3
+
2x
6
5
(
)
1
6
4
(
)
2
6
2
(
)
4
Any correct form for any term of
C 3
2x or
C 3
2x
or
C 3
2x
M1
1
2
4
5
4
2
Any two correct: t(2) =
6 3
(
2x
)
1
, t(3) =
15 3
(
2x
)
2
t(5) =
15 3
(
2x
)
4
A1
×
×
×
×
×
×
All 3 terms correct 2nd term = 2916x, 3rd term= 4860x2 , 5th term=2160x4 A1
4
2
2160
4860
x
x
Method using consecutive terms of a GP e.g
=
2
2916
x
4860
x
...
=
...
x
15
x
=
4
2
a
8
sin
θ
(ii) (a) Attempts to use
∞=
=
θ
S
1
5
1 2cos
−
r
−
8
1 cos
−
2
θ
= −
θ
5
1 2cos
2
−
8 16cos
θ
= −
5 5cos
θ
5cos
2
θ−
16cos
θ
+ =
3
0
1
(b)
Solves
5cos
2
θ−
16cos
θ
+
3
=
0
cos
θ
=
5
1
2
1
Attempts
2
ar
=
sin
θ
×
2cos
θ
=
1
−
××
2
5
5
48
=
125
(i) (a)
M1: For an attempt at using the binomial expansion to find one of the required terms.
Condone the omission of the brackets on the 2x terms. Allow using combination notation.
Don't be concerned over any labelling (2nd term). May be awarded from a full expansion
6
A1: Any two correct terms, simplified or not. The
C must now be numerical.
2
Don't be concerned by labelling and accept as part of a large list or expansion
A1: All three terms correct and simplified.
They may be part of a larger or complete expansion, which is fine. Look for the 2nd , 3rd and 5th terms
2
4
Accept the 3 terms with +'s between. So accept for example
2916
x
+
4860
x
+
2160
x
2
6
Note that the expansion can be done by (
)
6
6
3
+
2
x
=
3
1
+
x
The unsimplified versions of the terms are:
3
2
4
2
6 5 2
×
6 5 4 3 2
×××
6
6
6
t (2) =
3
×
6 3 x
t(3) =
3
×
and t(5) =
3
×
3 x
4 3 2 1 3 x
2
×××
(i)(b)
M1: Uses the fact that the 2nd, 3rd and 5th terms of their binomial expansion in (a) are consecutive terms of a
GP to set up an equation in x.
There are a few ways in which the equation could be set up so look carefully at what is written.
4
2
2
''2160
x
''
''4860
x
''
''4860
x
''
''2916 ''
x
The most common ways are
=
,
=
and
2 '
2
''2916 ''
4
''4 60
8
''
x
''2
160
'
'
'
' 8 0
4
6
'
x
x
x
2
''4860
x
''
4
2
''2160
''
=
×
''4860
''
. Condone slips e.g. 2196 for 2916
x
x
''2916 ''
x
The candidates must be using what they believe are the 2nd, 3rd and 5th terms of the binomial expansion
and not the 2nd, 3rd or 4th or any other combination.
If a list is given it must be the 2nd, 3rd and 5th terms in
+
+
2
+
3
+
4
+
....
where , , ,
a
bx
cx
dx
ex
dM1: It is dependent upon the first M and having the correct index on the powers of x.
It is for proceeding to an equation of the form ...
x =
...
15
A1:
x =
or exact equivalent. Condone an extra
x =
0
4
.........................................................................................................................................................................
If there are unlabelled/mislabelled terms in (a) that are used in (b) and you feel that they deserve more credit
than that afforded by the scheme, please use review. A common error is described below.
2
3
5x in (i) (a) and (b). We will treat as a SC only
It seems that some candidates are finding the terms in
x
,
x and
when appearing in both parts. FYI: These are
4860
2
, 4320
x and
3
576x . These values would score M1 in
5
x
(i)(a) for the
2x term and potentially M1, dM1 (when proceeding to ...x =...) in (i)(b)
...........................................................................................................................................................................
(ii)(a) Allow use of a different parameter, e.g.
θ
x
2
a
8
sin
θ
M1: Attempts to use
∞=
=
θ Condone slips only. The formula must be correct
S
1
5
1 2cos
−
r
−
2
2
dM1: Uses the identity
sin
θ
= −
1 cos
θto form an equation in just cosθ
A1*: Completes proof with sufficient working.
2
2
Condone one slip in notation. E.g.
sin
θ
sin
θ cos
cos
θbut the final line must be correct and
include the = 0
(ii)(b)
1
B1: States or uses
cos
θ
=
5
1
1
2
2
M1: Attempts
=
sin
θ
×
2cos
θ
=
1 ''
−
''
××
2 ''
''
ar
5
5
1
Alternatively finds the value of θ from their
cos
and uses this in
sin
2
2cos
θ
θ
=
ar =
θ
×
5
A1: 48
125 or exact equivalent. Eg. 0.384.
It cannot be scored by rounding a non exact answer or from a decimal value of θ
So, for example, answers of 48
125 achieved using a value of
θ
=
78.46….. don't score the final A1
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
10(ii)(b)[3]
Method using consecutive terms of a GP e.g
=
2
2916
x
4860
x
...
=
...
x
15
x
=
4
2
a
8
sin
θ
(ii) (a) Attempts to use
∞=
=
θ
S
1
5
1 2cos
−
r
−
8
1 cos
−
2
θ
= −
θ
5
1 2cos
2
−
8 16cos
θ
= −
5 5cos
θ
5cos
2
θ−
16cos
θ
+ =
3
0
1
(b)
Solves
5cos
2
θ−
16cos
θ
+
3
=
0
cos
θ
=
5
1
2
1
Attempts
2
ar
=
sin
θ
×
2cos
θ
=
1
−
××
2
5
5
48
=
125
(i) (a)
M1: For an attempt at using the binomial expansion to find one of the required terms.
Condone the omission of the brackets on the 2x terms. Allow using combination notation.
Don't be concerned over any labelling (2nd term). May be awarded from a full expansion
6
A1: Any two correct terms, simplified or not. The
C must now be numerical.
2
Don't be concerned by labelling and accept as part of a large list or expansion
A1: All three terms correct and simplified.
They may be part of a larger or complete expansion, which is fine. Look for the 2nd , 3rd and 5th terms
2
4
Accept the 3 terms with +'s between. So accept for example
2916
x
+
4860
x
+
2160
x
2
6
Note that the expansion can be done by (
)
6
6
3
+
2
x
=
3
1
+
x
The unsimplified versions of the terms are:
3
2
4
2
6 5 2
×
6 5 4 3 2
×××
6
6
6
t (2) =
3
×
6 3 x
t(3) =
3
×
and t(5) =
3
×
3 x
4 3 2 1 3 x
2
×××
(i)(b)
M1: Uses the fact that the 2nd, 3rd and 5th terms of their binomial expansion in (a) are consecutive terms of a
GP to set up an equation in x.
There are a few ways in which the equation could be set up so look carefully at what is written.
4
2
2
''2160
x
''
''4860
x
''
''4860
x
''
''2916 ''
x
The most common ways are
=
,
=
and
2 '
2
''2916 ''
4
''4 60
8
''
x
''2
160
'
'
'
' 8 0
4
6
'
x
x
x
2
''4860
x
''
4
2
''2160
''
=
×
''4860
''
. Condone slips e.g. 2196 for 2916
x
x
''2916 ''
x
The candidates must be using what they believe are the 2nd, 3rd and 5th terms of the binomial expansion
and not the 2nd, 3rd or 4th or any other combination.
If a list is given it must be the 2nd, 3rd and 5th terms in
+
+
2
+
3
+
4
+
....
where , , ,
a
bx
cx
dx
ex
dM1: It is dependent upon the first M and having the correct index on the powers of x.
It is for proceeding to an equation of the form ...
x =
...
15
A1:
x =
or exact equivalent. Condone an extra
x =
0
4
.........................................................................................................................................................................
If there are unlabelled/mislabelled terms in (a) that are used in (b) and you feel that they deserve more credit
than that afforded by the scheme, please use review. A common error is described below.
2
3
5x in (i) (a) and (b). We will treat as a SC only
It seems that some candidates are finding the terms in
x
,
x and
when appearing in both parts. FYI: These are
4860
2
, 4320
x and
3
576x . These values would score M1 in
5
x
(i)(a) for the
2x term and potentially M1, dM1 (when proceeding to ...x =...) in (i)(b)
...........................................................................................................................................................................
(ii)(a) Allow use of a different parameter, e.g.
θ
x
2
a
8
sin
θ
M1: Attempts to use
∞=
=
θ Condone slips only. The formula must be correct
S
1
5
1 2cos
−
r
−
2
2
dM1: Uses the identity
sin
θ
= −
1 cos
θto form an equation in just cosθ
A1*: Completes proof with sufficient working.
2
2
Condone one slip in notation. E.g.
sin
θ
sin
θ cos
cos
θbut the final line must be correct and
include the = 0
(ii)(b)
1
B1: States or uses
cos
θ
=
5
1
1
2
2
M1: Attempts
=
sin
θ
×
2cos
θ
=
1 ''
−
''
××
2 ''
''
ar
5
5
1
Alternatively finds the value of θ from their
cos
and uses this in
sin
2
2cos
θ
θ
=
ar =
θ
×
5
A1: 48
125 or exact equivalent. Eg. 0.384.
It cannot be scored by rounding a non exact answer or from a decimal value of θ
So, for example, answers of 48
125 achieved using a value of
θ
=
78.46….. don't score the final A1
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom

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