Mathematics · May / June 2022 · Paper 31 · Question 9
YMA01
As printed in the paper
The question as text
9. y O x M Figure 5 In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Figure 5 shows the curve with equation + x π < x < 3 π y = 1 2 cos – 2 + x 2 1 sin The point M, shown in Figure 5, is the minimum point on the curve.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
9(a)[4]
=
e
1.75
t−
10
or equivalent eg
9e
−
0.75
t
−
e
t
−
10
=
0
M1
Proceeds from their equation of the form
A
e
Ct D
±
−
B
e
Et F
±
=
0
or equivalent leading to a value or an
expression (which may be unsimplified) for t. If log work is seen it must be correct.
Eg 40
4ln9
or 10
ln9
+
+
will score this mark
7
1.75
9e
−
0.75
t
e
t
−
10
0
−
=
awrt 6.97 can also score this mark
This cannot be scored from an unsolvable equation.
awrt 6.97 or awrt 6.96 including exact unsimplified expressions such as 40
+
4ln9
or 10
A1
7
A1
awrt 11.9 (ms1) with all previous marks scored. Condone lack of units
Note: awrt 6.97 and awrt 11.9 with no calculus scores 0 marks
M1
Attempts to integrate the expression for v. Award for
12
t
±
A
e
(
t
−
10)
±
B
e
−
0.75
t
.
Do not accept eg
12
t
±
A
e
(
t
−
10) 1
+
±
B
e
−
0.75
t
+
1
Allow if their attempt to integrate appears in part (a) as long as it is a clear attempt at
integration.
Limits may have been substituted in.
12
12
t
−
e
(
t
−
10)
+
16e
−
0.75
t
12
t
−
e
(
t
−
10)
−−
A1
(with or without +C) and allow unsimplified eg
0.75
Do not withhold this mark if they write 12x but subsequently replace it with T later on.
M1
Substitutes T and 0 into their changed expression with three different terms in t , sets equal
100 and attempts to make 12T or T the subject.
Eg (
) (
)
12
T
−
e
T
−
10
+
16e
−
0.75
T
−−
e
−
10
+
16
=
100
12
T
=
...
Condone arithmetical slips in their rearrangement.
1 116 16e
(
)
T
−
0.75
T
e
T
−
10
e
−
10
A1*
=
−
+
−
with no errors including bracket errors/omissions.
12
Must have T = …
(c)
(i)
M1
Substitutes 10 into the iterative formula. The expression with 10 embedded is sufficient to score this
mark.
May be implied by awrt 9.7492 or awrt 9.7493. May also be implied by 9.7306
awrt 9.7493
A1
(ii) *Be aware that this can be found directly on a calculator which is not acceptable*
A1
9.7293 (seconds) (cannot be awarded without the method mark)
SC If they over round in (i) and (ii) and achieve 9.749 and 9.729 then award M1A0A1
d
y
(1 sin )( 2sin )
+
x
−
x
−
(1
+
2cos )cos
x
x
9(a)
=
d
x
(
)
2
1 sin
+
x
(1 sin )( 2sin )
+
x
−
x
−
(1 2cos )cos
+
x
x
=
0
−
2sin
x
−
cos
x
−
2
=
0
2sin
x
+
cos
x
= − *
2
(b)
Possible solutions from
R
sin(
x
+
α
) or
R
cos(
x
−
α
)
:
2sin
x
+
cos
x
= −
2
5sin(
x
+
0.464)
= − or 5 cos(
2
x −
1.107)
= −
2
or eg solutions from squaring
(2sin
x
+
cos )
x
2
=
4
cos (3cos
x
x
−
4sin )
x
=
0
3
5sin(
x
+
0.464)
= −
2 or
5 cos(
x
−
1.107)
= −
2
x
=
... or tan
x
=
x
=
...
4
awrt 3.79
x =
(a)
M1
Attempts to differentiate by the product rule or quotient rule and proceeds to:
A
(1 sin )(sin )
x
x
(1 2cos )cos
x
x
±
+
±
+
Quotient rule:
(Scored for the numerator only)
....
Product rule:
±
cos (1 2cos )(1 sin )
x
+
x
+
x
−
2
±
A
sin (1 sin )
x
+
x
−
1
Alternatively, they express 1 2cos
x
1
2cos
x
+
=
+
and proceed to either
1 sin
+
x
1 sin
+
x
1 sin
+
x
cos
x
A
sin (1 sin )
x
+
x
±
cos (2cos )
x
x
Quotient rule:
±
±
(
)
(
)
1 sin
+
x
2
1 sin
+
x
2
Product rule:
±
cos (1 sin )
x
+
x
−
2
±
A
cos
2
x
(1 sin )
+
x
−
2
±
B
sin (1 sin )
x
+
x
−
1
Condone arithmetical slips and invisible brackets.
d
y
(1 sin )( 2sin )
+
x
−
x
−
(1 2cos )cos
+
x
x
=
A1
or equivalent (which may be unsimplified)
d
x
(
)
2
1 sin
+
x
If is is via the product rule then it may appear as
d
y
=
−
cos (1
x
+
2cos )(1 sin )
x
+
x
−
2
−
2sin (1 sin )
x
+
x
−
1
or equivalent (which may be
d
x
unsimplified).
dM1
Sets the numerator of their quotient equal to zero, multiplies out the brackets and attempts to use the
±
sin
2
x
±
cos
2
x
= ± (which may be implied by their working) leading to an equation in
1
identity
sin x and cos x only. There must be no squared terms.
If it is via the product rule they will need to either combine their terms into a single fraction and then
set the numerator equal to zero (or multiply both sides by (
)
2
1 sin x
+
), and then attempt to use the
identity
±
sin
2
x
±
cos
2
x
= ± leading to an equation in sin x and cos x only. There must be no
1
squared terms.
Condone slips and invisible brackets. It is dependent on the first method mark.
A1*
Achieves 2sin
x
+
cos
x
= − with no errors or missing brackets around 1
2
+
2cos x
or 1 sin x
+
when needed in the main body of their solution.
(Do not penalise the fraction line not being quite long enough for the numerator.)
It is acceptable to proceed to a correct d
y
, multiply out the brackets and proceed directly from
d
x
−
2sin
x
−
2sin
2
x
−
cos
x
−
2cos
2
x
=
0
2sin
x
+
cos
x
= − to score full marks. You do not need
2
sin
2
x
cos
2
x
1
to explicitly see
+
= stated or substituted in.
Note if they only ever use the numerator of the quotient formula:
vu
'
−
uv
'
from the beginning
Maximum score is M1A0dM1A0*
(b) *Be aware this can be solved entirely using a calculator which is not acceptable*
Solution by expressing 2sin
x
+
cos
x
R
sin(
x
±
α
)
R
cos(
x
±
α
)
in the form
or
R =
2
2
+
1
2
=
5
M1
−
2
−
1
(
)
(
)
or (
)
(
)
M1
α
=
tan
−
1
±
1
α
=
tan
−
1
±
2
(or
cos
1
±
sin
1
±
)
2
5
5
" 5"sin(
x +
"0.464")
= − or " 5"cos(
2
x −
"1.107")
= − oe (or in another variable)
2
A1
dM1
Attempts to solve their equation and proceeds to find a value for x.
−
2
−
2
eg
x
=
π
−
sin
−
1
±
"0.464"
or
x
=
sin
−
1
±
"0.464"
" 5"
" 5"
You may need to check their angle on your calculator. Condone slips in their rearrangement.
It is dependent on both of the previous method marks.
A1
awrt 3.79 only
Solutions squaring both sides and using the identity
sin
2
x
+
cos
2
x
1
M1
Attempts to square both sides and multiplies out the brackets. Condone sign slips only on the brackets
they are multiplying out. Do not accept just squaring all the individual terms.
sin
2
cos
2
= ± and proceeds to a quadratic equation in either sin x or
1
M1
Attempts to use
±
x
±
x
cos x or an equation of the form
±
A
cos
2
x
±
B
cos sin
x
x
=
0
.
be done before squaring both sides. Eg look for
The use of the identity may
2sin
x
+
2
=
cos
x
2sin
x
+
2
=
1 sin
−
2
x
...
Condone sign slips only.
A1
3cos
2
x
−
4sin cos
x
x
=
0
oe or
5sin
2
x
+
8sin
x
+
3
=
0
oe or
5cos
2
x
+
4cos
x
=
0
oe
The = 0 may be implied by later work and terms do not need to be collected on one side.
dM1
Attempts to solve their quadratic equation in either sin x or cos x or an equation of the form
±
A
cos
2
x
±
B
cos sin
x
x
=
0
and proceeds to find a value for x. Do not allow for solving
cos
x =
0
Usual rules apply for solving a quadratic. They may even just state the roots from their calculator.
It is dependent on both of the previous method marks.
A1
awrt 3.79 only
The table below may help with following through a particular approach:
(2sin
x
+
cos )
x
2
=
4
(2sin
x
+
2)
2
=
cos
2
x
(cos
x
+
2)
2
=
M1
4sin
2
x
+
4sin cos
x
x
+
cos
2
x
=
4
4sin
2
x
+
8sin
x
+
4
=
cos
2
x
cos
2
x
+
4cos
x
+
M1
±
sin
2
x
±
cos
2
x
= ±
1
Attempts to use
A1
3cos
2
x
−
4sin cos
x
x
=
0
oe
5sin
2
x
+
8sin
x
+
3
=
0
oe
5cos
2
x
+
4cos
dM1
Eg cos (3cos
x
x
−
4sin )
x
=
0
Eg (5sin
x
+
3)(sin
x
+
1)
=
0
Eg cos (5cos
x
x
=
...
x
=
...
x
=
A1
awrt 3.79
It is possible that they may divide the equation by either sin x or cos x and use
1
+
tan
2
x
=
or
1 cot
+
2
x
=
cosec
2
x
resulting in the following equations (or equivalent
simplified rearrangements of
these) to solve:
3cot
2
x
=
4cot
x
cot (5cot
x
x
+
4cot cosec )
x
x
=
0
3cosec
2
x
+
8cosec
x
+
5
=
0
4tan
x =
3
4sec
x +
5
=
0
The mark scheme should be applied in the same way but looking for use of
±±
1
tan
2
x
= ±
sec
±±
1 cot
2
x
= ±
cosec
2
x
and proceeding to equations involving cot x, cosec x, tan x or sec x
Send to review if unsure.
Solutions using the t method substitution
x
2
t
1
−
t
2
Using
t
=
tan 2
sin
x
= +
cos
x
= +
1
t
2
1
t
2
±
2
t
M1
Attempts to express 2sin
x
+
cos
x
= − in terms of t. Attempts to use
2
sin
x
= ±±
and
1
t
2
±±
1
t
2
cos
x
= ±±
.
1
t
2
2
t
1
−
t
2
+
M1
Proceeds from an equation of the form
2
= −
2
to a 3TQ
1
+
t
2
1
+
t
2
4
t
1
−
t
2
+
= −
2
4
t
+ −
1
t
2
= −
2(1
+
t
2
)
...
1
+
t
2
1
+
t
2
t
2
4
t
3
0
A1
+
+
=
or equivalent. The = 0 may be implied by later work and the three terms do
need to be on the same side of the equation.
dM1
Correctly rearranges their equation to find a value for x
(
)
t
2
+
4
t
+
3
=
0
= −
t
3
x
=
2 arctan(" 3")
−
+
π
t = −
1
Ignore any reference to
It is dependent on both of the previous method marks.
A1
awrt 3.79
There may be other variants of these methods, but the mark scheme should still be able to be applied. If
in doubt send to review.
Answer only scores 0 marks
earson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
9(b)[5]
M1
Attempts to integrate the expression for v. Award for
12
t
±
A
e
(
t
−
10)
±
B
e
−
0.75
t
.
Do not accept eg
12
t
±
A
e
(
t
−
10) 1
+
±
B
e
−
0.75
t
+
1
Allow if their attempt to integrate appears in part (a) as long as it is a clear attempt at
integration.
Limits may have been substituted in.
12
12
t
−
e
(
t
−
10)
+
16e
−
0.75
t
12
t
−
e
(
t
−
10)
−−
A1
(with or without +C) and allow unsimplified eg
0.75
Do not withhold this mark if they write 12x but subsequently replace it with T later on.
M1
Substitutes T and 0 into their changed expression with three different terms in t , sets equal
100 and attempts to make 12T or T the subject.
Eg (
) (
)
12
T
−
e
T
−
10
+
16e
−
0.75
T
−−
e
−
10
+
16
=
100
12
T
=
...
Condone arithmetical slips in their rearrangement.
1 116 16e
(
)
T
−
0.75
T
e
T
−
10
e
−
10
A1*
=
−
+
−
with no errors including bracket errors/omissions.
12
Must have T = …
(c)
(i)
M1
Substitutes 10 into the iterative formula. The expression with 10 embedded is sufficient to score this
mark.
May be implied by awrt 9.7492 or awrt 9.7493. May also be implied by 9.7306
awrt 9.7493
A1
(ii) *Be aware that this can be found directly on a calculator which is not acceptable*
A1
9.7293 (seconds) (cannot be awarded without the method mark)
SC If they over round in (i) and (ii) and achieve 9.749 and 9.729 then award M1A0A1
d
y
(1 sin )( 2sin )
+
x
−
x
−
(1
+
2cos )cos
x
x
9(a)
=
d
x
(
)
2
1 sin
+
x
(1 sin )( 2sin )
+
x
−
x
−
(1 2cos )cos
+
x
x
=
0
−
2sin
x
−
cos
x
−
2
=
0
2sin
x
+
cos
x
= − *
2
(b)
Possible solutions from
R
sin(
x
+
α
) or
R
cos(
x
−
α
)
:
2sin
x
+
cos
x
= −
2
5sin(
x
+
0.464)
= − or 5 cos(
2
x −
1.107)
= −
2
or eg solutions from squaring
(2sin
x
+
cos )
x
2
=
4
cos (3cos
x
x
−
4sin )
x
=
0
3
5sin(
x
+
0.464)
= −
2 or
5 cos(
x
−
1.107)
= −
2
x
=
... or tan
x
=
x
=
...
4
awrt 3.79
x =
(a)
M1
Attempts to differentiate by the product rule or quotient rule and proceeds to:
A
(1 sin )(sin )
x
x
(1 2cos )cos
x
x
±
+
±
+
Quotient rule:
(Scored for the numerator only)
....
Product rule:
±
cos (1 2cos )(1 sin )
x
+
x
+
x
−
2
±
A
sin (1 sin )
x
+
x
−
1
Alternatively, they express 1 2cos
x
1
2cos
x
+
=
+
and proceed to either
1 sin
+
x
1 sin
+
x
1 sin
+
x
cos
x
A
sin (1 sin )
x
+
x
±
cos (2cos )
x
x
Quotient rule:
±
±
(
)
(
)
1 sin
+
x
2
1 sin
+
x
2
Product rule:
±
cos (1 sin )
x
+
x
−
2
±
A
cos
2
x
(1 sin )
+
x
−
2
±
B
sin (1 sin )
x
+
x
−
1
Condone arithmetical slips and invisible brackets.
d
y
(1 sin )( 2sin )
+
x
−
x
−
(1 2cos )cos
+
x
x
=
A1
or equivalent (which may be unsimplified)
d
x
(
)
2
1 sin
+
x
If is is via the product rule then it may appear as
d
y
=
−
cos (1
x
+
2cos )(1 sin )
x
+
x
−
2
−
2sin (1 sin )
x
+
x
−
1
or equivalent (which may be
d
x
unsimplified).
dM1
Sets the numerator of their quotient equal to zero, multiplies out the brackets and attempts to use the
±
sin
2
x
±
cos
2
x
= ± (which may be implied by their working) leading to an equation in
1
identity
sin x and cos x only. There must be no squared terms.
If it is via the product rule they will need to either combine their terms into a single fraction and then
set the numerator equal to zero (or multiply both sides by (
)
2
1 sin x
+
), and then attempt to use the
identity
±
sin
2
x
±
cos
2
x
= ± leading to an equation in sin x and cos x only. There must be no
1
squared terms.
Condone slips and invisible brackets. It is dependent on the first method mark.
A1*
Achieves 2sin
x
+
cos
x
= − with no errors or missing brackets around 1
2
+
2cos x
or 1 sin x
+
when needed in the main body of their solution.
(Do not penalise the fraction line not being quite long enough for the numerator.)
It is acceptable to proceed to a correct d
y
, multiply out the brackets and proceed directly from
d
x
−
2sin
x
−
2sin
2
x
−
cos
x
−
2cos
2
x
=
0
2sin
x
+
cos
x
= − to score full marks. You do not need
2
sin
2
x
cos
2
x
1
to explicitly see
+
= stated or substituted in.
Note if they only ever use the numerator of the quotient formula:
vu
'
−
uv
'
from the beginning
Maximum score is M1A0dM1A0*
(b) *Be aware this can be solved entirely using a calculator which is not acceptable*
Solution by expressing 2sin
x
+
cos
x
R
sin(
x
±
α
)
R
cos(
x
±
α
)
in the form
or
R =
2
2
+
1
2
=
5
M1
−
2
−
1
(
)
(
)
or (
)
(
)
M1
α
=
tan
−
1
±
1
α
=
tan
−
1
±
2
(or
cos
1
±
sin
1
±
)
2
5
5
" 5"sin(
x +
"0.464")
= − or " 5"cos(
2
x −
"1.107")
= − oe (or in another variable)
2
A1
dM1
Attempts to solve their equation and proceeds to find a value for x.
−
2
−
2
eg
x
=
π
−
sin
−
1
±
"0.464"
or
x
=
sin
−
1
±
"0.464"
" 5"
" 5"
You may need to check their angle on your calculator. Condone slips in their rearrangement.
It is dependent on both of the previous method marks.
A1
awrt 3.79 only
Solutions squaring both sides and using the identity
sin
2
x
+
cos
2
x
1
M1
Attempts to square both sides and multiplies out the brackets. Condone sign slips only on the brackets
they are multiplying out. Do not accept just squaring all the individual terms.
sin
2
cos
2
= ± and proceeds to a quadratic equation in either sin x or
1
M1
Attempts to use
±
x
±
x
cos x or an equation of the form
±
A
cos
2
x
±
B
cos sin
x
x
=
0
.
be done before squaring both sides. Eg look for
The use of the identity may
2sin
x
+
2
=
cos
x
2sin
x
+
2
=
1 sin
−
2
x
...
Condone sign slips only.
A1
3cos
2
x
−
4sin cos
x
x
=
0
oe or
5sin
2
x
+
8sin
x
+
3
=
0
oe or
5cos
2
x
+
4cos
x
=
0
oe
The = 0 may be implied by later work and terms do not need to be collected on one side.
dM1
Attempts to solve their quadratic equation in either sin x or cos x or an equation of the form
±
A
cos
2
x
±
B
cos sin
x
x
=
0
and proceeds to find a value for x. Do not allow for solving
cos
x =
0
Usual rules apply for solving a quadratic. They may even just state the roots from their calculator.
It is dependent on both of the previous method marks.
A1
awrt 3.79 only
The table below may help with following through a particular approach:
(2sin
x
+
cos )
x
2
=
4
(2sin
x
+
2)
2
=
cos
2
x
(cos
x
+
2)
2
=
M1
4sin
2
x
+
4sin cos
x
x
+
cos
2
x
=
4
4sin
2
x
+
8sin
x
+
4
=
cos
2
x
cos
2
x
+
4cos
x
+
M1
±
sin
2
x
±
cos
2
x
= ±
1
Attempts to use
A1
3cos
2
x
−
4sin cos
x
x
=
0
oe
5sin
2
x
+
8sin
x
+
3
=
0
oe
5cos
2
x
+
4cos
dM1
Eg cos (3cos
x
x
−
4sin )
x
=
0
Eg (5sin
x
+
3)(sin
x
+
1)
=
0
Eg cos (5cos
x
x
=
...
x
=
...
x
=
A1
awrt 3.79
It is possible that they may divide the equation by either sin x or cos x and use
1
+
tan
2
x
=
or
1 cot
+
2
x
=
cosec
2
x
resulting in the following equations (or equivalent
simplified rearrangements of
these) to solve:
3cot
2
x
=
4cot
x
cot (5cot
x
x
+
4cot cosec )
x
x
=
0
3cosec
2
x
+
8cosec
x
+
5
=
0
4tan
x =
3
4sec
x +
5
=
0
The mark scheme should be applied in the same way but looking for use of
±±
1
tan
2
x
= ±
sec
±±
1 cot
2
x
= ±
cosec
2
x
and proceeding to equations involving cot x, cosec x, tan x or sec x
Send to review if unsure.
Solutions using the t method substitution
x
2
t
1
−
t
2
Using
t
=
tan 2
sin
x
= +
cos
x
= +
1
t
2
1
t
2
±
2
t
M1
Attempts to express 2sin
x
+
cos
x
= − in terms of t. Attempts to use
2
sin
x
= ±±
and
1
t
2
±±
1
t
2
cos
x
= ±±
.
1
t
2
2
t
1
−
t
2
+
M1
Proceeds from an equation of the form
2
= −
2
to a 3TQ
1
+
t
2
1
+
t
2
4
t
1
−
t
2
+
= −
2
4
t
+ −
1
t
2
= −
2(1
+
t
2
)
...
1
+
t
2
1
+
t
2
t
2
4
t
3
0
A1
+
+
=
or equivalent. The = 0 may be implied by later work and the three terms do
need to be on the same side of the equation.
dM1
Correctly rearranges their equation to find a value for x
(
)
t
2
+
4
t
+
3
=
0
= −
t
3
x
=
2 arctan(" 3")
−
+
π
t = −
1
Ignore any reference to
It is dependent on both of the previous method marks.
A1
awrt 3.79
There may be other variants of these methods, but the mark scheme should still be able to be applied. If
in doubt send to review.
Answer only scores 0 marks
earson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.