Mathematics · May / June 2023 · Paper 11 · Question 6

YMA01

As printed in the paper

The question as text

6. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

6(a)[2]
A1: Fully correct expression in the required form. Depends on all previous marks.
y = … is not required. Apply isw once a correct expression is seen.
(b) The d
y
x = is not required throughout in part (b)
d
M1:
x
n
→
...
x
n
−
1
for at least one term where n ≠ 1.
A1ft: Correct differentiation of one of their terms where n ≠ 1.
Follow through their part (a). But not for kx →k
May be unsimplified.
Depends on both M marks in part (a).
A1: Correct derivative, with terms in simplest form. Apply isw once a correct expression is seen.
Score A0 if they include “+ c”
5(a)
1.5
4
2
−
1.5
2
∠
=
cos
−
1
=
...
or
∠
=
tan
−
1
=
...
AOB
AOB
4
1 5
.
or
π
1
.5
∠
AOB
=
−
s
in
−
1
= …
2
4
or
π
1.5
∠
AOB
=
−
tan
−
1
=
..
.
2
4
2
−
1.
5
2
= 1.186 *
(
)
=
4
×
1.186
=
4.744
l
(
)
6
4sin1.186
6
3.708...
2.29...
BC =
−
=
−
=
or
π
(
)
BC
=
6
−
4cos
−
1.186
=
6
−
3.708...
=
2.29...
2
or
(
)
12
−
55
=
6
−
4
2
−
1.5
2
=
6
−
3.708...
=
2.29...
=
BC
2
or
1.5
(
)
BC =
6
−
=
6
−
3.708...
=
2.29...
tan"0.384"
(
)
Perimeter is
4×
1.186...
+
6
−
4cos0.384...
+
1.5
+ +
6
4
=
..
= awrt 18.5 (m)
1
Area sector =
24
1.186...
2 ×
×
Area OBCD Examples:
1
or (
)
1
6 1.5
×
−
×
"3.708" 1.5
×
6
−
"3.708"
×
1.5
+
×
"3.708" 1.5
×
2
2
or (
1 6
)
1
+
6
−
"3.708"
×
1.5
or
6 1.5
×
−
×
1.5 4sin1.186
×
2
2
1
1
(
)
or
6 1.5
6
"3.708"
1.5
(= 6.219…)
××
+
−
×
2
2
1
×
4
2
×
1.18
6
+
6.21 . .
9 .
=
2
= awrt 15.7 (m2)
Notes:
Allow equivalent correct work in degrees.
Ignore any presence or absence of units throughout.
(a)
M1: For a complete and correct method to find angle AOB in radians or degrees. May attempt angle
π or equivalent
AOB directly using cosine, or angle BOD using sine then subtract angle BOD from
2
work using tan.
A1*: Obtains angle AOB = 1.186 with no errors.
Example in degrees:
1.
5
1 3
sin
∠
BOD
=
∠
BOD
=
sin
−
=
22.0
243
..
.
4
8
90
22.0243...
67.9756...
∠
AOB
=
−
=
π×
67.9756...
= 1.186 *
=
180
(b)
M1: Uses the correct arc length formula with r = 4 and 1.186. May be implied by awrt 4.74 or 4.75.
M1: Attempts the length of BC using an appropriate trig function and angle, or Pythagoras. May see
6
−
4sin 1.186
or 6
−
4cos"0.384"
or may use
6
−
4
2
−
1.5
2
. May be implied by awrt 2.3 provided
no incorrect working is seen.
ddM1: Depends on both previous marks. Adds the five lengths of the sides (or equivalent work)
to obtain the perimeter.
A1: Correct answer, accept awrt 18.5. No units required and ignore any given.
(c)
1
but may be implied by e.g. 1186
M1: Correct attempt at the sector area. Look for
2 ×
24
×
1.186...
1
π
awrt 9.5. Allow equivalent expressions in degrees e.g.
×
24
×
67.9756... 180
×
but not e.g.
2
1
24
67.9756
2 ×
×
M1: Correct attempt at the area of OBCD – which may be found as the difference of a rectangle and
triangle or sum of rectangle and triangle or a trapezium. See scheme for details.
The attempt to find BC, if used, must have involved the use of trigonometry or e.g. Pythagoras.
ddM1: Depends on both previous marks. Correct attempt at the full area, so adds sector area to
area of OBCD
A1: Correct answer, accept awrt 15.7. No units required and ignore any given.
Note that some candidates assume BC = 2 (i.e. 6 – 4) which greatly simplifies the question.
The maximum possible marks in (b) and (c) are: (b) M1M0ddM0A0 (c) M1M0ddM0A0
6(a)
2
1
1
1
1
−
=
−
×
−
×
+
r
r
2
r
r
r
r
r
r
2
1
=
2
+
−
2
r
2
r
(b)
1
1
3
2
2
−
=
×
3
2
2
3
2
2
3
2
2
+
+
−
3
2
2
−
=
=
3
−
2
2
)
(
2
3
2
−
2
2
(
)(
)
(b)
1
=
p
+
q
2
=
1
p
+
q
2
3
+
2
2
=
3
p
+
4
q
+
2
p
2
+
3
q
2
ALT
3
+
2
2
3
p
+
4
q
=
1
2
p
+
3
q
=
0
3
+
4
=
1
6
+
8
=
2
p
q
p
q
= −
2
3
−
8
=
1
=
3
q
p
p
2
+
3
=
0
6
+
9
=
0
p
q
p
q
1
=
3
−
2
2
3
+
2 2
(c)
2
1
1
3
+
2
2
−
=
3
+
2
2
+
−
2
3
+
2 2
3
+
2
2
=
3
+
2
2
+ −
3
2 2
−
2
=
...
(
=
4)
1
so
3
+
2
2
−
=
2
3
+
2 2
(c)
1
3
+
2
2
−
=
2
3
+
2 2
−=
1
2
3
+
2
2
Alt
3
+
2
2
(
)
(
)
2
2
+
2 2
=
4
3
+
2
2
4
+
8 2
+
8
=
12
+
8 2
✓ Hence true
Notes:
(a)
M1: Expands the bracket to obtain 3 or 4 terms with at least 2 correct which may be unsimplified.
Allow use of a different variable e.g. x for this mark.
A1: Correct simplified expansion in terms of r. Accept terms in any order.
2
1
Accept e.g.
2
+
−
2
−
2
and accept
2
+
−
2
. Do not isw and mark their final answer.
r
r
r
r
Note that if a correct simplified expression is seen and they then rewrite the expression
correctly in a different way then this mark should be awarded provided their expression is
correct.
(b)
M1: Correct process to rationalise the denominator, so look for multiplying numerator and
denominator by 3
−
2
2
or any multiple of this expression. No processing is required so just look
for the statement as shown above.
A1: Shows an intermediate line before obtaining 3
−
2
2
Examples of an acceptable intermediate line are:
3
−
2
2
3
−
2
2
3
−
2
2
3
−
2
2
3
−
2
2
,
,
,
,
(
)
(
)
2
9
−
8
2
9
−
6 2
+
6
2
−
8
1
3
2
−
2
2
9
−
6 2
+
6 2
−
2
2
Do not allow 3
−
2
2
as the final answer.
1
(b) ALT
1
M1: Sets
=
+
2,
multiplies up, compares rational and irrational parts to produce 2
p
q
3
+
2 2
equations in p and q.
A1: Solves simultaneously showing working and obtains 3
−
2 2
.
(c)
2
1
M1: Applies their result from (a) or the correct result to
3
+
2
2
−
3
+
2 2
dM1: Depends on the first mark. Applies their result from (b) and cancels
2 terms to achieve a
constant.
A1: Reaches the correct answer from correct work. There is no requirement to justify the positive
square root.
(c) ALT
M1: Multiplies equation through by 3
2
2
and applies
3
2
2
3
2
2
3
2
+
+
×
+
=
+
dM1: Depends on the first mark. Squares both sides. May or may not have cancelled the 2 before
squaring.
A1: Achieves similar expression for both sides and gives a (minimal) conclusion.
Note that there will be other methods for 6(c) e.g.:
1
3
+
2
2
−
1
3
+
2
2
−
=
3
+
2
2
3
+
2
2
M1: Attempts common denominator and applies
3
+
2
2
×
3
+
2
2
=
3
+
2
2
2
2
+
2
2
=
12
+
8 2
3
−
2
2
36
−
32
×
=
=
4
9
−
8
3
+
2
2
3
−
2
2
3
+
2
2
dM1: Squares and attempts to rationalise the denominator
1
So
3
+
2
2
−
=
2
*
3
2
2
+
A1*: Fully correct work
6(b)[2]
=
4
×
1.186
=
4.744
l
(
)
6
4sin1.186
6
3.708...
2.29...
BC =
−
=
−
=
or
π
(
)
BC
=
6
−
4cos
−
1.186
=
6
−
3.708...
=
2.29...
2
or
(
)
12
−
55
=
6
−
4
2
−
1.5
2
=
6
−
3.708...
=
2.29...
=
BC
2
or
1.5
(
)
BC =
6
−
=
6
−
3.708...
=
2.29...
tan"0.384"
(
)
Perimeter is
4×
1.186...
+
6
−
4cos0.384...
+
1.5
+ +
6
4
=
..
= awrt 18.5 (m)
6(c)[3]
1
Area sector =
24
1.186...
2 ×
×
Area OBCD Examples:
1
or (
)
1
6 1.5
×
−
×
"3.708" 1.5
×
6
−
"3.708"
×
1.5
+
×
"3.708" 1.5
×
2
2
or (
1 6
)
1
+
6
−
"3.708"
×
1.5
or
6 1.5
×
−
×
1.5 4sin1.186
×
2
2
1
1
(
)
or
6 1.5
6
"3.708"
1.5
(= 6.219…)
××
+
−
×
2
2
1
×
4
2
×
1.18
6
+
6.21 . .
9 .
=
2
= awrt 15.7 (m2)
Notes:
Allow equivalent correct work in degrees.
Ignore any presence or absence of units throughout.
(a)
M1: For a complete and correct method to find angle AOB in radians or degrees. May attempt angle
π or equivalent
AOB directly using cosine, or angle BOD using sine then subtract angle BOD from
2
work using tan.
A1*: Obtains angle AOB = 1.186 with no errors.
Example in degrees:
1.
5
1 3
sin
∠
BOD
=
∠
BOD
=
sin
−
=
22.0
243
..
.
4
8
90
22.0243...
67.9756...
∠
AOB
=
−
=
π×
67.9756...
= 1.186 *
=
180
(b)
M1: Uses the correct arc length formula with r = 4 and 1.186. May be implied by awrt 4.74 or 4.75.
M1: Attempts the length of BC using an appropriate trig function and angle, or Pythagoras. May see
6
−
4sin 1.186
or 6
−
4cos"0.384"
or may use
6
−
4
2
−
1.5
2
. May be implied by awrt 2.3 provided
no incorrect working is seen.
ddM1: Depends on both previous marks. Adds the five lengths of the sides (or equivalent work)
to obtain the perimeter.
A1: Correct answer, accept awrt 18.5. No units required and ignore any given.
(c)
1
but may be implied by e.g. 1186
M1: Correct attempt at the sector area. Look for
2 ×
24
×
1.186...
1
π
awrt 9.5. Allow equivalent expressions in degrees e.g.
×
24
×
67.9756... 180
×
but not e.g.
2
1
24
67.9756
2 ×
×
M1: Correct attempt at the area of OBCD – which may be found as the difference of a rectangle and
triangle or sum of rectangle and triangle or a trapezium. See scheme for details.
The attempt to find BC, if used, must have involved the use of trigonometry or e.g. Pythagoras.
ddM1: Depends on both previous marks. Correct attempt at the full area, so adds sector area to
area of OBCD
A1: Correct answer, accept awrt 15.7. No units required and ignore any given.
Note that some candidates assume BC = 2 (i.e. 6 – 4) which greatly simplifies the question.
The maximum possible marks in (b) and (c) are: (b) M1M0ddM0A0 (c) M1M0ddM0A0
6(a)
2
1
1
1
1
−
=
−
×
−
×
+
r
r
2
r
r
r
r
r
r
2
1
=
2
+
−
2
r
2
r
(b)
1
1
3
2
2
−
=
×
3
2
2
3
2
2
3
2
2
+
+
−
3
2
2
−
=
=
3
−
2
2
)
(
2
3
2
−
2
2
(
)(
)
(b)
1
=
p
+
q
2
=
1
p
+
q
2
3
+
2
2
=
3
p
+
4
q
+
2
p
2
+
3
q
2
ALT
3
+
2
2
3
p
+
4
q
=
1
2
p
+
3
q
=
0
3
+
4
=
1
6
+
8
=
2
p
q
p
q
= −
2
3
−
8
=
1
=
3
q
p
p
2
+
3
=
0
6
+
9
=
0
p
q
p
q
1
=
3
−
2
2
3
+
2 2
(c)
2
1
1
3
+
2
2
−
=
3
+
2
2
+
−
2
3
+
2 2
3
+
2
2
=
3
+
2
2
+ −
3
2 2
−
2
=
...
(
=
4)
1
so
3
+
2
2
−
=
2
3
+
2 2
(c)
1
3
+
2
2
−
=
2
3
+
2 2
−=
1
2
3
+
2
2
Alt
3
+
2
2
(
)
(
)
2
2
+
2 2
=
4
3
+
2
2
4
+
8 2
+
8
=
12
+
8 2
✓ Hence true
Notes:
(a)
M1: Expands the bracket to obtain 3 or 4 terms with at least 2 correct which may be unsimplified.
Allow use of a different variable e.g. x for this mark.
A1: Correct simplified expansion in terms of r. Accept terms in any order.
2
1
Accept e.g.
2
+
−
2
−
2
and accept
2
+
−
2
. Do not isw and mark their final answer.
r
r
r
r
Note that if a correct simplified expression is seen and they then rewrite the expression
correctly in a different way then this mark should be awarded provided their expression is
correct.
(b)
M1: Correct process to rationalise the denominator, so look for multiplying numerator and
denominator by 3
−
2
2
or any multiple of this expression. No processing is required so just look
for the statement as shown above.
A1: Shows an intermediate line before obtaining 3
−
2
2
Examples of an acceptable intermediate line are:
3
−
2
2
3
−
2
2
3
−
2
2
3
−
2
2
3
−
2
2
,
,
,
,
(
)
(
)
2
9
−
8
2
9
−
6 2
+
6
2
−
8
1
3
2
−
2
2
9
−
6 2
+
6 2
−
2
2
Do not allow 3
−
2
2
as the final answer.
1
(b) ALT
1
M1: Sets
=
+
2,
multiplies up, compares rational and irrational parts to produce 2
p
q
3
+
2 2
equations in p and q.
A1: Solves simultaneously showing working and obtains 3
−
2 2
.
(c)
2
1
M1: Applies their result from (a) or the correct result to
3
+
2
2
−
3
+
2 2
dM1: Depends on the first mark. Applies their result from (b) and cancels
2 terms to achieve a
constant.
A1: Reaches the correct answer from correct work. There is no requirement to justify the positive
square root.
(c) ALT
M1: Multiplies equation through by 3
2
2
and applies
3
2
2
3
2
2
3
2
+
+
×
+
=
+
dM1: Depends on the first mark. Squares both sides. May or may not have cancelled the 2 before
squaring.
A1: Achieves similar expression for both sides and gives a (minimal) conclusion.
Note that there will be other methods for 6(c) e.g.:
1
3
+
2
2
−
1
3
+
2
2
−
=
3
+
2
2
3
+
2
2
M1: Attempts common denominator and applies
3
+
2
2
×
3
+
2
2
=
3
+
2
2
2
2
+
2
2
=
12
+
8 2
3
−
2
2
36
−
32
×
=
=
4
9
−
8
3
+
2
2
3
−
2
2
3
+
2
2
dM1: Squares and attempts to rationalise the denominator
1
So
3
+
2
2
−
=
2
*
3
2
2
+
A1*: Fully correct work

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