Mathematics · Oct / Nov 2021 · Paper 31 · Question 6
YMA01
As printed in the paper
Show the answer and mark scheme
6(i)[4]
d
y
6
x
8
x
=
−
d
x
x
2
5
−
6
x
23
23
8
0
2
Stationary point when
−
x
=
⇒
x
=
⇒
x
=
only
4
2
2
5
x
−
(ii)(a)
d
y
4
24sin
cos
=
−
x
x
d
x
d
y
=
4 12sin 2
−
x
(b)
d
x
Maximum gradient = 16
(i)
(
)
Ax
A
M1
For differentiating
3ln
x
2
−
5
→
or
x
2
−
5
x −
2
5
d
y
6
x
6
x
8
x
8
x
A1
=
−
seen or implied by eg
=
d
x
x
2
5
x
2
5
−
−
d
y
Ax
dM1
For proceeding from
=
−
Bx
=
0
to a quadratic or cubic equation. Eg
d
x
x
2
−
5
6
x
−
8
x
3
+
40
x
=
0
oe
Do not be too concerned with the mechanics of their rearrangement.
23
For (
)
A1
x =
only provided dM1 has been scored. Solving the quadratic or cubic
2
with a calculator is acceptable. Condone recovery of slips provided the method is
correct.
23
x =
0
and
x = −
must be rejected if found or condone some minimal conclusion
2
eg stating
p
=
23
Mark (ii) (a) + (b) together
(ii) Way One
2
M1
For differentiating using the chain rule with
sin
x
→
...sin cos
x
x
x to proceed to d
y
dM1
.....and then using sin 2
x
=
2sin cos
x
=
A
+
B
sin 2
x
. Condone
d
x
A
+
B
sin
x provided their coefficient of sin cos
x
x has been halved or
the slip writing
writing 2sin cos x
d
y
(
)
A1
or
4
+ −
12 sin 2x
or 4 12 sin 2x
−
×
(or states A and B
=
4 12sin 2
−
x
d
x
following a correct method) Condone recovery of slips provided the method is correct.
A1
Maximum gradient = 16 (the previous A1 must have been scored)
(ii) Way Two
cos2
x
= ± ±
1 2sin
2
x
y
=
4
x
−
12sin
2
x
M1
Attempts to use
in an attempt to write
terms of cos2x
to proceed to d
y
dM1
...and then differentiates cos2
x
→
k
sin 2
x
=
A
+
B
sin 2
x
d
x
d
y
(
)
A1
=
4 12sin 2
−
x
or
4
+ −
12 sin 2x
or 4 12 sin 2x
−
×
(or states A and B
d
x
following a correct method) Condone recovery of slips provided the method is correct.
A1
Maximum gradient = 16 (the previous A1 must have been scored)
(Note some candidates may find the second derivative and set equal to 0 and then substitute in
their angle to find the maximum gradient.)
6(ii)[4]
d
y
6
x
8
x
=
−
d
x
x
2
5
−
6
x
23
23
8
0
2
Stationary point when
−
x
=
⇒
x
=
⇒
x
=
only
4
2
2
5
x
−
(ii)(a)
d
y
4
24sin
cos
=
−
x
x
d
x
d
y
=
4 12sin 2
−
x
(b)
d
x
Maximum gradient = 16
(i)
(
)
Ax
A
M1
For differentiating
3ln
x
2
−
5
→
or
x
2
−
5
x −
2
5
d
y
6
x
6
x
8
x
8
x
A1
=
−
seen or implied by eg
=
d
x
x
2
5
x
2
5
−
−
d
y
Ax
dM1
For proceeding from
=
−
Bx
=
0
to a quadratic or cubic equation. Eg
d
x
x
2
−
5
6
x
−
8
x
3
+
40
x
=
0
oe
Do not be too concerned with the mechanics of their rearrangement.
23
For (
)
A1
x =
only provided dM1 has been scored. Solving the quadratic or cubic
2
with a calculator is acceptable. Condone recovery of slips provided the method is
correct.
23
x =
0
and
x = −
must be rejected if found or condone some minimal conclusion
2
eg stating
p
=
23
Mark (ii) (a) + (b) together
(ii) Way One
2
M1
For differentiating using the chain rule with
sin
x
→
...sin cos
x
x
x to proceed to d
y
dM1
.....and then using sin 2
x
=
2sin cos
x
=
A
+
B
sin 2
x
. Condone
d
x
A
+
B
sin
x provided their coefficient of sin cos
x
x has been halved or
the slip writing
writing 2sin cos x
d
y
(
)
A1
or
4
+ −
12 sin 2x
or 4 12 sin 2x
−
×
(or states A and B
=
4 12sin 2
−
x
d
x
following a correct method) Condone recovery of slips provided the method is correct.
A1
Maximum gradient = 16 (the previous A1 must have been scored)
(ii) Way Two
cos2
x
= ± ±
1 2sin
2
x
y
=
4
x
−
12sin
2
x
M1
Attempts to use
in an attempt to write
terms of cos2x
to proceed to d
y
dM1
...and then differentiates cos2
x
→
k
sin 2
x
=
A
+
B
sin 2
x
d
x
d
y
(
)
A1
=
4 12sin 2
−
x
or
4
+ −
12 sin 2x
or 4 12 sin 2x
−
×
(or states A and B
d
x
following a correct method) Condone recovery of slips provided the method is correct.
A1
Maximum gradient = 16 (the previous A1 must have been scored)
(Note some candidates may find the second derivative and set equal to 0 and then substitute in
their angle to find the maximum gradient.)
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