Mathematics · Oct / Nov 2020 · Paper 41 · Question 6
YMA01
As printed in the paper
The question as text
6. A curve C has equation y = xsin x x > 0 y > 0
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[5]
y
=
x
sin
x
⇒
ln
y
=
sin ln
x
x
Differentiates 1 d
y
sin
x
=
+
ln cos
x
x
y
d
x
x
d
y
y
sin
x
=
+
y
ln cos
x
x
oe
d
x
x
Puts d
y
⇒ sin
x
0
6(b)[2]
x =
+
ln
x
cos
x
=
0
d
x
sin
x
ln
0
tan
ln
0*
x +
x
x
=
⇒
x
+
x
x
=
cos
(a)
B1:
y
=
x
sin
x
⇒
ln
y
=
sin ln
x
x
o.e. Condone this writtenlog
y
=
sin
x
log
x
ln
y
This may have been found via
log
y
sin
x
sin
x
=
→
=
ln
x
x
1
d
y
1
d
y
M1: For differentiation of
ln
y
→
×
or
log
y
→
×
e
y
d
x
y
d
x
M1: Attempts the product rule on sin ln
x
x If the rule is quoted it must be correct.
If candidates set
u
=
..., '
u
=
...,
v
=
..., '
v
=
...
it may be implied by their
vu
'
+
uv
'
If it is neither quoted nor implied score for sin
x
±
ln cos
x
x
x
A1: Correct differentiation to 1 d
sin
y
x
=
+
ln cos
x
x
y
d
x
x
A1: Achieves d
sin
or exact equivalent such as d
sin
ln
cos
y
y
x
y
x
+
x
x
=
+
y
ln cos
x
x
=
y
d
x
x
d
x
x
d
y
x
sin
x
sin
x
Accept this in terms of just x. For instance
=
+
x
sin
x
ln
x
cos
x
d
x
x
ISW after sight of a correct answer.
..........................................................................................................................................................................
Note that you will see candidates who write 1
sin
x
d
y
=
d
x
+
ln cos
x
x x
d
which can be marked in the same
y
x
way
.................................................................................................................................................................................
(b)
M1: Sets their d
y
x =
0
, divides by y or
x
sin x
to form a simplified equation in x.
d
This may be implied by d
y
y
sin
x
=
+
y
ln cos
x
x
=
0
⇒
sin
x
+
x
ln cos
x
x
=
0
o.e.
d
x
x
For this to be scored d
y
x must be in the form
d
y
sin
x
±
y
ln cos
x
x
...
sin
x
x
sin
x
sin
x
±
x
ln
x
cos
x
where ... could be 1
...
A1*: CSO Proceeds to the given answer of tan
x
+
x
ln
x
=
0
showing at least the intermediate line sin
x
x
ln
x
0
x +
=
cos
(
)
It can be implied by
sin
x
+
x
ln cos
x
x
=
0
÷
cos
x
⇒
tan
x
+
x
ln
x
=
0
y
=
x
sin
x
⇒
y
=
e
sin ln
x
x
6 (a)
d
y
y
=e
sin ln
x
x
→
=e
sin ln
x
x
×
...
Differentiates
d
x
d
sin
x
(
)
sin ln
x
x
=
+
ln cos
x
x
d
x
x
d
y
sin
x
=e
sin ln
x
x
×
+
ln cos
x
x
d
x
x
B1:
y
=
x
sin
x
⇒
y
=
e
sin ln
x
x
M1: Differentiates
e
sin ln
x
x
→
e
sin ln
x
x
×
...
where ... could be 1
M1: Attempts the product rule on sin ln
x
x. See main scheme on how to apply this mark.
sin
x
A2: Correct differentiation
=
+
ln cos
x
x
x
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