Mathematics · Oct / Nov 2023 · Paper 31 · Question 5
YMA01
As printed in the paper
The question as text
5. The curve C has equation y = ln x 2 k x ∈ x 2 k where k is a positive constant.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[3]
2
x
(
x
2
+
k
)
×
−
2 ln(
x
x
2
+
k
)
d
y
x
2
+
k
=
d
x
(
x
2
+
k
)
2
d
y
2
x
−
2 ln(
x
x
2
+
k
)
2 (1 ln(
x
−
x
2
+
k
))
=
=
d
x
(
x
2
+
k
)
2
(
x
2
+
k
)
2
5(b)[3]
x =
0
"1
" ln(
−
x
2
+
k
)
=
0
x
2
=
e
"1"
±
k
x
= ±
e
−
k
5(c)[1]
Upper limit is e or k < e
Notes
(a)
M1:
Attempts to apply the quotient rule. Score for an expression of the form
...
(
x
2
+
k
)
×
−
Qx
ln(
x
2
+
k
)
d
y
x
2
+
k
oe (may be simplified) where … is P or Px and P,Q > 0
=
2
2
d
x
(
x
+
k
)
May also attempt the product rule, in which case look for
...
(
)
−
2
(
)
−
1
−
Px x
2
+
k
ln(
x
2
+
k
)
+
x
2
+
k
×
where … is Q or Qx and P,Q > 0
x
2
+
k
M1:
Cancels
2x
+
k
and correctly takes a factor of (2)x from the numerator to achieve “B – ln(x2+k)”
(may be done in one step). They must have a factor x in both terms. Note this may be scored
following M0 e.g. if a numerator uv’ – vu’ is used, or if the denominator misses the square.
d
y
2 (1 ln(
x
−
x
2
+
k
))
A1:
=
from fully correct work. Both M’s must have been scored.
d
x
(
x
2
+
k
)
2
(b)
B1:
x =
0
M1:
Sets
"1" ln(
−
x
2
+
k
)
=
0
and proceeds to
x =
2
...
with correct undoing of ln.
(
A1ft:
x
= ±
e
−
k
or follow through
for a numerical B in their
B
−
ln
x
2
x
= ±
eB
−
k
(c)
B1ft: Upper limit is e or k < e stated. Follow on their (b) of form
x
( )
eB
k
B0 if k = e or
= ±
±
k
e
is given as the answer without statement about upper limit being e.
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