Mathematics · Oct / Nov 2024 · Paper 31 · Question 9
YMA01
As printed in the paper
The question as text
9. y P C l R O x Figure 5 Figure 5 shows a sketch of part of the curve C with equation y = f (x) where 2x f (x) = 6 4 x 2 x > – 1 x 2 1 2
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Show the answer and mark scheme
9(a)[3]
(
2
1
) (
12
4
)
2 6
2
4
2
x
+
×
x
+
−
x
+
x
−
f ( )
′
x
=
(
)
2
+
1
2
x
12
2
12
8
x
+
x
+
=
o.e
(
)
2
2
x
+
1
9(b)[3]
2
12 2
×
+
12 2
×+
8
16
At
2
f ( )
′
=
=
x =
x
(
)
2
5
2 2 1
×+
5
Full method of normal
(
)
y
−
6
= −
x
−
2
16
16
−
96
= −
5
+
10
16
+
5
=
106
*
y
x
y
x
9(c)[3]
1
5
Any correct value
=
3,
=
= −
A
B
or D
2
2
6
2
4
2
(
)(
2
1
)
Values of A, B and D
x
+
x
−
=
Ax
+
B
x
+
+
D
5
−
1
2
3
+
+
o.e
x
2
2
+
1
x
9(d)[5]
5
−
1
3
1
5
2
2
3
x
+
+
d
x
=
x
+
x
−
ln (2
x
+
1)
2
2
x
+
1
2
2
4
Area under curve =
2
2
3
1
5
3
1
5
3
1
1
1
5
5
2
2
x
+
x
−
ln (2
x
+
1)
=
×
2
+
×−
2
ln (5)
−
×
+
×
−
ln
2
2
4
2
2
4
2
3
2
3
4
3
1
3
2
5
Area of R =
1
−
1
106
2
3
+
+
d
+
××
6
−
2
x
x
2
2
x
+
1
2
5
1
3
Area of R = 964
5 ln3
−
15
4
(a)
M1: Condoning slips, this is for an attempt using
(
)
(
2
1
) (
)
6
2
4
2
x
+
×
Ax
+
B
−
C
x
+
x
−
the quotient rule to obtainMust be ‘−’ on the numerator
f ( )
′
x
=
(
)
2
+
1
2
x
B
the chain rule on 3f ( )
′
3
(
2
1
)
−
2
x
+
A
+
x
=
±
D
x
+
2
x
+
1
the product rule on ()(
(
)
(
)
−
1
(
)
6
2
+
4
−
2
2
+
1
−
1
f ( )
′
x
=
2
x
+
1
×
Ax
+
B
−
C
6
x
2
+
4
x
x
x
where A, B and C are > 0
A1: Correct (unsimplified) expression for f ( )
′
. Look for
x
(
)
(
2
1
) (
12
4
)
2 6
2
4
2
x
+
×
x
+
−
x
+
x
−
′
via the quotient rule
f ( )
x
=
(
)
2
2
+
1
x
) (
)
(
)
−
1
(
(
)
−
2
f ( )
′
x
=
2
x
+
1
×
12
x
+
4
+
6
x
2
+
4
x
−
2
×−
2 2
x
+
1
via the product rule
3
5 2
(
1
)
−
2
following division. Condone an incorrect value for A.
+
x
+
A1: Simplified answers such as (
)
4 3
2
+
3
+
2
x
x
12
2
+
12
+
8
x
x
(
)
2
,
or
3
+
5 2
+
1
−
following correct A.
x
(
2
1
)
2
2
x
+
4
+
4
+
1
x
x
ISW following a correct answer
(b)
M1: Substitutes
2
into their attempt at f ( )
′
.
x =
x
If f ( )
′
x
is incorrect then look for correctly embedded 2’s or a correct value.
dM1: Full method for finding the equation of the normal at (
)
2, 6 . Look for
y
−
6
= −
f (2)
5
A1*:
(
)
o.e with at least one correct intermediate line leading to 16
+
5
y
−
6
= −
x
−
2
y
16
5
106
5
106
If y
=
+
is used look for
6
2
You can allow
mx
c
= −
×
+
c
c
=
y
= −
x
+
16
16
16
16
the given answer. ISW following sight of correct answer. It must follow a correct derivative in (a)
(c)
1
5
B1: Any correct value for
=
3,
=
= −
.
A
B
or D
2
2
They may be embedded within an expression e.g. 3x in the quotient of a division sum
M1: Correct method for finding all three constants.
Via comparison/inspection look for
2
(
)(
)
Values of A, B and D
6
x
+
4
x
−
2
=
Ax
+
B
2
x
+
1
+
D
Via division look for a linear quotient and a constant remainder
5
−
1
1
5
A1:
2
o.e such as
which may be seen within an integral in (d).
3
+
+
3
+
−
x
x
2
2
x
+
1
2
2(2
x
+
1)
It is not just for values of A, B and D.
ISW following a correct answer (e.g. they may go on to 2
×but then the A1 ft in (d) would be unavailable)
(d)
D
M1: Attempts to integrate their
+
+
+ . Look for two terms in the correct form, one of which must
Ax
B
2
1
x
2
be the ln term. Award for
...
+
....
+
ln(2
+
1)
or ... ...
+
x
+
k
ln(2
x
+
1)
x
k
x
Also note that ln(4
x +
2)
and any multiple of (2
x +
1)
within the ln is also correct
k
3
1
5 ln (2
2
A1ft:
+
−
+
1)
o.e. but follow through on their values of A, B and D
x
x
x
2
2
4
dM1: A correct method of finding the area under the curve using limits of 1
3 and 2 either way around.
Dependent upon the previous M mark. Award if the intention is clear.
M1: For a correct method of finding the area of R. It is scored for adding the area under the curve to the
correct calculation for the area of the triangle. The integration of f(x) need not be correct but the limits
must be applied the correct way around. The area of the triangle may be done by integration. Look for
2
1
106
1
106
1
1
106
[
]
f( ) d
+
××
6
−
2
=
( )
2
+
××
6
−
2
=
−
+
××
6
−
x
x
I x
I
I
1
2
5
2
5
3
2
5
3
1
3
where (I x ) is their attempt at
f( ) d
x
x
A1: 964
5 ln3
−
15
4
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