Mathematics · Oct / Nov 2020 · Paper 31 · Question 9
YMA01
As printed in the paper
Show the answer and mark scheme
9(a)[4]
2
x
+
2
2
4
3
2
x
−
x
−
12
x
−
x
−
10
x
+
3
x
−
9
4
3
2
x
−
x
−
12
x
2
x
2
+
3
x
−
9
2
x
2
−
2
x
−
24
5
x
+
15
4
3
10
2
3
9
5(
3)
x
−
x
−
x
+
x
−
x
+
≡
x
2
+
2
+
x
2
+
x
−
12
(
x
−
4)(
x
+
3)
5
≡
x
2
+
2
+
P
=
2,
Q
=
5
or stating
(
x
−
4)
5
g'( )
x
=
2
x
−
9(b)[5]
(
4)
2
x
−
5
11
Subs
x =
2
into
g'(2)
=
2
×
2
−
=
2
4
(2
−
4)
11
7
m
=
g'(2)
=
(2, g(2))
=
2, 2
Uses
with
to form equation of tangent
4
7
11
11
(
2
)
2
y
−
=
x
−
⇒
y
=
x
−
2
4
4
5
1
∫
x
2
+
2
+
d
x
=
x
3
+
2 , 5ln
x
+
x
−
4
9(c)[5]
(
x
−
4)
3
2
1
8
(
)
Area R =
x
3
+
2
x
+
5ln
x
−
4
=
+
4
+
5ln 2
−
0
+
0
+
5ln 4
3
3
0
20
20
=
+
5ln 2
−
5ln 4
=
−
5ln 2
3
3
Alt (a)
x
4
−
x
3
−
10
x
2
+
3
x
−
9
≡
(
x
2
+
P
)(
x
2
−
x
−
12)
+
Q x
(
+
3)
Compare terms (OR sub in values) and solve simultaneously to find P and/or Q
2
ie
x
⇒
P
−
12
= −
10,
x
⇒−
P
+
Q
=
3,
const
⇒−
12 +3
P
Q
=
−
9
⇒
P
=
...
Q =
...
or
P
=
2,
Q
=
5
*
Award in the order shown here on ePEN.
(a)
M1: Divides to obtain a quadratic quotient and a linear remainder. May divide by (
x −
4)
and then by (
vice versa to reach these but must be a full process.
5
FYI: By x + 3 first gives
x
3
−
4
x
2
+
2
x
−
3
as quotient, followed by
x
2
+
2
+
x
−
4
x
−
4
x
+
3
(
x
+
3)
(
x
−
4
)
x
−
A1: Obtains a quotient of
x +
2
2
and a remainder of 5
x +
15
2
M1: Writes the given expression in the required form using
x
−
x
−
12
=
(
x
−
4
)(
x
+
3
)
or divides
their remainder term.
A1: Correct answer. (
)
"
2
2"
P
=
2,
Q
=
5
May be awarded following an incorrect
ax +
quadratic factor.
Alt:
M1: Multiplies though completely by the denominator and cancels the
x −
4
term.
M1: Complete process of comparing coefficients or substituting values to find a value for either P or Q
A1:
Either P = 2 or for showing Q = 5* (must have seen a correct equation for Q)
A1: Both P = 2 and showing Q = 5* Note that Q = 5 is given so it must be shown from correct work, not just stated.
Note M0M1A1A0 is possible if Q is assumed and factorisation of
x
2
−
x
−
12
is never seen.
(b)
Q
...
M1: For
→
x
−
4
(
)
x
−
4
2
5
A1: For
g'( )
x
=
2
x
−
. Note that this can be scored from an incorrect P.
(
x
−
4)
2
x =
2
M1: Attempts the gradient of C at the point where
dM1: Depends on previous M. A complete method of finding the equation of the tangent. If y
=
mx
must proceed as far as finding c.
A1:
y
=
2.75
x
−
2
or exact equivalent and isw.
x
4
x
3
10
x
2
3
x
9
−
−
+
−
Note: This may be attempted from the original function
g( )
x
=
x
2
+
x
−
12
M1: Scored for an attempt at the quotient rule A1 if correct and so on.
(
ax
3
+
bx
2
+
cx
+
d
)
(
x
2
+
x
−
12)
−
(
x
4
−
x
3
−
10
x
2
+
3
x
−
9)(
px
+
q
)
→
(
x
2
+
x
−
12
)
2
(c)
..
∫
M1: Attempts to integrate with
4) x
d
→
..ln
x
−
4
Condone ln(
x −
4)
(
x
−
5
1
∫
2
3
A1ft:
x
+
P
+
d
x
=
x
+
Px
+
5ln
x
−
4
following through on their P.
(
4)
3
x
−
dM1: Dependent on first M mark. Attempt the area of R using the limits 0 and 2 in their integrated function and
subtracting the correct way round (or recovered).
ddM1: Depends on both previous M’s. Scored for attempting to combine two log terms using correct log work
−
2
ln(
2)
ln( 4
)
ln
ln 2
Allow the method and final accuracy if
−
−
−
→
= −
is used (bod that modulus is meant)
4
−
Do not allow if ln(
−
a
)
→−
ln
a
is used.
Note: If a candidate gives correct values of m and c in (b) and of a and b in (c) but has not stated the answer in correct
form, then penalise only the first instance.
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