Mathematics · Oct / Nov 2024 · Paper 41 · Question 7
YMA01
As printed in the paper
The question as text
7. y R O 1 4 x Figure 4 Figure 4 shows a sketch of part of the curve with equation 3 x 1 y x 2 x 2
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[2]
7
3
x
−
1
...
=
3
+
or
...
−
x
+
2
x
+
2
x
+
2
3
x
1
7
−
=
3
−
x
+
2
x
+
2
7(b)[6]
7
2
( )
V
=
π
3
−
d
x
=
...
x
+
2
42
49
( )
(
)
=
π
9
−
+
d
x
=
... 42ln
−
x
+
2
+
...
x
+
2
(
)
2
x
+
2
or
42
49
49
( )
=
π
9
−
+
d
x
=
... ...
+
−
x
+
2
(
)
2
x
+
2
x
+
2
42
49
( )
(
)
=
π
9
−
+
d
x
=
... 42ln
−
x
+
2
+
...
x
+
2
(
)
2
x
+
2
and
42
49
49
( )
=
π
9
−
+
d
x
=
... ...
+
−
x
+
2
(
)
2
x
+
2
x
+
2
42
49
49
( )
(
)
=
π
9
−
+
d
x
=
9
x
−
42ln
x
+
2
−
x
+
2
(
2
)
2
x
+
2
x
+
49
4
π
49
49
✓
( )
(
)
( )
V
=
π
9
x
−
42ln
x
+
2
−
=
36
−
42ln6
−
−
9
−
42ln3
−
x
+
2
6
3
1
211
π
=
−
42ln 2
6
(a)
M1:
Obtains either A = 3 or
B = −
7
7
3
A1:
−
Just stating the values of A and B is insufficient but you may award this mark if it
x
+
2
−
7
7
is seen in (b). Allow
3
+
and
3
+ −
x
+
2
x
+
2
...
Condone transcription errors for the method marks e.g.
...ln
(
2
)
provided
(b)
→
x
−
x
+
2
no incorrect method is seen. Condone invisible brackets for the M marks
M1:
Uses their result from part (a) and applies the volume formula (with or without the π and
condone 2π ) and attempts to square the bracket. Condone if they just square the two terms
in the brackets or if their squaring is poor.
...
M1:
Integrates to obtain the correct form for one of the fractional terms i.e.
(
→
...ln
x
+
x
+
2
...
...
or
. This can be scored even if they have not squared any terms originally
→
(
)
x
+
2
2
x
+
2
if they have one of these terms of the required form. Condone invisible brackets for ln
...
dM1: Integrates to obtain the correct form for both fractional terms i.e.
→
...ln
(
x
+
2
)
and
x
+
2
...
...
→
. It is dependent on the previous method mark. Condone invisible brackets
(
2
)
2
x
+
2
x
+
for ln
x +
2
49
(
)
A1:
9
x
−
42ln
x
+
2
−
o.e. May be implied by further work.
x
+
2
M1:
Applies the limits 1 and 4 to an expression of the form
Ax
±
B
ln(
x
+
2)
±
C x
(
+
2)n
where
A B C n ≠
, , ,
0
and subtracts the correct way round. The expression with values embedded is
sufficient. Condone slips and invisible brackets.
π
211
A1:
42ln 2
−
6
Correct expression in the required form. The fraction does not need to be in simplest
form.
Alternative methods (including not using part (a)) requiring use of a substitutione.g.
u
=
7
or e.g.
u
=
. Watch out for these longer methods which can still score full marks.
x
+
2
2
3
x
−
1
42
49
M1:
If using
u
=
x
+
2
then
( )
d
x
( )
9
d
u
(They will need to divide
π
π
−
+
x
+
2
u
u
2
β
x
+
γ
their numerator by the denominator to achieve
α
+
). Condone slips.
(
x
+
2)
2
7
42
63
If using
u
=
then
( )
π
−
−
7du
Condone slips.
x
+
2
u
u
2
As above in the main scheme notes for one of the fractional terms i.e. ...
M1:
u →
...lnu
or
...
...
→
.
u
2
u
dM1: Integrates to obtain the correct form for both fractional terms
49
63
7
A1:
9
u
−
42ln
u
−
if using
u
=
x
+
2
or
42ln
u
+
−
7
u
if using
u
=
u
u
x
+
2
M1:
Applies the limits 1 and 4 to an expression of the required form (see above in main scheme
notes) if the expression is converted back to being in terms of x
or applies the limits 3 and 6 if in terms of u if using
u
=
x
+
2
or e.g. 7
7
7
,
6 if using
u
=
3
x
+
2
A1:
As above in main scheme notes.
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