Edexcel PURE 2024 October — Question 7

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicVolumes of Revolution
TypeRotation about x-axis: rational or reciprocal function
DifficultyStandard +0.3 This is a standard volumes of revolution question with algebraic fractions. Part (a) is routine partial fractions/algebraic division (A=3, B=-7). Part (b) requires squaring the expression and integrating term-by-term, which is methodical but straightforward for this topic. The integration involves standard forms and the final answer simplification is routine. Slightly easier than average due to the guided structure and standard techniques.
Spec1.02y Partial fractions: decompose rational functions1.08e Area between curve and x-axis: using definite integrals
7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fa121449-492f-4737-a9eb-a14a62ced47b-18_510_680_251_696} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} Figure 4 shows a sketch of part of the curve with equation $$y = \frac { 3 x - 1 } { x + 2 } \quad x > - 2$$
  1. Show that $$\frac { 3 x - 1 } { x + 2 } \equiv A + \frac { B } { x + 2 }$$ where \(A\) and \(B\) are constants to be found. The finite region \(R\), shown shaded in Figure 4, is bounded by the curve, the line with equation \(x = 4\), the \(x\)-axis and the line with equation \(x = 1\) This region is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
  2. Use the answer to part (a) and algebraic integration to find the exact volume of the solid generated, giving your answer in the form $$\pi ( p + q \ln 2 )$$ where \(p\) and \(q\) are rational constants.
Question 7:
AnswerMarks
7(a)3 x − 1 . .+ . 7
= 3 + or ... −
AnswerMarks
x + 2 x 2 x + 2M1
3 x − 1 7+
= 3 −
AnswerMarks
x + 2 x 2A1
(2)
AnswerMarks
(b)2
V ( ) 3 7 d x ...  =   −  =
AnswerMarks
x 2 +M1
 
 42 49
=() 9− + dx=...−42ln ( x+2 )+...
 x+2 ( x+2 )2 
 
or
 
4 2 4 9 4 9 
( ) 9 d x . . . . . .  = − + = + −
AnswerMarks
x 2 ( x 2 ) 2 x 2 + + +M1
 
4 2 4 9 
( ) 9 d x ... 4 2 l n ( x 2 ) ...  = − + = − + +
x 2 ( x 2 ) 2 + +
and
 
4 2 4 9 4 9 
( ) 9 d x . . . . . .  = − + = + −
AnswerMarks
x 2 ( x 2 ) 2 x 2 + + +dM1
 
4 2 4 9 4 9 
( ) 9 d x 9 x 4 2 l n ( x 2 )  = − + = − + −
AnswerMarks
x 2 ( x 2 ) 2 x 2 + + +A1
4 9 4 4 9 4 9      
V ( ) 9 x 4 2 l n ( x 2 ) ( ) 3 6 4 2 l n 6 9 4 2 l n 3   = − + − = − − − − −
x 2 6 3 +
AnswerMarks
1M1
2 1 1  
4 2 l n 2  = −
AnswerMarks
6A1
(6)

Total 8

M1: Integrates to obtain the correct form for one of the fractional terms i.e.
x
...
+ 2
→ ... l n ( 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 lnx+2
dM1: Integrates to obtain the correct form for both fractional terms i.e.
x
...
+ 2
→ ... l n ( x + 2 ) and
... ...
→ . It is dependent on the previous method mark. Condone invisible brackets
( x+2 )2 x+2
for lnx+2
49
A1: 9x−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 A x  B l n ( 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.
211 
A1:   −42ln2
 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 substitution e .g . u = x + 2
or e.g. u =
x
7
+ 2
. Watch out for these longer methods which can still score full marks.
M1: If using u = x + 2 then ( )
3
x
x 1
2
2
d x ( ) 9
4
u
2 4
u
9
2
d u  
 
+
− 

− + (They will need to divide
their numerator by the denominator to achieve
( x
x
2 ) 2
 
 +
+
+
). Condone slips.
If using u =
x
7
+ 2
then ( )
4
u
2 6
u
3
2
7 d u 

− − Condone slips.
...
M1: As above in the main scheme notes for one of the fractional terms i.e. →...lnu or
u
... ...
→ .
u2 u
dM1: Integrates to obtain the correct form for both fractional terms
49 63 7
A1: 9u−42lnu− if using u=x+2 or 42lnu+ −7u 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
3
,
7
6
if using u =
x
7
+ 2
PMT
A1: As above in main scheme notes.
Question 7:
--- 7(a) ---
7(a) | 3 x − 1 . .+ . 7
= 3 + or ... −
x + 2 x 2 x + 2 | M1
3 x − 1 7+
= 3 −
x + 2 x 2 | A1
(2)
(b) | 2
V ( ) 3 7 d x ...  =   −  =
x 2 + | M1
 
 42 49
=() 9− + dx=...−42ln ( x+2 )+...
 x+2 ( x+2 )2 
 
or
 
4 2 4 9 4 9 
( ) 9 d x . . . . . .  = − + = + −
x 2 ( x 2 ) 2 x 2 + + + | M1
 
4 2 4 9 
( ) 9 d x ... 4 2 l n ( x 2 ) ...  = − + = − + +
x 2 ( x 2 ) 2 + +
and
 
4 2 4 9 4 9 
( ) 9 d x . . . . . .  = − + = + −
x 2 ( x 2 ) 2 x 2 + + + | dM1
 
4 2 4 9 4 9 
( ) 9 d x 9 x 4 2 l n ( x 2 )  = − + = − + −
x 2 ( x 2 ) 2 x 2 + + + | A1
4 9 4 4 9 4 9      
V ( ) 9 x 4 2 l n ( x 2 ) ( ) 3 6 4 2 l n 6 9 4 2 l n 3   = − + − = − − − − −
x 2 6 3 +
1 | M1
2 1 1  
4 2 l n 2  = −
6 | A1
(6)
Total 8
M1: Integrates to obtain the correct form for one of the fractional terms i.e.
x
...
+ 2
→ ... l n ( 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 lnx+2
dM1: Integrates to obtain the correct form for both fractional terms i.e.
x
...
+ 2
→ ... l n ( x + 2 ) and
... ...
→ . It is dependent on the previous method mark. Condone invisible brackets
( x+2 )2 x+2
for lnx+2
49
A1: 9x−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 A x  B l n ( 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.
211 
A1:   −42ln2
 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 substitution e .g . u = x + 2
or e.g. u =
x
7
+ 2
. Watch out for these longer methods which can still score full marks.
M1: If using u = x + 2 then ( )
3
x
x 1
2
2
d x ( ) 9
4
u
2 4
u
9
2
d u  
 
+
− 


− + (They will need to divide
their numerator by the denominator to achieve
( x
x
2 ) 2
 
 +
+
+
). Condone slips.
If using u =
x
7
+ 2
then ( )
4
u
2 6
u
3
2
7 d u 

− − Condone slips.
...
M1: As above in the main scheme notes for one of the fractional terms i.e. →...lnu or
u
... ...
→ .
u2 u
dM1: Integrates to obtain the correct form for both fractional terms
49 63 7
A1: 9u−42lnu− if using u=x+2 or 42lnu+ −7u 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
3
,
7
6
if using u =
x
7
+ 2
PMT
A1: As above in main scheme notes.
7.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{fa121449-492f-4737-a9eb-a14a62ced47b-18_510_680_251_696}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}

Figure 4 shows a sketch of part of the curve with equation

$$y = \frac { 3 x - 1 } { x + 2 } \quad x > - 2$$
\begin{enumerate}[label=(\alph*)]
\item Show that

$$\frac { 3 x - 1 } { x + 2 } \equiv A + \frac { B } { x + 2 }$$

where $A$ and $B$ are constants to be found.

The finite region $R$, shown shaded in Figure 4, is bounded by the curve, the line with equation $x = 4$, the $x$-axis and the line with equation $x = 1$

This region is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.
\item Use the answer to part (a) and algebraic integration to find the exact volume of the solid generated, giving your answer in the form

$$\pi ( p + q \ln 2 )$$

where $p$ and $q$ are rational constants.
\end{enumerate}

\hfill \mbox{\textit{Edexcel PURE 2024 Q7}}
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