OCR Further Pure Core 2 2023 June — Question 4 4 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2023
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeApplied context: real-world solid
DifficultyChallenging +1.2 This is a straightforward volume of revolution problem requiring the standard formula V = π∫y²dx. The integrand simplifies to 1/(3x²-3x+1), which factors as 1/(3x-1)(x-1) and yields to partial fractions, followed by routine logarithmic integration. While it requires multiple techniques (partial fractions, logarithms), these are standard Further Maths procedures with no conceptual surprises, making it slightly above average difficulty.
Spec4.08d Volumes of revolution: about x and y axes4.08h Integration: inverse trig/hyperbolic substitutions

4 In this question you must show detailed reasoning. The region \(R\) is bounded by the curve with equation \(\mathrm { y } = \frac { 1 } { \sqrt { 3 \mathrm { x } ^ { 2 } - 3 \mathrm { x } + 1 } }\), the \(x\)-axis and the lines with equations \(x = \frac { 1 } { 2 }\) and \(x = 1\) (see diagram). The units of the axes are cm . \includegraphics[max width=\textwidth, alt={}, center]{7b2bfb4e-524f-4d1c-ae98-075c7fb404f9-3_778_1241_497_242} A pendant is to be made out of a precious metal. The shape of the pendant is modelled as the shape formed when \(R\) is rotated by \(2 \pi\) radians about the \(x\)-axis. Find the exact value of the volume of precious metal required to make the pendant, according to the model.

Question 4:
AnswerMarks
4DR
2
 1 
1 −
V = (3x2 −3x+1) 2 dx
1 
AnswerMarks Guidance
2 M1 3.3
aAccept squared out expression
Condone omission of dx
1 1 1
V d x d x   =  = 
3 x 2 3 x 1 3 x 2 x 1 − + − +
3
1 12
d x  = 
3 1 1 1  
x − − +
2 4 3
1 1
d x  = 
3 1 2 1  
x − +
AnswerMarks Guidance
2 1 2M1 2.2a
completed square form1
or d x  
2
1 1  
3 x − +
2 4
1
or  dx
2
 3 1
 3x−  +
 2  4
   
1
x −
 1 d x = 1 t a n − 1 2
 1  2 1 1 1
x − +
AnswerMarks Guidance
2 1 2 1 2 1 2A1 1.1
3
may be equivalent based on
AnswerMarks
their form and their substitutionCould see a substitution
 1 
eg u = 3 x − with d u = 3 d x
2
2 t a n 1 ( 3 ) t a n 1 ( 0 )  =  − − − 
3
2 2 3 
2   = =
3 3 9
2 3
so 2  cm3
AnswerMarks Guidance
9A1 3.4
[4]
Question 4:
4 | DR
2
 1 
1 −
V = (3x2 −3x+1) 2 dx
1 
2  | M1 | 3.3 | Using V b y 2 d x  =  with limits
a | Accept squared out expression
Condone omission of dx
1 1 1
V d x d x   =  = 
3 x 2 3 x 1 3 x 2 x 1 − + − +
3
1 12
d x  = 
3 1 1 1  
x − − +
2 4 3
1 1
d x  = 
3 1 2 1  
x − +
2 1 2 | M1 | 2.2a | Expressing the integral in
completed square form | 1
or d x  
2
1 1  
3 x − +
2 4
1
or  dx
2
 3 1
 3x−  +
 2  4
   
1
x −
 1 d x = 1 t a n − 1 2
 1  2 1 1 1
x − +
2 1 2 1 2 1 2 | A1 | 1.1 | 2 t a n 1 ( 3 ( 2 x 1 ) )  =  − − 
3
may be equivalent based on
their form and their substitution | Could see a substitution
 1 
eg u = 3 x − with d u = 3 d x
2
2 t a n 1 ( 3 ) t a n 1 ( 0 )  =  − − − 
3
2 2 3 
2   = =
3 3 9
2 3
so 2  cm3
9 | A1 | 3.4 | oe | Do not penalise missing units
[4]
4 In this question you must show detailed reasoning.
The region $R$ is bounded by the curve with equation $\mathrm { y } = \frac { 1 } { \sqrt { 3 \mathrm { x } ^ { 2 } - 3 \mathrm { x } + 1 } }$, the $x$-axis and the lines with equations $x = \frac { 1 } { 2 }$ and $x = 1$ (see diagram). The units of the axes are cm .\\
\includegraphics[max width=\textwidth, alt={}, center]{7b2bfb4e-524f-4d1c-ae98-075c7fb404f9-3_778_1241_497_242}

A pendant is to be made out of a precious metal. The shape of the pendant is modelled as the shape formed when $R$ is rotated by $2 \pi$ radians about the $x$-axis.

Find the exact value of the volume of precious metal required to make the pendant, according to the model.

\hfill \mbox{\textit{OCR Further Pure Core 2 2023 Q4 [4]}}