AQA Further AS Paper 1 2018 June — Question 9 6 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2018
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConic sections
TypeParabola area calculations
DifficultyStandard +0.3 This is a straightforward volumes of revolution question requiring integration of y² = 4x rotated about the x-axis, then solving for d given V = 1000. The setup is standard (π∫y²dx = π∫4x dx = 2πx²), requiring only basic integration and algebraic manipulation. While it's Further Maths content, it's a routine application of a core technique with no conceptual challenges or novel insights needed.
Spec4.08d Volumes of revolution: about x and y axes

  1. Sketch the graph of \(y^2 = 4x\) [1 mark] \includegraphics{figure_9a}
  2. Ben is using a 3D printer to make a plastic bowl which holds exactly \(1000\text{cm}^3\) of water. Ben models the bowl as a region which is rotated through \(2\pi\) radians about the \(x\)-axis. He uses the finite region enclosed by the lines \(x = d\) and \(y = 0\) and the curve with equation \(y^2 = 4x\) for \(y \geq 0\)
    1. Find the depth of the bowl to the nearest millimetre. [4 marks]
    2. What assumption has Ben made about the bowl? [1 mark]

Question 9:

AnswerMarks Guidance
9(a)Sketches correct parabola 1.2

AnswerMarks Guidance
9(b)(i)OLibmtaitisn sn o 𝜋𝜋t r�eq4u𝑥𝑥ir 𝑑𝑑e𝑥𝑥d for this mark.
Condone missing3.3 M1
2
volume=𝜋𝜋�𝑦𝑦 𝑑𝑑𝑥𝑥
0
𝑑𝑑
∴ 1000=𝜋𝜋�4𝑥𝑥 𝑑𝑑𝑥𝑥
0
2 𝑑𝑑
1000 4𝑥𝑥
=� �
𝜋𝜋 2 0
1000 2
=2𝑑𝑑 −0
𝜋𝜋
2
2𝜋𝜋𝑑𝑑 =1000
500
𝑑𝑑de=pt � h = 12.6 cm
𝜋𝜋
𝑑𝑑𝑥𝑥
Obtains and uses limits of and 0 (oe).
2
AnswerMarks Guidance
2𝑥𝑥 𝑑𝑑1.1b B1
Forms an equation of the form ( oe)
AnswerMarks Guidance
23.4 M1
𝑘𝑘𝑑𝑑 = volume
Correct depth to nearest millimetre.
Condone 126 or 12.6 without units.
NMS: 126 or 12.6 scores 4/4.
AnswerMarks Guidance
Using 1000000 mm3 leads to a correct answer of 399 mm for 4/4.3.2a A1

AnswerMarks Guidance
9(b)(ii)States appropriate assumption 3.5b
Total6
QMarking instructions AO
Question 9:
--- 9(a) ---
9(a) | Sketches correct parabola | 1.2 | B1
--- 9(b)(i) ---
9(b)(i) | OLibmtaitisn sn o 𝜋𝜋t r�eq4u𝑥𝑥ir 𝑑𝑑e𝑥𝑥d for this mark.
Condone missing | 3.3 | M1 | 𝑑𝑑
2
volume=𝜋𝜋�𝑦𝑦 𝑑𝑑𝑥𝑥
0
𝑑𝑑
∴ 1000=𝜋𝜋�4𝑥𝑥 𝑑𝑑𝑥𝑥
0
2 𝑑𝑑
1000 4𝑥𝑥
=� �
𝜋𝜋 2 0
1000 2
=2𝑑𝑑 −0
𝜋𝜋
2
2𝜋𝜋𝑑𝑑 =1000
500
𝑑𝑑de=pt � h = 12.6 cm
𝜋𝜋
𝑑𝑑𝑥𝑥
Obtains and uses limits of and 0 (oe).
2
2𝑥𝑥 𝑑𝑑 | 1.1b | B1
Forms an equation of the form ( oe)
2 | 3.4 | M1
𝑘𝑘𝑑𝑑 = volume
Correct depth to nearest millimetre.
Condone 126 or 12.6 without units.
NMS: 126 or 12.6 scores 4/4.
Using 1000000 mm3 leads to a correct answer of 399 mm for 4/4. | 3.2a | A1
--- 9(b)(ii) ---
9(b)(ii) | States appropriate assumption | 3.5b | B1 | The thickness of the plastic is negligible
Total | 6
Q | Marking instructions | AO | Mark | Typical solution
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $y^2 = 4x$
[1 mark]

\includegraphics{figure_9a}

\item Ben is using a 3D printer to make a plastic bowl which holds exactly $1000\text{cm}^3$ of water.

Ben models the bowl as a region which is rotated through $2\pi$ radians about the $x$-axis.

He uses the finite region enclosed by the lines $x = d$ and $y = 0$ and the curve with equation $y^2 = 4x$ for $y \geq 0$

\begin{enumerate}[label=(\roman*)]
\item Find the depth of the bowl to the nearest millimetre.
[4 marks]

\item What assumption has Ben made about the bowl?
[1 mark]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2018 Q9 [6]}}