AQA C3 — Question 6

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeRearrange to iterative form
DifficultyModerate -0.5 This is a standard C3 iteration question with routine tasks: showing a root exists by sign change, algebraic rearrangement (trivial), performing 2-3 iterations with a calculator, and drawing a cobweb diagram. All are textbook exercises requiring no problem-solving insight, making it slightly easier than average.
Spec1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

6 [Figure 1, printed on the insert, is provided for use in this question.]
The curve \(y = x ^ { 3 } + 4 x - 3\) intersects the \(x\)-axis at the point \(A\) where \(x = \alpha\).
  1. Show that \(\alpha\) lies between 0.5 and 1.0.
  2. Show that the equation \(x ^ { 3 } + 4 x - 3 = 0\) can be rearranged into the form \(x = \frac { 3 - x ^ { 3 } } { 4 }\).
    (1 mark)
    1. Use the iteration \(x _ { n + 1 } = \frac { 3 - x _ { n } { } ^ { 3 } } { 4 }\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\), giving your answer to two decimal places.
    2. The sketch on Figure 1 shows parts of the graphs of \(y = \frac { 3 - x ^ { 3 } } { 4 }\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis.
      (3 marks)

Question 6:
(a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Decreasing curve from \((-1,\ \pi)\) to \((1,\ 0)\)B1 Correct shape
End points \((-1,\ \pi)\) and \((1,\ 0)\) statedB1
(b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
Increasing curve; reflection of \(\cos^{-1}x\) in line \(y = \frac{\pi}{2}\)B1 Correct shape
End points \((-1,\ 0)\) and \((1,\ \pi)\) statedB1
I can see these are exam questions from AQA MPC3 June 2013, but I don't have access to the actual mark scheme document — the images shown contain only the question paper, not the mark scheme.
What I can do is provide worked solutions with likely mark allocations based on standard AQA marking conventions:
## Question 6:

**(a)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Decreasing curve from $(-1,\ \pi)$ to $(1,\ 0)$ | B1 | Correct shape |
| End points $(-1,\ \pi)$ and $(1,\ 0)$ stated | B1 | |

**(b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Increasing curve; reflection of $\cos^{-1}x$ in line $y = \frac{\pi}{2}$ | B1 | Correct shape |
| End points $(-1,\ 0)$ and $(1,\ \pi)$ stated | B1 | |

I can see these are exam questions from AQA MPC3 June 2013, but I don't have access to the actual mark scheme document — the images shown contain only the **question paper**, not the mark scheme.

What I can do is provide **worked solutions** with likely mark allocations based on standard AQA marking conventions:

---
6 [Figure 1, printed on the insert, is provided for use in this question.]\\
The curve $y = x ^ { 3 } + 4 x - 3$ intersects the $x$-axis at the point $A$ where $x = \alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ lies between 0.5 and 1.0.
\item Show that the equation $x ^ { 3 } + 4 x - 3 = 0$ can be rearranged into the form $x = \frac { 3 - x ^ { 3 } } { 4 }$.\\
(1 mark)
\item \begin{enumerate}[label=(\roman*)]
\item Use the iteration $x _ { n + 1 } = \frac { 3 - x _ { n } { } ^ { 3 } } { 4 }$ with $x _ { 1 } = 0.5$ to find $x _ { 3 }$, giving your answer to two decimal places.
\item The sketch on Figure 1 shows parts of the graphs of $y = \frac { 3 - x ^ { 3 } } { 4 }$ and $y = x$, and the position of $x _ { 1 }$.

On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of $x _ { 2 }$ and $x _ { 3 }$ on the $x$-axis.\\
(3 marks)
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA C3  Q6}}