AQA C3 2007 January — Question 8 7 marks

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
Year2007
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeSolve trigonometric equation via iteration
DifficultyStandard +0.3 This is a straightforward fixed point iteration question requiring: (a) reading coordinates from a standard inverse cosine graph, (b) simple substitution to verify a root lies in an interval, and (c) mechanical application of a given iteration formula. All steps are routine with no problem-solving insight needed, making it slightly easier than average.
Spec1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

8 The diagram shows the curve \(y = \cos ^ { - 1 } x\) for \(- 1 \leqslant x \leqslant 1\). \includegraphics[max width=\textwidth, alt={}, center]{6890a681-2b7f-4853-a5f0-f88b7b435367-4_492_698_1640_671}
  1. Write down the exact coordinates of the points \(A\) and \(B\).
  2. The equation \(\cos ^ { - 1 } x = 3 x + 1\) has only one root. Given that the root of this equation is \(\alpha\), show that \(0.1 \leqslant \alpha \leqslant 0.2\).
  3. Use the iteration \(x _ { n + 1 } = \frac { 1 } { 3 } \left( \cos ^ { - 1 } x _ { n } - 1 \right)\) with \(x _ { 1 } = 0.1\) to find the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to three decimal places.

Question 8:
Part (a)
AnswerMarks Guidance
AnswerMark Guidance
\(A(-1, \pi)\)B1
\(B\left(0, \frac{\pi}{2}\right)\)B1
Total: 2
Part (b)
AnswerMarks Guidance
AnswerMark Guidance
\(\cos^{-1}x - 3x - 1 = 0\)
\(f(0.1) = 0.17\) allow \(0.2, 0.1\)M1 Or comparing 'sides'
\(f(0.2) = -0.23\) allow \(-0.2\)
Change of sign \(\therefore\) rootA1
Total: 2
Part (c)
AnswerMarks Guidance
AnswerMark Guidance
\(x_1 = 0.1\)M1
\(x_2 = 0.1569 = 0.157\)A1
\(x_3 = 0.1378 = 0.138\)
\(x_4 = 0.144\)A1
Total: 3
# Question 8:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $A(-1, \pi)$ | B1 | |
| $B\left(0, \frac{\pi}{2}\right)$ | B1 | |
| **Total: 2** | | |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cos^{-1}x - 3x - 1 = 0$ | | |
| $f(0.1) = 0.17$ allow $0.2, 0.1$ | M1 | Or comparing 'sides' |
| $f(0.2) = -0.23$ allow $-0.2$ | | |
| Change of sign $\therefore$ root | A1 | |
| **Total: 2** | | |

## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x_1 = 0.1$ | M1 | |
| $x_2 = 0.1569 = 0.157$ | A1 | |
| $x_3 = 0.1378 = 0.138$ | | |
| $x_4 = 0.144$ | A1 | |
| **Total: 3** | | |

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8 The diagram shows the curve $y = \cos ^ { - 1 } x$ for $- 1 \leqslant x \leqslant 1$.\\
\includegraphics[max width=\textwidth, alt={}, center]{6890a681-2b7f-4853-a5f0-f88b7b435367-4_492_698_1640_671}
\begin{enumerate}[label=(\alph*)]
\item Write down the exact coordinates of the points $A$ and $B$.
\item The equation $\cos ^ { - 1 } x = 3 x + 1$ has only one root. Given that the root of this equation is $\alpha$, show that $0.1 \leqslant \alpha \leqslant 0.2$.
\item Use the iteration $x _ { n + 1 } = \frac { 1 } { 3 } \left( \cos ^ { - 1 } x _ { n } - 1 \right)$ with $x _ { 1 } = 0.1$ to find the values of $x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving your answers to three decimal places.
\end{enumerate}

\hfill \mbox{\textit{AQA C3 2007 Q8 [7]}}