| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve trigonometric equation via iteration |
| Difficulty | Standard +0.3 This is a straightforward fixed point iteration question requiring standard techniques: sketching arctan, drawing a line to show unique intersection, verifying bounds by substitution, and performing two iterations with a calculator. All steps are routine C3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Sketch of \(\tan^{-1}x\) with correct shape | B1 | shape |
| Asymptotes shown or stated, \(\frac{\pi}{2}\) seen | B1 | asymptotes (shown or stated) \(\left(\frac{\pi}{2}\right)\) seen |
| Answer | Marks | Guidance |
|---|---|---|
| Sketch of \(2x-1\) added to diagram | B1 | sketch of \(2x-1\) |
| Correct intersection shown | B1 | correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan^{-1}x - 2x + 1 = 0\) | ||
| \(f(0.8) = 0.07\), \(f(0.9) = -0.07\) | M1 | |
| Change of sign \(\therefore\) root | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \((x_1 = 0.8)\) | M1 | attempt at \(x_2\) |
| \(x_2 = 0.837(37)\ldots\) | A1 | for \(x_2\) |
| \(x_3 = 0.85\) | A1 | 3 |
## Question 7:
### Part (a):
Sketch of $\tan^{-1}x$ with correct shape | B1 | shape
Asymptotes shown or stated, $\frac{\pi}{2}$ seen | B1 | asymptotes (shown or stated) $\left(\frac{\pi}{2}\right)$ seen
**Total: 2 marks**
### Part (b)(i):
Sketch of $2x-1$ added to diagram | B1 | sketch of $2x-1$
Correct intersection shown | B1 | correct
**Total: 2 marks**
### Part (b)(ii):
$\tan^{-1}x - 2x + 1 = 0$ | |
$f(0.8) = 0.07$, $f(0.9) = -0.07$ | M1 |
Change of sign $\therefore$ root | A1 | 2 | allow $+$ve, $-$ve; A0 if $f(0.8)$, $f(0.9)$ wrong
**Total: 2 marks**
### Part (c):
$(x_1 = 0.8)$ | M1 | attempt at $x_2$
$x_2 = 0.837(37)\ldots$ | A1 | for $x_2$
$x_3 = 0.85$ | A1 | 3 | for $x_3$
**Total: 3 marks**
---7
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $y = \tan ^ { - 1 } x$.
\item \begin{enumerate}[label=(\roman*)]
\item By drawing a suitable straight line on your sketch, show that the equation $\tan ^ { - 1 } x = 2 x - 1$ has only one root.
\item Given that the root of this equation is $\alpha$, show that $0.8 < \alpha < 0.9$.
\end{enumerate}\item Use the iteration $x _ { n + 1 } = \frac { 1 } { 2 } \left( \tan ^ { - 1 } x _ { n } + 1 \right)$ with $x _ { 1 } = 0.8$ to find the value of $x _ { 3 }$, giving your answer to two significant figures.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2005 Q7 [9]}}