| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve exponential equation via iteration |
| Difficulty | Standard +0.3 This is a standard C3 fixed point iteration question with straightforward parts: showing a root exists in an interval using sign changes (routine), trivial algebraic rearrangement, and applying a given iteration formula twice with a calculator. All steps are mechanical with no problem-solving required, making it slightly easier than average. |
| Spec | 1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(1) = 3^1 - (10-1^3) = 3 - 9 = -6 < 0\) and \(f(2) = 3^2 - (10-2^3) = 9 - 2 = 7 > 0\), sign change \(\Rightarrow\) root between 1 and 2 | M1 A1 | Must show both values and conclude sign change |
| Answer | Marks | Guidance |
|---|---|---|
| \(3^x = 10 - x^3 \Rightarrow x^3 = 10 - 3^x \Rightarrow x = \sqrt[3]{10 - 3^x}\) | B1 | Shown convincingly |
| Answer | Marks | Guidance |
|---|---|---|
| \(x_2 = \sqrt[3]{10 - 3^1} = \sqrt[3]{7} = 1.913\) | B1 | |
| \(x_3 = \sqrt[3]{10 - 3^{1.913}} = \sqrt[3]{10 - 6.588...} = \sqrt[3]{3.412...} = 1.502\) | B1 | Accept \(x_3 = 1.501\) or \(1.502\) |
# Question 1:
## Part (a)
| $f(1) = 3^1 - (10-1^3) = 3 - 9 = -6 < 0$ and $f(2) = 3^2 - (10-2^3) = 9 - 2 = 7 > 0$, sign change $\Rightarrow$ root between 1 and 2 | M1 A1 | Must show both values and conclude sign change |
## Part (b)(i)
| $3^x = 10 - x^3 \Rightarrow x^3 = 10 - 3^x \Rightarrow x = \sqrt[3]{10 - 3^x}$ | B1 | Shown convincingly |
## Part (b)(ii)
| $x_2 = \sqrt[3]{10 - 3^1} = \sqrt[3]{7} = 1.913$ | B1 | |
| $x_3 = \sqrt[3]{10 - 3^{1.913}} = \sqrt[3]{10 - 6.588...} = \sqrt[3]{3.412...} = 1.502$ | B1 | Accept $x_3 = 1.501$ or $1.502$ |
---1 The curve $y = 3 ^ { x }$ intersects the curve $y = 10 - x ^ { 3 }$ at the point where $x = \alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ lies between 1 and 2 .
\item \begin{enumerate}[label=(\roman*)]
\item Show that the equation $3 ^ { x } = 10 - x ^ { 3 }$ can be rearranged into the form
$$x = \sqrt [ 3 ] { 10 - 3 ^ { x } }$$
\item Use the iteration $x _ { n + 1 } = \sqrt [ 3 ] { 10 - 3 ^ { x _ { n } } }$ with $x _ { 1 } = 1$ to find the values of $x _ { 2 }$ and $x _ { 3 }$, giving your answers to three decimal places.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2010 Q1 [5]}}