| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Show root in interval |
| Difficulty | Moderate -0.3 This is a standard C3 question covering routine sketching, sign change method, and fixed-point iteration. Part (a) requires basic exponential/linear sketches, (b) uses graph intersection reasoning, (c) applies straightforward sign change verification, and (d) involves calculator-based iteration. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02p Interpret algebraic solutions: graphically1.06a Exponential function: a^x and e^x graphs and properties1.06d Natural logarithm: ln(x) function and properties1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks |
|---|---|
| (a) Shape for \(y = 10 - x\) | B1 |
| Shape for \(y = e^x\) | B1 |
| Coordinates correct \((0, 10)\), \((10, 0)\) and \((0, 1)\) | B1 |
| (b) One solution as there is one point of intersection | B1 |
| (c) Substitute \(x = 2\) and \(x = 3\) into \(f(x) = e^x - 10 + x\) | M1 |
| \(f(2) = -0.61\), \(f(3) = 13.1\) | A1 |
| Both correct to 1 s.f., reason (change of sign) and conclusion (hence root) | A1 |
| (d) Substitutes \(x = 2\) into \(x_{n+1} = \ln(10 - x_n)\) | M1 |
| \(x_2 = 2.07794\), \(x_3 = 2.0695\), \(x_4 = 2.0707\) | A1, A1 |
(a) Shape for $y = 10 - x$ | B1
Shape for $y = e^x$ | B1
Coordinates correct $(0, 10)$, $(10, 0)$ and $(0, 1)$ | B1
(b) One solution as there is one point of intersection | B1
(c) Substitute $x = 2$ and $x = 3$ into $f(x) = e^x - 10 + x$ | M1
$f(2) = -0.61$, $f(3) = 13.1$ | A1
Both correct to 1 s.f., reason (change of sign) and conclusion (hence root) | A1
(d) Substitutes $x = 2$ into $x_{n+1} = \ln(10 - x_n)$ | M1
$x_2 = 2.07794$, $x_3 = 2.0695$, $x_4 = 2.0707$ | A1, A1
---\begin{enumerate}
\item (a) On the same diagram, sketch and clearly label the graphs with equations
\end{enumerate}
$$y = \mathrm { e } ^ { x } \quad \text { and } \quad y = 10 - x$$
Show on your sketch the coordinates of each point at which the graphs cut the axes.\\
(b) Explain why the equation $\mathrm { e } ^ { x } - 10 + x = 0$ has only one solution.\\
(c) Show that the solution of the equation
$$\mathrm { e } ^ { x } - 10 + x = 0$$
lies between $x = 2$ and $x = 3$\\
(d) Use the iterative formula
$$x _ { n + 1 } = \ln \left( 10 - x _ { n } \right) , \quad x _ { 1 } = 2$$
to calculate the values of $x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$.\\
Give your answers to 4 decimal places.
\hfill \mbox{\textit{Edexcel C3 2013 Q4 [9]}}