| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve trigonometric equation via iteration |
| Difficulty | Moderate -0.3 This is a standard C3 fixed-point iteration question with routine steps: showing a sign change, rearranging an equation algebraically, applying an iterative formula (calculator work), and verifying accuracy via sign change. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05h Reciprocal trig functions: sec, cosec, cot definitions and graphs1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(1.2) = 0.49166551...\), \(f(1.3) = -0.048719817...\) Sign change (and \(f(x)\) continuous) therefore root \(\alpha \in [1.2, 1.3]\) | M1A1 | M1: Evaluate both \(f(1.2)\) and \(f(1.3)\), at least one correct to awrt (or truncated) 1 sf. A1: Both values correct to 1 sf, sign change and conclusion. |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4\text{cosec}x - 4x + 1 = 0 \Rightarrow 4x = 4\text{cosec}x + 1\) | M1 | Attempt to make \(4x\) or \(x\) the subject. |
| \(\Rightarrow x = \text{cosec}x + \frac{1}{4} \Rightarrow x = \frac{1}{\sin x} + \frac{1}{4}\) | A1* | Candidate must rearrange to give required result. Must be clear \(f(x) = 0\). |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_1 = \frac{1}{\sin(1.25)} + \frac{1}{4}\) | M1 | Attempt to substitute \(x_0 = 1.25\). Can be implied by \(x_1 = \text{awrt } 1.3\) or \(x_1 = \text{awrt } 46°\). |
| \(x_1 = 1.303757858...\), \(x_2 = 1.286745793...\) | A1 | Both \(x_1 = \text{awrt } 1.3038\) and \(x_2 = \text{awrt } 1.2867\) |
| \(x_3 = 1.291744613...\) | A1 | \(x_3 = \text{awrt } 1.2917\) |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(1.2905) = 0.00044566695...\), \(f(1.2915) = -0.00475017278...\) Sign change (and \(f(x)\) continuous) therefore root \(\alpha \in (1.2905, 1.2915) \Rightarrow \alpha = 1.291\) (3 dp) | M1 | Choose suitable interval, e.g. \([1.2905, 1.2915]\) or tighter, at least one \(f(x)\) evaluated. |
| A1 | Both values correct to 1 sf, sign change and conclusion. | |
| (2) [9] |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(1.2) = 0.49166551...$, $f(1.3) = -0.048719817...$ Sign change (and $f(x)$ continuous) therefore root $\alpha \in [1.2, 1.3]$ | M1A1 | M1: Evaluate both $f(1.2)$ and $f(1.3)$, at least one correct to awrt (or truncated) 1 sf. A1: Both values correct to 1 sf, sign change and conclusion. |
| | **(2)** | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4\text{cosec}x - 4x + 1 = 0 \Rightarrow 4x = 4\text{cosec}x + 1$ | M1 | Attempt to make $4x$ or $x$ the subject. |
| $\Rightarrow x = \text{cosec}x + \frac{1}{4} \Rightarrow x = \frac{1}{\sin x} + \frac{1}{4}$ | A1* | Candidate must rearrange to give required result. Must be clear $f(x) = 0$. |
| | **(2)** | |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_1 = \frac{1}{\sin(1.25)} + \frac{1}{4}$ | M1 | Attempt to substitute $x_0 = 1.25$. Can be implied by $x_1 = \text{awrt } 1.3$ or $x_1 = \text{awrt } 46°$. |
| $x_1 = 1.303757858...$, $x_2 = 1.286745793...$ | A1 | Both $x_1 = \text{awrt } 1.3038$ and $x_2 = \text{awrt } 1.2867$ |
| $x_3 = 1.291744613...$ | A1 | $x_3 = \text{awrt } 1.2917$ |
| | **(3)** | |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(1.2905) = 0.00044566695...$, $f(1.2915) = -0.00475017278...$ Sign change (and $f(x)$ continuous) therefore root $\alpha \in (1.2905, 1.2915) \Rightarrow \alpha = 1.291$ (3 dp) | M1 | Choose suitable interval, e.g. $[1.2905, 1.2915]$ or tighter, at least one $f(x)$ evaluated. |
| | A1 | Both values correct to 1 sf, sign change and conclusion. |
| | **(2) [9]** | |
---3. $\mathrm { f } ( x ) = 4 \operatorname { cosec } x - 4 x + 1$, where $x$ is in radians.
\begin{enumerate}[label=(\alph*)]
\item Show that there is a root $\alpha$ of $\mathrm { f } ( x ) = 0$ in the interval [1.2,1.3].
\item Show that the equation $\mathrm { f } ( x ) = 0$ can be written in the form
$$x = \frac { 1 } { \sin x } + \frac { 1 } { 4 }$$
\item Use the iterative formula
$$x _ { n + 1 } = \frac { 1 } { \sin x _ { n } } + \frac { 1 } { 4 } , \quad x _ { 0 } = 1.25$$
to calculate the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$, giving your answers to 4 decimal places.
\item By considering the change of sign of $\mathrm { f } ( x )$ in a suitable interval, verify that $\alpha = 1.291$ correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2010 Q3 [9]}}