| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Standard +0.3 This is a straightforward fixed-point iteration question with standard parts: algebraic rearrangement (routine), applying an iteration formula (mechanical calculation), verifying a root using change of sign (standard technique), and a simple transformation. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| VIIIV SIHI NI JIIYM ION OC | VIIV SIHI NI JIIIM ION OC | VIIV SIHI NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2x^3 = \pm x \pm 20\) or \(x^3 = \frac{\pm x \pm 20}{2}\) \(\Rightarrow x = \sqrt[3]{\frac{\pm x \pm 20}{2}}\) | M1 | Correct order of operations including cube root. The "\(= 0\)" does not have to be seen initially and can be implied by e.g. \(2x^3 = \pm x \pm 20\) |
| \(x = \sqrt[3]{10 - \frac{1}{2}x}\) or \(x = \sqrt[3]{\left(10 - \frac{1}{2}x\right)}\) | A1 | Correct equation or exact equivalent e.g. \(x = \sqrt[3]{10 - 0.5x}\) or \(x = \sqrt[3]{-0.5x + 10}\) with no errors. Vinculum should encompass both terms, go beyond "−" or "+". \(x = \pm\sqrt[3]{\ldots}\) scores A0. Isw once correct answer obtained. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \sqrt[3]{a - bx} \Rightarrow x^3 = a - bx \Rightarrow x^3 + bx - a = 0 \Rightarrow 2x^3 + 2bx - 2a = 0 \Rightarrow a = \ldots, b = \ldots\) | M1 | Correct order of operations: cubes, collects to one side, multiplies by 2, then compares coefficients to establish values for \(a\) and \(b\) |
| \(\Rightarrow a = 10,\ b = \frac{1}{2}\) | A1 | Correct values, apply isw if necessary |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_2 = \sqrt[3]{10 - \frac{1}{2} \times 2.1}\) | M1 | Substitutes \(x_1 = 2.1\) into \(x_{n+1} = \sqrt[3]{a - bx_n}\) with numerical values of \(a\) and \(b\) to find \(x_2\). Can be implied by awrt 2.076 if \(a\) and \(b\) correct, otherwise may need to check. |
| \((x_2 =)\) awrt \(2.076\) \((x_3 =)\) awrt \(2.077\) | A1 | Correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(2.0765) = -0.016\ldots\) \(f(2.0775) = 0.011\ldots\) | M1 | Chooses suitable interval for \(x\) within \(2.077 \pm 0.0005\) and attempts to evaluate \(f(x) = 2x^3 + x - 20\) for both values, obtains at least one value correct to 1 sig fig (rounded or truncated) |
| Sign change (negative, positive), therefore root. | A1 | Both values correct awrt (or truncated) 1 sf, sign change, e.g. \(< 0, > 0\) or \(f(2.0765) \cdot f(2.0775) < 0\) or \(f(2.0765) < 0 < f(2.0775)\), and minimal conclusion e.g. therefore root. Allow tick, QED, hash, square box, smiley face etc. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.077\) | B1 | Cao |
# Question 1:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x^3 = \pm x \pm 20$ or $x^3 = \frac{\pm x \pm 20}{2}$ $\Rightarrow x = \sqrt[3]{\frac{\pm x \pm 20}{2}}$ | M1 | Correct order of operations including cube root. The "$= 0$" does not have to be seen initially and can be implied by e.g. $2x^3 = \pm x \pm 20$ |
| $x = \sqrt[3]{10 - \frac{1}{2}x}$ or $x = \sqrt[3]{\left(10 - \frac{1}{2}x\right)}$ | A1 | Correct equation or exact equivalent e.g. $x = \sqrt[3]{10 - 0.5x}$ or $x = \sqrt[3]{-0.5x + 10}$ with no errors. Vinculum should encompass both terms, go beyond "−" or "+". $x = \pm\sqrt[3]{\ldots}$ scores A0. Isw once correct answer obtained. |
## Part (a) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \sqrt[3]{a - bx} \Rightarrow x^3 = a - bx \Rightarrow x^3 + bx - a = 0 \Rightarrow 2x^3 + 2bx - 2a = 0 \Rightarrow a = \ldots, b = \ldots$ | M1 | Correct order of operations: cubes, collects to one side, multiplies by 2, then compares coefficients to establish values for $a$ and $b$ |
| $\Rightarrow a = 10,\ b = \frac{1}{2}$ | A1 | Correct values, apply isw if necessary |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_2 = \sqrt[3]{10 - \frac{1}{2} \times 2.1}$ | M1 | Substitutes $x_1 = 2.1$ into $x_{n+1} = \sqrt[3]{a - bx_n}$ with numerical values of $a$ and $b$ to find $x_2$. Can be implied by awrt 2.076 if $a$ and $b$ correct, otherwise may need to check. |
| $(x_2 =)$ awrt $2.076$ $(x_3 =)$ awrt $2.077$ | A1 | Correct values |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(2.0765) = -0.016\ldots$ $f(2.0775) = 0.011\ldots$ | M1 | Chooses suitable interval for $x$ within $2.077 \pm 0.0005$ and attempts to evaluate $f(x) = 2x^3 + x - 20$ for both values, obtains at least one value correct to 1 sig fig (rounded or truncated) |
| Sign change (negative, positive), therefore root. | A1 | Both values correct awrt (or truncated) 1 sf, sign change, e.g. $< 0, > 0$ or $f(2.0765) \cdot f(2.0775) < 0$ or $f(2.0765) < 0 < f(2.0775)$, and minimal conclusion e.g. therefore root. Allow tick, QED, hash, square box, smiley face etc. |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.077$ | B1 | Cao |
---1.
$$f ( x ) = 2 x ^ { 3 } + x - 20$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ can be rewritten as
$$x = \sqrt [ 3 ] { a - b x }$$
where $a$ and $b$ are positive constants to be determined.
\item Starting with $x _ { 1 } = 2.1$ use the iteration formula $x _ { n + 1 } = \sqrt [ 3 ] { a - b x _ { n } }$, with the numerical values of $a$ and $b$, to calculate the values of $x _ { 2 }$ and $x _ { 3 }$ giving your answers to 3 decimal places.
\item Using a suitable interval, show that 2.077 is a root of the equation $\mathrm { f } ( x ) = 0$ correct to 3 decimal places.
\item Hence state a root, to 3 decimal places, of the equation
$$2 ( x + 2 ) ^ { 3 } + x - 18 = 0$$
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VIIIV SIHI NI JIIYM ION OC & VIIV SIHI NI JIIIM ION OC & VIIV SIHI NI JIIYM ION OC \\
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\hfill \mbox{\textit{Edexcel C34 2019 Q1 [7]}}