| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 This is a structured multi-part question involving differentiation (using quotient/chain rule), algebraic manipulation to rearrange f'(x)=0, and applying a given iteration formula three times. All steps are routine A-level techniques with clear guidance, making it slightly easier than average despite multiple parts. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Show that \(f'(x) = \frac{6x^2 - x - 4\sqrt{x}}{2x\sqrt{x}}\) for \(f(x) = \frac{2x^2 - x}{\sqrt{x}} - 2\ln\left(\frac{x}{2}\right)\) | (4) | B1 Differentiates \(\ln\frac{x}{2} \to \frac{1}{x}\); M1 Correct method to differentiate \(\frac{2x^2 - x}{\sqrt{x}}\); A1 Correct differentiation of \(\frac{2x^2 - x}{\sqrt{x}}\); A1 Obtains \(\frac{dy}{dx} = \frac{6x^2 - x - 4\sqrt{x}}{2x\sqrt{x}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Show that x-coordinate of Q is solution of \(x = \sqrt{\frac{x}{6} + \frac{2\sqrt{x}}{3}}\) | (2) | M1 Sets \(6x^2 - x - 4\sqrt{x} = 0\) and writes line equivalent to \(x^2 = \frac{x + 4\sqrt{x}}{6}\); A1 Completely correct with all signs correct |
| Answer | Marks | Guidance |
|---|---|---|
| Taking \(x_0 = 0.8\), find values of \(x_1\), \(x_2\), \(x_3\) | (3) | c. M1 Attempt to substitute \(x_0 = 0.8\) into iterative formula; A1 Answer which rounds to \(x_1 = 0.854\); A1 Both answers which round to \(x_2 = 0.871\) and \(x_3 = 0.876\) |
**Part (a):**
Show that $f'(x) = \frac{6x^2 - x - 4\sqrt{x}}{2x\sqrt{x}}$ for $f(x) = \frac{2x^2 - x}{\sqrt{x}} - 2\ln\left(\frac{x}{2}\right)$ | (4) | B1 Differentiates $\ln\frac{x}{2} \to \frac{1}{x}$; M1 Correct method to differentiate $\frac{2x^2 - x}{\sqrt{x}}$; A1 Correct differentiation of $\frac{2x^2 - x}{\sqrt{x}}$; A1 Obtains $\frac{dy}{dx} = \frac{6x^2 - x - 4\sqrt{x}}{2x\sqrt{x}}$
**Part (b):**
Show that x-coordinate of Q is solution of $x = \sqrt{\frac{x}{6} + \frac{2\sqrt{x}}{3}}$ | (2) | M1 Sets $6x^2 - x - 4\sqrt{x} = 0$ and writes line equivalent to $x^2 = \frac{x + 4\sqrt{x}}{6}$; A1 Completely correct with all signs correct
OR Alternative working backwards: M1 Starts with answer and squares, multiplies each side by 6; A1 Completely correct $6x^2 - x - 4\sqrt{x} = 0$ and states $f(x) = 0$
**Part (c):**
Taking $x_0 = 0.8$, find values of $x_1$, $x_2$, $x_3$ | (3) | c. M1 Attempt to substitute $x_0 = 0.8$ into iterative formula; A1 Answer which rounds to $x_1 = 0.854$; A1 Both answers which round to $x_2 = 0.871$ and $x_3 = 0.876$
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cb92f7b6-2ba5-4703-9595-9ba8570fc52b-09_1152_1006_285_374}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a sketch of the curve $C$ with equation $y = \mathrm { f } ( x )$, where
$$f ( x ) = \frac { 2 x ^ { 2 } - x } { \sqrt { x } } - 2 \ln \left( \frac { x } { 2 } \right) , \quad x > 0$$
The curve has a minimum turning point at $Q$, as shown in Figure 4.\\
a. Show that $\mathrm { f } ^ { \prime } ( x ) = \frac { 6 x ^ { 2 } - x - 4 \sqrt { x } } { 2 x \sqrt { x } }$\\
b. Show that the $x$-coordinate of $Q$ is the solution of
$$x = \sqrt { \frac { x } { 6 } + \frac { 2 \sqrt { x } } { 3 } }$$
To find an approximation for the $x$-coordinate of $Q$, the iteration formula
$$x _ { n + 1 } = \sqrt { \frac { x _ { n } } { 6 } + \frac { 2 \sqrt { x _ { n } } } { 3 } }$$
is used.\\
c. Taking $x _ { 0 } = 0.8$, find the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$.
Give your answers to 3 decimal places.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q6 [9]}}