| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson with trigonometric or exponential functions |
| Difficulty | Standard +0.8 This is a substantial multi-part question requiring curve sketching, stationary point analysis with nature determination, inequality proofs involving logarithms, Newton-Raphson iteration, and synthesis of graphical reasoning. While individual parts use standard techniques, the question demands sustained mathematical reasoning across five connected parts, with part (ii) requiring insight about the relationship between the stationary point value and roots. This is more demanding than typical A-level questions but accessible to well-prepared students. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative1.09d Newton-Raphson method |
**Question 9:**
**(i)** $\frac{\text{d}y}{\text{d}x} = \frac{1}{x} - k = 0$ **M1** (Differentiate, put equal to 0 and attempt to solve)
$\Rightarrow x = \frac{1}{k}$ **A1** (aef)
$\frac{\text{d}^2y}{\text{d}x^2} = -\frac{1}{x^2}$ and at $x = \frac{1}{k}$, $\frac{\text{d}^2y}{\text{d}x^2} = -k^2 < 0$ and hence maximum **B1** (Clear conclusion required. Accept $-\frac{1}{(1/k)^2} < 0$)
$y = \ln\left(\frac{1}{k}\right) - 1 = -\ln k - 1$ **A1** (aef)
**(ii)** $-\ln k - 1 \geqslant 0$ **M1** ($y$ ordinate stated either $=$ or $\geqslant 0$ M1)
$k \leqslant \frac{1}{\text{e}}$ **A1** (AG)
**(iii)** $x_{n+1} = x_n - \frac{\ln x - \frac{1}{3}x}{\frac{1}{x} - \frac{1}{3}}$ **M1** (Correct formula)
$5,\ 4.570784343,\ 4.536651853,\ 4.536403668$ **A1** (Obtain at least 2 correct iterates including 5)
$4.54$ **A1** (Obtain answer correct to 3 s.f.)
**(iv)** If $k < 0$, $-kx$ adds a positive quantity onto $\ln x$ **B1**
If $k = 0$, the curve is $y = \ln x$ which has no maximum and continuously increases so crosses the $x$-axis once and has one root **B1**
Hence the function $y = \ln x - kx$ is continuously increasing so crosses the $x$-axis once only. **B1** (Award if the two cases are considered and an overall conclusion drawn)
**(v)** Should mention a maximum between $0 < k < \frac{1}{\text{e}}$ **B1** (Accept alternative arguments)
Conclude that the curve must cross the $x$-axis twice giving two real roots. **B1**
**Total: 12 marks**9 (i) Find the coordinates of the stationary point of the curve with equation
$$y = \ln x - k x , \text { where } k > 0 \text { and } x > 0$$
and determine its nature.\\
(ii) Hence show that the equation $\ln x - k x = 0$ has real roots if $0 < k \leqslant \frac { 1 } { \mathrm { e } }$.\\
(iii) In the particular case that $k = \frac { 1 } { 3 }$, the equation $\ln x - k x = 0$ has two roots, one of which is near $x = 5$.
Use the Newton-Raphson process to find, correct to 3 significant figures, the root of the equation $\ln x - \frac { 1 } { 3 } x = 0$ which is near $x = 5$.\\
(iv) Show that the equation $\ln x - k x = 0$ has one real root if $k \leqslant 0$.\\
(v) Explain why the equation $\ln x - k x = 0$ has two distinct real roots if $0 < k < \frac { 1 } { \mathrm { e } }$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2018 Q9 [12]}}