| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson with derivative given or simple |
| Difficulty | Standard +0.3 This is a multi-part question covering standard A-level calculus techniques: finding where functions are positive/increasing, locating stationary points, algebraic manipulation, proving no stationary points exist, and applying Newton-Raphson. All parts are routine applications of well-practiced methods with no novel insights required. The Newton-Raphson iteration is computational but straightforward. Slightly easier than average due to the guided structure and standard techniques. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| When \(t = 0\): \(y = \frac{2}{x^2}(x-1) = \frac{2}{x} - \frac{2}{x^2}\) | ||
| Attempt at differentiating: \(\frac{dy}{dx} = -\frac{2}{x^2} + \frac{4}{x^3} = \frac{1}{x^3}(4 - 2x)\) | M1 | |
| (a) \(y\) is positive if \(x > 1\); \(y\) is positive and increasing if \(1 < x < 2\) | B1 A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \(y\) is stationary when \(x = 2\). Consideration of the gradient either side, or the use of the second derivative, shows a maximum | B1 B1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Substitute \(t = 2\) and \(y = -2\) and show at least one line of manipulation to obtain \(x^3 + x - 1 = 0\) | B1 AG | [1] |
| (b) \(f = 3x^2 + 1\). This is positive definite \((\Rightarrow\) there are no roots) | B1 B1 AG | [2] |
| (c) Derive iterative formula: \(x_{n+1} = x_n - \frac{x_n^3 + x_n - 1}{3x_n^2 + 1} = \left(\frac{2x_n^3 + 1}{3x_n^2 + 1}\right)\) | M1 A1 | |
| Using \(x_0 = 1\), obtain a succession of approximations: 0.75, 0.68606, 0.68233, 0.68234 | M1 | |
| State \(V = 0.6823\) to 4 sf | A1 | [4] |
### (i)
When $t = 0$: $y = \frac{2}{x^2}(x-1) = \frac{2}{x} - \frac{2}{x^2}$ |
Attempt at differentiating: $\frac{dy}{dx} = -\frac{2}{x^2} + \frac{4}{x^3} = \frac{1}{x^3}(4 - 2x)$ | M1 |
**(a)** $y$ is positive if $x > 1$; $y$ is positive and increasing if $1 < x < 2$ | B1 A1 | [3]
*Note: Accept alternative convincing arguments.*
**(b)** $y$ is stationary when $x = 2$. Consideration of the gradient either side, or the use of the second derivative, shows a maximum | B1 B1 | [2]
### (ii)
**(a)** Substitute $t = 2$ and $y = -2$ and show at least one line of manipulation to obtain $x^3 + x - 1 = 0$ | B1 AG | [1]
**(b)** $f = 3x^2 + 1$. This is positive definite $(\Rightarrow$ there are no roots) | B1 B1 AG | [2]
**(c)** Derive iterative formula: $x_{n+1} = x_n - \frac{x_n^3 + x_n - 1}{3x_n^2 + 1} = \left(\frac{2x_n^3 + 1}{3x_n^2 + 1}\right)$ | M1 A1 |
Using $x_0 = 1$, obtain a succession of approximations: 0.75, 0.68606, 0.68233, 0.68234 | M1 |
State $V = 0.6823$ to 4 sf | A1 | [4]Let $y = (x - 1)\left(\frac{2}{x^2} + t\right)$ define $y$ as a function of $x$ ($x > 0$), for each value of the real parameter $t$.
\begin{enumerate}[label=(\roman*)]
\item When $t = 0$,
\begin{enumerate}[label=(\alph*)]
\item determine the set of values of $x$ for which $y$ is positive and an increasing function, [3]
\item locate the stationary point of $y$, and determine its nature. [2]
\end{enumerate}
\item It is given that $t = 2$ and $y = -2$.
\begin{enumerate}[label=(\alph*)]
\item Show that $x$ satisfies $f(x) = 0$, where $f(x) = x^3 + x - 1$. [1]
\item Prove that $f$ has no stationary points. [2]
\item Use the Newton-Raphson method, with $x_0 = 1$, to find $x$ correct to 4 significant figures. [4]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2010 Q7 [12]}}