| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | December |
| Marks | 16 |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Standard +0.3 This is a standard C3/C4 level question on numerical methods and function composition. Part (a) is algebraic manipulation, part (b) applies the Newton-Raphson formula (routine differentiation with product rule), part (c) is calculator work, and part (d) requires solving fg(x) = 2/13 which involves standard algebraic techniques. All parts follow predictable patterns with no novel insight required, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08i Integration by parts1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{e^{2x}} = \frac{x}{x^2 + 3} \Rightarrow x^2 + 3 = xe^{2x}\) | M1 | Equate expressions and cross-multiply (to remove fractions) |
| \(x^2 + 3 - xe^{2x} = 0\) | A1 | AG – sufficient working must be shown to indicate that result has been derived correctly |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dx}(xe^{2x}) = e^{2x} + 2xe^{2x}\) | M1 | Attempt at product rule for \(xe^{2x}\) – expression must be of the form \(\pm e^{2x}(1 \pm ke^{2x})\) |
| \(h'(x) = 2x - e^{2x} - 2xe^{2x}\) | A1 | |
| \(x_{n+1} = x_n - \frac{x_n^2 + 3 - x_ne^{2x_n}}{2x_n - e^{2x_n} - 2x_ne^{2x_n}}\) | M1* | Correct application of NR with their \(h'(x)\) |
| \(x_{n+1} = \frac{2x_n^2 - x_ne^{2x_n} - 2x_ne^{2x_n} - x_n^2 + 3x_ne^{2x_n}}{2x_n - e^{2x_n}(1 + 2x_n)}\) | M1dep* | Correctly combining as a single fraction and expanding any brackets in numerator |
| \(x_{n+1} = \frac{x_n^2 - 2x_n^2e^{2x_n} - 3}{2x_n - e^{2x_n}(1 + 2x_n)} = \frac{x_n^2(1 - 2e^{2x_n}) - 3}{2x_n - (1 + 2x_n)e^{2x_n}}\) | A1 | AG – sufficient working must be shown as the answer is given |
| Answer | Marks | Guidance |
|---|---|---|
| From graph eg \(x_1 = 1\) | M1 | Suitable starting value chosen |
| \(x_2 = 0.83195181..., \quad x_3 = 0.77546...\) | A1 | At least two correct applications of NR |
| \(x_4 = 0.77016..., \quad x_5 = 0.77011...\) | A1 | |
| \(\alpha = 0.770\) (correct to 3 dp) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(fg(x) = f(e^{-2x}) = \frac{e^{-2x}}{(e^{-2x})^2 + 3}\) | M1 | Attempt at \(fg(x)\) – need not be simplified |
| \(2e^{-4x} - 13e^{-2x} + 6 = 0\) | A1 | Correct equation – fractions must be removed and powers simplified |
| \(k = e^{-2x}\) | M1* | Substitute for \(e^{-2x}\) (or equivalent) |
| \((2k - 1)(k - 6) = 0\) | M1dep* | Attempt to solve resulting quadratic |
| \(e^{-2x} = \frac{1}{2} \Rightarrow x = -\frac{1}{2}\ln(\frac{1}{2})\) | A1 | www oe |
| \(e^{-2x} \neq 6\) it is given that \(x \geq 0\) | A1 | Correct statement or equivalent that \(e^{-2x}\) cannot be greater than 1 |
## Part (a)
$\frac{1}{e^{2x}} = \frac{x}{x^2 + 3} \Rightarrow x^2 + 3 = xe^{2x}$ | M1 | Equate expressions and cross-multiply (to remove fractions)
$x^2 + 3 - xe^{2x} = 0$ | A1 | AG – sufficient working must be shown to indicate that result has been derived correctly
## Part (b)
$\frac{d}{dx}(xe^{2x}) = e^{2x} + 2xe^{2x}$ | M1 | Attempt at product rule for $xe^{2x}$ – expression must be of the form $\pm e^{2x}(1 \pm ke^{2x})$
$h'(x) = 2x - e^{2x} - 2xe^{2x}$ | A1 |
$x_{n+1} = x_n - \frac{x_n^2 + 3 - x_ne^{2x_n}}{2x_n - e^{2x_n} - 2x_ne^{2x_n}}$ | M1* | Correct application of NR with their $h'(x)$
$x_{n+1} = \frac{2x_n^2 - x_ne^{2x_n} - 2x_ne^{2x_n} - x_n^2 + 3x_ne^{2x_n}}{2x_n - e^{2x_n}(1 + 2x_n)}$ | M1dep* | Correctly combining as a single fraction and expanding any brackets in numerator
$x_{n+1} = \frac{x_n^2 - 2x_n^2e^{2x_n} - 3}{2x_n - e^{2x_n}(1 + 2x_n)} = \frac{x_n^2(1 - 2e^{2x_n}) - 3}{2x_n - (1 + 2x_n)e^{2x_n}}$ | A1 | AG – sufficient working must be shown as the answer is given
## Part (c)
From graph eg $x_1 = 1$ | M1 | Suitable starting value chosen
$x_2 = 0.83195181..., \quad x_3 = 0.77546...$ | A1 | At least two correct applications of NR
$x_4 = 0.77016..., \quad x_5 = 0.77011...$ | A1 |
$\alpha = 0.770$ (correct to 3 dp) | A1 |
## Part (d)
**DR**
$fg(x) = f(e^{-2x}) = \frac{e^{-2x}}{(e^{-2x})^2 + 3}$ | M1 | Attempt at $fg(x)$ – need not be simplified
$2e^{-4x} - 13e^{-2x} + 6 = 0$ | A1 | Correct equation – fractions must be removed and powers simplified | Or equivalent
$k = e^{-2x}$ | M1* | Substitute for $e^{-2x}$ (or equivalent) | Alternatively: Factorise into two brackets containing $e^{-2x}$ M2
$(2k - 1)(k - 6) = 0$ | M1dep* | Attempt to solve resulting quadratic
$e^{-2x} = \frac{1}{2} \Rightarrow x = -\frac{1}{2}\ln(\frac{1}{2})$ | A1 | www oe
$e^{-2x} \neq 6$ it is given that $x \geq 0$ | A1 | Correct statement or equivalent that $e^{-2x}$ cannot be greater than 1\includegraphics{figure_5}
The functions f(x) and g(x) are defined for $x \geqslant 0$ by $\text{f}(x) = \frac{x}{x^2 + 3}$ and $\text{g}(x) = \text{e}^{-2x}$. The diagram shows the curves $y = \text{f}(x)$ and $y = \text{g}(x)$. The equation $\text{f}(x) = \text{g}(x)$ has exactly one real root $\alpha$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\alpha$ satisfies the equation $\text{h}(x) = 0$, where $\text{h}(x) = x^2 + 3 - x\text{e}^{2x}$. [2]
\item Hence show that a Newton-Raphson iterative formula for finding $\alpha$ can be written in the form
$$x_{n+1} = \frac{x_n^2(1 - 2\text{e}^{2x_n}) - 3}{2x_n - (1 + 2x_n)\text{e}^{2x_n}}.$$ [5]
\item Use this iterative formula, with a suitable initial value, to find $\alpha$ correct to 3 decimal places. Show the result of each iteration. [3]
\item In this question you must show detailed reasoning.
Find the exact value of $x$ for which $\text{fg}(x) = \frac{2}{13}$. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q5 [16]}}