| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2013 |
| Session | June |
| Marks | 4 |
| Topic | Taylor series |
| Type | Combining or manipulating standard series |
| Difficulty | Challenging +1.2 This question requires students to manipulate standard Maclaurin series (for ln(1+x) and ln(1-x)) and recognize the resulting series matches tanh^(-1)x from the formula booklet. While it involves multiple steps (expanding two logarithms, combining them, simplifying the series), the series are given in the formula booklet and the approach is relatively standard for Further Maths students. It's more challenging than routine differentiation or integration but doesn't require deep insight—just careful algebraic manipulation of known series. |
| Spec | 4.07e Inverse hyperbolic: definitions, domains, ranges4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n |
$\ln(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 + \ldots$ from the Formula Book
$\ln(1-x) = -x - \frac{1}{2}x^2 - \frac{1}{3}x^3 - \frac{1}{4}x^4 - \frac{1}{5}x^5 - \ldots$ **B1**
$\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) = \frac{1}{2}\{\ln(1+x) - \ln(1-x)\}$ **M1**
$= \frac{1}{2} \times 2\left\{x + \frac{1}{3}x^3 + \frac{1}{5}x^5 + \ldots\right\} = \tanh^{-1}x$ from the Formula Book **A1**
$\ln(1+x)$ valid for $-1 < x \leq 1$ and so $\ln(1-x)$ is valid for $-1 \leq x < 1$, so LHS valid for $-1 < x < 1$, which matches the range for RHS **B1**
**[4]**2 Use the standard Maclaurin series expansions given in the List of Formulae MF20 to show that
$$\frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right) \equiv \tanh ^ { - 1 } x \text { for } - 1 < x < 1$$
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2013 Q2 [4]}}