| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Trigonometric method of differences |
| Difficulty | Challenging +1.2 This is a structured multi-part question on inverse trigonometric functions and series. Part (i) is a standard geometric visualization (1 mark). Part (ii) requires applying the tangent addition formula to inverse tangents, which is a familiar technique (2 marks). Part (iii) is the most substantial, requiring students to recognize that tan^{-1}(2/n²) can be expressed as a difference of inverse tangents using the given substitution, then apply telescoping series—this requires insight but follows a guided path with 7 marks. While requiring multiple techniques and some problem-solving, the question provides significant scaffolding and uses well-known methods (method of differences, telescoping). It's moderately challenging for Further Maths but not exceptional. |
| Spec | 4.06b Method of differences: telescoping series4.08g Derivatives: inverse trig and hyperbolic functions |
\begin{enumerate}[label=(\roman*)]
\item By sketching a suitable triangle, show that $\tan^{-1} a + \tan^{-1} \left(\frac{1}{a}\right) = \frac{1}{4}\pi$, for $a > 0$. [1]
\item Given that $a$ and $b$ are positive and less than 1, express $\tan(\tan^{-1} a \pm \tan^{-1} b)$ in terms of $a$ and $b$. [2]
\item By letting $a = \frac{1}{n-1}$ and $b = \frac{1}{n+1}$, use the method of differences to prove that
$$\sum_{n=1}^{\infty} \tan^{-1} \left(\frac{2}{n^2}\right) = \frac{3}{4}\pi.$$ [7]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2015 Q13 [10]}}