CAIE FP1 2015 June — Question 4 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeTrigonometric method of differences
DifficultyChallenging +1.2 This is a standard method of differences question with inverse tan functions. The first part requires straightforward application of a given formula (tan(A-B)) with substitutions A = tan^(-1)(x+1) and B = tan^(-1)(x-1). The second part is a telescoping series where consecutive terms cancel, requiring students to recognize the pattern and write out the sum. While it involves Further Maths content (inverse trig and method of differences), the technique is routine and well-practiced in FP1, making it slightly above average difficulty but not requiring novel insight.
Spec1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs4.06b Method of differences: telescoping series
4 Use the formula for \(\tan ( A - B )\) in the List of Formulae (MF10) to show that $$\tan ^ { - 1 } ( x + 1 ) - \tan ^ { - 1 } ( x - 1 ) = \tan ^ { - 1 } \left( \frac { 2 } { x ^ { 2 } } \right)$$ Deduce the sum to \(n\) terms of the series $$\tan ^ { - 1 } \left( \frac { 2 } { 1 ^ { 2 } } \right) + \tan ^ { - 1 } \left( \frac { 2 } { 2 ^ { 2 } } \right) + \tan ^ { - 1 } \left( \frac { 2 } { 3 ^ { 2 } } \right) + \ldots .$$
Question 4:
AnswerMarks Guidance
Working/AnswerMarks Notes
\(A = \tan^{-1}(x+1)\); \(B = \tan^{-1}(x-1) \Rightarrow \tan(A-B) = \frac{(x+1)-(x-1)}{1+(x+1)(x-1)}\)M1A1
\(\tan(A-B) = \frac{2}{x^2} \Rightarrow A-B = \tan^{-1}\!\left(\frac{2}{x^2}\right)\)A1 AG
LHS \(= (\tan^{-1}2 - \tan^{-1}0) + (\tan^{-1}3 - \tan^{-1}1) + \ldots + (\tan^{-1}n - \tan^{-1}[n-2])\) \(+ (\tan^{-1}[n+1] - \tan^{-1}[n-1])\)M1A1
\(= \tan^{-1}[n+1] + \tan^{-1}n - \tan^{-1}1 - \tan^{-1}0 = \tan^{-1}[n+1] + \tan^{-1}n - \frac{1}{4}\pi\)A1A1 
## Question 4:

| Working/Answer | Marks | Notes |
|---|---|---|
| $A = \tan^{-1}(x+1)$; $B = \tan^{-1}(x-1) \Rightarrow \tan(A-B) = \frac{(x+1)-(x-1)}{1+(x+1)(x-1)}$ | M1A1 | |
| $\tan(A-B) = \frac{2}{x^2} \Rightarrow A-B = \tan^{-1}\!\left(\frac{2}{x^2}\right)$ | A1 | AG |
| LHS $= (\tan^{-1}2 - \tan^{-1}0) + (\tan^{-1}3 - \tan^{-1}1) + \ldots + (\tan^{-1}n - \tan^{-1}[n-2])$ $+ (\tan^{-1}[n+1] - \tan^{-1}[n-1])$ | M1A1 | |
| $= \tan^{-1}[n+1] + \tan^{-1}n - \tan^{-1}1 - \tan^{-1}0 = \tan^{-1}[n+1] + \tan^{-1}n - \frac{1}{4}\pi$ | A1A1 | |

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4 Use the formula for $\tan ( A - B )$ in the List of Formulae (MF10) to show that

$$\tan ^ { - 1 } ( x + 1 ) - \tan ^ { - 1 } ( x - 1 ) = \tan ^ { - 1 } \left( \frac { 2 } { x ^ { 2 } } \right)$$

Deduce the sum to $n$ terms of the series

$$\tan ^ { - 1 } \left( \frac { 2 } { 1 ^ { 2 } } \right) + \tan ^ { - 1 } \left( \frac { 2 } { 2 ^ { 2 } } \right) + \tan ^ { - 1 } \left( \frac { 2 } { 3 ^ { 2 } } \right) + \ldots .$$

\hfill \mbox{\textit{CAIE FP1 2015 Q4 [7]}}
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