| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2021 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Find exact trigonometric values |
| Difficulty | Challenging +1.2 This question requires applying the arctan addition formula and algebraic manipulation to verify given identities. Part (a) is straightforward substitution (n=1, n=2), while part (b) requires working backwards with the addition formula and simplifying a rational expression. The algebra is moderately involved but follows a clear path with no novel insight required—it's a standard application of a given formula with careful manipulation. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| If \(n=1\), \(n+1=2\) and \(n^2+n+1=3\) | E1 | 2.2a |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For \(n\) a positive integer, \(\frac{1}{n}\left(\frac{1}{n^2+n+1}\right) < 1\) | E1 | 2.3 |
| \(\arctan\left(\frac{1}{n+1}\right) + \arctan\left(\frac{1}{n^2+n+1}\right) = \arctan\left(\dfrac{\frac{1}{n+1}+\frac{1}{n^2+n+1}}{1 - \frac{1}{n+1}\cdot\frac{1}{n^2+n+1}}\right)\) | M1 | 3.1a — Use of arctan addition formula |
| \(\arctan\left(\dfrac{n^2+n+1+n+1}{(n^2+n+1)(n+1)-1}\right)\) | M1 | 1.1 — Clearing fractions within the fraction. Condone omission of arctan for this mark and condone 1 other error |
| \(\arctan\left(\dfrac{n^2+2n+2}{n^3+2n^2+2n}\right) = \arctan\left(\dfrac{n^2+2n+2}{n(n^2+2n+2)}\right) = \arctan\left(\frac{1}{n}\right)\) | A1 | 2.1 — Convincing completion (AG) |
| [4] |
## Question 14(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| If $n=1$, $n+1=2$ and $n^2+n+1=3$ | E1 | 2.2a |
| | **[1]** | |
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## Question 14(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For $n$ a positive integer, $\frac{1}{n}\left(\frac{1}{n^2+n+1}\right) < 1$ | E1 | 2.3 |
| $\arctan\left(\frac{1}{n+1}\right) + \arctan\left(\frac{1}{n^2+n+1}\right) = \arctan\left(\dfrac{\frac{1}{n+1}+\frac{1}{n^2+n+1}}{1 - \frac{1}{n+1}\cdot\frac{1}{n^2+n+1}}\right)$ | M1 | 3.1a — Use of arctan addition formula |
| $\arctan\left(\dfrac{n^2+n+1+n+1}{(n^2+n+1)(n+1)-1}\right)$ | M1 | 1.1 — Clearing fractions within the fraction. Condone omission of arctan for this mark and condone 1 other error |
| $\arctan\left(\dfrac{n^2+2n+2}{n^3+2n^2+2n}\right) = \arctan\left(\dfrac{n^2+2n+2}{n(n^2+2n+2)}\right) = \arctan\left(\frac{1}{n}\right)$ | A1 | 2.1 — Convincing completion (AG) |
| | **[4]** | |14
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\arctan \left( \frac { 1 } { n + 1 } \right) + \arctan \left( \frac { 1 } { n ^ { 2 } + n + 1 } \right) = \arctan \left( \frac { 1 } { n } \right) \Rightarrow \arctan \left( \frac { 1 } { 2 } \right) + \arctan \left( \frac { 1 } { 3 } \right) = \arctan 1 .$$
\item Use the arctan addition formula in line 23 to show that
$$\arctan \left( \frac { 1 } { n + 1 } \right) + \arctan \left( \frac { 1 } { n ^ { 2 } + n + 1 } \right) = \arctan \left( \frac { 1 } { n } \right) , \text { as given in line } 39 .$$
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2021 Q14 [5]}}