| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Mean value using inverse trig integral |
| Difficulty | Challenging +1.2 Part (a) requires computing a mean value using the given integral of sec x (straightforward application of the mean value formula with provided integration result). Part (b) requires arc length formula, finding dy/dx = tan x, then integrating √(1 + tan²x) = sec x using the given formula. While this involves multiple steps and the arc length concept, the integration is provided and the algebraic manipulation is standard for Further Maths students. This is moderately above average difficulty due to the arc length application and multi-step nature, but not exceptionally challenging. |
| Spec | 4.08a Maclaurin series: find series for function4.08e Mean value of function: using integral |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(MV = \dfrac{\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi}\sec x\, dx}{\frac{1}{3}\pi - \frac{1}{6}\pi} = \dfrac{\left[\ln(\sec x + \tan x)\right]_{\frac{1}{6}\pi}^{\frac{1}{3}\pi}}{\frac{1}{3}\pi - \frac{1}{6}\pi}\) | M1A1 | Uses formula for MV and integrates |
| \(= \frac{6}{\pi}\left\{\ln(2+\sqrt{3}) - \ln\!\left(\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right)\right\}\) | M1 | Substitutes limits |
| \(= \frac{6}{\pi}\ln\!\left(\frac{2+\sqrt{3}}{\sqrt{3}}\right)\) | A1 | (4) OE (1.47) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(y = -\ln\cos x \Rightarrow y' = \tan x\) | B1 | Differentiates wrt \(x\) |
| \(\sqrt{1+(y')^2} = \sec x\) | B1 | Finds \(\frac{ds}{dx}\) |
| \(s = \int_0^{\frac{\pi}{3}}\sec x\, dx = \left[\ln(\sec x + \tan x)\right]_0^{\frac{\pi}{3}}\) | M1 | Obtains arc length integral and integrates |
| \(= \ln(2+\sqrt{3})\) | A1 | (4) (1.32) |
## Question 6(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $MV = \dfrac{\int_{\frac{1}{6}\pi}^{\frac{1}{3}\pi}\sec x\, dx}{\frac{1}{3}\pi - \frac{1}{6}\pi} = \dfrac{\left[\ln(\sec x + \tan x)\right]_{\frac{1}{6}\pi}^{\frac{1}{3}\pi}}{\frac{1}{3}\pi - \frac{1}{6}\pi}$ | M1A1 | Uses formula for MV and integrates |
| $= \frac{6}{\pi}\left\{\ln(2+\sqrt{3}) - \ln\!\left(\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right)\right\}$ | M1 | Substitutes limits |
| $= \frac{6}{\pi}\ln\!\left(\frac{2+\sqrt{3}}{\sqrt{3}}\right)$ | A1 | **(4)** OE (1.47) |
## Question 6(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $y = -\ln\cos x \Rightarrow y' = \tan x$ | B1 | Differentiates wrt $x$ |
| $\sqrt{1+(y')^2} = \sec x$ | B1 | Finds $\frac{ds}{dx}$ |
| $s = \int_0^{\frac{\pi}{3}}\sec x\, dx = \left[\ln(\sec x + \tan x)\right]_0^{\frac{\pi}{3}}$ | M1 | Obtains arc length integral and integrates |
| $= \ln(2+\sqrt{3})$ | A1 | **(4)** (1.32) |
**Total: [8]**
---6 [In this question you may use, without proof, the formula $\int \sec x \mathrm {~d} x = \ln ( \sec x + \tan x ) + \operatorname { const }$.]
\begin{enumerate}[label=(\alph*)]
\item Let $y = \sec x$. Find the mean value of $y$ with respect to $x$ over the interval $\frac { 1 } { 6 } \pi \leqslant x \leqslant \frac { 1 } { 3 } \pi$.
\item The curve $C$ has equation $y = - \ln ( \cos x )$, for $0 \leqslant x \leqslant \frac { 1 } { 3 } \pi$. Find the arc length of $C$.
\end{enumerate}
\hfill \mbox{\textit{CAIE FP1 2013 Q6 [8]}}