| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Derivative of inverse trig function |
| Difficulty | Standard +0.3 This is a structured Further Maths question with clear scaffolding. Part (a)(i) is a standard derivation from first principles that appears in textbooks. Part (a)(ii) requires recognizing standard inverse trig integrals with simple substitutions. Part (b) involves routine polar-to-Cartesian conversions using given identities. While it's Further Maths content, the question guides students through each step methodically with no novel insights required, making it slightly easier than an average A-level question overall. |
| Spec | 1.07s Parametric and implicit differentiation4.08h Integration: inverse trig/hyperbolic substitutions4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| (b) \(r = \tan \theta \Rightarrow x = r \cos \theta = \frac{\sin \theta}{\cos \theta} \times \cos \theta = \sin \theta\) | M1 A1(ag) | Using \(x = r \cos \theta\) o.e. |
| \(\Rightarrow r^2 = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\sin^2 \theta}{1 - \sin^2 \theta} = \frac{x^2}{1-x^2}\) | M1 A1(ag) | Obtaining \(r^2\) in terms of \(x\) |
| \(r^2 = x^2 + y^2 \Rightarrow x^2 + y^2 = \frac{x^2}{1-x^2}\) | M1 | Obtaining \(y^2\) in terms of \(x\) |
| \(\Rightarrow y^2 = \frac{x^2}{1-x^2} - x^2\) | M1 | |
| \(\Rightarrow y^2 = \frac{x^2 - x^2(1-x^2)}{1-x^2} = \frac{x^3}{1-x^2}\) | A1(ag) | |
| \(\Rightarrow y = \frac{x^2}{\sqrt{1-x^2}}\) | A1(ag) | Ignore discussion of \(\pm\) |
| Asymptote \(x = 1\) | B1 [7] | Condone \(x = \pm 1\); \(x^2 = 1\) B0 |
**(b)** $r = \tan \theta \Rightarrow x = r \cos \theta = \frac{\sin \theta}{\cos \theta} \times \cos \theta = \sin \theta$ | M1 A1(ag) | Using $x = r \cos \theta$ o.e.
$\Rightarrow r^2 = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\sin^2 \theta}{1 - \sin^2 \theta} = \frac{x^2}{1-x^2}$ | M1 A1(ag) | Obtaining $r^2$ in terms of $x$
$r^2 = x^2 + y^2 \Rightarrow x^2 + y^2 = \frac{x^2}{1-x^2}$ | M1 | Obtaining $y^2$ in terms of $x$
$\Rightarrow y^2 = \frac{x^2}{1-x^2} - x^2$ | M1 |
$\Rightarrow y^2 = \frac{x^2 - x^2(1-x^2)}{1-x^2} = \frac{x^3}{1-x^2}$ | A1(ag) |
$\Rightarrow y = \frac{x^2}{\sqrt{1-x^2}}$ | A1(ag) | Ignore discussion of $\pm$
Asymptote $x = 1$ | B1 [7] | Condone $x = \pm 1$; $x^2 = 1$ B0
---1
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Differentiate the equation $\sin y = x$ with respect to $x$, and hence show that the derivative of $\arcsin x$ is $\frac { 1 } { \sqrt { 1 - x ^ { 2 } } }$.
\item Evaluate the following integrals, giving your answers in exact form.\\
(A) $\int _ { - 1 } ^ { 1 } \frac { 1 } { \sqrt { 2 - x ^ { 2 } } } \mathrm {~d} x$\\
(B) $\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \frac { 1 } { \sqrt { 1 - 2 x ^ { 2 } } } \mathrm {~d} x$
\end{enumerate}\item A curve has polar equation $r = \tan \theta , 0 \leqslant \theta < \frac { 1 } { 2 } \pi$. The points on the curve have cartesian coordinates $( x , y )$. A sketch of the curve is given in Fig. 1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{99f0c663-bb5b-4456-854c-df177f5d8349-2_493_796_1123_605}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
Show that $x = \sin \theta$ and that $r ^ { 2 } = \frac { x ^ { 2 } } { 1 - x ^ { 2 } }$.\\
Hence show that the cartesian equation of the curve is
$$y = \frac { x ^ { 2 } } { \sqrt { 1 - x ^ { 2 } } } .$$
Give the cartesian equation of the asymptote of the curve.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2012 Q1 [18]}}