| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2016 |
| Session | June |
| Marks | 17 |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Derivative of inverse trig function |
| Difficulty | Challenging +1.8 This is a substantial multi-part Further Maths question requiring: (i) proving the derivative of sec⁻¹x using implicit differentiation and the chain rule, (ii) integration by parts with inverse trig functions, and (iii) finding tangent coordinates and computing a complex area involving both integration and geometric components. While the techniques are standard for Further Maths (inverse trig derivatives, integration by parts, area calculations), the question requires careful execution across multiple steps and the final area calculation involves algebraic manipulation to reach a specific exact form. This is moderately challenging for Further Maths but follows predictable patterns. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.08i Integration by parts4.08g Derivatives: inverse trig and hyperbolic functions |
**Question 13** [3+4+6+4 marks]
**(i)(a)**
Let $y = \sec^{-1}x$, i.e. $\sec y = x$
$\Rightarrow \cos y = \frac{1}{x} \Rightarrow y = \cos^{-1}\!\left(\frac{1}{x}\right)$ B1
Then $\frac{d}{dx}\left(\sec^{-1}x\right) = \frac{d}{dx}\left(\cos^{-1}\frac{1}{x}\right)$
$= -\frac{1}{\sqrt{1-(1/x)^2}} \times \frac{-1}{x^2}$ M1 (Using MF20 and the Chain Rule)
$= \frac{1}{x\sqrt{x^2-1}}$ A1 **(AG)**
[Allow **M1 A1** for valid non-"deduced" approaches]
**[3]**
**(b)**
$\int \sec^{-1}x \cdot 1\,\mathrm{d}x$ M1 (Use of integration by "parts")
$= x\cdot\sec^{-1}x - \int x \cdot \frac{1}{x\sqrt{x^2-1}}\,\mathrm{d}x$ A1 A1
$= \left[x\cdot\sec^{-1}x - \cosh^{-1}x\right]$ A1 (Condone lack of "$+C$")
**[4]**
**(ii)(a)**
$\frac{1}{x\sqrt{x^2-1}} = \frac{1}{\sqrt{2}} \Rightarrow x^2(x^2-1)=2$ M1
$\Rightarrow x^4-x^2-2=(x^2-2)(x^2+1)=0$
$\Rightarrow x=\sqrt{2}$ and $y=\frac{1}{4}\pi$ A1 A1 (i.e. $P=(\sqrt{2},\frac{1}{4}\pi)$)
$\frac{\frac{1}{4}\pi}{\sqrt{2}-c} = \frac{1}{\sqrt{2}}$ M1A1 (or by $y-\frac{1}{4}\pi = \frac{1}{\sqrt{2}}(x-\sqrt{2})$ & $y=0$)
$c = \sqrt{2} - \frac{\pi\sqrt{2}}{4}$ A1 (i.e. $Q=\left(\sqrt{2}-\frac{\pi\sqrt{2}}{4},\ 0\right)$)
**[6]**
**(b)**
Area $\triangle = \frac{1}{2}\times\frac{\pi\sqrt{2}}{4}\times\frac{\pi}{4} = \frac{\pi^2\sqrt{2}}{32}$ B1
Area under curve $= \sqrt{2}\cdot\frac{\pi}{4} - \ln(1+\sqrt{2})$ B1 (using **(iii)**'s answer and the limits $(1,\sqrt{2})$)
Then $R = \frac{\pi^2\sqrt{2}}{32} - \frac{\pi\sqrt{2}}{4} + \ln(1+\sqrt{2})$ M1 (Difference in areas)
$= \ln(1+\sqrt{2}) - \frac{\pi(8-\pi)\sqrt{2}}{32}$ A1 **(AG)**
**[4]**13 (i) (a) Given that $x \geqslant 1$, show that $\sec ^ { - 1 } x = \cos ^ { - 1 } \left( \frac { 1 } { x } \right)$, and deduce that $\frac { \mathrm { d } } { \mathrm { d } x } \left( \sec ^ { - 1 } x \right) = \frac { 1 } { x \sqrt { x ^ { 2 } - 1 } }$.\\
(b) Use integration by parts to determine $\int \sec ^ { - 1 } x \mathrm {~d} x$.\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{5d526fd9-72f8-42b1-b156-fd4a0c764c82-4_670_1029_1073_596}
The diagram shows the curve $S$ with equation $y = \sec ^ { - 1 } x$ for $x \geqslant 1$. The line $L$, with gradient $\frac { 1 } { \sqrt { 2 } }$, is the tangent to $S$ at the point $P$ and cuts the $x$-axis at the point $Q$. The point $I$ has coordinates $( 1,0 )$.\\
(a) Determine the exact coordinates of $P$ and $Q$.\\
(b) The region $R$, shaded on the diagram, is bounded by the line segments $P Q$ and $Q I$ and the $\operatorname { arc } I P$ of $S$. Show that $R$ has area
$$\ln ( 1 + \sqrt { 2 } ) - \frac { \pi ( 8 - \pi ) \sqrt { 2 } } { 32 } .$$
{www.cie.org.uk} after the live examination series.
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\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2016 Q13 [17]}}