| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2016 |
| Session | Specimen |
| Marks | 6 |
| Topic | Hyperbolic functions |
| Type | Second derivative relations with hyperbolics |
| Difficulty | Challenging +1.2 This question requires knowledge of hyperbolic functions (a Further Maths topic) and involves finding a derivative, expressing it in terms of y, then recognizing the reverse relationship for integration. While it requires multiple steps and the connection between parts isn't immediately obvious, the differentiation is straightforward using chain rule, and the integration follows directly once the substitution is identified. It's harder than average due to the Further Maths content and the need to recognize the pattern, but the techniques themselves are standard. |
| Spec | 1.08h Integration by substitution4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| ALT.2 Set \(t^2=\tan\theta\), \(2t\,\mathrm{d}t=\sec^2\theta\,\mathrm{d}\theta\) M1 Full substitution; \(\int\frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}}\,\mathrm{d}\theta\) A1 \(=\int\sec\theta\,\mathrm{d}\theta = \ln | \sec\theta+\tan\theta | = \ln |
**(i)** $y=(\sinh x)^{\frac{1}{2}} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(\sinh x)^{-\frac{1}{2}}.\cosh x$ **OR** $y^2=\sinh x \Rightarrow 2y\frac{\mathrm{d}y}{\mathrm{d}x}=\cosh x$ **M1 A1**
$= \frac{\sqrt{1+y^4}}{2y}$ **A1**
**(ii)** $\int\frac{2y}{\sqrt{1+y^4}}\,\mathrm{d}y = \int 1\,\mathrm{d}x$ — By separating variables in **(i)**'s answer **M1**
$\Rightarrow x = \int\frac{2y}{\sqrt{1+y^4}}\,\mathrm{d}y$ **A1**
But $x = \sinh^{-1}y^2$ so $\int\frac{2t}{\sqrt{1+t^4}}\,\mathrm{d}x = \sinh^{-1}(t^2)+C$ — condone missing "$+C$" **A1**
**ALT.1** Set $t^2=\sinh\theta$, $2t\,\mathrm{d}t=\cosh\theta\,\mathrm{d}\theta$ **M1** Full substitution; $\int\frac{\cosh\theta}{\sqrt{1+\sinh^2\theta}}\,\mathrm{d}\theta$ **A1** $=\int 1\,\mathrm{d}\theta = \theta = \sinh^{-1}(t^2)$ **A1**
**ALT.2** Set $t^2=\tan\theta$, $2t\,\mathrm{d}t=\sec^2\theta\,\mathrm{d}\theta$ **M1** Full substitution; $\int\frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}}\,\mathrm{d}\theta$ **A1** $=\int\sec\theta\,\mathrm{d}\theta = \ln|\sec\theta+\tan\theta| = \ln|t^2+\sqrt{1+t^4}|$ **A1**
**Total: 6 marks**4 (i) Given that $y = \sqrt { \sinh x }$ for $x \geqslant 0$, express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $y$ only.\\
(ii) Hence or otherwise find $\int \frac { 2 t } { \sqrt { 1 + t ^ { 4 } } } \mathrm {~d} t$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2016 Q4 [6]}}