| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove inverse hyperbolic logarithmic form |
| Difficulty | Standard +0.3 This is a standard Further Maths question testing routine techniques with inverse hyperbolic functions. Part (i) requires substituting the definition of cosh and solving a quadratic in e^y (textbook method), part (ii) is straightforward differentiation using the chain rule or the result from (i), and part (iii) applies part (i) directly. While hyperbolic functions are a Further Maths topic, these are well-practiced procedures with clear pathways, making this slightly easier than average overall. |
| Spec | 4.07d Differentiate/integrate: hyperbolic functions4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow y = \ln x + \sqrt{x^2-1}\) | M1, A1, B1, A1 | Finding 3 term quadratic in \(e^y\); Correct solution. Condone ignoring -ve sign at this point; Including reason oe. Condone interchange of x and y but final ans must be correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{1}{x+\sqrt{x^2-1}} \times \frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}} = \frac{1}{\sqrt{x^2-1}}\) | M1, A1 | Alt: \(x = \cosh y \Rightarrow \frac{dy}{dx} = \frac{1}{\sinh y} = \frac{1}{\sqrt{x^2-1}}\) |
| (iii) \(x = \cosh^{-1}3 = \ln 3 + \sqrt{8} = -\ln 3 + \sqrt{8}\) oe | M1, A1, A1 | Use of \(\cosh^{-1}\); ft, -ve the first answer |
| [3] |
**(i)** $x = \cosh y = \frac{e^y + e^{-y}}{2} \Rightarrow e^y + e^{-y} = 2x$
$\Rightarrow e^{2y} - 2xe^y + 1 = 0$
$\Rightarrow e^y = \frac{2x \pm \sqrt{4x^2-4}}{2} = x \pm \sqrt{x^2-1}$
$\Rightarrow y = \ln x \pm \sqrt{x^2-1}$
Reject $-$ sign as principal value taken
$\Rightarrow y = \ln x + \sqrt{x^2-1}$ | M1, A1, B1, A1 | Finding 3 term quadratic in $e^y$; Correct solution. Condone ignoring -ve sign at this point; Including reason oe. Condone interchange of x and y but final ans must be correct
**(ii)** $y = \ln x + \sqrt{x^2-1} \Rightarrow \frac{dy}{dx} = \frac{1}{x+\sqrt{x^2-1}} \times \left(1 + \frac{2x}{2\sqrt{x^2-1}}\right)$
$= \frac{1}{x+\sqrt{x^2-1}} \times \frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}} = \frac{1}{\sqrt{x^2-1}}$ | M1, A1 | Alt: $x = \cosh y \Rightarrow \frac{dy}{dx} = \frac{1}{\sinh y} = \frac{1}{\sqrt{x^2-1}}$
**(iii)** $x = \cosh^{-1}3 = \ln 3 + \sqrt{8} = -\ln 3 + \sqrt{8}$ oe | M1, A1, A1 | Use of $\cosh^{-1}$; ft, -ve the first answer
| [3] |
---6 (i) Given that $y = \cosh ^ { - 1 } x$, show that $y = \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)$.\\
(ii) Show that $\frac { \mathrm { d } } { \mathrm { d } x } \left( \cosh ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 2 } - 1 } }$.\\
(iii) Solve the equation $\cosh x = 3$, giving your answers in logarithmic form.
\hfill \mbox{\textit{OCR FP2 2014 Q6 [9]}}