| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Second derivative relations with hyperbolics |
| Difficulty | Standard +0.8 Part (i) is a standard Further Maths derivation requiring knowledge of inverse hyperbolic functions and implicit differentiation. Part (ii) requires careful application of the chain rule, product rule, and algebraic manipulation to verify a second-order differential equation—this involves multiple differentiation steps and non-trivial simplification using hyperbolic identities. While systematic, it demands precision and is more challenging than typical A-level questions but remains a standard FP2 exercise. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08g Derivatives: inverse trig and hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Reasonable attempt to differentiate \(\sinh y = x\) to get \(dy/dx\) in terms of \(y\) | M1 | Allow \(\pm\cosh y\,dy/dx = 1\) |
| Replace \(\sinh y\) to get A.G. | A1 | Clearly use \(\cosh^2 - \sinh^2 = 1\) |
| SC: Attempt to differentiate \(y = \ln(x+\sqrt{x^2+1})\) using chain rule | M1 | |
| Clearly tidy to A.G. | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Reasonable attempt at chain rule | M1 | To give a product |
| Get \(dy/dx = a\sinh(a\sinh^{-1}x)/\sqrt{x^2+1}\) | A1 | |
| Reasonable attempt at product/quotient rule | M1 | Must involve sinh and cosh |
| Get \(d^2y/dx^2\) correctly in some form | A1\(\checkmark\) | From \(dy/dx = k\sinh(a\sinh^{-1}x)/\sqrt{x^2+1}\) |
| Substitute in and clearly get A.G. | A1 | |
| SC: Write \(\sqrt{x^2+1}\,dy/dx = k\sinh(a\sinh^{-1}x)\) or similar, then derive A.G. |
# Question 6:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Reasonable attempt to differentiate $\sinh y = x$ to get $dy/dx$ in terms of $y$ | M1 | Allow $\pm\cosh y\,dy/dx = 1$ |
| Replace $\sinh y$ to get A.G. | A1 | Clearly use $\cosh^2 - \sinh^2 = 1$ |
| SC: Attempt to differentiate $y = \ln(x+\sqrt{x^2+1})$ using chain rule | M1 | |
| Clearly tidy to A.G. | A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Reasonable attempt at chain rule | M1 | To give a product |
| Get $dy/dx = a\sinh(a\sinh^{-1}x)/\sqrt{x^2+1}$ | A1 | |
| Reasonable attempt at product/quotient rule | M1 | Must involve sinh and cosh |
| Get $d^2y/dx^2$ correctly in some form | A1$\checkmark$ | From $dy/dx = k\sinh(a\sinh^{-1}x)/\sqrt{x^2+1}$ |
| Substitute in and clearly get A.G. | A1 | |
| SC: Write $\sqrt{x^2+1}\,dy/dx = k\sinh(a\sinh^{-1}x)$ or similar, then derive A.G. | | |
---6 (i) Show that $\frac { \mathrm { d } } { \mathrm { d } x } \left( \sinh ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 2 } + 1 } }$.\\
(ii) Given that $y = \cosh \left( a \sinh ^ { - 1 } x \right)$, where $a$ is a constant, show that
$$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + x \frac { \mathrm {~d} y } { \mathrm {~d} x } - a ^ { 2 } y = 0$$
\hfill \mbox{\textit{OCR FP2 2010 Q6 [7]}}