| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Integrate using hyperbolic substitution |
| Difficulty | Standard +0.8 This is a structured Further Maths question on hyperbolic functions requiring multiple proofs and integration techniques. Part (i) is routine FP2 material, part (ii) is a standard derivation, but parts (iii) and (iv) require careful substitution work and manipulation of hyperbolic identities across multiple steps. The 18-mark total and need to connect hyperbolic substitutions with definite integral evaluation places this moderately above average difficulty, though it follows predictable FP2 patterns rather than requiring novel insight. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07f Inverse hyperbolic: logarithmic forms4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cosh u = \frac{e^u + e^{-u}}{2}\) | ||
| \(\Rightarrow 2\cosh^2 u = \frac{e^{2u} + 2 + e^{-2u}}{2}\) | B1 | \((e^u + e^{-u})^2 = e^{2u} + 2 + e^{-2u}\) |
| \(\Rightarrow 2\cosh^2 u - 1 = \frac{e^{2u} + e^{-2u}}{2}\) | B1 | \(\cosh 2u = \frac{e^{2u} + e^{-2u}}{2}\) |
| \(= \cosh 2u\) | B1 (ag) | Completion www |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \text{arsinh } y\) | ||
| \(\Rightarrow \sinh x = y\) | ||
| \(\Rightarrow y = \frac{e^x - e^{-x}}{2}\) | M1 | Expressing y in exponential form (\(\frac{1}{2}\) — must be correct) |
| \(\Rightarrow e^{2x} - 2ye^x - 1 = 0\) | ||
| \(\Rightarrow (e^x)^2 - y^2 - 1 = 0\) | ||
| \(\Rightarrow e^x - y = \sqrt{y^2 + 1}\) | ||
| \(\Rightarrow e^x = y \pm \sqrt{y^2 + 1}\) | M1 | Reaching \(e^x\) by quadratic formula or completing the square. Condone no ±. Or argument of ln must be positive |
| Take + because \(e^x > 0\) | B1 | |
| \(\Rightarrow x = \ln(y + \sqrt{y^2 + 1})\) | A1 (ag) | Completion www but independent of B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 2\sinh u \Rightarrow \frac{dx}{du} = 2\cosh u\) | M1 | \(\frac{dx}{du}\) and substituting for all elements |
| \(\int \sqrt{x^2 + 4} dx = \int \sqrt{4\sinh^2 u + 4} \times 2\cosh u \, du\) | A1 | Substituting for all elements correctly |
| \(= \int 4\cosh u \, du\) | ||
| \(= \int 2\cosh 2u + 2 \, du\) | M1 | Simplifying to an integrable form. Any form, e.g. \(\frac{e^u-\frac{1}{e^u} + 2u}\) |
| \(= \sinh 2u + 2u + c\) | A1 | Condone omission of + c throughout |
| \(= 2\sinh u \cosh u + 2u + c\) | ||
| \(= x\sqrt{1 + \frac{x^2}{4}} + 2\text{arsinh}\frac{x}{2} + c\) | M1 | Using double "angle" formula and attempt to express \(\cosh u\) in terms of x |
| \(= \frac{1}{2}\sqrt{x^2 + 4} + 2\text{arsinh}\frac{x}{2} + c\) | A1 (ag) | Completion www |
| Answer | Marks | Guidance |
|---|---|---|
| \(t + 2t + 5 = (t+1)^2 + 4\) | B1 | Completing the square |
| \(\int \sqrt{t^2 + 2t + 5} dt = \int \sqrt{(t+1)^2 + 4} dt\) | ||
| \(= \left[\frac{1}{2}\sqrt{x^2 + 4} + 2\text{arsinh}\frac{x}{2}\right]\) | M1 | Simplifying to an integrable form, by substituting \(x = t + 1\) s.o.i. complete alternative method. Correct limits consistent with their method seen anywhere |
| 5 marks total |
## Part (i)
$\cosh u = \frac{e^u + e^{-u}}{2}$ | |
$\Rightarrow 2\cosh^2 u = \frac{e^{2u} + 2 + e^{-2u}}{2}$ | B1 | $(e^u + e^{-u})^2 = e^{2u} + 2 + e^{-2u}$
$\Rightarrow 2\cosh^2 u - 1 = \frac{e^{2u} + e^{-2u}}{2}$ | B1 | $\cosh 2u = \frac{e^{2u} + e^{-2u}}{2}$
$= \cosh 2u$ | B1 (ag) | Completion www | | 3 marks total
## Part (ii)
$x = \text{arsinh } y$ | |
$\Rightarrow \sinh x = y$ | |
$\Rightarrow y = \frac{e^x - e^{-x}}{2}$ | M1 | Expressing y in exponential form ($\frac{1}{2}$ — must be correct)
$\Rightarrow e^{2x} - 2ye^x - 1 = 0$ | |
$\Rightarrow (e^x)^2 - y^2 - 1 = 0$ | |
$\Rightarrow e^x - y = \sqrt{y^2 + 1}$ | |
$\Rightarrow e^x = y \pm \sqrt{y^2 + 1}$ | M1 | Reaching $e^x$ by quadratic formula or completing the square. Condone no ±. Or argument of ln must be positive
Take + because $e^x > 0$ | B1 |
$\Rightarrow x = \ln(y + \sqrt{y^2 + 1})$ | A1 (ag) | Completion www but independent of B1 | | 4 marks total
## Part (iii)
$x = 2\sinh u \Rightarrow \frac{dx}{du} = 2\cosh u$ | M1 | $\frac{dx}{du}$ and substituting for all elements
$\int \sqrt{x^2 + 4} dx = \int \sqrt{4\sinh^2 u + 4} \times 2\cosh u \, du$ | A1 | Substituting for all elements correctly
$= \int 4\cosh u \, du$ | |
$= \int 2\cosh 2u + 2 \, du$ | M1 | Simplifying to an integrable form. Any form, e.g. $\frac{e^u-\frac{1}{e^u} + 2u}$
$= \sinh 2u + 2u + c$ | A1 | Condone omission of + c throughout
$= 2\sinh u \cosh u + 2u + c$ | |
$= x\sqrt{1 + \frac{x^2}{4}} + 2\text{arsinh}\frac{x}{2} + c$ | M1 | Using double "angle" formula and attempt to express $\cosh u$ in terms of x
$= \frac{1}{2}\sqrt{x^2 + 4} + 2\text{arsinh}\frac{x}{2} + c$ | A1 (ag) | Completion www | | 6 marks total
## Part (iv)
$t + 2t + 5 = (t+1)^2 + 4$ | B1 | Completing the square
$\int \sqrt{t^2 + 2t + 5} dt = \int \sqrt{(t+1)^2 + 4} dt$ | |
$= \left[\frac{1}{2}\sqrt{x^2 + 4} + 2\text{arsinh}\frac{x}{2}\right]$ | M1 | Simplifying to an integrable form, by substituting $x = t + 1$ s.o.i. complete alternative method. Correct limits consistent with their method seen anywhere | A1 |
| | 5 marks total
---\begin{enumerate}[label=(\roman*)]
\item Prove, from definitions involving exponentials, that
$$\cosh 2u = 2\cosh^2 u - 1.$$ [3]
\item Prove that $\arsinh y = \ln\left(y + \sqrt{y^2 + 1}\right)$. [4]
\item Use the substitution $x = 2\sinh u$ to show that
$$\int \sqrt{x^2 + 4} dx = 2\arsinh \frac{x}{2} + \frac{x}{2}\sqrt{x^2 + 4} + c,$$
where $c$ is an arbitrary constant. [6]
\item By first expressing $t^2 + 2t + 5$ in completed square form, show that
$$\int_{-1}^1 \sqrt{t^2 + 2t + 5} dt = 2\left(\ln(1 + \sqrt{2}) + \sqrt{2}\right).$$ [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2009 Q4 [18]}}