| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using double angle formulas |
| Difficulty | Challenging +1.2 This is a Further Maths question covering standard hyperbolic function techniques: deriving the inverse hyperbolic cosine formula (bookwork), applying it to a straightforward integral, and solving an equation using the double angle formula cosh(2x) = 2cosh²(x) - 1 to get a quadratic. While it requires multiple techniques and is inherently harder than single maths content, these are all standard FP2 procedures without requiring novel insight or complex problem-solving. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07e Inverse hyperbolic: definitions, domains, ranges4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\cosh y = x \Rightarrow x = \frac{1}{2}(e^y+e^{-y})\) | B1 | Using correct exponential definition |
| \(\Rightarrow (e^y)^2-2xe^y+1=0\) | M1 | Obtaining quadratic in \(e^y\) |
| \(\Rightarrow e^y = \frac{2x\pm\sqrt{4x^2-4}}{2} = x\pm\sqrt{x^2-1}\) | M1, A1 | Solving quadratic; \(x\pm\sqrt{x^2-1}\) |
| \(\Rightarrow y = \ln(x\pm\sqrt{x^2-1})\) | ||
| \((x+\sqrt{x^2-1})(x-\sqrt{x^2-1})=1\) | M1 | Validly attempting to justify \(\pm\) in printed answer |
| \(\Rightarrow y = \pm\ln(x+\sqrt{x^2-1})\) | A1 (ag) | |
| \(\text{arcosh}(x) = \ln(x+\sqrt{x^2-1})\) because this is the principal value | B1 | Reference to arcosh as a function, or correctly to domains/ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int_{\frac{4}{5}}^{1}\frac{1}{\sqrt{25x^2-16}}\,dx = \frac{1}{5}\int_{\frac{4}{5}}^{1}\frac{1}{\sqrt{x^2-\frac{16}{25}}}\,dx\) | ||
| \(= \frac{1}{5}\left[\text{arcosh}\left(\frac{5x}{4}\right)\right]_{\frac{4}{5}}^{1}\) | M1, A1A1 | arcosh alone, or any cosh substitution; \(\frac{1}{5}\), \(\frac{5x}{4}\) |
| \(= \frac{1}{5}\left(\text{arcosh}\left(\frac{5}{4}\right)-\text{arcosh}(1)\right)\) | ||
| \(= \frac{1}{5}\ln\left(\frac{5}{4}+\sqrt{\left(\frac{5}{4}\right)^2-1}\right)-0\) | M1 | Substituting limits and using (i) correctly at any stage (or using limits of \(u\) in logarithmic form); dependent on first M1 |
| \(= \frac{1}{5}\ln 2\) | A1 | |
| OR \(= \frac{1}{5}\left[\ln\left(x+\sqrt{x^2-\frac{16}{25}}\right)\right]_{\frac{4}{5}}^{1}\) | M | \(\ln\left(kx+\sqrt{k^2x^2+\ldots}\right)\); give M1 for \(\ln\left(k_1x+\sqrt{k_2^2x^2+\ldots}\right)\) |
| A1A | \(\frac{1}{5}\ln\left(x+\sqrt{x^2-\frac{16}{25}}\right)\) o.e. | |
| \(= \frac{1}{5}\ln\frac{8}{5}-\frac{1}{5}\ln\frac{4}{5} = \frac{1}{5}\ln 2\) | A |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(5\cosh x - \cosh 2x = 3\) | ||
| \(\Rightarrow 5\cosh x-(2\cosh^2 x-1)=3\) | M1 | Attempting to express \(\cosh 2x\) in terms of \(\cosh x\) |
| \(\Rightarrow 2\cosh^2 x - 5\cosh x+2=0\) | ||
| \(\Rightarrow (2\cosh x-1)(\cosh x-2)=0\) | M1 | Solving quadratic to obtain at least one real value of \(\cosh x\) |
| \(\Rightarrow \cosh x = \frac{1}{2}\) (rejected) | A1 | Or factor \(2\cosh x-1\) |
| or \(\cosh x = 2\) | A1 | |
| \(\Rightarrow x = \ln(2+\sqrt{3})\) | A1ft | F.t. \(\cosh x = k,\ k>1\) |
| \(x = -\ln(2+\sqrt{3})\) or \(\ln(2-\sqrt{3})\) | A1ft | F.t. other value. Max. A1A0 if additional real values quoted |
# Question 4:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cosh y = x \Rightarrow x = \frac{1}{2}(e^y+e^{-y})$ | B1 | Using correct exponential definition |
| $\Rightarrow (e^y)^2-2xe^y+1=0$ | M1 | Obtaining quadratic in $e^y$ |
| $\Rightarrow e^y = \frac{2x\pm\sqrt{4x^2-4}}{2} = x\pm\sqrt{x^2-1}$ | M1, A1 | Solving quadratic; $x\pm\sqrt{x^2-1}$ |
| $\Rightarrow y = \ln(x\pm\sqrt{x^2-1})$ | | |
| $(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})=1$ | M1 | Validly attempting to justify $\pm$ in printed answer |
| $\Rightarrow y = \pm\ln(x+\sqrt{x^2-1})$ | A1 (ag) | |
| $\text{arcosh}(x) = \ln(x+\sqrt{x^2-1})$ because this is the principal value | B1 | Reference to arcosh as a function, or correctly to domains/ranges |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_{\frac{4}{5}}^{1}\frac{1}{\sqrt{25x^2-16}}\,dx = \frac{1}{5}\int_{\frac{4}{5}}^{1}\frac{1}{\sqrt{x^2-\frac{16}{25}}}\,dx$ | | |
| $= \frac{1}{5}\left[\text{arcosh}\left(\frac{5x}{4}\right)\right]_{\frac{4}{5}}^{1}$ | M1, A1A1 | arcosh alone, or any cosh substitution; $\frac{1}{5}$, $\frac{5x}{4}$ |
| $= \frac{1}{5}\left(\text{arcosh}\left(\frac{5}{4}\right)-\text{arcosh}(1)\right)$ | | |
| $= \frac{1}{5}\ln\left(\frac{5}{4}+\sqrt{\left(\frac{5}{4}\right)^2-1}\right)-0$ | M1 | Substituting limits and using (i) correctly at any stage (or using limits of $u$ in logarithmic form); dependent on first M1 |
| $= \frac{1}{5}\ln 2$ | A1 | |
| OR $= \frac{1}{5}\left[\ln\left(x+\sqrt{x^2-\frac{16}{25}}\right)\right]_{\frac{4}{5}}^{1}$ | M | $\ln\left(kx+\sqrt{k^2x^2+\ldots}\right)$; give M1 for $\ln\left(k_1x+\sqrt{k_2^2x^2+\ldots}\right)$ |
| | A1A | $\frac{1}{5}\ln\left(x+\sqrt{x^2-\frac{16}{25}}\right)$ o.e. |
| $= \frac{1}{5}\ln\frac{8}{5}-\frac{1}{5}\ln\frac{4}{5} = \frac{1}{5}\ln 2$ | A | |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $5\cosh x - \cosh 2x = 3$ | | |
| $\Rightarrow 5\cosh x-(2\cosh^2 x-1)=3$ | M1 | Attempting to express $\cosh 2x$ in terms of $\cosh x$ |
| $\Rightarrow 2\cosh^2 x - 5\cosh x+2=0$ | | |
| $\Rightarrow (2\cosh x-1)(\cosh x-2)=0$ | M1 | Solving quadratic to obtain at least one real value of $\cosh x$ |
| $\Rightarrow \cosh x = \frac{1}{2}$ (rejected) | A1 | Or factor $2\cosh x-1$ |
| or $\cosh x = 2$ | A1 | |
| $\Rightarrow x = \ln(2+\sqrt{3})$ | A1ft | F.t. $\cosh x = k,\ k>1$ |
| $x = -\ln(2+\sqrt{3})$ or $\ln(2-\sqrt{3})$ | A1ft | F.t. other value. Max. A1A0 if additional real values quoted |
---4 (i) Given that $\cosh y = x$, show that $y = \pm \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)$ and that $\operatorname { arcosh } x = \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)$.\\
(ii) Find $\int _ { \frac { 4 } { 5 } } ^ { 1 } \frac { 1 } { \sqrt { 25 x ^ { 2 } - 16 } } \mathrm {~d} x$, expressing your answer in an exact logarithmic form.\\
(iii) Solve the equation
$$5 \cosh x - \cosh 2 x = 3$$
giving your answers in an exact logarithmic form.
\hfill \mbox{\textit{OCR MEI FP2 2011 Q4 [18]}}