| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using double angle formulas |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring proof using exponentials, solving a hyperbolic equation, and evaluating a hyperbolic integral. While each individual part uses standard FP2 techniques (exponential definitions, substitution, standard integrals), the combination of multiple hyperbolic concepts and the need for exact logarithmic answers elevates it slightly above average difficulty for Further Maths material. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sinh x = \dfrac{e^x-e^{-x}}{2} \Rightarrow \sinh^2 x = \dfrac{(e^x-e^{-x})^2}{4} = \dfrac{e^{2x}-2+e^{-2x}}{4}\) | B1 | \(e^{2x}-2+e^{-2x}\) |
| \(\Rightarrow 2\sinh^2 x + 1 = \dfrac{e^{2x}-2+e^{-2x}}{2}+1 = \dfrac{e^{2x}+e^{-2x}}{2} = \cosh 2x\) | B1 | Correct completion |
| \(\Rightarrow 2\sinh 2x = 4\sinh x\cosh x\) | B1 | Both correct derivatives |
| \(\Rightarrow \sinh 2x = 2\sinh x\cosh x\) | B1 | Correct completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\cosh 2x + 3\sinh x = 3\) | ||
| \(\Rightarrow 2(1+2\sinh^2 x)+3\sinh x = 3\) | M1 | Using identity |
| \(\Rightarrow 4\sinh^2 x + 3\sinh x - 1 = 0\) | A1 | Correct quadratic |
| \(\Rightarrow (4\sinh x-1)(\sinh x+1)=0\) | M1 | Solving quadratic |
| \(\Rightarrow \sinh x = \tfrac{1}{4},\ -1\) | A1 | Both |
| Use of \(\text{arsinh}\, x = \ln(x+\sqrt{x^2+1})\) | M1 | Must obtain at least one value of \(x\) |
| \(\Rightarrow x = \text{arsinh}(\tfrac{1}{4}) = \ln\!\left(\dfrac{1+\sqrt{17}}{4}\right)\) | A1 | Must evaluate \(\sqrt{x^2+1}\) |
| \(x = \text{arsinh}(-1) = \ln(-1+\sqrt{2})\) | A1 | |
| OR \(2e^{4x}+3e^{3x}-6e^{2x}-3e^x+2=0\) | ||
| \(\Rightarrow (2e^{2x}-e^x-2)(e^{2x}+2e^x-1)=0\) | M1A1 | Factorising quartic |
| \(\Rightarrow e^x=\dfrac{1\pm\sqrt{17}}{4}\) or \(-1\pm\sqrt{2}\) | M1A1 | Solving either quadratic |
| \(\Rightarrow x=\ln\!\left(\dfrac{1+\sqrt{17}}{4}\right)\) or \(\ln(-1+\sqrt{2})\) | M1A1A1 | Using ln (dependent on first M1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cosh t = \dfrac{5}{4} \Rightarrow \dfrac{e^t+e^{-t}}{2}=\dfrac{5}{4}\) | ||
| \(\Rightarrow 2e^{2t}-5e^t+2=0\) | M1 | Forming quadratic in \(e^t\) |
| \(\Rightarrow (2e^t-1)(e^t-2)=0\) | M1 | Solving quadratic |
| \(\Rightarrow e^t=\tfrac{1}{2},\ 2\) | A1 | |
| \(\Rightarrow t=\pm\ln 2\) | A1 (ag) | Convincing working |
| \(\displaystyle\int_4^5\dfrac{1}{\sqrt{x^2-16}}\,dx = \left[\text{arcosh}\,\dfrac{x}{4}\right]_4^5\) | B1 | |
| \(= \text{arcosh}\,\dfrac{5}{4} - \text{arcosh}\,1\) | M1 | Substituting limits |
| \(= \ln 2\) | A1 | A0 for \(\pm\ln 2\) |
| OR \(\displaystyle\int_4^5\dfrac{1}{\sqrt{x^2-16}}\,dx = \left[\ln(x+\sqrt{x^2-16})\right]_4^5\) | B1 | |
| \(= \ln 8 - \ln 4 = \ln 2\) | M1, A1 | Substituting limits |
# Question 4:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sinh x = \dfrac{e^x-e^{-x}}{2} \Rightarrow \sinh^2 x = \dfrac{(e^x-e^{-x})^2}{4} = \dfrac{e^{2x}-2+e^{-2x}}{4}$ | B1 | $e^{2x}-2+e^{-2x}$ |
| $\Rightarrow 2\sinh^2 x + 1 = \dfrac{e^{2x}-2+e^{-2x}}{2}+1 = \dfrac{e^{2x}+e^{-2x}}{2} = \cosh 2x$ | B1 | Correct completion |
| $\Rightarrow 2\sinh 2x = 4\sinh x\cosh x$ | B1 | Both correct derivatives |
| $\Rightarrow \sinh 2x = 2\sinh x\cosh x$ | B1 | Correct completion |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\cosh 2x + 3\sinh x = 3$ | | |
| $\Rightarrow 2(1+2\sinh^2 x)+3\sinh x = 3$ | M1 | Using identity |
| $\Rightarrow 4\sinh^2 x + 3\sinh x - 1 = 0$ | A1 | Correct quadratic |
| $\Rightarrow (4\sinh x-1)(\sinh x+1)=0$ | M1 | Solving quadratic |
| $\Rightarrow \sinh x = \tfrac{1}{4},\ -1$ | A1 | Both |
| Use of $\text{arsinh}\, x = \ln(x+\sqrt{x^2+1})$ | M1 | Must obtain at least one value of $x$ |
| $\Rightarrow x = \text{arsinh}(\tfrac{1}{4}) = \ln\!\left(\dfrac{1+\sqrt{17}}{4}\right)$ | A1 | Must evaluate $\sqrt{x^2+1}$ |
| $x = \text{arsinh}(-1) = \ln(-1+\sqrt{2})$ | A1 | |
| OR $2e^{4x}+3e^{3x}-6e^{2x}-3e^x+2=0$ | | |
| $\Rightarrow (2e^{2x}-e^x-2)(e^{2x}+2e^x-1)=0$ | M1A1 | Factorising quartic |
| $\Rightarrow e^x=\dfrac{1\pm\sqrt{17}}{4}$ or $-1\pm\sqrt{2}$ | M1A1 | Solving either quadratic |
| $\Rightarrow x=\ln\!\left(\dfrac{1+\sqrt{17}}{4}\right)$ or $\ln(-1+\sqrt{2})$ | M1A1A1 | Using ln (dependent on first M1) |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cosh t = \dfrac{5}{4} \Rightarrow \dfrac{e^t+e^{-t}}{2}=\dfrac{5}{4}$ | | |
| $\Rightarrow 2e^{2t}-5e^t+2=0$ | M1 | Forming quadratic in $e^t$ |
| $\Rightarrow (2e^t-1)(e^t-2)=0$ | M1 | Solving quadratic |
| $\Rightarrow e^t=\tfrac{1}{2},\ 2$ | A1 | |
| $\Rightarrow t=\pm\ln 2$ | A1 (ag) | Convincing working |
| $\displaystyle\int_4^5\dfrac{1}{\sqrt{x^2-16}}\,dx = \left[\text{arcosh}\,\dfrac{x}{4}\right]_4^5$ | B1 | |
| $= \text{arcosh}\,\dfrac{5}{4} - \text{arcosh}\,1$ | M1 | Substituting limits |
| $= \ln 2$ | A1 | A0 for $\pm\ln 2$ |
| OR $\displaystyle\int_4^5\dfrac{1}{\sqrt{x^2-16}}\,dx = \left[\ln(x+\sqrt{x^2-16})\right]_4^5$ | B1 | |
| $= \ln 8 - \ln 4 = \ln 2$ | M1, A1 | Substituting limits |
---4 (i) Prove, using exponential functions, that
$$\cosh 2 x = 1 + 2 \sinh ^ { 2 } x$$
Differentiate this result to obtain a formula for $\sinh 2 x$.\\
(ii) Solve the equation
$$2 \cosh 2 x + 3 \sinh x = 3$$
expressing your answers in exact logarithmic form.\\
(iii) Given that $\cosh t = \frac { 5 } { 4 }$, show by using exponential functions that $t = \pm \ln 2$.
Find the exact value of the integral
$$\int _ { 4 } ^ { 5 } \frac { 1 } { \sqrt { x ^ { 2 } - 16 } } \mathrm {~d} x$$
\hfill \mbox{\textit{OCR MEI FP2 2010 Q4 [18]}}