| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using sech/tanh identities |
| Difficulty | Standard +0.3 This is a straightforward application of the standard identity sech²x = 1 - tanh²x to convert to a quadratic in tanh x, then solve and use the inverse hyperbolic function formula. It's a routine Further Maths exercise requiring only direct substitution and algebraic manipulation, making it slightly easier than average overall but typical for its topic. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.06f Laws of logarithms: addition, subtraction, power rules4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(15(1-\tanh^2 x) + 7\tanh x = 13\) | M1 | Uses \(\text{sech}^2x = 1-\tanh^2x\) |
| \(15\tanh^2 x - 7\tanh x - 2 = 0\) | A1 | Correct 3 term quadratic, terms in any order |
| \((5\tanh x+1)(3\tanh x-2)=0 \Rightarrow \tanh x = -\frac{1}{5}, \frac{2}{3}\) | M1A1 | M1: Solves 3 term quadratic to obtain at least one value for \(\tanh x\); A1: Both correct values. If solved by formula accept \(\frac{7\pm13}{30}\) |
| \(x = \frac{1}{2}\ln\frac{2}{3},\ \frac{1}{2}\ln 5\) | A1, A1 | A1: One correct exact answer; A1: Both exact answers correct. Allow equivalents e.g. \(x=\frac{1}{2}\ln 2-\frac{1}{2}\ln 3,\ \ln\frac{\sqrt{6}}{3},\ \ln\sqrt{\frac{2}{3}},\ \ln\sqrt{5}\) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(15\left(\frac{2}{e^x+e^{-x}}\right)^2 + 7\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)=13\) | M1 | Substitutes correct exponential forms; equation may be rearranged before substitution; \(\frac{1}{2}\)s may have been cancelled |
| \(6e^{2x}-34+20e^{-2x}=0\) | A1 | Correct 3 term quadratic in \(e^{2x}\) |
| \(3e^{4x}-17e^{2x}+10=0\) | ||
| \((3e^{2x}-2)(e^{2x}-5)=0\) or \((3e^x-2e^{-x})(e^x-5e^{-x})=0 \Rightarrow e^{2x}=\frac{2}{3}\) or \(5\) | M1A1 | M1: Solves 3 term quadratic to obtain at least one value for \(e^{2x}\); A1: Both correct values |
| \(x=\frac{1}{2}\ln\frac{2}{3},\ \frac{1}{2}\ln 5\) | A1, A1 | A1: One correct answer; A1: Both answers correct. Allow e.g. \(x=\frac{1}{2}\ln 2-\frac{1}{2}\ln 3\) |
# Question 1:
$15\text{sech}^2x + 7\tanh x = 13$
| Working/Answer | Marks | Notes |
|---|---|---|
| $15(1-\tanh^2 x) + 7\tanh x = 13$ | M1 | Uses $\text{sech}^2x = 1-\tanh^2x$ |
| $15\tanh^2 x - 7\tanh x - 2 = 0$ | A1 | Correct 3 term quadratic, terms in any order |
| $(5\tanh x+1)(3\tanh x-2)=0 \Rightarrow \tanh x = -\frac{1}{5}, \frac{2}{3}$ | M1A1 | M1: Solves 3 term quadratic to obtain at least one value for $\tanh x$; A1: Both correct values. If solved by formula accept $\frac{7\pm13}{30}$ |
| $x = \frac{1}{2}\ln\frac{2}{3},\ \frac{1}{2}\ln 5$ | A1, A1 | A1: One correct exact answer; A1: Both exact answers correct. Allow equivalents e.g. $x=\frac{1}{2}\ln 2-\frac{1}{2}\ln 3,\ \ln\frac{\sqrt{6}}{3},\ \ln\sqrt{\frac{2}{3}},\ \ln\sqrt{5}$ etc |
**Alternative Using Exponentials:**
| Working/Answer | Marks | Notes |
|---|---|---|
| $15\left(\frac{2}{e^x+e^{-x}}\right)^2 + 7\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)=13$ | M1 | Substitutes correct exponential forms; equation may be rearranged before substitution; $\frac{1}{2}$s may have been cancelled |
| $6e^{2x}-34+20e^{-2x}=0$ | A1 | Correct 3 term quadratic in $e^{2x}$ |
| $3e^{4x}-17e^{2x}+10=0$ | | |
| $(3e^{2x}-2)(e^{2x}-5)=0$ or $(3e^x-2e^{-x})(e^x-5e^{-x})=0 \Rightarrow e^{2x}=\frac{2}{3}$ or $5$ | M1A1 | M1: Solves 3 term quadratic to obtain at least one value for $e^{2x}$; A1: Both correct values |
| $x=\frac{1}{2}\ln\frac{2}{3},\ \frac{1}{2}\ln 5$ | A1, A1 | A1: One correct answer; A1: Both answers correct. Allow e.g. $x=\frac{1}{2}\ln 2-\frac{1}{2}\ln 3$ |
**Total: 6 marks**
---\begin{enumerate}
\item Solve the equation
\end{enumerate}
$$15 \operatorname { sech } ^ { 2 } x + 7 \tanh x = 13$$
Give your answers in terms of simplified natural logarithms.\\
\hfill \mbox{\textit{Edexcel F3 2018 Q1 [6]}}