| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using sech/tanh identities |
| Difficulty | Standard +0.8 This is a Further Maths question requiring knowledge of hyperbolic identities (sech²x = 1 - tanh²x), leading to a quadratic in tanh x, then using the definition tanh x = (e^(2x) - 1)/(e^(2x) + 1) to solve for x. It involves multiple steps and manipulation of exponentials, but follows a standard method for this topic with no novel insight required. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5\dfrac{e^x - e^{-x}}{e^x + e^{-x}} + 7 = \dfrac{10}{e^x + e^{-x}}\) | B1 | The given equation correctly expressed in terms of exponentials in any form |
| \(5(e^{2x}-1)+7(e^{2x}+1)=10e^x\) | M1 | Attempt quadratic in \(e^x\) |
| \(12e^{2x}-10e^x+2=0\) | A1 | Correct quadratic |
| \(6e^{2x}-5e^x+1=0 \Rightarrow (3e^x-1)(2e^x-1)=0\) | M1 | Solves their 3TQ in \(e^x\) |
| \(x=\ln\!\left(\tfrac{1}{3}\right),\quad \ln\!\left(\tfrac{1}{2}\right)\) | A1 | Both correct (Allow \(-\ln 3\) and/or \(-\ln 2\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(50\tanh^2 x + 70\tanh x + 24 = 0\) | B1 | Correct quadratic in \(\tanh x\) |
| \(\tanh x = -\dfrac{4}{5},\quad \tanh x = -\dfrac{3}{5}\) | M1A1 | M1: Solves their 3TQ in \(\tanh x\); A1: Correct values |
| \(\dfrac{e^{2x}-1}{e^{2x}+1}=-\dfrac{4}{5} \Rightarrow e^{2x}=\dfrac{1}{9} \Rightarrow x=\ln\dfrac{1}{3}\) | M1A1 | M1: Uses correct exponential form of \(\tanh x\) to obtain a value for \(x\) at least once; A1: Both answers correct |
| \(\dfrac{e^{2x}-1}{e^{2x}+1}=-\dfrac{3}{5} \Rightarrow e^{2x}=\dfrac{1}{4} \Rightarrow x=\ln\dfrac{1}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(24\sinh^2 x + 50\sinh x + 24 = 0\) | B1 | Correct quadratic in \(\sinh x\) |
| \(\sinh x = -\dfrac{4}{3},\quad \sinh x = -\dfrac{3}{4}\) | M1A1 | M1: Solves their 3TQ in \(\sinh x\); A1: Correct values |
| \(\dfrac{e^x - e^{-x}}{e^x+e^{-x}}=-\dfrac{4}{3} \Rightarrow e^x=\dfrac{1}{3} \Rightarrow x=\ln\dfrac{1}{3}\) | M1A1 | M1: Uses correct exponential form of \(\sinh x\) to obtain a value for \(x\) at least once; A1: Both answers correct |
| \(\dfrac{e^x - e^{-x}}{e^x+e^{-x}}=-\dfrac{3}{4} \Rightarrow e^x=\dfrac{1}{2} \Rightarrow x=\ln\dfrac{1}{2}\) |
# Question 1: $5\tanh x + 7 = 5\text{sech}\, x$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5\dfrac{e^x - e^{-x}}{e^x + e^{-x}} + 7 = \dfrac{10}{e^x + e^{-x}}$ | B1 | The given equation correctly expressed in terms of exponentials in any form |
| $5(e^{2x}-1)+7(e^{2x}+1)=10e^x$ | M1 | Attempt quadratic in $e^x$ |
| $12e^{2x}-10e^x+2=0$ | A1 | Correct quadratic |
| $6e^{2x}-5e^x+1=0 \Rightarrow (3e^x-1)(2e^x-1)=0$ | M1 | Solves their 3TQ in $e^x$ |
| $x=\ln\!\left(\tfrac{1}{3}\right),\quad \ln\!\left(\tfrac{1}{2}\right)$ | A1 | Both correct (Allow $-\ln 3$ and/or $-\ln 2$) |
**Alternative 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $50\tanh^2 x + 70\tanh x + 24 = 0$ | B1 | Correct quadratic in $\tanh x$ |
| $\tanh x = -\dfrac{4}{5},\quad \tanh x = -\dfrac{3}{5}$ | M1A1 | M1: Solves their 3TQ in $\tanh x$; A1: Correct values |
| $\dfrac{e^{2x}-1}{e^{2x}+1}=-\dfrac{4}{5} \Rightarrow e^{2x}=\dfrac{1}{9} \Rightarrow x=\ln\dfrac{1}{3}$ | M1A1 | M1: Uses correct exponential form of $\tanh x$ to obtain a value for $x$ at least once; A1: Both answers correct |
| $\dfrac{e^{2x}-1}{e^{2x}+1}=-\dfrac{3}{5} \Rightarrow e^{2x}=\dfrac{1}{4} \Rightarrow x=\ln\dfrac{1}{2}$ | | |
**Alternative 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $24\sinh^2 x + 50\sinh x + 24 = 0$ | B1 | Correct quadratic in $\sinh x$ |
| $\sinh x = -\dfrac{4}{3},\quad \sinh x = -\dfrac{3}{4}$ | M1A1 | M1: Solves their 3TQ in $\sinh x$; A1: Correct values |
| $\dfrac{e^x - e^{-x}}{e^x+e^{-x}}=-\dfrac{4}{3} \Rightarrow e^x=\dfrac{1}{3} \Rightarrow x=\ln\dfrac{1}{3}$ | M1A1 | M1: Uses correct exponential form of $\sinh x$ to obtain a value for $x$ at least once; A1: Both answers correct |
| $\dfrac{e^x - e^{-x}}{e^x+e^{-x}}=-\dfrac{3}{4} \Rightarrow e^x=\dfrac{1}{2} \Rightarrow x=\ln\dfrac{1}{2}$ | | |
---\begin{enumerate}
\item Solve the equation
\end{enumerate}
$$5 \tanh x + 7 = 5 \operatorname { sech } x$$
Give each answer in the form $\ln k$ where $k$ is a rational number.\\
\hfill \mbox{\textit{Edexcel FP3 2014 Q1 [5]}}