Edexcel FP3 — Question 30 7 marks

Exam BoardEdexcel
ModuleFP3 (Further Pure Mathematics 3)
Marks7
PaperDownload PDF ↗
TopicHyperbolic functions
TypeFind stationary points of hyperbolic curves
DifficultyStandard +0.3 Part (a) is straightforward substitution and manipulation using the definition cosh(2x) = (e^(4x) + 1)/(2e^(2x)), which with x = ln k becomes routine algebra. Part (b) requires differentiating tanh(2x) using sech²(2x), setting f'(ln 2) = 0, and solving for p—standard calculus with hyperbolic functions. This is typical FP3 bookwork with no novel insight required, making it slightly easier than average.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions

  1. Show that, for \(x = \ln k\), where \(k\) is a positive constant, $$\cosh 2x = \frac{k^4 + 1}{2k^2}.$$ [3]
Given that \(f(x) = px - \tanh 2x\), where \(p\) is a constant,
  1. find the value of \(p\) for which \(f(x)\) has a stationary value at \(x = \ln 2\), giving your answer as an exact fraction. [4]
(Total 7 marks)

\begin{enumerate}[label=(\alph*)]
\item Show that, for $x = \ln k$, where $k$ is a positive constant,
$$\cosh 2x = \frac{k^4 + 1}{2k^2}.$$ [3]
\end{enumerate}

Given that $f(x) = px - \tanh 2x$, where $p$ is a constant,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the value of $p$ for which $f(x)$ has a stationary value at $x = \ln 2$, giving your answer as an exact fraction. [4]
\end{enumerate}

(Total 7 marks)

\hfill \mbox{\textit{Edexcel FP3  Q30 [7]}}