WJEC Further Unit 4 2024 June — Question 8 11 marks

Exam BoardWJEC
ModuleFurther Unit 4 (Further Unit 4)
Year2024
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeFind stationary points of hyperbolic curves
DifficultyChallenging +1.2 Part (a) is a standard derivation of the inverse hyperbolic sine derivative using implicit differentiation and the identity cosh²y - sinh²y = 1, requiring algebraic manipulation but following a well-established method. Part (b) requires implicit differentiation of a product, finding where dy/dx = 0, and proving uniqueness of the stationary point, likely involving showing a function is monotonic. While this involves multiple techniques and some problem-solving, it's a typical Further Maths question on hyperbolic functions without requiring particularly novel insight—moderately above average difficulty.
Spec4.07d Differentiate/integrate: hyperbolic functions4.08g Derivatives: inverse trig and hyperbolic functions

  1. By writing \(y = \sinh^{-1}(4x + 3)\) as \(\sinh y = 4x + 3\), show that \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4}{\sqrt{16x^2 + 24x + 10}}\). [5]
  2. Show that the graph of \(e^{-3x} \cdot y = \sinh 2x\) has only one stationary point. [6]

Question 8:
AnswerMarks
811
Question 8:
8 | 11
\begin{enumerate}[label=(\alph*)]
\item By writing $y = \sinh^{-1}(4x + 3)$ as $\sinh y = 4x + 3$, show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4}{\sqrt{16x^2 + 24x + 10}}$. [5]

\item Show that the graph of $e^{-3x} \cdot y = \sinh 2x$ has only one stationary point. [6]
\end{enumerate}

\hfill \mbox{\textit{WJEC Further Unit 4 2024 Q8 [11]}}