OCR FP2 — Question 9 12 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeFind stationary points of hyperbolic curves
DifficultyStandard +0.3 This is a straightforward Further Maths question on hyperbolic functions. Part (i) is routine algebraic manipulation using standard definitions. Part (ii) requires differentiation, solving a quadratic equation, and using the second derivative test—all standard techniques with no novel insight required. While it's Further Maths content, the execution is mechanical and follows textbook methods.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07b Hyperbolic graphs: sketch and properties4.07d Differentiate/integrate: hyperbolic functions

9
  1. Using the definitions of \(\cosh x\) and \(\sinh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), prove that $$\sinh 2 x = 2 \sinh x \cosh x$$
  2. Show that the curve with equation $$y = \cosh 2 x - 6 \sinh x$$ has just one stationary point, and find its \(x\)-coordinate in logarithmic form. Determine the nature of the stationary point.

9 (i) Using the definitions of $\cosh x$ and $\sinh x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$, prove that

$$\sinh 2 x = 2 \sinh x \cosh x$$

(ii) Show that the curve with equation

$$y = \cosh 2 x - 6 \sinh x$$

has just one stationary point, and find its $x$-coordinate in logarithmic form. Determine the nature of the stationary point.

\hfill \mbox{\textit{OCR FP2  Q9 [12]}}