Year: 2020
Paper: 3
Question Number: 2
Course: UFM Pure
Section: Hyperbolic functions
No solution available for this problem.
In spite of the change to criteria for entering the paper, there was still a very healthy number of candidates, and the vast majority handled the protocols for the online testing very well. Just over half the candidates attempted exactly six questions, and whilst about 10% attempted a seventh, hardly any did more than seven. With 20% attempting five questions, and 10% attempting only four, overall, there were very few candidates not attempting the target number. There was a spread of popularity across the questions, with no question attracting more than 90% of candidates and only one less than 10%, but every question received a good number of attempts. Likewise, there was a spread of success on the questions, though every question attracted at least one perfect solution.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The curve $C$ has equation $\sinh x + \sinh y = 2k$, where $k$ is a positive constant.
\begin{questionparts}
\item Show that the curve $C$ has no stationary points and that $\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 0$ at the point $(x,y)$ on the curve if and only if
\[ 1 + \sinh x \sinh y = 0. \]
Find the co-ordinates of the points of inflection on the curve $C$, leaving your answers in terms of inverse hyperbolic functions.
\item Show that if $(x,y)$ lies on the curve $C$ and on the line $x + y = a$, then
\[ \mathrm{e}^{2x}(1 - \mathrm{e}^{-a}) - 4k\mathrm{e}^x + (\mathrm{e}^a - 1) = 0 \]
and deduce that $1 < \cosh a \leqslant 2k^2 + 1$.
\item Sketch the curve $C$.
\end{questionparts}
This was both the fourth most popular and successful question being attempted by 84% with a mean score of about 55%. Most generally performed much better in parts (i) and (ii) than in in (iii). In part (i), most successfully showed that were no stationary points and obtained the given result. Likewise, generally they found the points of inflection although a few struggled to do so. In part (ii), almost all candidates obtained the required equation and then noticed that it was a quadratic in e^x. Then they usually noticed that the discriminant being non-negative gave the higher bound for cosh a. A surprising number seemed not to notice there was a strict lower bound to deduce, and, as a consequence, did not subsequently appreciate that a was non-zero. Given the amount of information obtained in parts (i) and (ii), there was frequently a reluctance to apply this to part (iii). For example, although stationary points usually appeared, points of inflection often did not. Even fewer candidates used (ii) to deduce that the graph has to lie between the lines x + y = 0 and x + y = cosh⁻¹(2k² + 1). It is expected that candidates should observe that the graph is symmetrical in the line = x, that the two bounding lines should be labelled with their equations, and that the coordinates of the intercepts with the coordinate axes, the points of inflection and the point where it touches x + y = cosh⁻¹(2k² + 1) should be written in on the sketch.