| Exam Board | SPS |
|---|---|
| Module | SPS FM Pure (SPS FM Pure) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Topic | Differentiating Transcendental Functions |
| Type | Show gradient condition leads to equation |
| Difficulty | Challenging +1.2 This is a Further Maths differentiation question requiring quotient rule, chain rule, and trigonometric manipulation. Part (a) involves standard calculus techniques to find stationary points, leading to a straightforward simplification. Part (b) requires understanding how transformations affect turning points. While multi-step, the techniques are routine for FM students with no novel insight required. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
\includegraphics{figure_5}
Figure 5 shows a sketch of the curve with equation $y = f(x)$, where
$$f(x) = \frac{4\sin 2x}{e^{\sqrt{2}x-1}}, \quad 0 \leq x \leq \pi$$
The curve has a maximum turning point at $P$ and a minimum turning point at $Q$ as shown in Figure 5.
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$ coordinates of point $P$ and point $Q$ are solutions of the equation
$$\tan 2x = \sqrt{2}$$
[4]
\item Using your answer to part (a), find the $x$-coordinate of the minimum turning point on the curve with equation
$$y = 3 - 2f(x)$$
[2]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Pure 2021 Q14 [6]}}