| Exam Board | SPS |
|---|---|
| Module | SPS FM (SPS FM) |
| Year | 2020 |
| Session | September |
| Marks | 8 |
| Topic | Stationary points and optimisation |
| Type | Determine constant from stationary point condition |
| Difficulty | Challenging +1.2 This is a moderately challenging Further Maths question requiring understanding of inflection points (finding f''(x) and showing it equals zero with sign change) and using the distance formula with roots of a cubic. Part (a) is routine calculus, but part (b) requires solving for k by finding roots of f(x), using the distance formula AB = √[(x₂-x₁)² + (y₂-y₁)²] = √2|x₂-x₁| (since both points are on x-axis), then working backwards through integration and the constraint. The multi-step nature and algebraic manipulation elevate this above average difficulty. |
| Spec | 1.07f Convexity/concavity: points of inflection1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
\includegraphics{figure_5}
Figure 5 shows a sketch of the curve $C$ with equation $y = f(x)$.
The curve $C$ crosses the $x$-axis at the origin, $O$, and at the points $A$ and $B$ as shown in Figure 5.
Given that
$$f'(x) = k - 4x - 3x^2$$
where $k$ is constant.
\begin{enumerate}[label=(\alph*)]
\item show that $C$ has a point of inflection at $x = -\frac{2}{3}$ [3]
Given also that the distance $AB = 4\sqrt{2}$
\item find, showing your working, the integer value of $k$. [5]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM 2020 Q8 [8]}}