| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Second derivative test justification |
| Difficulty | Moderate -0.8 This is a straightforward application of standard differentiation techniques: find dy/dx, set to zero, solve the quadratic, then use the second derivative test. All steps are routine procedures covered early in A-level calculus with no problem-solving insight required, making it easier than average but not trivial since it requires correct execution of multiple steps. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dy}{dx} = 3x^2 - 2x - 16\) | M1A1 | M1: differentiates to achieve \(Ax^2 + Bx + C\) where \(A,B,C \neq 0\); A1: correct derivative |
| \(3x^2 - 2x - 16 = 0 \Rightarrow x = \ldots\) | M1 | Sets \(\frac{dy}{dx} = 0\) and attempts to solve quadratic |
| \(x = \frac{8}{3}, -2\) | A1 | Exact values; do not accept 2.67 but \(2.\dot{6}\) acceptable |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{d^2y}{dx^2} = Cx + D\) | M1 | Attempts to differentiate \(\frac{dy}{dx}\) to achieve \(Cx + D\) where \(C,D \neq 0\) |
| \(\frac{d^2y}{dx^2} = 6x - 2\) | A1 | Correct second derivative with correct values at both \(x\) values, correct conclusions for both |
| When \(x = \frac{8}{3}\): \(\frac{d^2y}{dx^2} = 14 > 0 \Rightarrow\) min | dM1 | Substitutes \(x\) values into \(\frac{d^2y}{dx^2}\) and draws consistent conclusion; dependent on previous M |
| When \(x = -2\): \(\frac{d^2y}{dx^2} = -14 < 0 \Rightarrow\) max | A1 | Must be clear which point corresponds to each conclusion; accept "concave down" for max, "concave up" for min |
# Question 2:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 3x^2 - 2x - 16$ | M1A1 | M1: differentiates to achieve $Ax^2 + Bx + C$ where $A,B,C \neq 0$; A1: correct derivative |
| $3x^2 - 2x - 16 = 0 \Rightarrow x = \ldots$ | M1 | Sets $\frac{dy}{dx} = 0$ and attempts to solve quadratic |
| $x = \frac{8}{3}, -2$ | A1 | Exact values; do not accept 2.67 but $2.\dot{6}$ acceptable |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d^2y}{dx^2} = Cx + D$ | M1 | Attempts to differentiate $\frac{dy}{dx}$ to achieve $Cx + D$ where $C,D \neq 0$ |
| $\frac{d^2y}{dx^2} = 6x - 2$ | A1 | Correct second derivative with correct values at both $x$ values, correct conclusions for both |
| When $x = \frac{8}{3}$: $\frac{d^2y}{dx^2} = 14 > 0 \Rightarrow$ min | dM1 | Substitutes $x$ values into $\frac{d^2y}{dx^2}$ and draws consistent conclusion; dependent on previous M |
| When $x = -2$: $\frac{d^2y}{dx^2} = -14 < 0 \Rightarrow$ max | A1 | Must be clear which point corresponds to each conclusion; accept "concave down" for max, "concave up" for min |
---2. A curve has equation
$$y = x ^ { 3 } - x ^ { 2 } - 16 x + 2$$
\begin{enumerate}[label=(\alph*)]
\item Using calculus, find the $x$ coordinates of the stationary points of the curve.
\item Justify, by further calculus, the nature of all of the stationary points of the curve.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2021 Q2 [7]}}