| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Second derivative test justification |
| Difficulty | Moderate -0.3 This is a standard C2 differentiation and integration question with routine procedures: finding stationary points by setting dy/dx=0, using the second derivative test, and calculating area under a curve. All steps are textbook exercises requiring no problem-solving insight, though the multi-part structure and area calculation involving multiple regions adds slight complexity beyond the most basic questions. |
| 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/dx1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dy}{dx} = 3x^2 - 16x + 20\) | M1, A1 | |
| \(3x^2 - 16x + 20 = 0\) | dM1, A1 | \((3x - 10)(x - 2) = 0\) |
| (b) \(\frac{d^2y}{dx^2} = 6x - 16\) | M1 | At \(x = 2\), \(\frac{d^2y}{dx^2} = \ldots\) |
| (c) \(\int(x^3 - 8x^2 + 20x)dx = \frac{x^4}{4} - \frac{8x^3}{3} + \frac{20x^2}{2} + (C)\) | M1, A1, A1 | (3 marks) |
| (d) \(4 - \frac{64}{3} + 40\) | M1 | \(\left(= \frac{68}{3}\right)\) |
| A: \(x = 2\): | B1 | \(y = 8 - 32 + 40 = 16\) (May be scored elsewhere) |
| Area of \(\triangle = \frac{1}{2}\left(\frac{10}{3} - 2\right) \times 16\) | M1 | \(\left(\frac{1}{2}(x_B - x_A) \times y_A\right)\) |
| Shaded area \(= \frac{68}{3} + \frac{32}{3} = \frac{100}{3}\) | M1, A1 | \(\left(= 33\frac{1}{3}\right)\) (5 marks) |
| (14 marks) | (a) The second M is dependent on the first, and requires an attempt to solve a 3 term quadratic. (b) M1: Attempt second differentiation and substitution of one of the \(x\) values. A1ft: Requires correct second derivative and negative value of the second derivative, but fit from their \(x\) value. (c) All 3 terms correct: M1 A1 A1. Two terms correct: M1 A1 A0. One power correct: M1 A0 A0. (d) Limits M1: Substituting their lower \(x\) value into a 'changed' expression. Area of triangle M1: Fully correct method. Alternative for the triangle (finding an equation for the straight line then integrating) requires a fully correct method to score the M mark. Final M1: Fully correct method (beware valid alternatives!). |
**(a)** $\frac{dy}{dx} = 3x^2 - 16x + 20$ | M1, A1 | |
| --- | --- | --- |
| $3x^2 - 16x + 20 = 0$ | dM1, A1 | $(3x - 10)(x - 2) = 0$ | $x = \ldots, \frac{10}{3}$ and $2$ (4 marks) |
| **(b)** $\frac{d^2y}{dx^2} = 6x - 16$ | M1 | At $x = 2$, $\frac{d^2y}{dx^2} = \ldots$ | A1ft | $-4$ (or $< 0$, or both), therefore maximum (2 marks) |
| **(c)** $\int(x^3 - 8x^2 + 20x)dx = \frac{x^4}{4} - \frac{8x^3}{3} + \frac{20x^2}{2} + (C)$ | M1, A1, A1 | (3 marks) |
| **(d)** $4 - \frac{64}{3} + 40$ | M1 | $\left(= \frac{68}{3}\right)$ | | |
| A: $x = 2$: | B1 | $y = 8 - 32 + 40 = 16$ (May be scored elsewhere) | | |
| Area of $\triangle = \frac{1}{2}\left(\frac{10}{3} - 2\right) \times 16$ | M1 | $\left(\frac{1}{2}(x_B - x_A) \times y_A\right)$ | $\left(= \frac{32}{3}\right)$ | |
| Shaded area $= \frac{68}{3} + \frac{32}{3} = \frac{100}{3}$ | M1, A1 | $\left(= 33\frac{1}{3}\right)$ (5 marks) |
| | (14 marks) | **(a)** The second M is dependent on the first, and requires an attempt to solve a 3 term quadratic. **(b)** M1: Attempt second differentiation and substitution of one of the $x$ values. A1ft: Requires correct second derivative and negative value of the second derivative, but fit from their $x$ value. **(c)** All 3 terms correct: M1 A1 A1. Two terms correct: M1 A1 A0. One power correct: M1 A0 A0. **(d)** Limits M1: Substituting their lower $x$ value into a 'changed' expression. Area of triangle M1: Fully correct method. Alternative for the triangle (finding an equation for the straight line then integrating) requires a fully correct method to score the M mark. Final M1: Fully correct method (beware valid alternatives!). |10.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{29c7baa1-6929-448a-a756-319ea75dffa7-14_636_956_285_513}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve with equation $y = x ^ { 3 } - 8 x ^ { 2 } + 20 x$. The curve has stationary points $A$ and $B$.
\begin{enumerate}[label=(\alph*)]
\item Use calculus to find the $x$-coordinates of $A$ and $B$.
\item Find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at $A$, and hence verify that $A$ is a maximum.
The line through $B$ parallel to the $y$-axis meets the $x$-axis at the point $N$.\\
The region $R$, shown shaded in Figure 3, is bounded by the curve, the $x$-axis and the line from $A$ to $N$.
\item Find $\int \left( x ^ { 3 } - 8 x ^ { 2 } + 20 x \right) \mathrm { d } x$.
\item Hence calculate the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2006 Q10 [14]}}