| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Determine nature of stationary points |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing standard C2 differentiation techniques: finding first and second derivatives of a polynomial, locating turning points by solving dy/dx=0, and classifying them using the second derivative test. All steps are routine applications of basic rules with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation needed. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dy}{dx} = 6x^2 - 10x - 4\) | M1 A1 | (2 marks) |
| (b) \(6x^2 - 10x - 4 = 0\) | M1 | |
| \(2(3x+1)(x-2) = 0\) | M1 | |
| \(x = 2\) or \(-\frac{1}{3}\) | A1 (both \(x\) values) | |
| Points are \((2, -10)\) and \(\left(-\frac{1}{3}, 2\frac{23}{27}\right.\) or \(\frac{71}{27}\) or 2.70 or better) | A1 (both \(y\) values) | (4 marks) |
| (c) \(\frac{d^2y}{dx^2} = 12x - 10\) | M1 A1 | (2 marks) |
| (d) \(x=2 \Rightarrow \frac{d^2y}{dx^2} = (14) \geq 0 \therefore [(2, -10)]\) is a Min | M1 | |
| \(x = -\frac{1}{4} \Rightarrow \frac{d^2y}{dx^2} = (-14) < 0 \therefore \left[\left(-\frac{1}{4}, \frac{23}{27}\right)\right]\) is a Max | A1 | (2 marks) |
(a) $\frac{dy}{dx} = 6x^2 - 10x - 4$ | M1 A1 | (2 marks)
(b) $6x^2 - 10x - 4 = 0$ | M1
$2(3x+1)(x-2) = 0$ | M1
$x = 2$ or $-\frac{1}{3}$ | A1 (both $x$ values)
Points are $(2, -10)$ and $\left(-\frac{1}{3}, 2\frac{23}{27}\right.$ or $\frac{71}{27}$ or 2.70 or better) | A1 (both $y$ values) | (4 marks)
(c) $\frac{d^2y}{dx^2} = 12x - 10$ | M1 A1 | (2 marks)
(d) $x=2 \Rightarrow \frac{d^2y}{dx^2} = (14) \geq 0 \therefore [(2, -10)]$ is a Min | M1
$x = -\frac{1}{4} \Rightarrow \frac{d^2y}{dx^2} = (-14) < 0 \therefore \left[\left(-\frac{1}{4}, \frac{23}{27}\right)\right]$ is a Max | A1 | (2 marks)
**Guidance:**
- (a) M1 for some correct attempt to differentiate $x^n \rightarrow x^{n-1}$
- (b) 1st M1 for setting their $\frac{dy}{dx} = 0$; 2nd M1 for attempting to solve 3TQ but it must be based on their $\frac{dy}{dx}$. NO marks for answers only in part (b)
- (c) M1 for attempting to differentiate their $\frac{dy}{dx}$
- (d) M1 for one correct use of their second derivative or a full method to determine the nature of one of their stationary points. A1: both correct ($=14$ and $=-14$) are not required
**Total: 10 marks**
---7. The curve $C$ has equation
$$y = 2 x ^ { 3 } - 5 x ^ { 2 } - 4 x + 2$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Using the result from part (a), find the coordinates of the turning points of $C$.
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item Hence, or otherwise, determine the nature of the turning points of $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2006 Q7 [10]}}