| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.3 This is a standard stationary points question requiring product rule differentiation of an exponential-polynomial product, solving a quadratic, then using the second derivative test. While it involves multiple steps (6 marks total), each technique is routine for C3 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dy}{dx} = e^{2x}(2x-4)\) | M1 | Product rule; allow 1 slip |
| \(+ (x^2-4x-2)2e^{2x}\) | A1 | |
| \(\frac{dy}{dx} = e^{2x}(2x-4+2x^2-8x-4)\) | M1 | Factorising \(e^{2x}(ax^2+6x+0)\) |
| \(e^{2x}(2x^2-6x-8)\) | A1 | or \(x^2-3x-4=0\) |
| \((x-4)(x+1)=0\) | m1 | Solving 3 term quadratic; dependent on both M marks |
| \(x=4,-1\) | A1 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{d^2y}{dx^2} = e^{2x}.2+(2x-4)2e^{2x}+(x^2-4x-2)4e^{2x}+2e^{2x}(2x-4)\) | M1 | Product rule from their \(\frac{dy}{dx}\) in form \(e^{2x}(\text{quadratic})\) giving \(e^{2x}(4x^2-8x-22)\) |
| Full correct expression | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x=4: y''=e^8(10)>0 \therefore\) MIN | M1 | Their 2 \(x\)'s in their \(\frac{d^2y}{dx^2}\); only of form \(e^{2x}(\text{quadratic})\) |
| \(x=-1: y''=e^{-2}(-10)<0 \therefore\) MAX | A1 | 2 |
# Question 6:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dy}{dx} = e^{2x}(2x-4)$ | M1 | Product rule; allow 1 slip |
| $+ (x^2-4x-2)2e^{2x}$ | A1 | |
| $\frac{dy}{dx} = e^{2x}(2x-4+2x^2-8x-4)$ | M1 | Factorising $e^{2x}(ax^2+6x+0)$ |
| $e^{2x}(2x^2-6x-8)$ | A1 | or $x^2-3x-4=0$ |
| $(x-4)(x+1)=0$ | m1 | Solving 3 term quadratic; dependent on both M marks |
| $x=4,-1$ | A1 | 6 | And no extras e.g. $x=0$ |
## Part (b)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{d^2y}{dx^2} = e^{2x}.2+(2x-4)2e^{2x}+(x^2-4x-2)4e^{2x}+2e^{2x}(2x-4)$ | M1 | Product rule from their $\frac{dy}{dx}$ in form $e^{2x}(\text{quadratic})$ giving $e^{2x}(4x^2-8x-22)$ |
| Full correct expression | A1 | 2 | |
## Part (b)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $x=4: y''=e^8(10)>0 \therefore$ MIN | M1 | Their 2 $x$'s in their $\frac{d^2y}{dx^2}$; only of form $e^{2x}(\text{quadratic})$ |
| $x=-1: y''=e^{-2}(-10)<0 \therefore$ MAX | A1 | 2 | CSO Both correct; allow values either side of $y$ or $y'$ |
---6 A curve has equation $y = \mathrm { e } ^ { 2 x } \left( x ^ { 2 } - 4 x - 2 \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of the $x$-coordinate of each of the stationary points of the curve.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item Determine the nature of each of the stationary points of the curve.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2009 Q6 [10]}}