| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.3 This is a straightforward application of the product rule with exponential and polynomial functions, followed by solving a quadratic equation to find stationary points. The 'show that' part guides students through the differentiation, making it slightly easier than average. The algebra is routine and the quadratic factorization is manageable, placing this just above a typical textbook exercise. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y' = e^{-4x}(2x+2) - 4e^{-4x}(x^2+2x-2)\) | M1 | \(y' = Ae^{-4x}(ax+b) \pm Be^{-4x}(x^2+2x-2)\) where \(A\) and \(B\) are non-zero constants |
| All correct | A1 | |
| \(= e^{-4x}(2x+2-4x^2-8x+8)\) | or \(-4x^2e^{-4x} - 6xe^{-4x} + 10e^{-4x}\) | |
| \(= 2e^{-4x}(5-3x-2x^2)\) | A1 | AG; all correct with no errors, \(2^{nd}\) line (OE) must be seen. Condone incorrect order on final line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = x^2e^{-4x} + 2xe^{-4x} - 2e^{-4x}\) | ||
| \(y' = -4x^2e^{-4x} + 2xe^{-4x} + 2x(-4e^{-4x}) + 2e^{-4x} + 8e^{-4x}\) | (M1) | \(Ax^2e^{-4x} + Bxe^{-4x} + Cxe^{-4x} + De^{-4x} + Ee^{-4x}\) |
| All correct | (A1) | |
| \(= -4x^2e^{-4x} - 6xe^{-4x} + 10e^{-4x}\) | ||
| \(= 2e^{-4x}(5-3x-2x^2)\) | (A1) | AG; all correct with no errors, \(3^{rd}\) line (OE) must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-(2x+5)(x-1)(=0)\) | M1 | OE Attempt at factorisation \((\pm 2x \pm 5)(\pm x \pm 1)\) or formula with at most one error |
| \(x = -\dfrac{5}{2},\ 1\) | A1 | Both correct and no errors. SC \(x=1\) only scores M1A0 |
| \(x=1,\ y=e^{-4}\) | m1 | For \(y = ae^b\) attempted |
| A1F | Either correct, follow through only from incorrect sign for \(x\) | |
| \(x=-\dfrac{5}{2},\ y=e^{10}\!\left(-\dfrac{3}{4}\right)\) | A1 | CSO 2 solutions only. Note: withhold final mark for extra solutions. Note: approximate values only for \(y\) can score m1 only |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y' = e^{-4x}(2x+2) - 4e^{-4x}(x^2+2x-2)$ | M1 | $y' = Ae^{-4x}(ax+b) \pm Be^{-4x}(x^2+2x-2)$ where $A$ and $B$ are non-zero constants |
| All correct | A1 | |
| $= e^{-4x}(2x+2-4x^2-8x+8)$ | | or $-4x^2e^{-4x} - 6xe^{-4x} + 10e^{-4x}$ |
| $= 2e^{-4x}(5-3x-2x^2)$ | A1 | AG; all correct with no errors, $2^{nd}$ line (OE) must be seen. Condone incorrect order on final line |
**Alternative method:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = x^2e^{-4x} + 2xe^{-4x} - 2e^{-4x}$ | | |
| $y' = -4x^2e^{-4x} + 2xe^{-4x} + 2x(-4e^{-4x}) + 2e^{-4x} + 8e^{-4x}$ | (M1) | $Ax^2e^{-4x} + Bxe^{-4x} + Cxe^{-4x} + De^{-4x} + Ee^{-4x}$ |
| All correct | (A1) | |
| $= -4x^2e^{-4x} - 6xe^{-4x} + 10e^{-4x}$ | | |
| $= 2e^{-4x}(5-3x-2x^2)$ | (A1) | AG; all correct with no errors, $3^{rd}$ line (OE) must be seen |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-(2x+5)(x-1)(=0)$ | M1 | OE Attempt at factorisation $(\pm 2x \pm 5)(\pm x \pm 1)$ or formula with at most one error |
| $x = -\dfrac{5}{2},\ 1$ | A1 | Both correct and no errors. SC $x=1$ only scores M1A0 |
| $x=1,\ y=e^{-4}$ | m1 | For $y = ae^b$ attempted |
| | A1F | Either correct, follow through only from incorrect sign for $x$ |
| $x=-\dfrac{5}{2},\ y=e^{10}\!\left(-\dfrac{3}{4}\right)$ | A1 | CSO 2 solutions only. Note: withhold final mark for extra solutions. Note: approximate values only for $y$ can score m1 only |
---1 A curve has equation $y = \mathrm { e } ^ { - 4 x } \left( x ^ { 2 } + 2 x - 2 \right)$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \mathrm { e } ^ { - 4 x } \left( 5 - 3 x - 2 x ^ { 2 } \right)$.
\item Find the exact values of the coordinates of the stationary points of the curve.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2010 Q1 [8]}}