| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Solve equation involving derivatives |
| Difficulty | Standard +0.3 This is a structured multi-part question testing standard C3 techniques: differentiating exponentials, finding stationary points, and using substitution to solve an exponential equation. All steps are routine and well-signposted, requiring only methodical application of learned procedures with no novel insight. Slightly easier than average due to extensive scaffolding. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.07e Second derivative: as rate of change of gradient1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = e^{2x} - 10e^x + 12x\); \(\frac{dy}{dx} = 2e^{2x} - 10e^x + 12\) | B1 B1 | \(2e^{2x}\); remaining terms correct, no extras; 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = 4e^{2x} - 10e^x\) | B1F | ft 1 slip; 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| \(2e^{2x} - 10e^x + 12 = 0\); \(e^{2x} - 5e^x + 6 = 0\) | B1 | AG (be convinced); 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| \(z^2 - 5z + 6 = 0\); \(z = 2, 3\); \(z = 2, e^x = 2\); \(x = \ln 2\); \(z = 3, e^x = 3\); \(x = \ln 3\) | M1 M1 A1 | use of \(z = e^x\) oe; finding \(e^x\) = their 2,3; all correct AG; SC: verification ln 2 (B1), ln 3 (B1); 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \ln 2\): \(y = e^{2\ln 2} - 10e^{\ln 2} + 12\ln 2\) or \(2^2 - 10 \times 2 + 12\ln 2 = 4 - 20 + 12\ln 2 = -16 + 12\ln 2\); \(x = \ln 3\): \(y = e^{2\ln 3} - 10e^{\ln 3} + 12\ln 3 = 9 - 30 + 12\ln 3 = -21 + 12\ln 3\) | M1 A1 A1 | either substitution of their \(x = \ln 2\) (\(e^x = 2\)) or their \(x = \ln 3\) (\(e^x = 3\)); 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \ln 2\): \(\frac{d^2y}{dx^2} = 4e^{2\ln 2} - 10e^{\ln 2} = 16 - 20 = -4\); \(\therefore\) maximum; \(x = \ln 3\): \(\frac{d^2y}{dx^2} = 4e^{2\ln 3} - 10e^{\ln 3} = 36 - 30 = 6\); \(\therefore\) minimum | M1 A1 A1 | use of; in either of their \(e^x = 2,3\) into their \(\frac{d^2y}{dx^2}\); CSO; CSO; 3 marks total |
### 5(a)(i)
$y = e^{2x} - 10e^x + 12x$; $\frac{dy}{dx} = 2e^{2x} - 10e^x + 12$ | B1 B1 | $2e^{2x}$; remaining terms correct, no extras; 2 marks total
### 5(a)(ii)
$\frac{d^2y}{dx^2} = 4e^{2x} - 10e^x$ | B1F | ft 1 slip; 1 mark total
### 5(b)(i)
$2e^{2x} - 10e^x + 12 = 0$; $e^{2x} - 5e^x + 6 = 0$ | B1 | AG (be convinced); 1 mark total
### 5(b)(ii)
$z^2 - 5z + 6 = 0$; $z = 2, 3$; $z = 2, e^x = 2$; $x = \ln 2$; $z = 3, e^x = 3$; $x = \ln 3$ | M1 M1 A1 | use of $z = e^x$ oe; finding $e^x$ = their 2,3; all correct AG; SC: verification ln 2 (B1), ln 3 (B1); 3 marks total
### 5(b)(iii)
$x = \ln 2$: $y = e^{2\ln 2} - 10e^{\ln 2} + 12\ln 2$ or $2^2 - 10 \times 2 + 12\ln 2 = 4 - 20 + 12\ln 2 = -16 + 12\ln 2$; $x = \ln 3$: $y = e^{2\ln 3} - 10e^{\ln 3} + 12\ln 3 = 9 - 30 + 12\ln 3 = -21 + 12\ln 3$ | M1 A1 A1 | either substitution of their $x = \ln 2$ ($e^x = 2$) or their $x = \ln 3$ ($e^x = 3$); 3 marks total
### 5(b)(iv)
$x = \ln 2$: $\frac{d^2y}{dx^2} = 4e^{2\ln 2} - 10e^{\ln 2} = 16 - 20 = -4$; $\therefore$ maximum; $x = \ln 3$: $\frac{d^2y}{dx^2} = 4e^{2\ln 3} - 10e^{\ln 3} = 36 - 30 = 6$; $\therefore$ minimum | M1 A1 A1 | use of; in either of their $e^x = 2,3$ into their $\frac{d^2y}{dx^2}$; CSO; CSO; 3 marks total
**Total for Question 5:** 13 marks5
\begin{enumerate}[label=(\alph*)]
\item A curve has equation $y = \mathrm { e } ^ { 2 x } - 10 \mathrm { e } ^ { x } + 12 x$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(2 marks)
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.\\
(1 mark)
\end{enumerate}\item The points $P$ and $Q$ are the stationary points of the curve.
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinates of $P$ and $Q$ are given by the solutions of the equation
$$\mathrm { e } ^ { 2 x } - 5 \mathrm { e } ^ { x } + 6 = 0$$
(1 mark)
\item By using the substitution $z = \mathrm { e } ^ { x }$, or otherwise, show that the $x$-coordinates of $P$ and $Q$ are $\ln 2$ and $\ln 3$.
\item Find the $y$-coordinates of $P$ and $Q$, giving each of your answers in the form $m + 12 \ln n$, where $m$ and $n$ are integers.
\item Using the answer to part (a)(ii), determine the nature of each stationary point.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2006 Q5 [13]}}