| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| 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 C3 question combining product rule differentiation, finding stationary points, algebraic manipulation, and iterative methods. Part (a) requires routine application of product rule and solving f'(x)=0. Parts (b)-(d) involve straightforward rearrangement and calculator-based iteration. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = 50x^2e^{2x} + 50xe^{2x}\) | M1A1 | Uses \(vu'+uv'\). If rule quoted it must be correct. Accept unsimplified forms e.g. \(25x^2 \times 2e^{2x} + 50x \times e^{2x}\). No marks unless differentiation seen. |
| Sets \(f'(x)=0\) to give \(x=-1\) and \(x=0\) (or one coordinate) | dM1A1 | Sets \(f'(x)=0\), factorises/cancels \(e^{2x}\). Dependent on first M1. |
| Obtains \((0,-16)\) and \((-1,\ 25e^{-2}-16)\) | A1 (CSO) | Both coordinates exact and correct as pairs. Accept \(x=-1,0\); \(y=\frac{25}{e^2}-16,\ -16\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(25x^2e^{2x}-16=0 \Rightarrow x^2 = \frac{16}{25}e^{-2x} \Rightarrow x = \pm\frac{4}{5}e^{-x}\) | B1* | Show that question — all elements must be seen. Must state \(f(x)=0\), show intermediate correct line, then reach given answer. Condone minus sign appearing on final line only. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Subs \(x_0 = 0.5\) into \(x = \frac{4}{5}e^{-x} \Rightarrow x_1 =\) awrt 0.485 | M1A1 | M1 for substituting \(x_0=0.5\). Can be implied by \(x_1 = \frac{4}{5}e^{-0.5}\) or awrt 0.49 |
| \(x_2 =\) awrt 0.492, \(x_3 =\) awrt 0.489 | A1 | Mark as second and third values given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\alpha = 0.49\) | B1 | |
| \(f(0.485) = -0.487,\ f(0.495) = (+)0.485\), sign change and deduction | B1 | Justify by: calculating \(f(0.485)\) and \(f(0.495)\) to awrt 1sf or 1dp showing sign change with minimal conclusion; or stating iteration is oscillating; or continued iteration to \(x_4 =\) awrt 0.491 with all values rounding to 0.49 |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = 50x^2e^{2x} + 50xe^{2x}$ | M1A1 | Uses $vu'+uv'$. If rule quoted it must be correct. Accept unsimplified forms e.g. $25x^2 \times 2e^{2x} + 50x \times e^{2x}$. No marks unless differentiation seen. |
| Sets $f'(x)=0$ to give $x=-1$ and $x=0$ (or one coordinate) | dM1A1 | Sets $f'(x)=0$, factorises/cancels $e^{2x}$. Dependent on first M1. |
| Obtains $(0,-16)$ and $(-1,\ 25e^{-2}-16)$ | A1 (CSO) | Both coordinates exact and correct as pairs. Accept $x=-1,0$; $y=\frac{25}{e^2}-16,\ -16$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $25x^2e^{2x}-16=0 \Rightarrow x^2 = \frac{16}{25}e^{-2x} \Rightarrow x = \pm\frac{4}{5}e^{-x}$ | B1* | Show that question — all elements must be seen. Must state $f(x)=0$, show intermediate correct line, then reach given answer. Condone minus sign appearing on final line only. |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Subs $x_0 = 0.5$ into $x = \frac{4}{5}e^{-x} \Rightarrow x_1 =$ awrt 0.485 | M1A1 | M1 for substituting $x_0=0.5$. Can be implied by $x_1 = \frac{4}{5}e^{-0.5}$ or awrt 0.49 |
| $x_2 =$ awrt 0.492, $x_3 =$ awrt 0.489 | A1 | Mark as second and third values given |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha = 0.49$ | B1 | |
| $f(0.485) = -0.487,\ f(0.495) = (+)0.485$, sign change and deduction | B1 | Justify by: calculating $f(0.485)$ and $f(0.495)$ to awrt 1sf or 1dp showing sign change with minimal conclusion; **or** stating iteration is oscillating; **or** continued iteration to $x_4 =$ awrt 0.491 with all values rounding to 0.49 |
---4.
$$\mathrm { f } ( x ) = 25 x ^ { 2 } \mathrm { e } ^ { 2 x } - 16 , \quad x \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Using calculus, find the exact coordinates of the turning points on the curve with equation $y = \mathrm { f } ( x )$.
\item Show that the equation $\mathrm { f } ( x ) = 0$ can be written as $x = \pm \frac { 4 } { 5 } \mathrm { e } ^ { - x }$
The equation $\mathrm { f } ( x ) = 0$ has a root $\alpha$, where $\alpha = 0.5$ to 1 decimal place.
\item Starting with $x _ { 0 } = 0.5$, use the iteration formula
$$x _ { n + 1 } = \frac { 4 } { 5 } \mathrm { e } ^ { - x _ { n } }$$
to calculate the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$, giving your answers to 3 decimal places.
\item Give an accurate estimate for $\alpha$ to 2 decimal places, and justify your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2013 Q4 [11]}}