| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard product rule application and stationary point analysis. While it requires multiple steps (differentiate twice, solve for stationary points, apply second derivative test), each step follows routine procedures with no conceptual challenges. The product rule with exponential is a core C3 skill, and the equation dy/dx = 0 factors easily to x(x+2)e^x = 0. Slightly easier than average due to its predictable structure and clean algebra. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07j 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 |
|---|---|---|
| (a) \(\frac{dy}{dx} = x^2e^x + 2xe^x\) | M1, A1, A1 | (3 marks) |
| (b) If \(\frac{dy}{dx} = 0\), \(e^x(x^2 + 2x) = 0\) (setting (a) = 0) | M1 | |
| \([e^x \neq 0]\) \(x(x + 2) = 0\) or \(x = 0\) | A1 | |
| \(x = -2\), \(y = 4e^{-2}\) (\(= 0.54...\)) | A1√ | (3 marks) |
| (c) \(\frac{d^2y}{dx^2} = x^2e^x + 2xe^x + 2xe^x + 2e^x\) \(\left[= (x^2 + 4x + 2)e^x\right]\) | M1, A1 | (2 marks) |
| (d) \(x = 0\): \(\frac{d^2y}{dx^2} > 0\) (=2) | M1 | |
| \(x = -2\): \(\frac{d^2y}{dx^2} < 0\) \(\left[= -2e^{-2} (= -0.270...)\right]\) | ||
| M1: Evaluate, or state sign of, candidate's (c) for at least one of candidate's \(x\) values from (b) | ||
| \(\therefore\) minimum | ∴ maximum | A1 (cso) |
| Answer | Marks |
|---|---|
| - Sketch curve |
**(a)** $\frac{dy}{dx} = x^2e^x + 2xe^x$ | M1, A1, A1 | (3 marks) |
**(b)** If $\frac{dy}{dx} = 0$, $e^x(x^2 + 2x) = 0$ (setting (a) = 0) | M1 | |
$[e^x \neq 0]$ $x(x + 2) = 0$ or $x = 0$ | A1 | |
$x = -2$, $y = 4e^{-2}$ ($= 0.54...$) | A1√ | (3 marks) |
**(c)** $\frac{d^2y}{dx^2} = x^2e^x + 2xe^x + 2xe^x + 2e^x$ $\left[= (x^2 + 4x + 2)e^x\right]$ | M1, A1 | (2 marks) |
**(d)** $x = 0$: $\frac{d^2y}{dx^2} > 0$ (=2) | M1 | |
$x = -2$: $\frac{d^2y}{dx^2} < 0$ $\left[= -2e^{-2} (= -0.270...)\right]$ | | |
M1: Evaluate, or state sign of, candidate's (c) for at least one of candidate's $x$ values from (b) | | |
$\therefore$ minimum | ∴ maximum | A1 (cso) | (2 marks) |
**Alt.(d)** For M1:
- Evaluate, or state sign of, $\frac{dy}{dx}$ at two appropriate values – on either side of at least one of their answers from (b) or
- Evaluate $y$ at two appropriate values – on either side of at least one of their answers from (b) or
- Sketch curve | | |
**Notes:**
- (a) Generous M for attempt at $f(x)g'(x) + f'(x)g(x)$. 1st A1 for one correct, 2nd A1 for the other correct. **Note that $x^2e^x$ on its own scores no marks**
- (b) 1st A1 ($x = 0$) may be omitted, but for 2nd A1 **both sets of coordinates needed**; f.t only on candidate's $x = -2$
- (c) M1 requires complete method for candidate's (a), result may be unsimplified for A1
- (d) A1 is cso; $x = 0$, min, and $x = -2$, max and no incorrect working seen., or (in alternative) sign of $\frac{dy}{dx}$ either side correct, or values of $y$ appropriate to t.p. Need only consider the quadratic, as may assume $e^x > 0$. If all marks gained in (a) and (c), and correct $x$ values, give M1A1 for correct statements with no working
---3. A curve $C$ has equation
$$y = x ^ { 2 } \mathrm { e } ^ { x }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, using the product rule for differentiation.
\item Hence find the coordinates of the turning points of $C$.
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item Determine the nature of each turning point of the curve $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2007 Q3 [10]}}