| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Topic | Tangents, normals and gradients |
| Type | Find stationary points |
| Difficulty | Moderate -0.3 This is a straightforward differentiation question using the product rule, followed by solving dy/dx = 0. Part (i) is routine calculus with the answer given, and part (ii) requires only basic algebraic manipulation to find x = 0 and x = 2, then substitution. Slightly easier than average due to the scaffolding provided. |
| Spec | 1.07q Product and quotient rules: differentiation |
(i) Attempt product rule — M1; Obtain $2xe^{-x}$ — A1; Obtain $\pm x^2 e^{-x}$ — M1; Obtain $xe^{-x}(2-x)$ — A1 **AG** **[4]**
(ii) Set equal to zero and solve — M1; At least two correct $x$ or $y$ values — A1; $(0, 0)$ and $(2, 4e^{-2})$ — A1 **[3]**
**Total: [7]**7 It is given that $y = x ^ { 2 } \mathrm { e } ^ { - x }$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = x \mathrm { e } ^ { - x } ( 2 - x )$.\\
(ii) Hence find the exact coordinates of the stationary points on the curve $y = x ^ { 2 } \mathrm { e } ^ { - x }$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2013 Q7 [7]}}