| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Topic | Tangents, normals and gradients |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a straightforward stationary points question requiring differentiation of a product (polynomial × exponential), setting the derivative to zero, and solving a quadratic. The product rule and exponential differentiation are standard A-level techniques, and the algebra is clean. Slightly easier than average due to its routine nature and clear structure. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
Use the product rule on given $f(x)$ or $x^2 e^{-x}$ and obtain a two term expression [M1]
Obtain $2xe^{-x}$ [A1]
Obtain $-(x^2-3)e^{-x}$ or $-x^2e^{-x}$ [A1]
Obtain and solve their 3 term quadratic $= 0$ [M1]
Obtain $x = -1, 3$ [A1]
Obtain $y = -2e, 6e^{-3}$ [A1]
State $\pm e^{-x} \neq 0$ or equivalent [B1]
**Total: [7]**9 A curve has equation $y = \left( x ^ { 2 } - 3 \right) \mathrm { e } ^ { - x }$. Find the exact coordinates of the stationary points of the curve.
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2015 Q9 [7]}}