| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2012 |
| Session | June |
| Marks | 5 |
| Topic | Differentiating Transcendental Functions |
| Type | Find stationary points - polynomial/exponential products |
| Difficulty | Moderate -0.8 This is a straightforward product rule differentiation followed by solving a simple inequality. Part (i) requires routine application of the product rule with an exponential function, and part (ii) only needs setting the derivative positive and solving a linear inequality. Both parts are standard textbook exercises requiring no problem-solving insight, making this easier than average. |
| Spec | 1.07o Increasing/decreasing: functions using sign of dy/dx1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Attempt product rule: \(\frac{dy}{dx} = 2e^{-2x} - 2(2x-3)e^{-2x}\) | M1, A1 | |
| \(= (8-4x)e^{-2x}\) | A1 [3] | |
| (ii) \(\frac{dy}{dx} \geq 0\) seen | B1 | |
| \(y\) is increasing when \(x \leq 2\) | B1 [2] | [5] |
**(i)** Attempt product rule: $\frac{dy}{dx} = 2e^{-2x} - 2(2x-3)e^{-2x}$ | M1, A1 |
$= (8-4x)e^{-2x}$ | A1 [3] |
**(ii)** $\frac{dy}{dx} \geq 0$ seen | B1 |
$y$ is increasing when $x \leq 2$ | B1 [2] | [5]Let $y = (2x - 3)e^{-2x}$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dy}{dx}$, giving your answer in the form $e^{-2x}(ax + b)$, where $a$ and $b$ are integers. [3]
\item Determine the set of values of $x$ for which $y$ is increasing. [2]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2012 Q7 [5]}}