| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Topic | Integration by Substitution |
| Type | Finding stationary points after integration |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard A-level techniques. Part (i) is a routine substitution integral with clear guidance (u = x²), and part (ii) requires product rule differentiation and solving e^(-x²/2) = 1/x, which reduces to finding where x² = 2. Both parts are textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \(du = 2xdx\) or equivalent used | M1 | |
| Substitute to obtain \(\int \frac{1}{2}e^{-\frac{1}{u}} du\) | A1 | |
| Obtain \(\left[-e^{-\frac{u}{2}}\right]\) | A1 | |
| Evaluate: 0.5 | WWW | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = -1 \times e^{-\frac{1}{x^3}} + x \times (-x) \times e^{-\frac{1}{x^2}}\) | M1 A1 | |
| Equate to zero and find at least one point | M1 | |
| Stationary points \(\left(1, e^{-0.5}\right), \left(-1, -e^{-0.5}\right)\) | A1 | [4] |
**Part (i)**
$du = 2xdx$ or equivalent used | M1
Substitute to obtain $\int \frac{1}{2}e^{-\frac{1}{u}} du$ | A1
Obtain $\left[-e^{-\frac{u}{2}}\right]$ | A1
Evaluate: 0.5 | WWW | A1 | [4]
*SC: For 0.5 without working – B2*
**Part (ii)**
$\frac{dy}{dx} = -1 \times e^{-\frac{1}{x^3}} + x \times (-x) \times e^{-\frac{1}{x^2}}$ | M1 A1
Equate to zero and find at least one point | M1
Stationary points $\left(1, e^{-0.5}\right), \left(-1, -e^{-0.5}\right)$ | A1 | [4]\begin{enumerate}[label=(\roman*)]
\item Using the substitution $u = x^2$, or otherwise, find the numerical value of
$$\int_0^{\sqrt{\ln 4}} xe^{-\frac{1}{2}x^2} \, dx.$$ [4]
\item Determine the exact coordinates of the stationary points of the curve $y = xe^{-\frac{1}{2}x^2}$. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2011 Q6 [8]}}