Standard +0.3 This is a straightforward application of the quotient rule followed by solving dy/dx = 0. The differentiation is routine (exponential and polynomial terms), and finding the stationary point requires basic algebraic manipulation and calculator use. Slightly above average difficulty due to the quotient rule with exponential, but still a standard textbook exercise.
2 The curve with equation \(y = \frac { \mathrm { e } ^ { - 2 x } } { 1 - x ^ { 2 } }\) has a stationary point in the interval \(- 1 < x < 1\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and hence find the \(x\)-coordinate of this stationary point, giving the answer correct to 3 decimal places.
Equate derivative to zero and obtain a 3 term quadratic in \(x\)
M1
Obtain a correct 3-term equation e.g. \(2x^2 + 2x - 2 = 0\) or \(x^2 + x = 1\)
A1
From correct work only
Solve and obtain \(x = 0.618\) only
A1
From correct work only
Total
5
**Question 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct quotient rule or correct product rule | M1 | |
| Obtain correct derivative in any form | A1 | $\dfrac{dy}{dx} = \dfrac{-2e^{-2x}(1-x^2)+2xe^{-2x}}{(1-x^2)^2}$ |
| Equate derivative to zero and obtain a 3 term quadratic in $x$ | M1 | |
| Obtain a correct 3-term equation e.g. $2x^2 + 2x - 2 = 0$ or $x^2 + x = 1$ | A1 | From correct work only |
| Solve and obtain $x = 0.618$ only | A1 | From correct work only |
| **Total** | **5** | |
2 The curve with equation $y = \frac { \mathrm { e } ^ { - 2 x } } { 1 - x ^ { 2 } }$ has a stationary point in the interval $- 1 < x < 1$. Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and hence find the $x$-coordinate of this stationary point, giving the answer correct to 3 decimal places.\\
\hfill \mbox{\textit{CAIE P3 2019 Q2 [5]}}