Standard +0.3 This is a straightforward application of the product rule to find dy/dx, setting it equal to zero, and solving for x. The exponential never equals zero, so students only need to solve a linear equation. The calculation is routine with no conceptual challenges beyond standard differentiation technique.
Equate first derivative to zero and solve for \(x\)
DM1
Solution must come from linear terms
Obtain \(x = -\frac{1}{2}\) only
A1
Obtain \(4e^{\frac{1}{4}}\) or exact equivalent only
A1
Total: 5
## Question 5:
| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate using the product rule | \*M1 | Must have $u$ and $v$ correct in a correct formula with $\dfrac{du}{dx} = 2$ and $\dfrac{dv}{dx} = me^{-\frac{1}{2}x}$ |
| Obtain correct $2e^{-\frac{1}{2}x} - \frac{1}{2}e^{-\frac{1}{2}x}(2x+5)$ | A1 | OE |
| Equate first derivative to zero and solve for $x$ | DM1 | Solution must come from linear terms |
| Obtain $x = -\frac{1}{2}$ only | A1 | |
| Obtain $4e^{\frac{1}{4}}$ or exact equivalent only | A1 | |
| **Total: 5** | | |
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5 Find the exact coordinates of the stationary point of the curve with equation $y = \mathrm { e } ^ { - \frac { 1 } { 2 } x } ( 2 x + 5 )$.\\
\hfill \mbox{\textit{CAIE P2 2019 Q5 [5]}}