Standard +0.3 This is a straightforward stationary points question requiring differentiation of exponential functions (standard P2 content), solving a quadratic equation in e^x, and using the second derivative test. All steps are routine with no problem-solving insight needed, making it slightly easier than average.
3 The equation of a curve is \(y = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - 5 \mathrm { e } ^ { x } + 4 x\). Find the exact \(x\)-coordinate of each of the stationary points of the curve and determine the nature of each stationary point.
Equate derivative to zero and carry out recognisable solution method for a quadratic in \(e^x\)
M1
Obtain \(e^x = 1\) or \(e^x = 4\)
A1
Obtain \(x = 0\) and \(x = \ln 4\)
A1
Use an appropriate method for determining nature of at least one stationary point
M1
\[\left(\frac{d^2y}{dx^2} = 2e^{2x} - 5e^x, \text{ when } x = 0, \frac{d^2y}{dx^2} = -(3), x = \ln 4, \frac{d^2y}{dx^2} = +(12)\right)\]
Conclude maximum at \(x = 0\) and minimum at \(x = \ln 4\) (no errors seen)
A1
[6]
Obtain derivative $e^{2x} - 5e^x + 4$ | B1 |
Equate derivative to zero and carry out recognisable solution method for a quadratic in $e^x$ | M1 |
Obtain $e^x = 1$ or $e^x = 4$ | A1 |
Obtain $x = 0$ and $x = \ln 4$ | A1 |
Use an appropriate method for determining nature of at least one stationary point | M1 |
$$\left(\frac{d^2y}{dx^2} = 2e^{2x} - 5e^x, \text{ when } x = 0, \frac{d^2y}{dx^2} = -(3), x = \ln 4, \frac{d^2y}{dx^2} = +(12)\right)$$ | |
Conclude maximum at $x = 0$ and minimum at $x = \ln 4$ (no errors seen) | A1 | [6]
---
3 The equation of a curve is $y = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - 5 \mathrm { e } ^ { x } + 4 x$. Find the exact $x$-coordinate of each of the stationary points of the curve and determine the nature of each stationary point.
\hfill \mbox{\textit{CAIE P2 2013 Q3 [6]}}