Standard +0.3 This is a straightforward application of the quotient rule (or rewriting as a product) to find dy/dx, setting it to zero to find stationary points, and using the second derivative test. The algebra is clean with the k² term, and determining nature is standard. Slightly above average difficulty due to the parameter k and requiring both first and second derivatives, but still a routine calculus exercise.
9 A curve has equation \(y = \frac { k ^ { 2 } } { x + 2 } + x\), where \(k\) is a positive constant. Find, in terms of \(k\), the values of \(x\) for which the curve has stationary points and determine the nature of each stationary point.
Sub their \(x\) value with \(k\) in it into \(\frac{d^2y}{dx^2}\)
When \(x = -2 + k\), \(\frac{d^2y}{dx^2} = \left(\frac{2}{k}\right)\) which is \((> 0)\) min
A1
Only 1 of bracketed items needed for each
When \(x = -2 - k\), \(\frac{d^2y}{dx^2} = \left(-\frac{2}{k}\right)\) which is \((<0)\)
A1
but \(\frac{d^2y}{dx^2}\) and \(x\) need to be correct.
max
[8]
$\frac{dy}{dx} = -k^2(x + 2)^{-2} + 1 = 0$ | M1A1 | Attempt differentiation & set to zero
$x + 2 = \pm k$ | DM1 | Attempt to solve
$x = -2 \pm k$ | A1 | cao
$\frac{d^2y}{dx^2} = 2k^2(x + 2)^{-3}$ | M1 | Attempt to differentiate again
| | | Sub their $x$ value with $k$ in it into $\frac{d^2y}{dx^2}$
When $x = -2 + k$, $\frac{d^2y}{dx^2} = \left(\frac{2}{k}\right)$ which is $(> 0)$ min | A1 | Only 1 of bracketed items needed for each
When $x = -2 - k$, $\frac{d^2y}{dx^2} = \left(-\frac{2}{k}\right)$ which is $(<0)$ | A1 | but $\frac{d^2y}{dx^2}$ and $x$ need to be correct.
max | | [8]
9 A curve has equation $y = \frac { k ^ { 2 } } { x + 2 } + x$, where $k$ is a positive constant. Find, in terms of $k$, the values of $x$ for which the curve has stationary points and determine the nature of each stationary point.
\hfill \mbox{\textit{CAIE P1 2013 Q9 [8]}}