C3 June 2015 Q9
9. Given that \(k\) is a negative constant and that the function \(\mathrm{f}(x)\) is defined by\[\mathrm{f}(x) = 2 - \frac{(x - 5k)(x - k)}{x^2 - 3kx + 2k^2}, \qquad x \geqslant 0\]
Justify your answer.
(2)| Scheme | Marks |
|---|---|
| \(x^2 - 3kx + 2k^2 = (x - 2k)(x - k)\) | B1 |
| \(2 - \dfrac{(x - 5k)(x - k)}{(x - 2k)(x - k)} = 2 - \dfrac{(x - 5k)}{(x - 2k)} = \dfrac{2(x - 2k) - (x - 5k)}{(x - 2k)}\) | M1 |
| \(= \dfrac{x + k}{(x - 2k)}\) | A1* |
| (3) |
Notes
B1: For seeing \(x^2 - 3kx + 2k^2 = (x - 2k)(x - k)\) anywhere in the solution
M1: For writing as a single term or two terms with the same denominator
Score for \(2 - \dfrac{(x - 5k)}{(x - 2k)} = \dfrac{2(x - 2k) - (x - 5k)}{(x - 2k)}\) or
\(2 - \dfrac{(x - 5k)(x - k)}{(x - 2k)(x - k)} = \dfrac{2(x - 2k)(x - k) - (x - 5k)(x - k)}{(x - 2k)(x - k)}\) \(\left(= \dfrac{x^2 - k^2}{x^2 - 3kx + 2k^2}\right)\)
A1*: Proceeds without any errors (including bracketing) to \(= \dfrac{x + k}{(x - 2k)}\)
| Scheme | Marks |
|---|---|
| Applies \(\dfrac{vu' - uv'}{v^2}\) to \(y = \dfrac{x + k}{x - 2k}\) with \(u = x + k\) and \(v = x - 2k\) | |
| \(\Rightarrow \mathrm{f}'(x) = \dfrac{(x - 2k) \times 1 - (x + k) \times 1}{(x - 2k)^2}\) | M1, A1 |
| \(\Rightarrow \mathrm{f}'(x) = \dfrac{-3k}{(x - 2k)^2}\) | A1 |
| (3) |
Notes
M1: Applies \(\dfrac{vu' - uv'}{v^2}\) to \(y = \dfrac{x + k}{x - 2k}\) with \(u = x + k\) and \(v = x - 2k\).
If the rule it is stated it must be correct. It can be implied by \(u = x + k\) and \(v = x - 2k\) with their \(u', v'\) and \(\dfrac{vu' - uv'}{v^2}\)
If it is neither stated nor implied only accept expressions of the form \(\mathrm{f}'(x) = \dfrac{x - 2k - x \pm k}{(x - 2k)^2}\)
The mark can be scored for applying the product rule to \(y = (x + k)(x - 2k)^{-1}\) If the rule it is stated it must be correct. It can be implied by \(u = x + k\) and \(v = (x - 2k)^{-1}\) with their \(u', v'\) and \(vu' + uv'\)
If it is neither stated nor implied only accept expressions of the form \(\mathrm{f}'(x) = (x - 2k)^{-1} \pm (x + k)(x - 2k)^{-2}\)
Alternatively writes \(y = \dfrac{x + k}{x - 2k}\) as \(y = 1 + \dfrac{3k}{x - 2k}\) and differentiates to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{A}{(x - 2k)^2}\)
A1: Any correct form (unsimplified) form of \(\mathrm{f}'(x)\).
\(\mathrm{f}'(x) = \dfrac{(x - 2k) \times 1 - (x + k) \times 1}{(x - 2k)^2}\) by quotient rule
\(\mathrm{f}'(x) = (x - 2k)^{-1} - (x + k)(x - 2k)^{-2}\) by product rule
and \(\mathrm{f}'(x) = \dfrac{-3k}{(x - 2k)^2}\) by the third method
A1: cao \(\mathrm{f}'(x) = \dfrac{-3k}{(x - 2k)^2}\). Allow \(\mathrm{f}'(x) = \dfrac{-3k}{x^2 - 4kx + 4k^2}\)
As this answer is not given candidates you may allow recovery from missing brackets
| Scheme | Marks |
|---|---|
| If \(\mathrm{f}'(x) = \dfrac{-Ck}{(x - 2k)^2} \Rightarrow \mathrm{f}(x)\) is an increasing function as \(\mathrm{f}'(x) > 0\), | M1 |
| \(\mathrm{f}'(x) = \dfrac{-3k}{(x - 2k)^2} > 0\) for all values of \(x\) as \(\dfrac{\text{negative} \times \text{negative}}{\text{positive}} = \text{positive}\) | A1 |
| (2) | |
| (8 marks) |
Notes
Note that this is B1 B1 on e pen. We are scoring it M1 A1
M1: If in part (b) \(\mathrm{f}'(x) = \dfrac{-Ck}{(x - 2k)^2}\), look for f(\(x\)) is an increasing function as \(\mathrm{f}'(x)\) / gradient \(> 0\)
Accept a version that states as \(k < 0 \Rightarrow -Ck > 0\) hence increasing
If in part (b) \(\mathrm{f}'(x) = \dfrac{(+)Ck}{(x - 2k)^2}\), look for f(\(x\)) is an decreasing function as \(\mathrm{f}'(x)\) / gradient \(< 0\)
Similarly accept a version that states as \(k < 0 \Rightarrow (+)Ck < 0\) hence decreasing
A1: Must have \(\mathrm{f}'(x) = \dfrac{-3k}{(x - 2k)^2}\) and give a reason that links the gradient with its sign.
There must have been reference to the sign of both numerator and denominator to justify the overall positive sign.