June 2018 Paper 1 Q2
2. A curve \(C\) has equation
\[y = x^2 - 2x - 24\sqrt{x}, \qquad x \gt 0\]| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 2 - 12x^{-\frac{1}{2}}\) | M1 A1 | 1.1b 1.1b |
| (ii) \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 + 6x^{-\frac{3}{2}}\) | B1ft | 1.1b |
| (3) |
Notes
(a)(i) M1: Differentiates to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = Ax + B + Cx^{-\frac{1}{2}}\)
A1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 2 - 12x^{-\frac{1}{2}}\) (Coefficients may be unsimplified)
(a)(ii) B1ft: Achieves a correct \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) for their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (Their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) must have a negative or fractional index)
| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(x = 4\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2 \times 4 - 2 - 12 \times 4^{-\frac{1}{2}} = \ldots\) | M1 | 1.1b |
| Shows \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and states "hence there is a stationary point" oe | A1 | 2.1 |
| (2) |
Notes
M1: Substitutes \(x = 4\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and attempts to evaluate. There must be evidence \(\left.\dfrac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=4} = \ldots\)
Alternatively substitutes \(x = 4\) into an equation resulting from \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) Eg. \(\dfrac{36}{x} = (x - 1)^2\) and equates
A1: There must be a reason and a minimal conclusion. Allow ✓, QED for a minimal conclusion
Shows \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and states "hence there is a stationary point" oe
Alt Shows that \(x = 4\) is a root of the resulting equation and states "hence there is a stationary point"
All aspects of the proof must be correct including a conclusion
| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(x = 4\) into their \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 + 6 \times 4^{-\frac{3}{2}} = (2.75)\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2.75 \gt 0\) and states "hence minimum" | A1ft | 2.2a |
| (2) | ||
| (7 marks) |
Notes
M1: Substitutes \(x = 4\) into their \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) and calculates its value, or implies its sign by a statement such as when \(x = 4 \Rightarrow \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} \gt 0\). This must be seen in (c) and not labelled (b). Alternatively calculates the gradient of \(C\) either side of \(x = 4\) or calculates the value of \(y\) either side of \(x = 4\).
A1ft: For a correct calculation, a valid reason and a correct conclusion. Ignore additional work where candidate finds \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) left and right of \(x = 4\). Follow through on an incorrect \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) but it is dependent upon having a negative or fractional index. Ignore any references to the word convex. The nature of the turning point is "minimum".
Using the gradient look for correct calculations, a valid reason…. goes from negative to positive, and a correct conclusion …minimum.