June 2019 Paper 1 Q8
8 In this question you must show detailed reasoning.
Show that the only stationary point on the graph of \(y = x^2 - 4\sqrt{x}\) is a minimum point at \((1, -3)\). [7]
| Scheme | Marks | AO |
|---|---|---|
| DR | ||
| \(y = x^2 - 4x^{\frac{1}{2}} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 4 \times \tfrac{1}{2}x^{-\frac{1}{2}}\) | M1 | 2.1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) gives \(x^{\frac{3}{2}} = 1 \Rightarrow x = 1\) | M1 | 3.1a |
| There is only one solution to this so there is only one stationary point on the graph | A1 | 2.2a |
| When \(x = 1,\ y = 1 - 4 = -3\) | B1 | 2.1 |
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 - 2\left(-\tfrac{1}{2}\right)x^{-\frac{3}{2}}\) | M1 | 1.1a |
| When \(x = 1,\ \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 3 > 0\) | M1 | 1.1a |
| So the stationary point is a minimum point | A1 | 2.2a |
| [7] |
Notes
M1: Uses fractional power in an attempt to differentiate
M1: Attempt to solve their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)
A1: Must obtain \(x = 1\) from correct working and indicate that this is the only stationary point
Allow SC1 for verifying that \(x = 1\) gives \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)
B1: From correct working seen (AG)
M1: Attempt to find second derivative
M1: Substituting into their expression
A1: Conclusion from correct working (AG)
Alternative for final three marks
| Scheme | Marks | AO |
|---|---|---|
| Attempt to evaluate \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) at a point \(x \neq 1\) | M1 | |
| Attempt to evaluate \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) at a point the other side of \(x = 1\) | M1 | |
| Correct conclusion from correct values | A1 |
As there is only one stationary point, allow for similarly evaluating \(y\) and comparing with −3