FP1 June 2018 Q2
2. \[\mathrm{f}(x) = \frac{3}{2}x^2 + \frac{4}{3x} + 2x - 5, \qquad x < 0\]
The equation \(\mathrm{f}(x) = 0\) has a single root \(\alpha\).
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(-3) = 2.05555555\ldots\) \(\mathrm{f}(-2.5) = -1.15833333\ldots\) | M1 |
| Sign change oe (and \(\mathrm{f}(x)\) is continuous) therefore a root \(\alpha\) {exists in the interval \([-3, -2.5]\).} | A1 |
| (2) |
Notes
M1: Attempt both of \(\mathrm{f}(-3) =\) awrt 2.1 or trunc 2 or 2.0 or \(\dfrac{37}{18}\) and \(\mathrm{f}(-2.5) =\) awrt \(-1.2\) or trunc \(-1.1\) or \(-\dfrac{139}{120}\)
A1: Both \(\mathrm{f}(-3) =\) awrt 2.1 and \(\mathrm{f}(-2.5) =\) awrt \(-1.2\), sign change and ‘root’ or ‘\(\alpha\)’. Any errors award A0.
| Scheme | Marks |
|---|---|
| \(\mathrm{f}'(x) = 3x - \dfrac{4}{3x^2} + 2\) | M1 A1 A1 |
| \(\alpha = -3 - \left(\dfrac{\text{"}2.055\ldots\text{"}}{\text{"}-7.148\ldots\text{"}}\right)\) | M1 |
| \(= -2.71243523\ldots\) or \(-\dfrac{1047}{386}\) or \(-2\dfrac{275}{386}\) | A1 |
| (5) |
Notes
M1: \(\dfrac{3}{2}x^2 \to \pm Ax\) or \(\dfrac{4}{3x} \to \pm Bx^{-2}\) or \(2x - 5 \to 2\). Calculus must be seen for this to be awarded.
A1: At least two terms differentiated correctly
A1: Correct derivative.
M1: Correct application of Newton-Raphson using their values from calculus.
A1: Exact value or awrt \(-2.712\)
| Scheme | Marks |
|---|---|
| \(\dfrac{-2.5 - \alpha}{\text{"}1.158\ldots\text{"}} = \dfrac{\alpha - -3}{\text{"}2.055\ldots\text{"}}\) or \(\dfrac{\alpha - -3}{\text{"}2.055\ldots\text{"}} = \dfrac{-2.5 - -3}{\text{"}2.055\ldots\text{"} + \text{"}1.158\ldots\text{"}}\) | M1 |
| \(\alpha = -3 + \left(\dfrac{\text{"}2.055\ldots\text{"}}{\text{"}2.055\ldots\text{"} + \text{"}1.158\ldots\text{"}}\right)(0.5)\) or \(\alpha = -3 + \left(\dfrac{\text{"}2.055\ldots\text{"}}{\text{"}3.213\ldots\text{"}}\right)(0.5)\) or \(\alpha = \left(\dfrac{(-2.5)(\text{"}2.055\ldots\text{"}) - 3(\text{"}1.158\ldots\text{"})}{\text{"}2.055\ldots\text{"} + \text{"}1.158\ldots\text{"}}\right)\) | dM1 |
| \(= -2.68020743\ldots\) or \(-\dfrac{3101}{1157}\) or \(-2\dfrac{787}{1157}\) | |
| \(= -2.680\ (3\text{ dp})\) | A1 cao |
| (3) | |
| (10 marks) |
Notes
M1: A correct linear interpolation statement \(\dfrac{-2.5 + \alpha}{\text{"}1.158\ldots\text{"}} = \dfrac{-\alpha - -3}{\text{"}2.055\ldots\text{"}}\) with correct signs, provided \(\alpha\) sign changed at the end. Do not award until \(\alpha\) is seen.
dM1: Achieves a correct linear interpolation statement with correct signs for \(\alpha = \ldots\) dependent on the previous method mark.
A1 cao: \(-2.680\): only penalise accuracy once in (b) and (c), but must be to at least 3sf.
ALT (c)
| Scheme | Marks |
|---|---|
| The gradient of the line between \((-3, 2.055\ldots)\) and \((-2.5, -1.158\ldots)\) is \(\dfrac{2.055\ldots - -1.158\ldots}{-3 - -2.5} = -6.427\ldots\) | |
| Equation of the line joining the points \(y - 2.055\ldots = -6.427\ldots(x - -3)\) | M1 |
| At \(y = 0\), \(0 - 2.055\ldots = -6.427\ldots(x - -3)\) | dM1 |
| \(\Rightarrow x = -2.680\) | A1 cao |
| (3) |
M1: Correct attempt to find the equation of a line between the two points.
dM1: Subs \(y = 0\) in their line and achieves \(x = \ldots\)
A1 cao: \(-2.680\): only penalise accuracy once in (b) and (c), but must be to at least 3sf.