FP1 June 2013 Q3
3. Given that \(x = \dfrac{1}{2}\) is a root of the equation \[2x^3 - 9x^2 + kx - 13 = 0, \qquad k \in \mathbb{R}\] find
(a) the value of \(k\), (3)
(b) the other 2 roots of the equation. (4)
| Scheme | Marks |
|---|---|
| Ignore part labels and mark part (a) and part (b) together. | |
| \(\mathrm{f}\left(\dfrac{1}{2}\right) = 2\left(\dfrac{1}{2}\right)^3 - 9\left(\dfrac{1}{2}\right)^2 + k\left(\dfrac{1}{2}\right) - 13\) Attempts f(0.5) | M1 |
| \(\left(\dfrac{1}{4}\right) - \left(\dfrac{9}{4}\right) + \left(\dfrac{k}{2}\right) - 13 = 0 \Rightarrow k = \ldots\ldots\) Sets f(0.5) = 0 and leading to \(k = \) | dM1 |
| \(k = 30\) cao | A1 |
| (3) |
Notes
Alternative using long division:
| Scheme | Marks |
|---|---|
| \(2x^3 - 9x^2 + kx - 13 \div (2x - 1)\) \(= x^2 - 4x + \dfrac{1}{2}k - 2\) (Quotient) Remainder \(\dfrac{1}{2}k - 15\) Full method to obtain a remainder as a function of \(k\) | M1 |
| \(\dfrac{1}{2}k - 15 = 0\) Their remainder = 0 | dM1 |
| \(k = 30\) | A1 |
Alternative by inspection:
| Scheme | Marks |
|---|---|
| \((2x - 1)(x^2 - 4x + 13) = 2x^3 - 9x^2 + 30x - 13\) First M for \((2x - 1)(x^2 + bx + c)\) or \(\left(x - \dfrac{1}{2}\right)(2x^2 + bx + c)\) Second M1 for \(ax^2 + bx + c\) where (\(b = -4\) or \(c = 13\)) or (\(b = -8\) or \(c = 26\)) | M1dM1 |
| \(k = 30\) | A1 |
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = (2x - 1)(x^2 - 4x + 13)\) or \(\left(x - \dfrac{1}{2}\right)(2x^2 - 8x + 26)\) M1: \((x^2 + bx \pm 13)\) or \((2x^2 + bx \pm 26)\) Uses inspection or long division or compares coefficients and \((2x - 1)\) or \(\left(x - \dfrac{1}{2}\right)\) to obtain a quadratic factor of this form. | M1 |
| \(x^2 - 4x + 13\) or \(2x^2 - 8x + 26\) A1 \((x^2 - 4x + 13)\) or \((2x^2 - 8x + 26)\) seen | A1 |
| \(x = \dfrac{4 \pm \sqrt{4^2 - 4 \times 13}}{2}\) or equivalent Use of correct quadratic formula for their 3TQ or completes the square. | M1 |
| \(x = \dfrac{4 \pm 6\mathrm{i}}{2} = 2 \pm 3\mathrm{i}\) oe | A1 |
| (4) | |
| Total 7 |